• Commentary
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Contents
1. Sets
2. Relations and Functions
3. Trigonometric Functions
4. Principle of Mathematical Induction
5. Complex Numbers and Quadratic Equations
6. Linear Inequalities
7. Permutations and Combinations
8. Binomial Theorem
9. Sequences and Series
10. Straight Lines and Pair of Straight Lines
11. Conic Sections
12. Limits and Derivatives
13. Mathematical Reasoning
14. Statistics
15. Probability
16. Relations and Functions
17. Inverse Trigonometric Functions
18. Matrices
19. Determinants
20. Continuity and Differentiability
21. Applications of Derivatives
22. Integrals
23. Applications of Integrals
24. Differential Equations
25. Vector Algebra
26. Three Dimensional Geometry
27. Probability
28. Properties of Triangles
Mock Test 1 with Solutions
Mock Test 2 with Solutions

##### Citation preview

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EBD_8344

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EBD_8344

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EBD_8344

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EBD_8344

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1 Sets TOPIC Ć 1.

2.

Set A has m elements and set B has n elements. If the total number of subsets of A is 112 more than the total number of subsets of B, then the value of m×n is ______. [Sep. 06, 2020 (I)] Let S = {1, 2, 3, ... , 100}. The number of non-empty subsets A of S such that the product of elements in A is even is : [Jan. 12, 2019 (I)]

(

3.

5.

7.

8.

)

50 50 (a) 2100 - 1 (b) 2 2 - 1 (c) 250 – 1 (d) 250 + 1 Let S = {x Î R : x ³ 0 and

2| (a) (b) (c) (d) 4.

6.

Sets, Types of Sets, Disjoint Sets, Complementary Sets, Subsets, Power Set, Cardinal Number of Sets, Operations on Sets

x - 3 | + x ( x - 6) + 6 = 0 . Then S : contains exactly one element. contains exactly two elements. contains exactly four elements. is an empty set.

æ1ö If f(x) + 2f ç ÷ = 3x , x ¹ 0 and èxø S = {x Î R : f(x) = f(–x)}; then S: (a) contains exactly two elements. (b) contains more than two elements. (c) is an empty set. (d) contains exactly one element. Let P = {q : sinq – cosq = cosq =

(b) Q Ë P (c) P = Q

(Y, Z) that can formed such that Y Í X , Z Í X and Y Ç Z is empty is : [2012] (a) 52 (b) 35 (c) 25 (d) 53 If A, B and C are three sets such that A Ç B = A Ç C and [2009] A È B = A È C , then (a) A = C (b) B = C (c) A Ç B = f

[2018]

TOPIC n 9.

[2016]

2 cosq} and Q = {q : sinq +

2 sinq} be two sets. Then: [Online April 10, 2016]

(a) P Ì Q and Q - P ¹ f

A relation on the set A = {x : |x| < 3, x ÎZ }, where Z is the set of integers is defined by R = {(x, y) : y = |x|, x ¹ – 1}. Then the number of elements in the power set of R is: [Online April 12, 2014] (a) 32 (b) 16 (c) 8 (d) 64 Let X ={1,2,3,4,5}. The number of different ordered pairs

10.

11.

(d) A = B

Venn Diagrams, Algebraic Operations on Sets, De Morgan’s Law, Number of Elements in Different Sets

A survey shows that 73% of the persons working in an office like coffee, whereas 65% like tea. If x denotes the percentage of them, who like both coffee and tea, then x cannot be : [Sep. 05, 2020 (I)] (a) 63 (b) 36 (c) 54 (d) 38 A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. If x% of the people read both the newspapers, then a possible value of x can be : [Sep. 04, 2020 (I)] (a) 29 (b) 37 (c) 65 (d) 55 Let

50

n

i =1

i =1

U X i = U Yi = T , where each Xi contains 10 elements

and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi's and exactly 6 of sets Yi's, then n is equal to [Sep. 04, 2020 (II)] (a) 15 (b) 50 (c) 45 (d) 30

(d) P Ë Q

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12.

13.

14.

15.

Mathematics

Let X = {n Î N: l £ n £ 50}. If A = {n Î X: n is a multiple of 2} and B = {n Î X: n is a multiple of 7}, then the number of elements in the smallest subset of X containing both A and B is __________. [Jan. 7, 2020 (II)]

16.

2 Let Z be the set of integers. If A = {xÎZ : 2(x + 2) ( x – 5x + 6) = 1} and B = {x Î Z : –3 < 2x – 1< 9}, then the number of subsets of the set A × B, is : [Jan. 12, 2019 (II)] (a) 215 (b) 218 (c) 212 (d) 210

In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is: [Jan. 10, 2019 (II)] (a) 102 (b) 42 (c) 1 (d) 38 f ¹ A Ç B Í C . Then Let A, B and C be sets such that which of the following statements is not true ? [April 12, 2019 (II)] (a) B Ç C ¹ f (b) If ( A - B) Í C , then A Í C

17.

(c) (C È A) Ç (C È B) = C (d) If (A - C) Í B , then A Í B

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Sets

1.

(28) 2m = 112 + 2n Þ 2m - 2n = 112 m -n

Þ 2 (2 n

5.

- 1) = 2 (2 - 1) 4

3

\ m = 7, n = 4 Þ mn = 28 2.

3.

7.

Case-II: x Î[9, ¥] 2( x - 3) + x - 6 x + 6 = 0

8.

2 -x x

f (x) = f (- x) Þ x2 = 2

(b) Q

B = ( B Ç A) È B

= ( A È B) Ç ( C È B) = ( A È C) Ç ( B È C)

æ 1ö (a) f (x) + 2f ç ÷ = 3x ....(1) è xø 3 æ 1ö f ç ÷ + 2f (x) = ....(2) è xø x Adding (1) and (2)

Þ f (x) =

(b) A = {x : |x | < 3, x Î Z } A = {–2, – 1, 0, 1, 2} R = {(x, y) : y = |x |, x ¹ -1} R = {(–2, 2), (0, 0), (1, 1), (2, 2)} R has four elements Number of elements in the power set of R = 24 = 16 (b) Let X = {1, 2, 3, 4, 5} n(x) = 5 Each element of x has 3 options. Either in set Y or set Z or none. (Q Y Ç Z = f) So, number of ordered pairs = 35

= ( A Ç C) È B

Since x Î[9, ¥] \ x = 16 Hence, x = 4 & 16

1 æ1ö Þ f (x) + f ç ÷ = x + ...(3) x x è ø Substracting (1) from (2) æ1ö 3 Þ f (x) - f ç ÷ = - 3x ...(4) èxø x On adding (3) and (4)

2 sinq

\ P= Q 6.

Þ x - 8 x + 12 = 0 Þ x = 4, 2 Þ x = 16, 4

4.

2 cosq

Þ sinq + cosq =

2(3 - x ) + x - 6 x + 6 = 0

Þ x - 4 x = 0 Þ x = 16,0

Þ sinq = cosq +

Þ ( 2 - 1) sinq = cosq

(b) Case-I: x Î[0,9]

Since x Î[0,9] \ x=4

2 cosq

æ 2 -1 ö = ( 2 + 1) cosq = ç ÷ cosq è 2 -1 ø

(b) Q Product of two even number is always even and product of two odd numbers is always odd. \ Number of required subsets = Total number of subsets – Total number of subsets having only odd numbers = 2100 – 250 = 250(250 – 1)

(c) sinq – cosq =

= ( A Ç B) È C = ( A Ç C) È C =C 9.

(b) Given, n(C ) = 73, n(T ) = 65, n(C Ç T ) = x \ 65 ³ n(C Ç T ) ³ 65 + 73 - 100 Þ 65 ³ x ³ 38 Þ x ¹ 36.

10.

(d) Let n(U ) = 100, then n(A) = 63, n(B) = 76 n( A Ç B ) = x Now, n( A È B ) = n( A) + n( B) - n( A Ç B) £ 100

-2 2 2 -x = +xÞx= x x x

= 63 + 76 - x £ 100 Þ x ³ 139 - 100 Þ x ³ 39

or x = 2, - 2

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Mathematics

Q = {6n: n Î N, 1 £ n £ 23} – P

Q n( A Ç B) £ n( A)

Þ n(Q) = 19

Þ x £ 63

R = {15n: n Î N, 1 £ n £ 9} – P

\ 39 £ x £ 63

11.

(d)

50

n

i =1

i =1

Þ n(R) = 5

U X i = U Yi = T

S = {10n: n Î N, 1 £ n £ 14} – P Þ n(S) = 10

Q

n( X i ) = 10, n (Yi ) = 5

So,

U X i = 500, U Yi = 5n

50

n(T) = 70 – n(P) – n(Q) – n(S) = 70 – 33 = 37 n(V) = 46 – n(P) – n(Q) – n(R) = 46 – 28 = 18

n

i =1

n(W) = 28 – n(P) – n(R) – n(S) = 28 – 19 = 9

i =1

Þ Number of required students

500 5n Þ = Þ n = 30 20 6

12.

= 140 – (4 + 19 + 5 + 10 + 37 + 18 + 9) = 140 – 102 = 38

(29) From the given conditions,

15.

n(A) = 25, n(B) = 7 and n(A Ç B) = 3

In (3) option,

n(A È B) = n(A) + n(B) – n(A Ç B)

If A = C then A – C = f

= 25 + 7 – 3 = 29 13.

(d) (1), (2) and (4) are always correct.

Clearly, f Í B but A Í B is not always true.

(a) Let x Î A, then Q 2( x+2)( x2 -5 x+6) = 1 Þ (x + 2)(x – 2)(x – 3) = 0 x = –2, 2, 3

16.

(a)

A

B 25

20 8

A = {–2, 2, 3} Then, n(A) = 3 Let x Î B, then –3 < 2x – 1 < 9

% of people who reads A only = 25 – 8 = 17%

–1 < x < 5 and x Î Z

% of people who read B only = 20 – 8 = 12%

\ B = {0, 1, 2, 3, 4}

n(B) = 5

= 17 × 0.3 = 5.1%

n(A ´ B) = 3 ´ 5 = 15

Hence, Number of subsets of A ´ B = 2

= 12 × 0.4 = 4.8%

15

(d)

= 8 × 0.5 = 4%

Maths T Q

Physics

W

S

P R

V

17.

= 5.1 + 4.8 + 4 = 13.9% (c) n(P) = 25% n(C) = 15%

n ( P ¢ È C ¢ ) =65% Þ n(P È C)¢ = 65%

n ( P È C ) = 35% P = {30, 60, 90, 120} Þ n(P) = 4

n ( P Ç C) = n ( P ) + n (C) - n ( P È C) 25 + 15 – 35 = 5% x × 5% = 2000 x = 40,000

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2 Relations and Functions TOPIC Ć 1.

The domain of the definition of the function

6.

1 + log10 ( x3 - x ) is: 4 - x2 (a) (–1, 0) È (1, 2) È (3, ¥) (b) (–2, –1) È (–1, 0) È (2, ¥) (c) (–1, 0) È (1, 2) È (2, ¥) (d) (1, 2) È (2, ¥) The range of the function f ( x) =

Let R1 and R2 be two relations defined as follows :

R1 = {(a, b) Î R 2 : a 2 + b 2 Î Q} and R2 = {(a, b) Î R 2 : a 2 + b2 Ï Q} , where Q is the set of all rational numbers. Then : [Sep. 03, 2020 (II)] (a) Neither R1 nor R2 is transitive. (b) R2 is transitive but R1 is not transitive. (c) R1 is transitive but R2 is not transitive. (d) R1 and R2 are both transitive. 2.

5.

Relations, Domain, Codomain and Range of a Relation, Functions, Domain, Codomain and Range of a Function

| x | +5 ö The domain of the function f ( x) = sin -1 æç ÷ is è x2 +1 ø (-¥, - a] È [a, ¥]. Then a is equal to :

f ( x) = (a) R

3.

4.

17 2

(b)

17 - 1 1 + 17 (d) (c) 2 2

Let f : R ® R be defined by f ( x ) = the range of f is :

x , x Î R. Then 1 + x2 [Jan. 11, 2019 (I)]

é 1 1ù (a) ê - , ú ë 2 2û

(b) R – [–1, 1]

é 1 1ù (c) R - ê - , ú ë 2 2û

(d) (–1, 1) – {0}

(c) R – {0}

(d) [– 1, 1]

1

The domain of the function f ( x) =

8.

(a) (0, ¥ ) (b) (– ¥ , 0) (c) (– ¥ , ¥ ) – {0} (d) (– ¥ , ¥ ) Domain of definition of the function f ( x) =

17 +1 2

If R = {(x, y) : x, y Î Z, x2 + 3 y 2 £ 8} is a relation on the set of integers Z, then the domain of R–1 is : [Sep. 02, 2020 (I)] (a) {–2, –1, 1, 2} (b) {0, 1} (c) {–2, –1, 0, 1, 2} (d) {–1, 0, 1}

(b) (– 1, 1)

[Online May 7, 2012]

7.

[Sep. 02, 2020 (I)] (a)

x , x Î R , is 1+ x

[April. 09, 2019 (II)]

3 4 - x2

is

+ log10 ( x 3 - x) , is

(a) ( -1,0) È (1,2) È ( 2, ¥)

(b) (a, 2)

(c) ( -1,0) È ( a,2)

(d) (1,2) È (2, ¥)

TOPIC n

9.

x -x

[2011]

[2003]

Even and Odd Functions, Explicit and Implicit Functions, Greatest Integer Function, Periodic Functions, Value of a Function, Equal Functions, Algebraic Operations on Functions

Let [t] denote the greatest integer £ t. Then the equation in x, [x]2 + 2[x + 2] – 7 = 0 has : [Sep. 04, 2020 (I)] (a) exactly two solutions (b) exactly four integral solutions (c) no integral solution (d) infinitely many solutions

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10.

11.

12.

13.

14.

Mathematics

Let f (x) be a quadratic polynomial such that f (–1) + f (2) = 0. If one of the roots of f (x) = 0 is 3, then its other root lies in : [Sep. 02, 2020 (II)] (a) (–1, 0) (b) (1, 3) (c) (–3, –1) (d) (0, 1)

15.

Then f(x) at x = -

x[ x] , 1 + x2 where [x] denotes the greatest integer £ x. Then the range of f is: [Jan. 8, 2020 (II)]

Let f (1, 3) ® R be a function defined by f (x) =

æ 2 3ö æ 3 4 ö (a) çè , ÷ø È çè , ÷ø 5 5 4 5

æ 2 1 ö æ 3 4ö (b) çè , ÷ø È çè , ø÷ 5 2 5 5

æ 2 4ö (c) çè , ÷ø 5 5

æ 3 4ö (d) çè , ÷ø 5 5

æ 2x ö 1- x ö If f(x) = loge æç , |x| < 1, then f ç ÷ is equal to : ÷ è 1+ x2 ø è 1+ x ø [April 8, 2019 (I)] (a) 2f(x) (b) 2f(x2) (c) (f(x)) 2 (d) –2f(x) Let f(x) = ax (a > 0) be written as f(x) = f1(x) + f2(x), where f1(x) is an even function and f2(x) is an odd function. Then f1(x + y) + f1(x – y) equals : [April. 08, 2019 (II)] (a) 2f1(x) f1(y) (b) 2f1(x + y) f1(x – y) (c) 2f1(x)f2(y) (d) 2f1(x + y) f2(x – y)

é 1 3n ù n , where [n] denotes the greatest Let f ( n ) = ê + ë 3 100 úû 56

integer less than or equal to n. Then

å f ( n) is equal to:

n=1

(a) 56

(b) 689

Let f be an odd function defined on the set of real numbers such that for x ³ 0, f(x) = 3 sin x + 4 cos x.

(a)

11p is equal to: [Online April 11, 2014] 6

3 +2 3 2

3 (b) - + 2 3 2

3 3 -2 3 (d) - - 2 3 2 2 A real valued function f (x) satisfies the functional equation f (x – y) = f (x) f (y) – f (a – x) f (a + y) where a is a given constant and f (0) = 0, f (2a – x) is equal to [2005] (a) – f (x) (b) f (x) (c) f (a) + f (a – x) (d) f (– x) The graph of the function y = f(x) is symmetrical about the line x = 2, then [2004]

(c)

16.

17.

18.

(a)

f ( x ) = - f (- x)

(b) f (2 + x) = f (2 - x)

(c)

f ( x ) = f ( - x)

(d) f ( x + 2) = f ( x - 2)

If f : R ® R satisfies f ( x + y ) = f ( x) + f ( y ) , for all x, n

y Î R and f(1) = 7, then S f (r ) is r=1

(a)

7 n (n + 1) 2

(b)

(c)

7 (n + 1) 2

(d) 7 n + ( n + 1)

7n 2

[Online April 19, 2014] (c) 1287 (d) 1399

[2003]

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Relations and Functions

1.

Þ 1 ³ 4y2

1/ 4 (a) For R1 let a = 1 + 2, b = 1 - 2, c = 8

Þ |y| £

bR1c Þ b2 + c 2 = (1 - 2)2 + (81/ 4 )2 = 3 Î Q

1 1 Þ - £ y£ 2 2

aR1c Þ a 2 + c2 = (1 + 2)2 + (81/ 4 )2 = 3 + 4 2 Ï Q

é 1 1ù Þ The range of f is ê - , ú . ë 2 2û

\ R1 is not transitive. For R2 let a = 1 + 2, b = 2, c = 1 - 2

5.

aR2b Þ a + b = (1 + 2) + ( 2) = 5 + 2 2 Ï Q 2

2

2

2

bR2 c Þ b 2 + c 2 = ( 2)2 + (1 - 2)2 = 5 - 2 2 Ï Q aR2c Þ a 2 + c2 = (1 + 2) 2 + (1 - 2) 2 = 6 Î Q | x | +5 ö (c) Q f ( x) = sin -1 æç è x 2 + 1 ÷ø \ -1 £

| x | +5 x2 + 1

£1

6.

Þ| x | +5 £ x2 + 1

[Q x 2 + 1 ¹ 0]

x 1+ x which is not defined for x = – 1

æ 1 - 17 ö æ 1 + 17 ö Þ ç| x | ÷ ç| x | - 2 ÷ ³ 0 2 è øè ø

x 1- x which is not defined for x = 1 Thus f(x) defined for all values of R except 1 and – 1 Hence, range = (– 1, 1).

If x < 0, | x | = – x Þ f ( x ) =

æ ö 1 + 17 ö é1 + 17 Þ x Î ç -¥, Èê , ¥÷ ÷ 2 ø ë 2 è ø

3.

1 + 17 2

(b)

f ( x) =

8.

(a)

f ( x) =

\ R = {(1, 1), (2, 1), (1, - 1), (0, 1), (1, 0)}

4.

(a) f(x) =

+ log10 ( x 3 - x)

x ¹ ± 4 and - 1 < x < 0 or 1 < x < ¥

x Let, y = 1 + x2

Þ yx2 – x + y = 0 Þ x =

3 4 - x2

4 - x 2 ¹ 0; x 3 - x > 0;

x , x ÎR 1 + x2

Þ 1 – 4y2 ³ 0

1 , f (x) is define if | x | – x > 0 x -x Þ | x | > x, Þ x < 0 Hence domain of f (x) is (– ¥, 0)

7.

(d) Since, R = {(x, y) : x, y Î Z, x2 + 3 y 2 £ 8}

Þ DR-1 = {-1, 0, 1}

x (b) f ( x) = 1 + x , x Î R If x > 0, | x | = x Þ f ( x ) =

Þ x2 - | x | -4 ³ 0

\a =

(c) To determine domain, denominator ¹ 0 and x3 – x > 0 i.e., 4 – x2 ¹ 0 x ¹ ±2 ...(1) and x (x – 1) (x + 1) > 0 – + – + x Î (– 1, 0) È (1, ¥) ...(2) Hence domain is intersection of (1) & (2). i.e., x Î (–1, 0) È (1, 2) È (2, ¥)

\ R2 is not transitive. 2.

1 2

aR1b Þ a 2 + b2 = (1 + 2)2 + (1 - 2) 2 = 6 Î Q

1 ± 1- 4 y2 2

+

+ –1

0

1

\ D = ( -1, 0) È (1, ¥) -

{ 4}

D = ( -1, 0) È (1, 2) È (2, ¥).

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9.

Mathematics

(d) The given equation

2 æ 1 + x2 - 2 x ö æ 1- x ö ÷ = log çç = log 2 ç 1+ x ÷ ÷ è ø è 1+ x + 2x ø

[ x]2 + 2[ x] + 4 - 7 = 0 Þ [ x]2 + 2[ x] - 3 = 0

æ 1- x ö = 2 log ç ÷ = 2f (x) è 1+ x ø

Þ [ x]2 + 3[ x] - [ x] - 3 = 0 Þ ([ x] + 3)([ x] - 1) = 0 Þ [ x] = 1 or –3

13.

Þ x Î[ -3, - 2) È [1, 2) \ The equation has infinitely many solutions. 10.

æ a x + a- x ö æ a x - a-x ö ÷+ç ÷ f (x) = ax = çç ÷ ç ÷ 2 2 è ø è ø

(a) Let f ( x) = ax 2 + bx + c Given : f ( -1) + f (2) = 0

where f1(x) =

a - b + c + 4a + 2b + c = 0 Þ 5a + b + 2c = 0

Þb=

f2(x) =

11.

(b)

æ a x+ y + a - x - y ö æ a x - y + a - x + y ÷+ç = çç ÷ ç 2 2 è ø è

c -6 and a = 3 = 5 a

-2 Î ( -1, 0) 5

ì x ïï 2 ; x Î (1, 2) x +1 f ( x) í ï 2 x ; x Î[2,3) îï x 2 + 1

14.

1é x y a (a + a - y ) + a - x (a y + a - y ) ù û 2ë

=

(a x + a - x )(a y + a - y ) = 2f1(x) . f1(y) 2

é 1 3n ù n (d) Let f (n) = ê + ë 3 100 úû where [n] is greatest integer function,

So,

æ 2 1 ö æ 6 4ù y Îç , ÷ Èç , ú è 5 2 ø è 10 5 û

å

n =1

f (n) = 1 (23) + 1 (24) + ... + 1 (55) + 2(56)

= (23 + 24 + ... + 55) + 112 =

æ 2 1 ö æ 3 4ù Þ y Îç , ÷ È ç , ú è 5 2 ø è 5 5û

33 [46 + 32] + 112 2

33 (78) + 112 = 1399. 2 (c) Given f be an odd function f (x) = 3 sin x + 4 cos x Now,

=

æ1- x ö (a) f (x) = log ç ÷ ,|x| 0 Þ 2[sin q + cos q - 1] > 0

éæ p p öù = 4 êç cos 2 - sin 2 ÷ ú 8 8 øû ëè

1 pö æ Þ sin q + cos q > 1 Þ sin ç q + ÷ > è 4ø 2

éæ 4 p 4p 2p 2 p öù êç sin 8 + cos 8 + sin 8 cos 8 ÷ ú è øû ë

pé æ p pö ù = cos ê 4 ç1 - sin 2 cos 2 ÷ - 3ú 4ë è 8 8ø û

9.

2 sin a 1 = and 2 cos a 7

Þ

2 sin b 1 = 2 10

12.

p æ p 3p ö æ pö Î ç , ÷ Þ q Î ç 0, ÷ è 2ø 4 è4 4ø

(a) Let the height of the tower be h and distance of the foot of the tower from the point A is d. By the diagram, Q

1 - cos 2 b 1 = 2 10

B

30°

h

30 m 45°

1 1 tan a = and sin b = 7 10

tan b =

\

Þ q+

1 é 1ù 1 1- ú = ê 2 2ë û 2 2

(1)

\

(a) Let f ( x, y ) = x + y - 1 Given (1, 2) and (sin q, cos q) are lies on same side.

p p p p - 4sin 6 - 3cos 4 + 3sin 4 8 8 8 8

éæ p p öæ p p öù -3 êç cos2 - sin 2 ÷ç cos 2 + sin 2 ÷ ú 8 8 øè 8 8 øû ëè

...(ii)

pö p 1æ 1 1 1 - cos ÷ = - cos and L= ç 2è 2 8ø 2 2 2 8

1 2 x 2

pé p pù + sin 3 ê3sin - 4sin 3 ú 8ë 8 8û

=

...(i)

and L – M = - cos

3

= 4cos6

1 p p = cos = 8 4 2

p 8 From equations (i) and (ii),

x

Maximum value of sinq is 1 at q =

2 (d) L + M = 1 - 2sin

1 3

A tan 45 =

1 2 2 tan b 3 tan 2b = = 3 =3= 2 1 8 4 1 - tan b 1 9 9

P

d h =1 d

h=d

2.

tan 30 =

...(i) h - 30 d

3(h - 30) = d

...(ii)

tan a + tan 2b tan(a + 2b) = 1 - tan a tan 2b

Put the value of h from (i) to (ii),

1 3 4 + 21 + 7 4 = = 28 = 1 1 3 25 1- . 7 4 28

d=

3d = d + 30 3 30 3 = 15 3 3 -1

(

)

(

3 + 1 = 15 3 + 3

)

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Trigonometric Functions

13.

(b) cos2 10 – cos10 cos50 + cos2 50

2

12 æ5ö cos (a – b) = 1 - ç ÷ = 13 è 13 ø

æ 1 + cos 20° ö æ 1 + cos100° ö 1 =ç ÷+ç ÷ - (2cos10° cos50°) 2 2 è ø è ø 2

5 12 Now, tan 2a = tan ((a + b) + (a – b))

Þ tan (a – b) =

1 1 = 1 + (cos 20° + cos100°) - [cos60° + cos 40°] 2 2

æ 1ö 1 = ç1 - ÷ + [cos20° + cos100° - cos 40°] è 4ø 2

14.

=

3 1 + [2cos 60° ´ cos 40° - cos 40°] 4 2

=

3 4

4 5 + 3 12 = 63 tan(a + b) + tan(a - b) = = 4 5 16 1 - tan(a + b).tan(a - b) 1- . 3 12

17.

sin4 a + 4 cos4 b + 2 = 4 2 sin a × cos b, a, b Î [0, p] Then, by A.M., G.M. ineqality; A.M. ³ G.M.

(a)

5 15º

(d) Q The given equation is

sin 4 a + 4cos 4 b + 1 + 1 ³ sin 4 a × 4 cos 4 b × 1 ×1 4

(

10

)

1 4

sin4a + 4cos4b + 1 + 1 ³ 4 2 sin a× |cos b |

5

Inequality still holds when cosb < 0 but L.H.S. is positive than cosb > 0, then L.H.S. = R.H.S

15º d By the diagram, tan15 =

(

)

5 3 +1 5 5 Þd = = d tan15 3 -1

(

5 4+2 3

=

15.

\

2

Þ a=

) =5 2+ 3 ( )

\

1 sin3A 4 \ sin10 sin50 sin70 = sin10 sin(60 – 10)

1 sin30 4

1 1 sin230 = 4 16 (b) Q a + b and a – b both are acute angles.

16.

2

cos (a + b) = tan (a + b) =

3 4 æ3ö , then sin (a + b) = 1 - ç ÷ = 5 5 è5ø 4 3

And sin (a – b) =

5 , then 13

p p and b = 2 4

cos (a + b) – cos (a – b)

= –sinb – sinb = -2sin 18.

Þ sin10 sin30 sin50 sin70 =

1 4

æp ö æp ö = cos çè + b÷ø - cos çè - bø÷ 2 2

(a) Q sin(60 + A).sin(60 – A) sinA =

sin(60 + 10) =

sin4 a = 1 and cos4 b =

p =- 2 4

1 k k (a) fk(x) = (sin x + cos x ) k 1 4 4 f4(x) = [sin x + cos x] 4 1é (sin 2 x) 2 ù 2 2 2 ú = 4 ê(sin x + cos x ) 2 ë û 1 é (sin 2 x )2 ù ú = 4 ê1 2 ë û 1 6 6 f6(x) = [sin x + cos x ] 6

1é 3 2 2 2 2ù = ê(sin x + (cos x) - (sin x) ú 6ë 4 û

EBD_8344

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Mathematics

1é 3 2ù = ê1 - (sin 2 x) ú 6ë 4 û Now f4(x) – f(6)(x) = = 19.

1 1 (sin 2 x )2 1 - + (sin 2 x )2 4 6 8 8

2 ìïæ 17 1ö 17 üï 1 ³ -4 íç cos 2 x - ÷ - ý ³ è 4 4ø 16 ïþ 2 îï

1 12

(a) A = cos

p p p p .cos 3 ... cos 10 .sin 10 22 2 2 2

p p p pö 1æ = çè cos 2 .cos 3 ... cos 9 sin 9 ÷ø 2 2 2 2 2 =

1æ p pö 1 p ç cos 2 .sin 2 ÷ø = 9 sin 2 28 è 2 2 2

22.

1 512 (a) We have 5 tan2 x – 5 cos2 x = 2 (2 cos2 x –1 ) + 9 Þ 5 tan2 x – 5 cos2 x = 4 cos2 x –2 + 9 Þ 5 tan2 x = 9 cos2 x + 7 Þ 5 (sec2 x – 1) = 9 cos2 x + 7 Let cos2 x = t 5 Þ - 9t - 12 = 0 t Þ 9t2 + 12t – 5 = 0 Þ 9t2 + 15t – 3t – 5 = 0 Þ (3t – 1) (3t + 5) = 0

2 ìïæ 1ö 17 üï - 4 íç cos 2 x - ÷ - ý è 4ø 16 ïþ îï 2 0 < cos x < 1

-

1 1 3 £ cos2 x - £ 4 4 4 2

1ö 9 æ 0 £ ç cos2 x - ÷ £ 4ø 16 è

1 2 17 2 15 - = 4 4 4

(b) Let cos a + cos b =

3 2

a+b a -b 3 cos = 2 2 2

...(i)

1 2

a +b a -b 1 cos = 2 2 2 On dividing (ii) by (i), we get

Þ 2sin

...(ii)

æ a + bö 1 tan ç è 2 ÷ø = 3 a+b Þ 2q = a + b 2 Consider sin 2q + cos 2q = sin (a + b) + cos (a + b)

Given : q =

2 1 16 8 7 3 + 9 + = = = 1 1 10 10 5 1+ 1+ 9 9

2

cos2 x 1 1 ïü ïì - 4 ícos 4 x -1 + - ý 2 16 16 þï îï

m=

and sin a + sin b =

æ1ö 1 cos 2x = 2 cos2 x – 1 = 2 ç ÷ – 1 = – 3 è3ø

21.

17 4

Þ 2 cos

1 5 Þ t = as t ¹ – . 3 3

7 æ 1ö cos 4x = 2 cos2 2x – 1 = 2 ç - ÷ - 1 = 9 è 3ø 1 (b) 4 + sin2 2x – 2 cos4 x 2 4 + 2 (1 – cos2 x) cos2 x – 2 cos4 x

M=

M–m=

=

20.

2

17 æ 2 1 ö 17 9 17 £ ç cos x - ÷ - £ 16 è 4 ø 16 16 16

23.

(b) cosecq =

p-q p+q , sin q = p+q p-q 2

2 pq æ p - qö = cos q = ± 1 - sin 2 q = 1 - ç è p + q ø÷ ( p + q) æ p qö cot ç + ÷ = è 4 2ø

p q q cot cot - 1 cot - 1 4 2 2 = p q q cot + cot cot + 1 4 2 2

q q cos - sin 2 2 = q q cos + sin 2 2 On rationali ing denominator, we get

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Trigonometric Functions

q qöæ q qö æ cos - sin cos + sin ç 2 2÷ ç 2 2÷ ç q q÷ç q q÷ çè cos + sin ÷ø çè cos + sin ÷ø 2 2 2 2

=

cos q q q q 2q sin + cos + 2sin cos 2 2 2 2

(d)

pq = p

- a 2 + b2 £ a cos q + b sin q £ a 2 + b 2

q p

A = sin 2 x + cos 4 x

\ – 2 £ cos q + sin q £ 2 Hence max. value of p + q is 28.

= sin 2 x + cos 2 x(1 - sin 2 x )

cos(a + b) =

sin(a - b) =

Þ

1 1 Þ 1 + sin 2 x = 2 4

3 , 4

...(i)

3 4 1 + tan x 2 Þ 3 tan x + 8 tan x + 3 = 0 2

\ tan x = for

(a)

-8 ± 64 - 36 -4 ± 7 =6 3

p < x < p , tan x < 0 2

\ tan x =

29.

=-

-4 - 7 3

u 2 = a2 + b2 + 2

tan 2a = tan [ (a + b) + (a - b)]

26.

3 5 + tan(a + b) + tan(a - b) 56 = = 4 12 = 1 - tan(a + b ) tan(a - b ) 1 - 3 . 5 33 4 12 (b) Given that

3 2 Þ 2 [cos (b – g ) + cos ( g – a) + cos (a – b)] + 3 = 0 Þ 2 [cos (b – g ) + cos ( g – a) + cos (a – b)] + sin2 a + cos2 a + sin2 b + cos2 b + sin2 g + cos2 a = 0 2 2 2 Þ [sin a + sin b + sin g + 2 sin a sin b + 2 sin b sin g + 2 sin g sina ] + [cos2a + cos2 b + cos2 g + 2cosa cosb + 2 cos b cos g + 2cos g cos a] = 0

cos (b – g ) + cos ( g – a) + cos (a – b) = -

2

p 7 " n ³ 1

(b) an < 7 " n ³ 1

(c) an < 4 " n ³ 1

(d) an < 3 " n ³ 1

EBD_8344

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1.

2.

Mathematics

(a) Þ & \ (b)

P(n) = n2 – n + 41 P(3) = 9 – 3 + 41 = 47 (prime) P(5) = 25 – 5 + 41 = 61 (prime) P(3) and P(5) are both prime i.e., true. S(K) = 1+3+5+...+(2K – 1) = 3 + K2

Adding 2K + 1 on both sides Þ 1 + 3 + 5.... + (2 K - 1) + 2 K + 1 = 3 + K 2 + 2K + 1 = 3 + (K + 1)2 = S ( K + 1)

\ S ( K ) Þ S ( K + 1)

S (1) :1 = 3 + 1, which is not true

Q S (1) is not true. \ P.M.I cannot be applied Let S(K) is true, i.e. S(K) : 1 + 3 + 5.... + (2 K - 1) = 3 + K 2

3.

(b) For n = 1, a1 = Then am + 1 =

7 < 7. Let am < 7.

7 + am

Þ a2m + 1 = 7 + am < 7 + 7 < 14. Þ am + 1 < 14 < 7; So, by the principle of mathematical induction an < 7, " n.

www.jeebooks.in M-25

5

Complex Numbers and Quadratic Equations TOPIC Ć

1.

If

Integral Powers of lota, Algebraic Operations of Complex Numbers, Conjugate, Modulus and Argument or Amplitude of a Complex Number

[Jan. 7, 2020 (II)]

-1 æ 4 ö (a) p - tan çè ÷ø 3

(a)

4ö (d) tan çè ÷ø 3

If the four complex numbers z, z , z - 2Re( z ) and z - 2Re( z ) represent the vertices of a square of side 4 units in the Argand plane, then |z| is equal to : [Sep. 05, 2020 (I)] (c) 2 2

(b) 4

8.

(d) 2

4.

5.

æ -1 + i 3 ö The value of çç [Sep. 05, 2020 (II)] ÷÷ is : è 1- i ø (a) – 215 (b) 215 i (c) – 215 i (d) 65

æ1+ iö If ç è 1 - i ÷ø

m/2

æ1+ iö =ç è i - 1÷ø

9.

= 1, (m, n Î N ) , then the

(a)

3

(b) 2 3

(c)

3 2

(d)

1 3

10

(c)

(d)

7

8

2z - n = 2i - 1 for 2z + n [April 12, 2019 (II)]

Let z Î C with Im(z) = 10 and it satisfies

The equation z - i = z - 1 , i = -1 , represents:

1 (a) a circle of radius . 2 (b) the line through the origin with slope 1. (c) a circle of radius 1. (d) the line through the origin with slope – 1.

greatest common divisor of the least values of m and n is _________. [Sep. 03, 2020 (I)] If z1, z2 are complex numbers such that Re(z1) = |z1 – 1|,

2

(b)

[April 12, 2019 (I)]

n/3

p Re(z2) = |z2 – 1| and arg( z1 - z2 ) = , then Im( z1 + z2 ) is 6 equal to : [Sep. 03, 2020 (II)]

17 2

[Jan. 9, 2020 (II)]

some natural number n. Then : (a) n = 20 and Re(z) = –10 (b) n = 40 and Re(z) = 10 (c) n = 40 and Re(z) = –10 (d) n = 20 and Re(z) = 10

30

3.

If z be a complex number satisfying |Re(z)| + |Im(z)| = 4, then |z| cannot be:

-1 æ

(c) - tan çè ÷ø 4

(a) 4 2

7.

-1 æ 3 ö (b) p - tan çè ÷ø 4

-1 æ 3 ö

z -i =1 z + 2i

Let z be a complex number such that

5 and z = . Then the value of |z + 3i| is : 2 [Jan. 9, 2020 (I)] 7 15 (a) 10 (b) (c) (d) 2 3 2 4

3 + i sin q , q Î [0, 2p], is a real number, then an argument 4 - i cos q

of sinq + icosq is:

2.

6.

10.

If a > 0 and = equal to : 1 3 (a) - - i 5 5

(c)

1 3 - i 5 5

2 (1 + i )2 , has magnitude , then is 5 a -i [April 10, 2019 (I)]

3 1 (b) - - i 5 5 1 3 (d) - + i 5 5

EBD_8344

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11.

Mathematics

If z and w are two complex numbers such that zw = 1 and arg(z) – arg(w) =

12.

p , then: 2

(c) zw = -i

1- i (d) z w = 2

[April 09, 2019 (II)] (b) 4 Im (w) > 5 (d) 5 Im (w) < 1

z -a (a Î R ) is a purely imaginary number and | | = 2, z +a then a value of a is : [Jan. 12, 2019 (I)]

1 (d) 2 2 Let 1 and 2 be two complex numbers satisfying | 1| = 9 and | 2 | – |3|–|4i||=|4. Then the minimum value of | 1 – 2| is : [Jan. 12, 2019 (II)]

(b) 1

(b)

(c) 1

20.

21.

(d) 2

(b) | | =

(d) Im( ) = 0

Then the sum of the elements in A is: (b) p

(c)

3p 4

[Online April 11, 2015] (c) –2

(d) –5

If z is a complex number such that z ³ 2, then the minimum

1 : 2

[2014]

(a) is strictly greater than

5 2

(b) is strictly greater than

3 5 but less than 2 2

22.

5 2 (d) lie in the interval (1, 2) For all complex numbers of the form 1 + ia, a Î R , if 2 = x + iy, then [Online April 19, 2014] (a) y2 – 4x + 2 = 0 (b) y2 + 4x – 4 = 0 (c) y2 – 4x – 4 = 0 (d) y2 + 4x + 2 = 0

23.

Let ¹ – i be any complex number such that

(c) is equal to

5 2

-i is a +i

purely imaginary number.

ì æ p ö 3 + 2isin q ü is purely imaginary ý . Let A= íqÎ ç - , p ÷ : 2 1 2isin q ø î è þ

5p 6

is :

(b) –4

value of z +

[Jan. 10 2019 (II)]

(a)

(lmz )5

[Jan. 11, 2019 (II)]

3 1 2 2 that 3 | 1 | = 4 | 2 |. If = 2 + 3 then: 2 1

17.

-1 æ 1 ö (b) sin ç ÷ è 3ø

p p (d) 3 6 If is a non-real complex number, then the minimum

(a) –1

5 5 34 41 (a) (b) (c) (d) 3 4 3 4 Let 1 and 2 be any two non- ero complex numbers such

1 17 (c) | | = 2 2

2 + 3isin q is purely imaginary, is: 1 - 2isin q [2016]

lmz 5

)

(a) Re( ) = 0

(d) equal to R

A value ofqfor which

value of

-1 .

Then | | is equal to :

1 + (1 – 8a) z is a purely 1– z

(c)

(c)

2

1ü ý 4þ

-1 æ 3 ö (a) sin çç 4 ÷÷ è ø

Let be a complex number such that + = 3 + i

( where i =

16.

19.

If

(a) 0 15.

ì 1 (c) í0, , – î 4

5 + 3z Let z Î C be such that |z| < 1. If w = 5(1 - z ) , then :

(a) 2 14.

-1 + i 2

(b) z w =

The set of all a Î R, for which w =

imaginary number, for all z Î C satisfying |z| = 1 and Re z ¹ 1, is [Online April 15, 2018] (a) {0} (b) an empty set

[April 10, 2019 (II)]

(a) zw = i

(a) 5 Re (w) > 4 (c) 5 Re (w) > 1 13.

18.

[Jan. 9 2019 (I)] (d)

2p 3

Then + (a) (b) (c) (d)

1 is:

[Online April 12, 2014]

ero any non- ero real number other than 1. any non- ero real number. a purely imaginary number.

www.jeebooks.in M-27

24.

If 1, 2 and 3, 4 are 2 pairs of complex conugate numbers, then æ arg ç 1 è 4

æ 2 ö ö ÷ + arg ç ÷ equals: ø è 3 ø

25.

31.

[Online April 9, 2014]

26.

= 1}

(c)

{ : ¹ 1}

(b)

{ : = }

(d)

{ :

= 1, ¹ 1}

If z is a complex n umber of unit modulus and

æ 1+ z ö argument q, then arg ç equals: è 1 + z ÷ø (a) –q 27.

p –q 2

(b)

(c) q

(d) p – q

p 3 [Online April 25, 2013] Statement 1 is true Statement 2 is true; Statement 2 is a correct explanation for Statement 1. Statement 1 is false; Statement 2 is true Statement 1 is true, Statement 2 is false. Statement 1 is true; Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.

(b) (c) (d)

æ 1+ z Let a = Im ç ç 2iz è

2

33.

34.

[Online April 23, 2013]

36.

Z2 Z1

2Z1 + 3Z 2 is a purely imaginary number, then 2Z - 3Z is equal to: 1 2

(a) 2

(b) 5

(c) 3

[Online April 9, 2013] (d) 1

2

(b) 2 z1 + z2 2

z1 z2

2

)

2

Statement 2: Z1 + Z 2 £ Z1 + Z 2 [Online May 7, 2012] (a) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1. (b) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1. (c) Statement 1 is true, Statement 2 is false. (d) Statement 1 is false, Statement 2 is true. The number of complex numbers such that |z – 1| = |z + 1| = |z – i| equals [2010] (a) 1 (b) 2 (c) ¥ (d) 0 The conugate of a complex number is

–1 i –1

(b)

1 i +1

(c)

–1 i +1

1

ö ÷ , where z is any non- ero complex ÷ ø

If Z1 ¹ 0 and Z2 be two complex numbers such that

(

)

Statement 2: |Z| = |W|, implies arg Z – arg W = p. (a) Statement 1 is true, Statement 2 is false. (b) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1. (c) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1. (d) Statement 1 is false, Statement 2 is true. Let Z1 and Z2 be any two complex number.

(a)

The set A = {a : | z | = 1 and z ¹ ±1 } is equal to: (a) (– 1, 1) (b) [– 1, 1] (c) [0, 1) (d) (– 1, 0] 29.

is equal to [Online May 26, 2012]

complex number is

35.

number.

2

Statement 1: Z1 - Z 2 ³ Z1 - Z 2

Let z satisfy| z | = 1 and = 1– z . Statement 1 : z is a real number.

(a)

28.

32.

[2013]

Statement 2 : Principal argument of is

+ z1 - z2

Statement 1:If arg Z + arg W = p, then Z = -W .

w - w = k (1 - ) , for some real number k, is

{ :

2

(d) z1 + z2 Let Z and W be complex numbers such that |Z| = |W|, and arg Z denotes the principal argument of Z. [Online May 19, 2012] (c)

(b)

(a)

z1 + z2

(a) 2 ( z1 + z2

[Online April 11, 2014]

p 3p (c) (d) p 2 2 Let w (Im w ¹ 0) be a complex number. Then the set of all complex number satisfying the equation

(a) 0

30.

1 then that i –1 [2008]

(d) æx

1 i –1

If z = x - i y and z 3 = p + iq, then çè p + q ø÷ ( p + q ) is equal to [2004] (a) –2 (b) –1 (c) 2 (d) 1 Let and w be complex numbers such that z + i w = 0 and arg zw = p. Then arg equals [2004] (a)

5p 4

(b)

p 2

(c)

3p 4

(d)

2

2

p 4

x

37.

æ1+ i ö ÷ = 1 then è1- i ø

If ç (a) (b) (c) (d)

x = 2n + 1 , where n is any positive integer x = 4n , where n is any positive integer x = 2n , where n is any positive integer x = 4n + 1 , where n is any positive integer..

[2003]

EBD_8344

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38.

Mathematics

If z and w are two non- ero complex numbers such that

p zw = 1 and Arg ( z ) - Arg (w ) = , then zw is equal to 2 [2003] (a) – 1 (b) 1 (c) – i (d) i 39. If | – 4 | < | – 2 |, its solution is given by [2002] (a) Re(z) > 0 (b) Re(z) < 0 (c) Re(z) > 3 (d) Re(z) > 2 40. z and w are two non ero complex numbers such that | z | = | w| and Arg + Arg w = p then equals [2002] (a) w (b) – w (c) w (d) – w

TOPIC n

41.

Rotational Theorem, Square Root of a Complex Number, Cube Roots of Unity, Geometry of Complex Numbers, De-moiver’s Theorem, Powers of Complex Numbers

(a) line, y = –x (c) line, y = x

2 (b) straight line whose slope is - . 3

(c) straight line whose slope is

(d) circle whose diameter is 47.

-1 , then z lies on the: [Sep. 06, 2020 (II)] (b) imaginary axis (d) real axis

-1 + i 3 , then a + b is equal to : 2 [Sep. 04, 2020 (II)] (b) 24 (c) 33 (d) 57

2p 2p ö æ 1 + sin + i cos ç 9 9 ÷ is : The value of ç 2p 2p ÷ - i cos ÷ ç 1 + sin è 9 9 ø

1 (1 - i 3) 2

(b)

1 (c) - ( 3 - i ) 2

44.

1 ( 3 - i) 2

1 (d) - (1 - i 3) 2

(b) -2 6

(c) 6 100

45.

50.

51.

The imaginary part of (3 + 2 -54)1/ 2 - (3 - 2 -54)1/ 2 can be : [Sep. 02, 2020 (II)] (a) - 6

(d)

6 100

2k 3k -1 + i 3 . If a = (1 + a) å a and b = å a , 2 k =0 k =0 then a and b are the roots of the quadratic equation: [Jan. 8, 2020 (II)] (a) x2 + 101x + 100 = 0 (b) x2 – 102 x + 101 = 0 (c) x2 – 101x + 100 = 0 (d) x2 + 102x +101 = 0

Let a =

x + iy 1 ö æ i = -1 , where x and y are real Let ç -2 - i ÷ = 3 ø 27 è numbers then y – x equals : [Jan. 11, 2019 (I)] (a) 91 (b) – 85 (c) 85 (d) – 91

(

)

5

æ 3 iö æ 3 iö Let z = çç + ÷÷ + çç - ÷÷ . If R(z) and I(z) 2ø è 2 2ø è 2 respectively denote the real and imaginary parts of z, then: [Jan. 10, 2019 (II)] (a) I(z) = 0 (b) R(z) > 0 and I(z) > 0 (c) R(z) < 0 and I(z) > 0 (d) R(z) = – (c) n

[Sep. 02, 2020 (I)] (a)

5 . 2

5

49.

3

43.

3 . 2

If z = 3 + i ( i = -1 ) , then (1 + iz + z5 + iz8)9 is equal 2 2 to: [April 08, 2019 (II)] (a) 0 (b) 1 (c) (– 1 + 2i)9 (d) – 1 3

48.

If a and b are real numbers such that (2 + a ) = a + ba ,

(a) 9

[Jan. 7, 2020 (I)]

æ 1 3ö (a) circle whose centre is at èç - , - ø÷ . 2 2

4

where a =

æ z -1 ö If Re çè ÷ = 1, where z = x + iy, then the point (x, y) lies 2z + i ø on a:

Let z = x + iy be a non- ero complex number such that z2 = i |z|2, where i =

42.

46.

52.

æ1+ i 3 ö The least positive integer n for which çç ÷÷ = 1, is è1 – i 3 ø

[Online April 16, 2018] (a) 2 (b) 6 (c) 5 (d) 3 The point represented by 2 + i in the Argand plane moves 1 unit eastwards, then 2 units northwards and finally from there 2 2 units in the south–westwards direction. Then its new position in the Argand plane is at the point represented by : [Online April 9, 2016] (a) 1 + i (b) 2 + 2i (c) –2 – 2i (d) –1 – i A complex number is said to be unimodular if | | = 1. 1 - 2 2 Suppose 1 and 2 are complex numbers such that 2 - 1 2

is unimodular and 2 is not unimodular. Then the point 1 lies on a: [2015]

www.jeebooks.in M-29

(a) circle of radius 2. (b) circle of radius 2. (c) straight line parallel to x-axis (d) straight line parallel to y-axis. 53.

TOPIC Đ

2 If z ¹ 1 and z is real, then the point represented by the

z -1

complex number z lies : [2012] (a) either on the real axis or on a circle passing through the origin. (b) on a circle with centre at the origin (c) either on the real axis or on a circle not passing through the origin. (d) on the imaginary axis. 54.

If w( ¹ 1) is a cube root of unity, and (1 + w )7 = A + Bw.

55.

Then (A, B) equals [2011] (a) (1, 1) (b) (1, 0) (c) (–1, 1) (d) (0, 1) If | z + 4 | £ 3, then th e maximum value of [2007] | z + 1 | is (a) 6 (b) 0 (c) 4 (d) 10

56.

If w =

z 1 z- i 3

and | w | = 1, then lies on

p 2

(b) – p

(c) 0

(d)

(c) – 1, 1 – 2 w , 1 – 2 w

2

(d) – 1, 1 + 2 w , 1 + 2 w

2

1

(a) 2 (b) 3 (c) 1 (d) 4 If a and b are the roots of the equation 2x(2x + 1) = 1, then b is equal to: [Sep. 06, 2020 (II)] (a) 2a(a + 1) (b) –2a(a + 1) (c) 2a(a – 1) (d) 2a2 The product of the roots of the equation 9x2 – 18| x | + 5 = 0, is : [Sep. 05, 2020 (I)] (a)

64.

5 9

(b)

25 81

(c)

5 27

(d)

25 9

If a and b are the roots of the equation, 7 x 2 - 3 x - 2 = 0, a 1- a

2

+

b 1 - b2

is equal to : [Sep. 05, 2020 (II)]

-p 2

(a) 65.

[2005]

(a) –1, –1 + 2 w , – 1 – 2 w2 (b) –1, – 1, – 1

60.

63.

If the cube roots of unity are 1, w , w 2 then the roots of the equation ( x –1)3 + 8 = 0, are

59.

62.

(b) a circle (d) a parabola

| z1 + z2 | = | z1 | + | z2 | , then arg z1 – arg z2 is equal to [2005]

58.

1

æ a 3 ö 8 æ b3 ö 8 Then the value of ç ÷ + ç ÷ is: [Sep. 06, 2020 (I)] è b5 ø è a5 ø

If z1 and z2 are two non- ero complex numbers such that

(a)

If a and b be two roots of the equation x2 – 64x + 256 = 0.

the the value of

(a) an ellipse (c) a straight line 57.

[2005]

61.

Solutions of Quadratic Equations, Sum and Product of Roots, Nature of Roots, Relation Between Roots and Co-efficients, Formation of an Equation with Given Roots.

If | z 2 - 1|=| z |2 +1, then lies on [2004] (a) an ellipse (b) the imaginary axis (c) a circle (d) the real axis The locus of the centre of a circle which touches the circle | z – z1 | = a and | – 2 | = b externally (z, z1 & z2 are complex numbers) will be [2002] (a) an ellipse (b) a hyperbola (c) a circle (d) none of these

66.

27 32

(b)

1 24

(c)

3 8

(d)

27 16

2z + i , z = x + iy and k > 0. If the curve z - ki represented by Re(u) + Im(u) = 1 intersects the y-axis at the points P and Q where PQ = 5, then the value of k is : [Sep. 04, 2020 (I)] (a) 3/2 (b) 1/2 (c) 4 (d) 2

Let u =

Let l ¹ 0 be in R. If a and b are roots of the equation, x 2 - x + 2l = 0 and a and g are the roots of the equation, bg is equal to : l [Sep. 04, 2020 (II)] (a) 27 (b) 18 (c) 9 (d) 36 If a and b are the roots of the equation x2 + px + 2 = 0 and 3 x 2 - 10 x + 27 l = 0, then

67.

1 1 and are the roots of the equation 2x2 + 2qx + 1 = 0, b a 1 öæ 1 öæ 1 öæ 1ö æ then ç a - ÷ ç b - ÷ ç a + ÷ ç b + ÷ is equal to : a øè b øè b øè aø è

[Sep. 03, 2020 (I)]

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Mathematics

(a)

9 (9 + q 2 ) 4

(b)

9 (9 - q 2 ) 4

9 9 (9 + p 2 ) (9 - p 2 ) (d) 4 4 The set of all real values of l for which the quadratic

77.

(c)

68.

78.

equations, (l 2 + 1) x 2 - 4lx + 2 = 0 always have exactly one root in the interval (0, 1) is : [Sep. 03, 2020 (II)] (a) (0, 2) (b) (2, 4] (c) (1, 3] (d) (–3, –1) 69.

Let a and b be the roots of the equation, 5 x 2 + 6 x - 2 = 0.

79.

Let p, q Î R. If 2 – 3 is a root of the quadratic equation, x2 + px + q = 0, then: [April 9, 2019 (I)] (a) p2 – 4q + 12 = 0 (b) q2 – 4p – 16 = 0 (c) q2 + 4p + 14 = 0 (d) p2 – 4q – 12 = 0 If m is chosen in the quadratic equation (m2 + 1) x2 – 3x + (m2 + 1)2 = 0 such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is: [April 09, 2019 (II)] (a) 10 5 (b) 8 3 (c) 8 5 (d) 4 3 The sum of the solutions of the equation

If Sn = an + bn , n = 1, 2,3, ..., then : [Sep. 02, 2020 (I)] (a) 6S6 + 5S5 = 2S4 70.

71.

x -2 + x

(b) 6S6 + 5S5 + 2S4 = 0

(c) 5S6 + 6S5 = 2S4 (d) 5S6 + 6S5 + 2S4 = 0 The number of real roots of the equation, e4x + e3x – 4e2x + ex + 1 = 0 is: [Jan. 9, 2020 (I)] (a) 1 (b) 3 (c) 2 (d) 4 The least positive value of ‘a’ for which the equation, 33 = 2a has real roots is _______. 2 [Jan. 8, 2020 (I)] If the equation, x2 + bx + 45 = 0 (b Î R) has conugate

80.

73.

74.

complex roots and they satisfy |z + l| = 2 10 , then: [Jan. 8, 2020 (I)] (b) b2 + b = 72 (a) b2 – b = 30 (c) b2–b = 42 (d) b2 + b = 12 Let a and b be the roots of the equation x2 – x – l = 0. If pk = (a)k + (b)k, k ³ l, then which one of the following statements is not true ? [Jan. 7, 2020 (II)] (a) p3 = p5 – p4 (b) P5 = 11 (c) (p1 + p2 + p3 + p4 + p5) = 26 (d) p5 = p2 × p3 Let a and b be two real roots of the equation (k +1) tan2x

81.

12

(a)

82.

83.

(b)

2 (sin q + 8)12

212 26 (d) 6 (sin q - 8) (sin q + 8)12 The number of real roots of the equation (c)

76.

5 + 2 x - 1 = 2 x (2 x - 2) is:

(a) 3

84.

(b) 2

[April 10, 2019 (II)] (c) 4

(d) 1

1 = 1, is : l

[Jan. 12, 2019 (I)] (b) 4 - 3 2

(c) –2 + 2 (d) 4 - 2 3 If one real root of the quadratic equation 81x2 + kx + 256 = 0 is cube of the other root, then a value of k is : [Jan. 11, 2019 (I)] (a) –81 (b) 100 (c) 144 (d) – 300 Consider the quadratic equation (c – 5)x2 – 2cx + (c – 4) = 0, c ¹ 5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is: [Jan. 10, 2019 (I)] (a) 18 (b) 12 (c) 10 (d) 11 The value of l such that sum of the squares of the roots of the quadratic equation, x2 + (3 – l)x + 2 = l has the least value is: [Jan. 10, 2019 (II)]

15 4 (b) 1 (c) (d) 2 8 9 Let a and b be two roots of the equation x2 + 2x + 2 = 0, then a15 + b15 is equal to: [Jan. 9, 2019 (I)] (a) – 256 (b) 512 (c) – 512 (d) 256 The number of all possible positive integral values of a for which the roots of the quadratic equation, 6x2 – 11x + a = 0 are rational numbers is: [Jan. 09, 2019 (II)] (a) 3 (b) 2 (c) 4 (d) 5 (a)

12

2 (sin q - 4)12

æaö the least value of n for which ç ÷ = 1 is : èbø [April 8, 2019 (I)] (a) 2 (b) 5 (c) 4 (d) 3 If l be the ratio of the roots of the quadratic equation in x, 3m2x2 + m(m – 4)x + 2 = 0, then the least value of m for

(a) 2 - 3

(a) 10 2 (b) 10 (c) 5 (d) 5 2 If a and b are the roots of the quadratic equation, x2 + x sin q

a12 + b12 æ pö – 2sinq = 0 0, qÎ ç 0, ÷ , then is (a -12 + b-12 )(a - b)24 è 2ø equal to : [April 10, 2019 (I)]

[April 8, 2019 (I)] (a) 9 (b) 12 (c) 4 (d) 10 If a and b be the roots of the equation x2 – 2x + 2 = 0, then

which l +

– 2 . ltan x = (1 – k), where k(¹ –1) and l are real numbers. If tan2(a + b) = 50, then a value of l is: [Jan. 7, 2020 (I)] 75.

)

x - 4 + 2 = 0, (x > 0) is equal to:

n

2x2 + (a – 10)x + 72.

(

85.

86.

www.jeebooks.in M-31

87.

88.

If both the roots of the quadratic equation x2 – mx + 4 = 0 are real and distinct and they lie in the interval [1, 5], then m lies in the interval: [Jan. 09, 2019 (II)] (a) (– 5, – 4) (b) (4, 5) (c) (5, 6) (d) (3, 4) Let z0 be a root of the quadratic equation, x2 + x + 1 = 0. If z = 3 + 6i z081 – 3i z093, then arg z is equal to: [Jan. 09, 2019 (II)]

97.

2 (x 2 - 5x + 5) x + 4x -60 = 1 is :

98.

1 1 1 + = are equal in x+ p x+q r

magnitude but opposite in sign, then the sum of squares of these roots is equal to. [Online April 16, 2018] (b) p2 + q2 (a) p2 + q2 + r2

p2 + q2 (c) (d) 2 If an angle A of a D ABC satisfies 5 cos A + 3 = 0, then the roots of the quadratic equation, 9x2 + 27x + 20 = 0 are. [Online April 16, 2018] (a) sin A, sec A (b) sec A, tan A (c) tan A, cos A (d) sec A, cot A If tan A and tan B are the roots of the quadratic equation, 3x2 – 10x – 25 = 0 then the value of 3 sin2 (A + B) – 10 sin (A + B). cos (A + B) – 25 cos2 (A + B) is [Online April 15, 2018] (a) 25 (b) – 25 (c) – 10 (d) 10 If f (x) is a quadratic expression such that f (a) + f (b) = 0, and – 1 is a root of f (x) = 0, then the other root of f (x) = 0 is [Online April 15, 2018] 2 (p2 + q2)

90.

91.

92.

(a) 93.

5 8

(b) -

8 5

(c)

5 8

(d)

8 5

[2018]

94.

(a) 0 (b) 1 (c) 2 (d) – 1 If, for a positive integer n, the quadratic equation,

95.

x(x + 1) + (x + 1) (x + 2) + ..... + (x + n - 1 ) (x + n) = 10n has two consecutive integral solutions, then n is equal to: [2017] (a) 11 (b) 12 (c) 9 (d) 10 The sum of all the real values of x satisfying the equation

(

96.

)

(a) 6 (b) 5 (c) 3 If x is a solution of the equation,

(d) – 4

3 1 (b) (c) 2 2 (d) 2 4 2 Let a and b be the roots of equation x2 – 6x – 2 = 0. If an =

(a) 99.

an – bn, for n ³ 1, then the value of

a10 - 2a 8 is equal to: 2a 9

[2015] (a) 3 (b) – 3 (c) 6 (d) – 6 100. If the two roots of the equation, (a – 1)(x4 + x2 + 1) + (a + 1)(x2 + x + 1)2 = 0 are real and distinct, then the set of all values of ‘a’ is : [Online April 11, 2015]

æ 1ö (a) ç 0, ÷ è 2ø

æ 1 ö æ 1ö (b) ç - , 0 ÷ È ç 0, ÷ è 2 ø è 2ø

æ 1 ö (c) ç - , 0 ÷ (d) (– ¥, –2) È (2, ¥) è 2 ø 101. If 2 + 3i is one of the roots of the equation 2x3 – 9x2 + kx – 13 = 0, k Î R, then the real root of this equation : [Online April 10, 2015] (a) exists and is equal to –

1 . 2

1 . 2 (c) exists and is equal to 1. (d) does not exist. 102. If a Î R and the equation

(b) exists and is equal to

If a, b Î C are the distinct roots, of the equation x 2 - x + 1 = 0 , then a101 + b107 is equal to :

[2016]

1ö æ 2x + 1 - 2x - 1 = 1, ç x ³ ÷ , then 4x 2 - 1 is equal è 2ø to : [Online April 10, 2016]

p p p (a) (b) (c) (d) 0 4 6 3 89. Let p, q and r be real numbers (p ¹ q, r ¹ 0), such that the roots of the equation

(a) p(b) = 11 (b) p(b) = 19 (c) p(–2) = 19 (d) p(–2) = 11 The sum of all real values of x satisfying the equation

( x -1) x 2 +5x -50 2 = 1 is : [Online April 9, 2017] (a) 16 (b) 14 (c) – 4 (d) – 5 Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x–1 and it leaves remainder 6 when divided by x+1; then : [Online April 8, 2017]

- 3 ( x - [ x ]) + 2 ( x - [ x ]) + a 2 = 0 2

(where [x] denotes the greatest integer £ x ) has no integral solution, then all possible values of a lie in the interval: [2014] (a)

( -2, -1)

(b)

( -¥, -2 ) È ( 2, ¥ )

(c)

( -1, 0 ) È ( 0,1)

(d)

(1, 2 )

3x 2 + x + 5 = x - 3 , where x is real, has; [Online April 19, 2014] (a) no solution (b) exactly one solution (c) exactly two solution (d) exactly four solution

103. The equation

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104. The sum of the roots of the equation, x2 + |2x – 3| – 4 = 0, is: [Online April 12, 2014] (d) - 2

(a) 2 (b) – 2 (c) 2 105. If a and b are roots of the equation,

x 2 - 4 2 kx + 2e 4 lnk - 1 = 0 for some k, and a2 + b2 = 66, then a3 + b3 is equal to: [Online April 11, 2014] = + + ( )= + (a) 248 2 (b) 280 2 (c) -32 2 (d) -280 2

1

1

106. If

and

a

b

are the roots of the equation,

ax2 + bx + 1 = 0 (a ¹ 0, a, b, Î R), then the equation,

(

) (

)

x x + b3 + a 3 - 3abx = 0 as roots : [Online April 9, 2014] 3

(a) a

3

(c)

ab and ab

2

and b

2

(b) ab (d)

a

-

1

3 2

2

and a

1

2b

3

and b- 2

107. If p an d q are non- ero real numbers and a3 + b3 = - p, ab = q, then a quadratic equation whose

roots are

a 2 b2 is : , b a

(a) px2 – qx + p2 = 0 (c) px2 + qx + p2 = 0

[Online April 25, 2013] (b) qx2 + px + q2 = 0 (d) qx2 – px + q2 = 0

108. If a and b are roots of the equation x 2 + px +

3p = 0, 4

such that | a - b |= 10, then p belongs to the set : [Online April 22, 2013] (a) {2, – 5} (b) {– 3, 2} (c) {– 2, 5} (d) {3, – 5} 109. If a complex number z statisfies the equation

z + 2 | z + 1| +i = 0 , then | z | is equal to : [Online April 22, 2013] (a) 2 (b) 3 (c) 5 (d) 1 110. Let p, q, r Î R and r > p > 0. If the quadratic equation px2 + qx + r = 0 has two complex roots a and b, then |a| + |b| is [Online May 19, 2012] (a) equal to1 (b) less than 2 but not equal to 1 (c) greater than 2 (d) equal to 2 111. If the sum of the square of the roots of the equation x2 – (sina – 2) x – (1 + sina) = 0 is least, then a is equal to [Online May 12, 2012] (a)

p 6

(b)

p 4

(c)

p 3

(d)

p 2

112. The value of k for which the equation (k – 2)x2 + 8x + k + 4 = 0 has both roots real, distinct and negative is [Online May 7, 2012] (a) 6 (b) 3 (c) 4 (d) 1 113. Let for a ¹ a1 ¹ 0, f ( x ) = ax 2 + bx + c, g ( x) = a1 x 2 + b1 x + c1an 1

=

+ c1and p ( x ) = f ( x ) - g ( x ) .

If p ( x ) = 0 only for x = -1 and p (– 2) = 2, then the value of p (b) is : [2011 RS] (a) 3 (b) 9 (c) 6 (d) 18 114. Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4,3). Rahul made a mistake in writing down coefficient of x to get roots (3,2). The correct roots of equation are : [2011 RS] (a) 6, 1 (b) 4, 3 (c) – 6, – 1 (d) – 4, – 3 115. Let a, b be real and z be a complex number. If z2 + az + b = 0 has two distinct roots on the line Re z =1, then it is necessary that : [2011] (b) b = 1

(a) b Î (-1, 0)

(c) b Î (1, ¥) (d) b Î (0,1) 116. If a and b are the roots of the equation x2 – x + 1 = 0, then a2009 + b2009 = [2010] (a) –1 (b) 1 (c) 2 (d) –2 117. If the roots of the equation bx2 + cx + a = 0 be imaginary, then for all real values of x, the expression 3b2x2 + 6bcx + 2c2 is : [2009] (a) less than 4ab (b) greater than – 4ab (c) 1ess than – 4ab (d) greater than 4ab 118. If the difference between the roots of the equation x2 + ax + 1 = 0 is less than values of a is

5 , then the set of possible [2007]

(a) (3, ¥) (b) (- ¥, - 3) (c) (– 3, 3) (d) (-3, ¥) 119. All the values of m for which both roots of the equation x 2 - 2mx + m 2 - 1 = 0 are greater than – 2 but less than 4, lie in the interval [2006] (a) -2 < m < 0 (b) m > 3 (c) -1 < m < 3 (d) 1 < m < 4 120. If the roots of the quadratic equation

x 2 + px + q = 0 are tan30 and tan15 , respectively, then the value of 2 + q – p is (a) 2 (b) 3 (c) 0 (d) 1

[2006]

121. If z 2 + z + 1 = 0 , where is complex number, then the value 2

2

of

1ö æ 2 1ö æ 3 1ö æ çè z + ÷ø + çè z + 2 ÷ø + çè z + 3 ÷ø z z z

2

+ ......... + æç z 6 + è

1ö ÷ z6 ø

2

is

[2006] (a) 18

(b) 54

(c) 6

(d) 12

-

www.jeebooks.in M-33

p æ Pö 122. In a triangle PQR, Ð R = . If tan ç ÷ and è 2ø 2

æ Qö – tan ç ÷ are the roots of ax 2 + bx + c = 0, a ¹ 0 then è 2ø [2005] (a) a = b + c (c) b = c

(b) c = a + b (d) b = a + c

131. Difference between the corresponding roots of x2+ax+b=0 [2002] and x2+bx+a=0 is same and a ¹ b, then (a) a + b + 4 = 0 (b) a + b – 4 = 0 (c) a – b – 4 = 0 (d) a – b + 4 = 0 132. If a ¹ b but a2 = 5a – 3 and b2 = 5b – 3 then the equation having a/b and b/a as its roots is [2002] (a) 3x2 – 19x + 3 = 0 (b) 3x2 + 19x – 3 = 0 (d) x2 – 5x + 3 = 0. (c) 3x2 – 19x – 3 = 0

123. If the roots of the equation x 2 – bx + c = 0 be two consecutive integers, then b 2 – 4c equals (a) – 2 (b) 3 (c) 2 (d) 1

[2005]

124. If one root of the equation x 2 + px + 12 = 0 is 4, while the equation x 2 + px + q = 0 has equal roots , then the value of ‘q’ is [2004] (a) 4

(b) 12

(1 - p)

125. If

(c) 3

(d)

is a root of quadratic equation [2004]

(a) –1, 2 (b) –1, 1 (c) 0, –1 (d) 0, 1 126. The number of real solutions of the equation x 2 - 3 x + 2 = 0 is

[2003]

(a) 3 (b) 2 (c) 4 (d) 1 127. The value of 'a' for which one root of the quadratic equation 2

2

(a - 5a + 3) x + (3a - 1) x + 2 = 0 is twice as large as the other is [2003]

(a) -

1 3

(b)

2 3

(c) -

2 3

(d)

1 3

128. Let Z1 and Z 2 be two roots of the equation Z 2 + aZ + b = 0 , Z being complex. Further , assume that

the origin, Z1 and Z 2 form an equilateral triangle. Then [2003] (a) a 2 = 4b

133. If 5, 5r, 5r2 are the lengths of the sides of a triangle, then r cannot be equal to: [Jan. 10, 2019 (I)]

3 5 7 3 (b) (c) (d) 4 4 4 2 134. Let a, b Î R, a ¹ 0 be such that the equation, ax2 – 2bx + 5 = 0 has a repeated root a, which is also a root of the equation, x2 – 2bx – 10 = 0. If b is the other root of this equation, then a2 + b2 is equal to :[Jan. 9, 2020 (II)] (a) 25 (b) 26 (c) 28 (d) 24 135. If lÎ R is such that the sum of the cubes of the roots of the equation, x2 + (2 – l) x + (10 – l) = 0 is minimum, then the magnitude of the difference of the roots of this equation is [Online April 15, 2018] (a)

49 4

x 2 + px + (1 - p) = 0 then its root are

TOPIC Ė

Condition for Common Roots, Maximum and Minimum value of Quadratic Equation, Quadratic Expression in two Variables, Solution of Quadratic Inequalities.

(a) 20 (b) 2 5 (c) 2 7 (d) 4 2 136. If | z – 3 + 2i | £ 4 then the difference between the greatest value and the least value of | z | is [Online April 15, 2018] (a)

(b) 2 13

(d) 4 + 13

(c) 8

(a) 2 (b) 3 (c) 3 (d) 2 138. If non- ero real numbers b and c are such that min f(x) > max g(x), where f(x) = x2 + 2bx + 2c2 and g(x) = – x2 – 2cx + b2 (x Î R);

(b) a 2 = b

(c) a 2 = 2b (d) a 2 = 3b 129. If p and q are the roots of the equation x2 + px + q = 0, then [2002] (a) p = 1, q = –2 (b) p = 0, q = 1 (c) p = –2, q = 0 (d) p = – 2, q = 1 130. Product of real roots of the equation t2 x2 + | x | + 9 = 0 [2002] (a) is always positive (b) is always negative (c) does not exist (d) none of these

13

137. If the equations x2 + bx – 1 = 0 and x2 + x + b = 0 have a common root different from –1, then |b| is equal to : [Online April 9, 2016]

then

c lies in the interval: b

[Online April 19, 2014]

æ 1ö (a) ç 0, ÷ è 2ø

é1 1 ö (b) ê , ÷ ë2 2 ø

é 1 ù 2ú (c) ê ë 2, û

(d)

(

2, ¥

)

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Mathematics

139. If equations ax2 + bx + c = 0 ( a, b, c Î R, a ¹ 0 ) and 2x2 + 3x + 4 = 0 have a common root, then a : b : c equals: [Online April 9, 2014] (a) 1 : 2 : 3 (b) 2 : 3 : 4 (c) 4 : 3 : 2 (d) 3 : 2 : 1 140. If the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0, a,b,c Î R, have a common root, then a : b : c is [2013] (a) 1 : 2 : 3 (b) 3 : 2 : 1 (c) 1 : 3 : 2 (d) 3 : 1 : 2 141. The least integral value a of x such that

x-5 x 2 + 5x - 14

143. If z -

(a) 1

(d) a £ -2

4 = 2 , then the maximum value of | | is equal to : z

(b) 4

(c) 3

145. If x is real, the maximum value of

>0,

satisfies : [Online April 23, 2013] (a) a2 + 3a – 4 = 0 (b) a2 – 5a + 4 = 0 (c) a2 – 7a + 6 = 0 (d) a2 + 5a – 6 = 0 142. The values of ‘a’ for which one root of the equation x2 – (a + 1 ) x + a2 + a – 8 = 0 exceeds 2 and the other is lesser than 2, are given by : [Online April 9, 2013] (a) 3 < a < 10 (b) a ³ 10 (c) -2 < a < 3

144. The quadratic equations x 2 – 6x + a = 0 and x2 – cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is [2009]

(a)

1 4

(b) 41

(d) 2 3x 2 + 9 x + 17 3x 2 + 9 x + 7

(c) 1

5 + 1 (b) 2

(c) 2 + 2

(d)

17 7

2 2 146. If both the roots of the quadratic equation x - 2 kx + k + k – 5 = 0 are less than 5, then k lies in the interval

[2005] (a) (5, 6]

(b) (6, ¥ )

(c) (– ¥ , 4) (d) [4, 5]

147. The value of a for which the sum of the squares of the roots of the equation x2 – (a – 2) x – a – 1 = 0 assume the least value is [2005] (a) 1 (b) 0 (c) 3 (d) 2

[2009] (a)

(d)

is [2006]

3 +1

www.jeebooks.in M-35

1.

(b) Let z = z=

3 + i sin q , after rationalising 4 - i cos q

æ (1 + i )2 ö Þç ÷ è 2 ø

(3 + i sin q) (4 + i cos q) ´ (4 - i cos q) (4 + i cos q)

3 4

5.

D(z–2Re(z))

(b) Let z1 = x1 + iy1 and z2 = x2 + iy2

Þ ( x1 - 1) 2 + y12 = x12

æ 4ö æ4ö = p + tan -1 ç - ÷ = p - tan -1 ç ÷ è 3ø è3ø (c)

=1

Q | z1 - 1 | = Re( z1 )

æ cos q ö arg(sinq + icosq) = p + tan–1 ç sin q ÷ è ø

2.

n /3

m (least) = 8, n (least) = 12 GCD (8, 12) = 4.

Þ tanq = –

3cosq + 4sinq = 0

æ (1 + i )2 ö =ç ÷ è -2 ø

Þ i m / 2 = ( -i ) n / 3 = 1

As z is purely real Þ

m/2

Þ y12 - 2 x1 + 1 = 0

...(i)

| z2 - 1|= Re( z2 ) Þ ( x2 - 1)2 + y22 = x22

C(z –2Re(z ))

Þ y22 - 2 x2 + 1 = 0

...(ii)

From eqn. (i) – (ii),

y12 - y22 - 2( x1 - x2 ) = 0 æ x -x ö Þ y1 + y2 = 2 ç 1 2 ÷ è y1 - y2 ø

A(z)

Q arg( z1 - z2 ) =

B(z)

Then, | z - z | = 4 Þ | 2iy | = 4 Þ | y | = 2 Also, | z - ( z - 2 Re( z )) | = 4

Þ

y1 - y2 1 = x1 - x2 3

Þ

2 1 = y1 + y2 3

Þ | 2 Re( z ) | = 4 Þ | 2 x | = 4 Þ | x | = 2 \| z | = x + y = 4+ 4 = 2 2 2

2p

3.

30

æ -1 + 3i ö \ çç ÷÷ è 1- i ø

30

æ 2p p ö ö æ ç + ÷i = ç 2eè 3 4 ø ÷ ç ÷ è ø

=2

15

4.

i

(c) Q -1 + 3i = 2 × e 3 and 1 - i = 2 × e

p - i ×e 2

m/2

ip 4

é y1 - y2 2 ù = ê From, ú x1 - x2 y1 + y2 û ë

\ y1 + y2 = 2 3 Þ lm( z1 + z2 ) = 2 3 6.

(b) Let z = x + iy

z -i = 1 Þ x2 + (y – 1)2 z + 2i

= x2 + (y + 2)2 Þ –2y + 1 = 4y + 4

= -2 × i.

æ 1+ i ö =ç è i - 1÷ø

-

Then,

15

(4)

æ 1+ i ö Given that ç è 1 - i ÷ø

p 6

æ y - y2 ö p Þ tan -1 ç 1 = è x1 - x2 ÷ø 6

Let z = x + iy Q Length of side of square = 4 units

2

Þ

6y = – 3 Þ y = -

Q

|z| =

n /3

=1

...(iii)

1 2

5 25 Þ x2 + y2 = 2 4

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x2 =

\

z = x + iy

| + 3i| =

7.

24 =6 4

Þ

Since, it is given that | z | = Þ z= ±

6-

i 2

Then, from equation (i), 2 2 = 5 1 + a2

25 49 6+ = 4 4

Now, square on both side; we get

7 Þ |z + 3i| = 2 (c) z = x + iy |x| + |y| = 4

Þ 1 + a2 = 10 Þ a = ± 3 Since, it is given that a > 0 Þ a = 3 Then, z =

z = x2 + y 2

=

Minimum value of |z| = 2 2 Maximum value of |z| = 4

11.

So, | | can’t be 7 . (c) Let Re (z) = x i.e., z = x + 10i

2i (3 + i ) -1 + 3i = 10 5

(2x – n) + 20i = (2i – 1) ((2x + n) + 20i)

-1 3 - i 5 5

(c) Given | zw | = 1

...(i)

æzö p and arg ç ÷ = èwø 2

...(ii)

\

2z – n = (2i – 1) (2z + n)

é ù æzö êQ Re ç w ÷ = 0 ú è ø ë û

z z + =0 w w

Þ zw = - z w

On comparing real and imaginary parts, – (2x + n) – 40 = 2x – n and 20 = 4x + 2n – 20

from equation (i), zzww = 1 [using zz =| z |2 ]

Þ 4x = – 40 and 40 = – 40 + 2n

( z w)2 = -1 Þ z w = ± i

Þ x = – 10 and n = 40

from equation (ii), - arg( z ) - arg w =

Hence, Re(z) = – 10 9.

(1 + i )2 1 + i 2 + 2i 2i = = 3-i 3-i a -i

Hence, z =

z Î éë 8, 16 ùû

8.

2 5

(b) Given equation is, | z – 1 | = | z – i | Þ

(x – 1)2 + y2 = x2 + (y – 1)2

Þ

1 – 2x = 1 – 2y Þ x – y = 0

[Here, z = x + iy]

z=

a2 + 1

=

2

4(1 + a 2 ) 2 2

(1 + a )

=

2 1 + a2

5 ( w - 1) 3 + 5w

Þ 25 ( ww - w - w + 1) < 9 + 25ww + 15w + 15w

a2 + 1

æ -2 ö æ 2a ö ç 2 ÷ +ç ÷ = è a + 1 ø è a2 + 1 ø

Þ|z|=

5 + 3z Þ 5w - 5wz = 5 + 3z 5 - 5z

Q | z | < 1, \5|w – 1| < |3 + 5w|

2ai - 2

2

|z|=

w=

Þ 5w - 5 = z ( 3 + 5w) Þ z =

(1 + i )2 a + i ´ 10. (a) z = a -i a +i (1 - 1 + 2i )(a + i )

Hence, zw = -i 12. (c)

Hence, locus is straight line with slope 1.

p -p - arg( z w) = 2 2

(Q z

4 + 4a 2 2

(a + 1)

2

...(i)

Þ 16 < 40w + 40w Þ w + w > Þ Re ( w) >

2

2 2 Þ 2Re(w) > 5 5

1 5

=zz

)

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13. (a) Let t =

z-a z+a

3z1 2 z2 3r1 i (q -f ) 2 r2 i (f -q ) + e z = 2 z + 3 z = 2r e 3 r1 2 1 2

Q

t is purely imaginary number.

\

t + t =0

Þ

z -a z -a + =0 z +a z +a

Þ

=

2 3 ´ [cos(q - f) - i sin(q - f)] 3 4

(z – a)( z + a) + ( z – a)(z + a) = 0 – a2 + z z

Þ

zz

Þ

z z – a2 = 0

Þ

|z|2 – a2 = 0

Þ

a2 = 4

Þ

a= ±2

3 4 ´ (cos(q – f) + i sin(q – f)) + 2 3

1ö 1ö æ æ z = çè 2 + ÷ø cos(q - f) + i èç 2 - ø÷ sin(q - f) 2 2

– a2 = 0

\

14. (a) | 1 | = 9, | 2 – 3 – 4i | = 4 2 lies on a circle with centre C2(3, 4) and radius r2 = 4 So, minimum value of | 1 – 2 | is ero at point of contact (i.e. A)

17.

25 9 cos2 (q - f) + sin 2 (q - f) 4 4

16 9 3 5 cos 2 (q - f) + Þ £ |z| £ 2 2 4 4

=

1 lies on a circle with centre C1(0, 0) and radius r1 = 9

| z|=

(d) Suppose z =

3 + 2i sin q 1 - 2i sin q

Since, z is purely imaginary, then z + z = 0 Þ

3 + 2i sin q 3 - 2i sin q + =0 1 - 2i sin q 1 + 2i sin q

Þ

(3 + 2i sin q)(1 + 2i sin q) + (3 - 2i sin q)(1 - 2i sin q) 1 + 4sin 2 q

=0 sin2 q =

Þ

p p 2p q= - , , 3 3 3

15. (b) Since, |z| + z = 3 + i Let z = a + ib, then |z| + z = 3 + i Þ

a 2 + b 2 + a + ib = 3 + i

Compare real and imaginary coefficients on both sides b = 1,

a2 + b2 + a = 3

Now, the sum of elements in A = -

a2 + 1 = a2 + 9 – 6a 4 6a = 8 Þ a = 3

Then,

p p 2p 2p + + = 3 3 3 3

18. (a) Q |z| = 1 & Re z ¹ 1 Suppose z = x + iy Þ x2 + y2 = 1 .....(i) Now, w =

a 2 +1 = 3 – a

3 3 Þ sin q = 4 2

Þ

1 + (1 – 8a ) z 1– z

Þ w=

1 + (1 – 8a ) ( x + iy) 1 – ( x + iy )

Þ w=

1 + (1 – 8a) ( x + iy ) )((1 – x) + iy) 1 – ( x + iy))((1 – x) + iy)

2

|z| =

16 5 æ 4ö çè 3 ÷ø +1 = 9 + 1 = 3

16. (none) Let z1 = r1eiq and z2 = r2eif 3|z1| = 4|z2| Þ 3r1 = 4r2

é (1 + x (1 – 8a ) (1 – x) – (1 – 8a) y 2 ù û Þ w= ë (1 – x )2 + y 2 +i

[ (1 + x (1 – 8a)) y – (1 – 8a) y (1 – x)] (1 – x )2 + y 2

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As, w is purely imaginary. So, é(1 + x (1 – 8a ) )(1 – x) – (1 – 8a) y 2 ù ë û Re w = =0 (1 – x)2 + y 2 Þ (1 – x) + x (1 – 8a) (1 – x) = (1 – 8a) y2 Þ (1 – x) + x (1 – 8a) – x2 (1 – 8a) = (1 – 8x)y2 Þ (1 – x) + x (1 – 8a) = 1 – 8a [From (i), x2 + y2 = 1] Þ 1 – 8a = 1 Þa=0 \ a Î {0} 19. (b) Rationali ing the given expression (2 + 3isin q)(1 + 2isin q) 1 + 4sin 2 q

For the given expression to be purely imaginary, real part of the above expression should be equal to ero. 1 = 0 Þ sin 2 q = 3 1 + 4 sin 2 q 1 1 æ ö Þ sin q = ± Þ q = sin -1 ç ÷ è 3ø 3 20. (b) Let = reiq

Im 5 Consider

(Im )

r (sin q ) 5

=

sin 5 q

5

sin q 5

5 cosec4 q

20sin 3 q

+

sin q 5

– 20 cosec2

5sin q sin 5 q

q + 16

Im 5 (Im )5

is – 4.

21. (d) We know minimum value of |Z1 + Z2| is | |Z1| – |Z2||. Thus minimum value of Z +

£ Z+

1 1 £| Z | + 2 2

Since, | Z |³ 2 therefore

2-

1 1 1 < Z + < 2+ 2 2 2

Þ

3 1 5 < Z+ < 2 2 2

x2 + y2 -1 x + ( y + 1) 2

x 2 + ( y + 1) 2

sin q

minimum value of

x 2 + ( y + 1) 2

x2 + y2 -1

5

x 2 - 2ix ( y + 1) + xi ( y - 1) + y 2 - 1

2

-

2 xi x + ( y + 1) 2 2

for pure imaginary, we have

16sin5 q - 20sin3 q + 5sin q

16sin5 q

= =

=

z-i is purely imaginary means its real part is ero. z+i

=

(Q eiq = cos q + i sinq)

sin 5q

= 4a 2 + 4 - 4a 2 - 4 = 0 = R.H.S. Hence, y2 + 4x – 4 = 0 23. (c) Let z = x + iy

=

r5 (sin 5q)

=

5

2 2 LHS : y 2 + 4 x - 4 = (2a ) + 4(1 - a ) - 4

x + iy - i x + i ( y - 1) x - i ( y + 1) ´ = x + i ( y + 1) x - i ( y + 1) x + iy + i

2 - 6 sin 2 q

Þ

22. (b) Let z = 1+ ia, a Î R z2 = (1 + ia) (1 + ia) x + iy = (1 + 2ia – a2) On comparing real and imaginary parts, we get x = 1 – a2, y = 2a Now, consider option (b), which is y2 + 4x – 4 = 0

1 1 is | Z | 2 2

=0

Þ Þ

x2 + y2 = 1 (x + iy) (x – iy) = 1

Þ

x + iy =

and

1 = x – iy z

z+

1 =z x - iy

1 = ( x + iy ) + ( x - iy ) = 2 x z

1ö æ çè z + ÷ø is any non- ero real number z æz ö æz ö 24. (a) Consider arg ç 1 ÷ + arg ç 2 ÷ è z4 ø è z3 ø

= arg( z1 ) - arg( z4 ) + arg( z2 ) - arg( z3 ) = (arg( z1 ) + arg( z2 )) - (arg( z3 ) + arg( z4 ))

æ z2 = z1 &ö given ç è z4 = z3 ÷ø

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= (arg( z1 ) + arg( z1 )) - (arg( z3 ) + arg( z3 ))

ìalso (arg( z1 ) = - arg( z1 )ü í ý îarg( z3 ) = - arg( z3 ) þ

28. (a) Let z = x + iy Þ z2 = x2 – y2 + 2ixy Now,

= (arg( z1 ) - arg( z1 )) - (arg( z3 ) - arg( z3 ))

1 + z 2 1 + x 2 - y 2 + 2ixy ( x 2 - y 2 + 1) + 2ixy = = 2iz 2i ( x + iy ) 2ix - 2 y ( x2 - y 2 + 1) + 2ixy -2 y - 2ix = ´ -2 y + 2ix -2 y - 2ix

=0–0=0 25. (d) Consider the equation

=

w - wz = k (1 - z ), k Î R

a=

w - wz Clearly z ¹ 1 and is purely real 1- z

Þ w - wz - wz + wzz = w - wz - wz + wzz 2 2 Þ w+w| z| = w+w| z|

Þ ( w - w)(| z |2 ) = w - w Þ |z|2 = 1 (QIm w ¹ 0) Þ |z| = 1 and z ¹ 1 \ The required set is {z : |z| = 1, z ¹ 1 } 26. (c) Given | | = 1, arg z = q 1 z

æ 1+ z ö æ 1+ z ö = arg \ arg ç = arg (z) = q. ç 1÷ è 1 + z ÷ø çè 1 + ÷ø z

27. (b) Let z = x + iy, z = x – iy Now, z = 1 – z Þ x + iy = 1 – (x – iy) 1 Þ 2x = 1 Þ x = 2 Now, | z | = 1 Þ x2 + y2 = 1 Þ y2 = 1 – x2

Þ

3 y =± 2

Now, tan q =

y (q is the argument) x

3 1 ¸ 2 2 = 3

(+ve since only principal argument)

=

p 3 Hence, z is not a real number So, statement-1 is false and 2 is true.

Þ

q = tan -1 3 =

x ( x 2 + y 2 + 1) 2( x 2 + y 2 )

x2 + y 2 = 1

Þ x2 + y2 = 1

w - wz w - wz = 1- z 1- z

Þ z=

2( x 2 + y 2 )

Since, | z | = 1 Þ

w - wz w - wz \ = 1- z 1- z

Þ

y ( x 2 + y 2 - 1) + x ( x 2 + y 2 + 1)i

x (1 + 1) =x 2 ´1 Also z ¹ 1 Þ x + iy ¹ 1 \ A = (– 1, 1) 29. (d) Let z1 = 1 + i and z2 = 1 – i

\ a=

z2 1 - i (1 - i ) (1 - i ) = = =-i z1 1 + i (1 + i ) (1 - i )

æz ö 2 + 3ç 2 ÷ 2 z1 + 3z2 è z1 ø = 2 - 3i = 2 z1 - 3z2 æ z ö 2 + 3i 2-3ç 2 ÷ è z1 ø 2 z1 + 3z2 2 - 3i 2 - 3i = = 2 z1 - 3z2 2 + 3i 2 + 3i 4+9 = =1 4+9 30. (b) z1 + z2 2 + z1 - z2 2

= z1 2 + z2 2

2

2

+ 2 z1 z2 + z1 + z2

= 2 z1 + 2 z2

2

é êQ ë

2

z1 | z |ù = 1 ú z2 | z2 | û

- 2 z1 z2

2 2 = 2 éê z1 + z2 ùú ë û

31. (a) Let |Z|= |W| = r Þ Z= reiq, W = reif where q + f = p \

–if W = re Now, Z = rei(p – f) = reip × e–if = – re–if

= –W Thus, statement-1 is true but statement-2 is false. 32. (b) Statement - 1 and 2 both are true. It is fundamental property. But Statement - 2 is not correct explanation for Statement - 1.

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33. (a) Let z = x + iy

38. (a)

Arg( z w) = arg( z ) + arg(w)

z - 1 = z + 1 Þ ( x -1)2 + y2 = ( x + 1)2 + y2

Þ x=0

p 2 [Q arg ( z ) = - arg( )]

Þ Re z = 0

= - arg( z ) + arg w = -

z - 1 = z - i Þ ( x - 1)2 + y 2 = x2 + ( y - 1)2

Þ x= y 2

2

2

z + 1 = z - i Þ ( x + 1) + y = x + ( y - 1)

\ z w = -1 39. (c) Given that | z – 4 | < | z – 2 | Let z = x + iy Þ | (x – 4) + iy) | < | (x – 2) + iy | Þ (x – 4)2 + y2 < (x – 2)2 + y2 Þ x2 – 8x + 16 < x2 – 4x + 4 Þ 12 < 4x Þ x > 3 Þ Re(z) > 3 40. (b) Let | z | = | w | = r \ z = reiq, w = reif where q + f = p.

2

Þ x = –y Only (0, 0) will satisfy all conditions. Þ Number of complex number z = 1 34. (c)

| z w |=| z || w |=| z || w |=| zw |= 1 [Q z = z ]

æ 1 ö 1 1 –1 çè i – 1÷ø = (i - 1) = – i – 1 = i + 1 1

\ z = rei(p–f) = reip . e–if = –re–if = – w .

35. (a) Given that z 3 = p + iq

[Q eip = –1 and w = re–if]

Þ z = p3 + (iq )3 + 3 p (iq )( p + iq )

41. (c) Let z = x + iy

Þ x - iy = p3 - 3 pq 2 + i (3 p 2 q - q 3 )

Q z 2 = i | z |2

Comparing both side, we get

\ x 2 - y 2 + 2ixy = i( x 2 + y 2 )

\ x = p3 - 3 pq 2 Þ

x = p 2 - 3q2 p

Þ x 2 - y 2 = 0 and 2xy = x2 + y 2

...(i)

Þ ( x - y )( x + y ) = 0 and ( x - y)2 = 0 Þx=y

y and y = q - 3 p q Þ = q 2 - 3 p 2 ...(ii) q 3

2

Adding (i) and (ii), we get

\

42. (a) Given that, a =

æ x yö x y + = -2 p 2 - 2q2 \ ç + ÷ ( p 2 + q 2 ) = -2 è p qø p q

36. (c) Given that arg zw = p Þ arg z + arg w = p

\ (2 + w)4 = a + bw Þ (4 + w2 + 4w)2 = a + bw Þ (w2 + 4(1 + w)) 2 = a + bw ...(i)

z + iw = 0 Þ z = -iw Replace i by –i, we get p \ z = iw Þ arg z = + arg w 2 p Þ arg z = + p - arg z (from (i)) 2 3p \ arg = 4 37. (b) Given that

Þ (w 2 - 4w 2 ) 2 = a + bw [Q1 + w = -w 2 ] Þ ( -3w 2 )2 = a + bw Þ 9w 4 = a + bw

x

x é (1 + i )2 ù æ 1+ i ö ú =1 çè ÷ø = 1 Þ ê 2 1- i êë 1 - i úû

æ 1 + i 2 + 2i ö x + ç ÷ = 1 Þ (i ) = 1; \ x = 4n ; n Î I 1 1 + è ø

(Q w3 = 1)

Þ 9w = a + bw On comparing, a = 0, b = 9 Þ a + b = 0 + 9 = 9.

43. (c)

x

-1 + 3i =w 2

5p 5p ö æ 1 + cos + i sin ç 18 18 ÷ ç 5p 5p ÷ ç 1 + cos - i sin ÷ è 18 18 ø

3

5p 5p 5p ö æ 2cos 2 + i 2sin × cos ç 36 36 36 ÷ =ç 5p 5p ÷ 2 5p - i 2sin × cos ÷ ç 2cos è 36 36 36 ø

3

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Complex Numbers and Quadratic Equations 3

5p 5p ö æ 6 cos + i sin ç 36 36 ÷ = æ cos 5p + i sin 5p ö =ç ç ÷ 5p 5p ÷ è 36 36 ø - i sin ÷ ç cos è ø 36 36

5p ö 5p ö 5p 5p æ æ = cos ç 6 ´ ÷ + i sin ç 6 ´ ÷ = cos + i sin è è 36 ø 36 ø 6 6 =44. (b) Let

3 1 1 + i = - ( 3 - i) 2 2 2 3 + 2 -54 = 3 + 6 6i 3 + 6 6i = a + ib

Þ a 2 - b2 = 3 and ab = 3 6 Þ a 2 + b2 = (a 2 - b 2 ) 2 + 4a 2b 2 = 15 So, a = ±3 and b = ± 6

47. (d)

æ 1 3 i 3 ö + = -i ç - + i ÷ = -iw ç ÷ 2 2 è 2 2 ø

where w is imaginary cube root of unity. Now, (1 + i + 5 + i 8)9 = (1 + w – iw2 + iw2)9 = (1 + w)9 = (– w2)9 = – w18 = – 1 (Q 1 + w + w2 = 0) 48. (a) –(6 + i)3 = x + iy Þ –[216 + i3 + 18i(6 + i)] = x + iy Þ –[216 – i +108i – 18] = x + iy Þ –216 + i – 108i + 18 = x + iy Þ –198 – 107i = x + iy Þ x = – 198, y = –107 Þ y – x = –107 + 198 = 91 5

æ 3 iö æ 3 iö + ÷ +ç - ÷ 49. (a) z = ç è 2 2ø è 2 2ø

5

5

p pö æ p pö æ = ç cos + i sin ÷ + ç cos - i sin ÷ è 6 6ø è 6 6ø

3 + 6 6 i = ± (3 + 6 i)

Similarly, 3 - 6 6 i = ± (3 - 6 i)

5

lm ( 3 + 6 6i - 3 - 6 6i ) = ±2 6 45. (b) Let a = w, b = 1 + w3 + w6 + ..... = 101 a = (1 + w) (1 + w2 + w4 + ..... w198 + w200) = (1 + w)

Þ

a =

(

1 - (w2 )101

)

(w + 1)(w202 - 1)

=

2

1- w

(1 + w)(1 - w) 1 - w2

(w2 - 1)

=1

Required equation = x2 – (101 + 1)x + (101) × 1 = 0 Þ

x2 – 102x + 101 = 0

46. (d) Q

z = x + iy

5

5

æ ip ö æ -i p ö p 6 6 = ç e ÷ + ç e ÷ = 2 cos 6 = 3 è ø è ø

Þ

I(z) = 0, Re(z) = 3

æ1+ i 3 ö 50. (d) Let l = çç ÷÷ . è1– i 3 ø æ1 + i 3 ö æ1 + i 3 ö \ l = çç ÷÷ ´ çç ÷÷ è1 – i 3 ø è1 + i 3 ø

æ – 2 + i2 3 ö æ 1 – i 3 ö =ç ÷÷ = çç ÷÷ ç 4 è ø è –2 ø æ1 + i 3 ö æ1 – i 3 ö Also, l = çç ÷÷ ÷÷ ´ çç è1 – i 3 ø è1 – i 3 ø

æ z - 1 ö ( x - 1) + iy ç 2 z + i ÷ = 2( x + iy ) + i è ø

æ ö æ –2 ö 4 =ç ÷=ç ÷ è – 2 – i2 3 ø è 1 + i 3 ø

( x - 1) + iy 2 x - (2 y + 1)i = ´ 2 x + (2 y + 1)i 2 x - (2 y + 1)i

æ z + 1 ö 2 x( x - 1) + y (2 y + 1) Re ç =1 ÷= è 2z + i ø (2 x ) 2 + (2 y + 1) 2 2

2 2 æ 1ö æ 3ö 5ö æ ÷÷ . Þ ç x + ÷ + ç y + ÷ = çç 2 4 4 è ø è ø è ø

3

æ1+ i 3 ö æ1+ i 3 ö æ1 + i 3 ö æ1+ i 3 ö Now, ç ÷ =ç ÷´ç ÷´ç ÷ è1 – i 3 ø è1 – i 3 ø è1 – i 3 ø è1 – i 3 ø æ1+ i 3 ö æ – 2 ö æ1 – i 3 ö =ç ÷´ç ÷ =1 ÷´ç è1 – i 3 ø è1 + i 3 ø è – 2 ø

\ least positive integer n is 3.

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Mathematics

54. (a) (1 + w)7 = A + Bw (–w2)7 = A + Bw (Q w14 = w12.w2 = w2) – w2 = A + Bw 1 + w = A + Bw Þ A = 1, B = 1. 55. (a) |z + 1 | = | z + 4 – 3 | £ | z + 4 | + | –3 | £ | 3 | + | – 3|

(3,3)

51. (a)

2 2 (1,1)

Final

(3,1)

(2,1) 1

position

Þ | z + 1 | £ 6 Þ | z + 1|max = 6

56. (c) Given that w = So new position is at the point 1 + i 52. (a) Þ

1 - 2 2 =1 2 - 1 2

Þ |w|=

2

1 - 2 2 = 2 - 1 2

2

Þ

( 1 - 2 2 )( 1 - 2 2 ) = (2 - 1 2 )(2 - 1 2 )

Þ

( 1 - 2 2 )( 1 - 2 2 ) = (2 - 1 2 )(2 - 1 2 )

Þ

( 1 1) - 2 1 2 - 2 1 2 + 4 2 2

2

1 + 4 2

Þ

1 + 4 2

2

2 2

= 4 + 1 – 4 – 1

2

2

2 2

57. (c)

1

Q

2 ¹ 1

\

1

2

Þ

1 = 2

Þ 53. (a)

é z1 z ù = 1ú êQ z2 û ë z2

| z1 + z2 | = | z1 | + | z2 | Þ z1 and z2 are collinear

and are to the same side of origin; hence arg z1 – arg z2 = 0.

=0

(z

|z| =1 1 |z- i| 3

æ 1ö Þ distance of from origin and point ç 0, ÷ is same è 3ø hence lies on bisector of the line oining points (0, 0) and (0, 1/3). Hence lies on a straight line.

2

2

1 z- i 3

1 Þ z = z- i 3

= 4 - 2 1 2 - 2 1 2 + 1 1 2 2 Þ

z

2

)(

- 4 1 - z2

2

)=0

58. (c) Q ( x - 1)3 + 8 = 0 Þ ( x - 1) = (-2) (1)1/ 3

Þ x – 1 = – 2 or -2w or - 2w 2 or x = – 1 or 1 – 2 w or 1 – 2 w2 .

=4

59. (b) Given that | z 2 - 1|=| z |2 +1 Þ| z 2 - 1|2 = ( zz + 1) 2 2

[Q z = ]

Point 1 lies on circle of radius 2. z2 z 2 é æ z1 ö z1 ù = = ú êQ z - 1 z - 1 ë çè z2 ÷ø z2 û

2 2 Þ zzz - z = z. z . z - z

Þ z 2 .z - z 2 = z 2 .z - z 2

Þ z 2 ( z - z ) - ( z - z )( z + z ) = 0

(

)

2 Þ (z - z ) z -(z + z ) = 0

Either z - z = 0 or z 2 - ( z + z ) = 0 Either z = z Þ real axis or z 2 = z + z Þ zz - z - z = 0

Þ(z2 -1)(z 2 -1) = (zz +1)2 (Qz1 - z2 = z1 - 2) Þ 2 2 - 2 - 2 + 1 = 2 2 + 2 + 1 Þ z 2 + 2 zz + z 2 = 0

Þ ( z + z )2 = 0 Þ z = - z

Þ z is purely imaginary 60. (b) Let the circle be |z – z0| = r. Then according to given conditions |z0 – z1| = r + a ...(i) |z0 – z2|= r + b ...(ii) Subtract (ii) from (i) we get |z0 – z1| – |z0 – z2| = a – b. \ Locus of centre z0 is |z – z1| –|z – z2| = a – b, which represents a hyperbola.

represents a circle passing through origin.

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Imaginary part of u

61. (a) Q a + b = 64, ab = 256

a 3/8

b3/8

a +b

64 =2 5/8 5/8 5/8 8 5/8 32 b a (ab) (2 ) 62. (b) Let a and b be the roots of the given quadratic equation, +

=

64

=

=

2x2 + 2x - 1 = 0

...(i)

Then, a + b = -

1 Þ -1 = 2a + 2b 2

-2 x( y - K ) + x (2 y + 1) x 2 + ( y - K )2

Q Re(u ) + Im(u ) = 1

Þ 2 x2 + 2 y 2 - 2 Ky + y - K - 2 xy + 2 Kx + 2 xy + x = x 2 + y 2 + K 2 - 2 Ky Since, the curve intersect at y-axis

[Q a is root of

and 4a 2 + 2a - 1 = 0 eq. (i)]

= Im(u ) =

\x = 0

Þ y 2 + y - K ( K + 1) = 0

Þ 4a 2 + 2a + 2a + 2b = 0 Þ b = -2a (a + 1)

Let y1 and y2 are roots of equations if x = 0 Q y1 + y2 = -1

63. (b) Let | x | = y then

9 y 2 - 18 y + 5 = 0

y1 y2 = -( K 2 + K )

Þ 9 y 2 - 15 y - 3 y + 5 = 0

\ ( y1 - y2 )2 = (1 + 4 K 2 + 4 K )

Þ (3 y - 1)(3 y - 5) = 0

Given PQ = 5 Þ| y1 - y2 |= 5

Þ y=

1 5 1 5 or Þ| x |= or 3 3 3 3

Roots are ±

3 x 2 - 10 x + 27l = 0

25 81 64. (d) Let a and b be the roots of the quadratic equation

= =

a

+

\ 3a 2 - 10a + 27l = 0 3a 2 - 3a + 6l = 0

b

Þ a + b = 1 Þ 3l +

1 - b2

a - ab(a + b) + b

2 3

2 1 = 1 Þ l = and 3 9

ag = 9l Þ 3l × g = 9l Þ g = 3 \

(a + b) - ab(a + b) 1 - (a + b) 2 + 2ab + (ab) 2

bg = 18 l

67. (d)

3 2 3 + ´ 27 7 7 7 = = -2 4 16 9 + 1- + 2 ´ 49 7 49

a ×b = 2 and a + b = - p also

1 1 + = -q a b

Þ p = 2q

2( x + iy ) + i 2 x + i (2 y + 1) u= = ( x + iy ) - ki x + i( y - k )

Real part of u = Re(u ) =

...(ii)

Now, ab = 2l Þ 3l ×b = 2l Þ b =

1 - (a 2 + b 2 ) + (ab) 2

65. (d)

...(i)

\ On subtract, we get a = 3l

7 x2 - 3x - 2 = 0 3 -2 \ a + b = , ab = 7 7 1 - a2

as K > 0, \ K = 2 66. (b) Since a is common root of x 2 - x + 2l = 0 and

1 5 and ± 3 3

\ Product =

Now,

Þ 4 K 2 + 4 K - 24 = 0 Þ K = 2 or – 3

2 x + ( y - K )(2 y + 1) 2

1ö æ 1ö æ 1ö æ 1ö æ Now çè a - a ÷ø èç b - bø÷ èç a + bø÷ çè b + a ÷ø é ù 1 a bùé 1 = êab + - - ú êab + + 2ú ab b a û ë ab ë û

x 2 + ( y - K )2

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=

9 é 5 a2 + b2 ù 9 2 ê ú = [5 - ( p - 4)] 2 ë2 2 û 4

9 = (9 - p 2 ) [Q a 2 + b2 = (a + b) 2 - 2ab] 4 68. (c) The given quadratic equation is

(l + 1) x - 4lx + 2 = 0 2

2

Q One root is in the interval (0, 1) \ f (0) f (1) £ 0

Þ 2(l 2 + 1 - 4l + 2) £ 0 Þ 2(l 2 - 4l + 3) £ 0

71. (8) Since, 2x2 + (a – 10) x + \

D³0

Þ

æ 33 ö (a - 10)2 - 4(2) ç - 2a ÷ ³ 0 è 2 ø

Þ (a – 10)2 – 4(33 – 4a) ³ 0 Þ a2 – 4a – 32 ³ 0 Þ (a – 8) (a + 4) ³ 0 Þ a£–4Èa³8 Þ aÎ (– ¥, – 4] È [8, ¥) 72. (a) Let z = a ± ib be the complex roots of the equation So, sum of roots = 2a = – b and Product of roots = a2 + b2 = 45 (a + 1)2 + b2 = 40

(l - 1)(l - 3) £ 0 Þ l Î[1, 3] But at l = 1, both roots are 1 so l ¹ 1 \ l Î (1, 3] 69. (c) Since, a and b are the roots of the equaton 5x2 + 6x – 2 = 0 Then, 5a 2 + 6a - 2 = 0 , 5b 2 + 6b - 2 = 0 5a 2 + 6a = 2

5S6 + 6S5 = 5(a 6 + b6 ) + 6(a 5 + b5 )

Given, | z + 1|= 2 10 Þ (a + 1)2 – a2 = – 5 [Q b2 = 45 – a2] Þ 2a + 1 = – 5 Þ 2a = – 6 2 Hence, b = 6 and b – b = 30 73. (d) a5 = 5a + 3 b5 = 5b + 3 p5 = 5(a + b) + 6 = 5(1) + 6 [Q from x2 – x – 1 = 0, a + b =

= (5a 4 + 6a 5 ) + (5b6 + 6b5 ) = a 4 (5a 2 + 6a ) + b 4 (5b 2 + 6b) = 2(a 4 + b4 ) = 2S4 70. (a) Let ex = t Î (0, ¥) Given equation t4 + t3 – 4t2 + t + 1 = 0 Þ Þ

tan a × tan b =

\ Þ Þ

(k + 1) tan 2 x - 2l tan x + ( k - 1) = 0

tan a + tan b =

æ 2 1 ö æ 1ö çt + 2 ÷ + çt + t ÷ - 4 = 0 ø t ø è è

y2 + y – 6 = 0 y = – 3, 2

1 Þ y=2 Þ t + = 2 t Þ ex + e–x = 2 x = 0, is the only solution of the equation

Hence, there only one solution of the given equation.

-b = 1] a

p5 = 11 and p5 = a2 + b2 = a + 1 + b + 1 p2 = 3 and p3 = a3 + b3 = 2a + 1 + 2b + 1 = 2(1) + 2 = 4 p2 × p3 = 12 and p5 = 11 Þ p5 ¹ p2 × p3 74. (b)

1 1 t2 + t – 4 + + 2 = 0 t t

1 Let t + = y t 2 (y – 2) + y – 4 = 0 y2 + y – 6 = 0

33 = 2a has real roots, 2

2l k +1

k -1 k +1

[Sum of roots] [Product of roots]

2l k + 1 = 2l = l tan(a + b) = k -1 2 2 1k +1

tan 2 (a + b) =

l2 = 50 2

l = 10. 75. (b) Given equation is, x2 + x sin q – 2 sin q = 0 a + b = – sin q and ab = – 2 sin q

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(a12 + b12 )a12b12 (a12 + b12 )(a - b) 24

=

80. (c) The given quadratic equation is x2 – 2x + 2 = 0

(ab)12 (a - b) 24

Then, the roots of the this equation are

\ | a - b |= (a + b)2 - 4ab = sin 2 q + 8sin q 12

\

(ab)

( a - b)

24

=

12

(2sin q) 12

12

sin q(sin q + 8)

=

12

2

Now,

a 1 - i (1 - i ) 2 = = =i b 1 + i 1 - i2

or

a a 1 - i (1 - i ) 2 = = = – i So, = ±i 2 b b 1+ i 1- i

(sin q + 8)12

76. (d) Let 2x – 1 = t 5 + | t | = (t + 1) (t – 1) Þ | t | = t2 – 6 When t > 0, t2 – t – 6 = 0 Þ t = 3 or – 2 t = – 2 (reected) When t < 0, t2 + t – 6 = 0 Þ t = – 3 or 2 (both reected) \ 2x – 1 = 3 Þ 2x = 4 Þ x = 2 77. (d) Since 2 - 3 is a root of the quadratic equation x2 + px + q = 0

Þ x 2 + px + q = [ x - (2 - 3)[ x - (2 + 3)] = x 2 - (2 + 3)x - (2 - 3)x + (22 - ( 3) 2 )

Þ

81. (b) Let roots of the quadratic equation are a, b.

a a b 1 and l + = 1 Þ + = 1 b b a l

(a + b) 2 - 2ab = 1...(i) ab

\

Now, by comparing p = –4, q = 1 Þ p2 – 4q – 12 = 16 – 4 – 12 = 0 78. (c) Sum of roots =

m2 + 1

Now, a + b = 3, ab = 1 Þ | -a - b | = 5

(

a+b=

m(4 - m) 4 - m 2 and ab = = 2 m 3 3m 3m 2

Put these values in eq (1),

3

Q sum of roots is greatest. \ m = 0 Hence equation becomes x2 – 3x + 1 = 0

)

a3 - b3 = (a - b) a 2 + b2 + ab = 5 ( 9 - 1) = 8 5

x =a given equation will become:

79. (d) Let

æ 4 - mö çè ÷ 3m ø 2 m 3 2

Þ

| a – 2 | + a2 – 4a + 4 – 2 = 0 | a – 2 | + (a – 2)2 – 2 = 0

2

=3

(m – 4)2 = 18 Þ m = 4 ± 18

Therefore, least value is 4 – 18 = 4 - 3 2 82. (d) Let a and b be the roots of the equation, 81x2 + kx + 256 = 0

| a – 2 | + a (a – 4) + 2 = 0 Þ Þ

n must be a multiple of 4.

Hence, the required least value of n = 4.

The quadratic equation is, 3m2x2 + m(m – 4) x + 2 = 0

= x 2 - 4x + 1

\

n

æaö Now, ç ÷ = 1 Þ (±i)n = 1 èbø

Given, l =

\ 2 + 3 is the other root

1

Given (a) 3 = b Þ a = b3 256 81

Let | a – 2| = y (Clearly y ³ 0) Þ y + y2 – 2 = 0

Q

Product of the roots =

Þ Þ

\

(a)(b) =

Þ

4 64 æ 4ö b4 = ç ÷ Þ b = Þ a = è 3ø 3 27

y = 1 or – 2 (reected) | a – 2 | = 1 Þ a = 1, 3

When

x = 1Þx=1

x = 3Þx=9 Hence, the required sum of solutions of the equation = 10 When

2 ± -4 =1± i 2

256 81 4

Q Sum of the roots = -

k 81

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a+ b=-

\

k k 4 64 =Þ + 81 3 27 81

Þ k = –300 83. (d) Consider the quadratic equation (c – 5) x2 – 2cx + (c – 4) = 0 Now, f(0).f(3) > 0 and f(0).f(2) < 0 Þ (c – 4) (4c – 49) > 0 and (c – 4) (c – 24) < 0 Þ

æ 49 ö c Î(–¥, 4) È ç , ¥ ÷ and c Î (4, 24) è 4 ø

Þ

æ 49 ö c Î ç , 24 ÷ è 4 ø

\ S = {13, 14, ..., 23} 84. (d) The given quadratic equation is x2 + (3 – l) x + 2 = l Sum of roots = a + b = l – 3 Product of roots = ab = 2 – l a2 + b2 = (a + b)2 – 2ab = (l – 3)2 – 2 (2– l) = l2 – 4l + 5 = (l – 2)2 + 1 For least (a2 + b2) l = 2. 85. (a) Consider the equation x2 + 2x + 2 = 0

-2 ± 4 - 8 = -1 ± i 2 Let a = –1 + i, b = –1 – i a15 + b15 = (–1 + i)15 + (–1 – i)15

x=

æ =ç è

÷ ø

( 2)

15

=

= =

(

3p 15 -i ö 2e 4

÷ ø

- i 45p ù é i 45 p êe 4 + e 4 ú êë úû

2 ) .2 cos 15

-2 2

= -2

æ +ç è

45p ( )15 3p = 2 .2cos 4 4

( 2)

15

( 2)

14

\

Discriminant D must be perfect square number.

D = (–11)2 – 4 × 6 × a = 121 – 24a must be a perfect square Hence, possible values for a are a = 3, 4, 5. \ 3 positive integral values are possible. 87. (b) Given quadratic equation is: x2 – mx + 4 = 0 Both the roots are real and distinct. So, discriminant B2 – 4AC > 0.

æ 49 ö Integral values in the interval ç , 24 ÷ are 13, 14, ..., 23. 4 è ø

3p 15 i ö 2e 4

86. (a) The roots of 6x2 – 11x + a = 0 are rational numbers.

\

m2 – 4 × 1 × 4 > 0

\

(m – 4) (m + 4) > 0

\

m Î (–฀, –4) È (4, ฀)

Since, both roots lies in [1, 5] -m Î (1, 5) 2

\

-

Þ

m Î (2, 10)

...(ii)

And 1 × (1 – m + 4) > 0 Þ m < 5 \

m Î (–฀, 5)

...(iii)

And 1 × (25 – 5m + 4) > 0 Þ m < \

29 ö æ m Îç - ฀ ÷ è 5ø

29 5

...(iv)

From (i), (ii), (iii) and (iv), m Î (4, 5) 88. (a) Q z0 is a root of quadratic equation x2 + x + 1 = 0 \

z0 = w or w2 Þ z03 = 1

\

z = 3 + 6i z081 – 3i z093

= 3 + 6i((z0)3)27 – 3i((z0)3)31 = 3 + 6i – 3i = 3 + 3i \

89. (b)

æ 3ö p = arg(z) tan–1 çè 3÷ø 4

1 1 1 + = x+ p x+q r

x+ p+x+q

( x + p) ( x + q) = -256

...(i)

=

1 r

(2x + p + q) r = x2 + px + qx + pq x2 + (p + q – 2r) x + pq – pr – qr = 0 Let a and b be the roots.

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\ a + b = – (p + q – 2r) .... (i) & ab = pq – pr – qr .... (ii) Q a = – b (given) \ in eq. (1), we get Þ – (p + q – 2r) = 0 ... (iii) Now, a2 + b2 = (a + b)2 – 2ab = (– (p + q –2r))2 – 2 (pq –pr – qr) .... (from (i) and (ii)) = p2 + q2 + 4r2 + 2pq – 4pr – 4qr – 2pq + 2pr + 2qr = p2 + q2 + 4r2 – 2pr – 2qr = p2 + q2 + 2r (2r – p – q) ... (from (iii)) = p2 + q2 + 0 = p2 + q2 90. (b) Here, 9x2 + 27x + 20 = 0 – b ± b – 4ac

and f (2) = 6 – 3a As, f (1) + f (2) = 0 Þ 2 – 2a + 6 – 3a = 0 Þ a =

8 5 93. (b) a, b are roots of x2 – x + 1 = 0 Therefore, the other root is

\

a = -w and b = -w2

where w is cube root of unity \ a101 + b107 = (–w)101 + (–w)107 = –[w2 + w] = –[–1] = 1 n

94. (a) We have,

2

\ x=

Þ x=

n

å

2a 2

Þ

4 5 ,– 3 3

\ sec A =

3 5

1 5 =– cos A 3

Here, A is an obtuse angle. \ tan A = – sec 2 A – 1 = –

4 . 3

Hence, roots of the equation are sec A and tan A. 91. (b) As tan A and tan B are the roots of 3x2 – 10x – 25 = 0, tan A + tan B So, tan (A + B) = = 1 – tan A tan B

10 3 = 10 / 3 = 5 25 28 / 3 14 1+ 3

Now, cos2 (A + B) = – 1 + 2 cos2 (A + B) =

1 – tan 2 ( A + B ) 1 + tan 2 ( A + B )

Þ cos 2 ( A + B ) =

196 221

\ 3sin2 (A + B) – 10 sin (A + B) cos (A + B) – 25 cos2 (A + B) = cos2 (A + B) [3 tan2 (A + B) – 10 tan (A + B) – 25] =

(x + r - 1)(x + r) = 10n

r =1

(x 2 + xr + (r –1)x + r 2 - r) = 10n n

Þ

2´9

Given, cos A = –

å

r =1

– 27 ± 27 – 4 ´ 9 ´ 20

Þ x=–

8 5

75 – 700 – 4900 196 5525 196 ´ =– ´ = – 25 196 221 196 221

92. (d) If a and – 1 are the roots of the polynomial, then we get f (x) = x2 + (1 – a) x – a. \ f (1) = 2 – 2a

å

(x 2 + (2r - 1)x + r(r - 1)) = 10n

r =1 nx2 + {1 + 3 + 5 + .... + (2n – 1) }x

+ {1.2 + 2.3 +.... + (n – 1) n} = 10 n (n - 1) n(n + 1) = 10n Þ nx2 + n2 x + 3 n 2 - 31 =0 Þ x2 + nx + 3 Let a and a + 1 be its two solutions (Q it has two consequtive integral solutions) Þ a + (a + 1) = – n -n - 1 ...(i) Þ a= 2 n 2 - 31 Also a (a+1) = ...(ii) 3 Putting value of (i) in (ii), we get 2 æ n + 1ö æ 1 - n ö n - 31 -ç = ÷ ç ÷ è 2 øè 2 ø 3

Þ 95. (c) Þ Þ

n2 = 121 Þ n = 11 (x – 1) (x2 + 5x – 50) = 0 (x – 1) (x + 10) (x – 5) = 0 x = 1, 5, –10 Sum = – 4 96. (c) Let p (x) = ax2 + bx + c Q p (0) = 1 Þ c = 1 Also, p (1) = 4 & p ( -1) = 6 Þ Þ Þ

a+ b+1= 4&a –b+1=6 a +b=3&a–b=5 a = 4 & b = –1 p (x) = 4x2 – x + 1 p (b) = 16 – 2 + 1 = 15 p (–2) = 16 + 2 + 1 = 19

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(x 2 - 5 x + 5) x

97. (c)

2

+ 4x - 60

=1

Case I x2 – 5x + 5 = 1 and x2 + 4x – 60 can be any real number Þ x = 1, 4 Case II x2 – 5x + 5 = –1 and x2 + 4x – 60 has to be an even number Þ x = 2, 3 where 3 is reected because for x = 3, x2 + 4x – 60 is odd. Case III x2 – 5x + 5 can be any real number and x2 + 4x – 60 = 0 Þ x = –10, 6 Þ Sum of all values of x = –10 + 6 + 2 + 1 + 4 = 3 2x + 1 - 2 x -1 = 1

98. (a)

.....(i)

Þ

2x + 1 + 2x – 1 – 2 4 x 2 - 1 = 1

Þ Þ Þ

4x – 1 = 2 4 x 2 - 1 16x2 – 8x + 1 = 16x2 – 4 8x = 5

Þ

5 x = which satisfies equation (i) 8

So,

100. (b) (a – 1) (x4 + x2 + 1) + (a + 1) (x2 + x + 1)2 = 0 Þ (a – 1) (x2 + x + 1) (x2 – x + 1) + (a + 1) (x2 + x + 1)2 = 0 Þ (x2 + x + 1) [(a – 1) (x2 – x + 1) + (a + 1) (x2 + x + 1)] = 0 Þ (x2 + x + 1) (ax2 + x + a) = 0 For roots to be distinct and real, a ¹ 0 and 1 – 4a2 > 0 Þ a ¹ 0 and a2

a

1 + b a

1

1 a b

3 for x > , x 2 + 2 x - 3 - 4 = 0 2 x2 + 2x – 7 = 0

x=

b= -

-2 ± 4 + 28 -2 ± 4 2 = -1 ± 2 2 = 2 2

Here x = 2 2 - 1

3ü ì í2 2 - 1 < ý 2þ î

1 Þ a = ab a

a+ b

)

x( x + b3 ) + (a3 - 3abx) = 0 Þ

x2 + (b3 - 3ab) x + a3 = 0

(

)

3

+3

(

ab

)(

)

a + b ù + (ab)3/ 2 = 0 ûú

Þ x 2 - éëa3/2 + b3/2 + 3 ab ( a + b ) - 3 ab ( a + b ) ùû x + (ab)3/2 = 0

2± 4+ 4 2± 2 2 = = 1± 2 x= 2 2

Þ

x2 - (a 3/ 2 + b3/ 2 ) x + a 3/ 2b3/ 2 = 0

Roots of this equation are a 3/ 2 , b3/ 2 107. (b) Given a3 + b3 = – p and ab = q

3ü ì í(1 - 2) < ý 2þ î

Sum of roots : (2 2 - 1) + (1 - 2) = 105. (d)

æ a + bö b ÷ = -a = ç ab ø è

x2 + é - a + b ëê

3 2 x2 – 2x + 3 – 4 = 0 Þ x2 – 2x – 1 = 0

Here x = 1 - 2

be the roots of ax 2 + bx + 1 = 0

b

Putting values of a and b, we get

for x
0 Þ k2 + 2k – 8 < 16 Þ k2 + 6k – 4k – 24 < 0 Þ (k + 6) (k – 4) < 0 Þ –6 1 ) Hence, |a| + |b| is greater than 2. 111. (d) Given equation is x2 – (sina – 2)x – (1 + sina) = 0 Let x1 and x2 be two roots of quadratic equation. \ x1 + x2 = sina – 2 and x1x2 = – (1 + sina) (x1 + x2)2 = (sina – 2)2 = sin2a + 4 – 4 sina Þ

x12 + x22 = sin 2 a + 4 - 4sin a - 2 x1 x2

= sin2a + 4 – 4 sina + 2 (1 + sina) = sin2a – 2 sina + 6 ...(i)

p 2

Hence, a =

Þ p2 – 4 ×

(x + iy) +

p p p p , a = , a = and a = in (i) one by one 2 6 4 3

k+4 > 0 Þk>–4 k -2 k -2 > 0 Þk>2 k+4

Roots are real so, – 6 < k < 4 So, 6 and 4 are not correct. Since, k > 2, so 1 is also not correct value of k. \k=3 113. (d) p (x) = 0 Þ

f ( x ) = g ( x)

Þ

ax2 + bx + c = a1x 2 + b1 x + c1

Þ

(a - a1 ) x2 + (b - b1 ) x + (c - c1 ) = 0.

It has only one solution, x = – 1 Þ

b - b1 = a - a1 + c - c1

Sum of roots

Þ

...(i)

-(b - b1 ) = -1 - 1 (a - a1 )

b - b1 =1 2 (a - a1 )

ÞÞ b - b1 = 2 ( a - a1 ) ...

...(ii)

Now p (– 2) = 2 Þ f (– 2) – g (– 2) = 2 Þ 4a – 2b + c – 4a1 + 2b1 – c1 = 2 Þ 4 (a – a1) – 2 (b – b1) + (c – c1) = 2 ...(iii) From equations, (i), (ii) and (iii) a - a1 = c - c1 =

1 ( b - b1 ) = 2 2

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Now, p ( 2) = f ( 2) - g (2)

Þ 12 b2 (3 c2 – 2 c2+ y ) ³ 0 [Q b2 ³ 0] Þ c2 + y ³ 0

= 4 ( a - a1 ) + 2 ( b - b1 ) + ( c - c1 )

But from eqn. (i), c2 < 4ab or – c2 > – 4ab \ we get y ³ – c2 > – 4ab

= 8 + 8 + 2 = 18 114. (a) Let the correct equation be

Þ y > – 4 ab

ax 2 + bx + c = 0 Now, Sachin’s equation

118. (c) Let a and b are roots of the equation x2 + ax + 1 = 0

ax 2 + bx + c ' = 0 Given that, roots found by Sachin’s are 4 and 3 Þ -

b =7 a

a + b = – a and ab = 1

....(i)

Given that | a - b | < 5 Þ

Rahul’s equation, ax 2 + b ' x + c = 0 Given that roots found by Rahul’s are 3 and 2

(a + b)2 - 4ab < 5

(Q

c =6 Þ ...(ii) a From (i) and (ii), roots of the correct equation

x 2 - 7 x + 6 = 0 are 6 and 1. 115. (c) Since both the roots of given quadratic equation lie in the line Re z = 1 i.e., x = 1, hence real part of both the roots are 1. Let both roots be 1 + ia and 1 – ia Product of the roots, 1 + a2 = b

(a - b) 2 = (a + b ) 2 - 4ab

)

a 2 - 4 < 5 Þ a2 – 4 < 5 Þ a2 – 9 < 0 Þ a2 < 9 Þ – 3 < a < 3 Þ a Î (–3, 3) Þ

119. (c) Given equation is x 2 - 2mx + m 2 - 1 = 0 Þ ( x - m) 2 - 1 = 0 Þ ( x - m + 1)( x - m - 1) = 0 Þ x = m - 1, m + 1

Q a +1 ³ 1

m – 1 > –2 and m + 1 < 4

2

Þ m > - 1 and m < 3 Þ -1 < m < 3

\b ³ 1 Þ Q b Î (1, ¥) 116. (b)

Þ y ³ – c2

x2 - x + 1 = 0 Þ x =

x=

1± 3 i 2

a=

1 3 +i = -w2 2 2

120. (b) Given that x 2 + px + q = 0

1± 1- 4 2

Sum of roots = tan30 + tan15 = – p Product of roots = tan30 . tan15 = q

tan 45° =

Þ – p = 1- q Þ q - p = 1 \ 2+ q - p = 3

1 i 3 b= = -w 2 2

121. (d)

a2009 + b2009 = (-w2 )2009 + (-w)2009

= -w2 - w = 1 117. (b) Given that roots of the equation bx2 + cx + a = 0 are imaginary \ c2 – 4ab < 0 Let y = 3b2x2 + 6 bc x + 2c2 Þ 3b2x2 + 6 bc x + 2c2 – y = 0

As x is real, D ³ 0 Þ 36 b2c2 – 12 b2 (2c2 – y ) ³ 0

-p tan30° + tan15° Þ =1 1 - tan30°.tan15° 1 - q

z 2 + z + 1 = 0 Þ = w or w 2

So, z + [Q w3 = 1]

....(i)

1 = w + w 2 = -1 z é 1 ù 2 2 êëQ z = w and 1 + w + w = 0 úû

z2 + z3 +

1 z2 1 z

3

= w 2 + w = -1, = w3 + w3 = 2

[Q w3 = 1]

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Mathematics

z4 +

1 z

4

= -1, z 5 +

1 z

5

Þ ( x - 2)( x - 1) = 0

= -1

Þ x = 1, 2 Þ x = ±1, ±2

1 =2 z6 \ The given sum = 1+1 + 4 + 1 + 1 + 4 = 12 and z 6 +

æ Pö æ Qö 122. (b) tan ç ÷ , tan ç ÷ are the roots of ax 2 + bx + c = 0 è 2ø è 2ø b æ Pö æ Qö tan ç ÷ + tan ç ÷ = è 2ø è 2ø a

\ No.of solution = 4 127. (b) Let one roots of given equation be a \ Second roots be 2a then a + 2a = 3a =

Þ a=

æ Pö æ Qö c tan ç ÷ × tan ç ÷ = è 2ø è 2ø a

(

1 - 3a a - 5a + 3 2

1 - 3a

)

2

3 a - 5a + 3

....(i)

2 and a.2a = 2a 2 = 2 a - 5a + 3

æ Pö æ Qö tan ç ÷ + tan ç ÷ è 2ø è 2ø æ P Qö = tan ç + ÷ = 1 è 2 2ø æ Pö æ Qö 1 - tan ç ÷ tan ç ÷ è 2ø è 2ø

é 1 (1 - 3a)2 ù 2 ú= 2 \ 2ê 2 2 9 ëê (a - 5a + 3) ûú a - 5a + 3

[from (i)] pù é êëQ P + Q = 2 úû

b b a-c a =1 Þ Þ - = c a a 1a

(1 - 3a) 2 (a 2 - 5a + 3)

=9

Þ 9a 2 - 6a + 1 = 9 a 2 - 45a + 27

-

Þ 39 a = 26 Þ a =

Þ –b=a–c Þ c=a+b 123. (d) Let a , a + 1 be roots Then a + a + 1 = b = sum of roots a (a + 1) = c = product of roots \ b 2 - 4c = (2a + 1)2 - 4a (a + 1) = 1 . 2

124. (d) Given that 4 is a root of x + px + 12 = 0

2 3

128. (d) Given that Z 2 + aZ + b = 0 ; Z1 + Z 2 = - a & Z1Z 2 = b 0, Z1, Z 2 form an equilateral triangle \ 02 + Z12 + Z 2 2 = 0.Z1 + Z1.Z 2 + Z 2 .0 (for an equilateral triangle,

Þ 16 + 4 p + 12 = 0 Þ p = -7

Z12 + Z 22 + Z32 = Z1Z 2 + Z 2 Z 3 + Z3Z1 )

Now, the equation x 2 + px + q = 0

2 2 Þ Z1 + Z 2 = Z1Z 2

has equal roots. \D =0

Þ ( Z1 + Z 2 ) 2 = 3Z1Z 2

p 2 49 = 4 4 125. (c) Let the second root be a. 2 Þ p - 4q = 0 Þ q =

Then a + (1 - p ) = - p Þ a = -1 Also a.(1 - p) =1 - p Þ (a - 1)(1 - p) = 0 Þ p =1 [Q a = -1] \ Roots are a = -1 and 1 - p = 0 126. (c) Given that

\ a 2 = 3b 129. (a) p + q = – p Þ q = 2p and pq = q Þ q (p – 1) = 0 Þ q = 0 or p = 1. If q = 0, then p = 0. or p = 1, then q = –2. c 9 = > 0, " t Î R a t2 \ Product of real roots is always positive.

130. (a) Product of real roots =

x 2 - 3 x + 2 = 0 Þ| x |2 -3 | x | +2 = 0

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131. (a) Let a and b are r oots of th e equation x2 + ax + b = 0 and g and d be the roots of the equation x2 + bx + a = 0 respectively. \ a + b = –a, ab = b and g + d = –b, g d = a. Given |a – b| = |g – d| Þ (a – b)2 = (g – d)2 Þ (a+ b)2 – 4ab = (g + d)2 – 4gd Þ a2 – 4b = b2 – 4a Þ (a2 – b2) + 4(a – b)= 0 Þ a+ b+4=0 (Q a ¹ b) 132. (a) Given that a2 = 5a – 3 and b2 = 5b – 3; Þ a & b are roots of equation, x2 = 5x – 3 or x2 – 5x + 3 = 0 \ a + b = 5 and ab = 3 Thus, the equation whose roots are

a b and is b a

For x2 – 2bx – 10 = 0 a + b = 2b and ab = – 10

...(ii) ...(iii)

b is also root of x2 – 2bx – 10 = 0 a Þ b2 – 2ab2 – 10a2 = 0 By eqn. (i) Þ 5a – 10a2 – 10a2 = 0 Þ 20a2 = 5a a=

1 5 a = and b2 = 4 4 a2 = 20 and b2 = 5 Now, a2 + b2 = 5 + 20 = 25 135. (b) Let, the roots of the equation, x2 + (2 – l) x + (10 – l) = 0 are a and b. Also roots of the given equation are

Þ

æ a b ö ab =0 x2 - x ç + ÷ + è b aø ab

l – 2 ± 4 – 4 l + l 2 – 40 + 4l l – 2 ± l 2 – 36 = 2 2

æ a 2 + b2 ö Þ x2 - x ç ÷ +1 = 0 è ab ø

The magnitude of the difference of the roots is

or 3x2 – 19x +3 = 0

So, a3 + b3 =

P 5r2

133. (c) Q

(l – 2)3 3 (l – 2) (l 2 – 36) + 4 4

(l – 2) (4l 2 – 4l – 104) = (l – 2) (l 2 – l – 26) = f (l ) 4 As f (l) attains its minimum value at l = 4. Therefore, the magintude of the difference of the roots is

=

5r R

5

DPQR is possible if 5 + 5r > 5r2

|i

Þ

1 + r > r2

Þ

r2 – r – 1 < 0

Þ

æ 1 5öæ 1 5ö çr - 2 + 2 ÷ çr - 2 - 2 ÷ < 0 è øè ø

Þ

æ - 5 + 1 5 + 1ö r Îç , 2 ÷ø è 2

20 |= 2 5

136. (b) | z – (3 – 2i) | £ 4 represents a circle whose centre is (3, – 2) and radius = 4. | z | = | z – 0 | represents the distance of point ‘z’ from origin (0, 0) y

R 4

x

O

7 æ - 5 + 1 5 + 1ö 7 Ï , \ r¹ 4 çè 2 2 ÷ø 4

Q

G

2b a

Þ a=

Þ

b2 = 5a

S

b a

2 and product of roots = a =

(a ¹ 0)

(3 , –2 ) 4

134. (a) ax2 – 2bx + 5 = 0, If a and a are roots of equations, then sum of roots 2a =

l 2 – 36

2

5 5 b Þ 2= a a a ...(i)

Suppose RS is the normal of the circle passing through origin ‘O’ and G is its center (3, – 2). Here, OR is the least distance and OS is the greatest distance OR = RG – OG and OS = OG + GS ...(i)

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Mathematics

As, RG = GS = 4 OG =

139. (b) Let a, b be the common roots of both the equations.

32 + ( - 2) 2 = 9 + 4 = 13

For first equation ax 2 + bx + c = 0 , we have

From (i), OR = 4 - 13 and OS = 4 + 13

(

) (

So, required difference = 4 + 13 - 4 - 13

)

= 13 + 13 = 2 13 a.b = 137. (c)

x 2 + bx - 1 = 0 common root x2 + x + b = 0 -

- x=

b +1 b -1

b +1 Put x = in equation b -1

-b a

a+b= c a

...(ii)

For second equation 2 x 2 + 3 x + 4 = 0 , we have -3 2

a+b=

...(iii)

2 ...(iv) 1 Now, from (i) & (iii) & from (ii) & (iv)

a.b=

æ b + 1 ö æ b +1 ö ç ÷ +ç ÷ +b =0 è b - 1 ø è b -1 ø

-b -3 = a 2

(b + 1)2 + (b + 1) (b – 1) + b (b – 1)2 = 0 b2 + 1 + 2b + b2 – 1 + b (b2 – 2b + 1) = 0 2b2 + 2b + b3– 2b2 + b = 0 b3 + 3b = 0 b(b2 + 3) = 0 b2 = – 3

b 3/ 2 = a 1

2

...(i)

c 2 = a 1

3 &c=2 2 putting these values in first equation, we get

Therefore on comparing we get a = 1, b =

3 x + 2 = 0 or 2 x 2 + 3x + 4 = 0 2 from this, we get a = 2, b = 3; c = 4 or a : b : c = 2 : 3 : 4 140. (a) Given equations are x2 + 2x + 3 = 0 …(i) ax2 + bx + c = 0 …(ii) Roots of equation (i) are imaginary roots in order pair. According to the question (ii) will also have both roots same as (i). Thus x2 +

b = ± 3i |b| = 3 138. (d) We have 2 2 f (x) = x + 2bx + 2c

and g(x) = - x 2 - 2cx + b2 , ( x Î R ) Þ f (x) = ( x + b)2 + 2c2 - b2 2 2 2 and g(x) = -( x + c) + b + c Now, fmin = 2c2 – b2 and gmax = b2 + c2 Given : min f (x) > max g(x)

Þ 2c 2 - b 2 > b 2 + c 2 Þ c 2 > 2b 2 Þ |c| > | b | 2

a b c = = = l (say) 1 2 3 Þ a = l, b = 2l, c = 3l Hence, required ratio is 1 : 2 : 3

Þ

|c| c > 2Þ > 2 |b| b

Þ

c Î ( 2, ¥) . b

x-5

> 0 Þ x2 + 5x – 14 < x – 5 x + 5x - 14 Þ x2 + 4x – 9 < 0 Þ a = – 5, – 4, – 3, – 2, – 1, 0, 1 a = – 5 does not satisfy any of the options a = – 4 satisfy the option (a) a2 + 3a – 4 = 0

141. (a)

2

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142. (c) x2 – (a + 1) x + a2 + a – 8 = 0 Since roots are different, therefore D > 0 Þ (a + 1)2 – 4(a2 + a – 8) > 0 Þ (a – 3) (3a + 1) < 0 There are two cases arises. Case I. a – 3 > 0 and 3a + 1 < 0

z

=

£

z

Þ

z

Þ

æ 2 + 20 ö çç z – ÷÷ 2 è ø

Þ

(z

2

(-

Þ

z

)

5 +1 £ max

+

5) (1 + 5)

(

Q x is real

x=5

)

z £

D³0

0

– (1 - 5) £ 0

(1 -

3x 2 ( y - 1) + 9 x( y - 1) + 7 y - 17 = 0

4 4 + z z

æ 2 - 20 ö çç z – ÷÷ £ 0 2 è ø

)(z

+ –¥

Þ

z-

–2 z -4£ 0

– (1 + 5)

3 x2 + 9 x + 7

Y axis

4 z

£ 2+

Þ

3 x 2 + 9 x + 17

Þ ( y - 1)( y - 41) £ 0 Þ 1 £ y £ 41 \ Max value of y is 41 – + + 8 –8 1 41 146. (c) Given that both roots of quadratic equation are less than 5 then (i)

4 =2 z

4 4 + z z

z-

y=

81( y - 1) 2 - 4 ´ 3( y - 1)(7 y - 17) ³ 0

11 3 Hence, no solution in this case Case II : a – 3 < 0 and 3a + 11 > 0 11 Þ a < 3 and a > 3 11 \ 0 Þ k2 – 9k + 20 > 0 Þ k (k – 4) –5(k – 4) > 0 Þ (k – 5) (k – 4) > 0 – + + ¥ –¥ 4 5 Þ k Î ( – ¥ , 4 ) U (5, ¥ ) (iii)

Sum of roots 0 and m ¹

Case 1: when y > 2 y2 – y + 2 = y – 1 + y – 2 D < 0 [ \ Equation not satisfy.]

Case 2: when 1 £ y £ 2 y2 – y2 + 2 = y – 1 – y + 2

7.

(3) Let the given quadratic expression (1 + 2m)x2 – 2(1 + 3m)x + 4(1 + m), is positive for all x Î R, then 1 + 2m > 0

y2 – y + 1 = 0 Q

é 1 ù êëQ 2 is not an integer úû

Þ number of integral values of m are infinitely many.

y2 – 3y + 5 = 0 Q

1 2

D < 0 [ \ Equation not satisfy.]

Case 3: when y £ 1 y2 – y + 2 = – y + 1 – y + 2

...(i)

D -

=1 ...(i)

For x = 1

1 2

31 + 41 > 51 For x = 3

m Î ( 3 - 2 3, 3 + 2 3 )

33 + 43 = 91 < 53

Then, integral values of m = {0, 1, 2, 3, 4, 5, 6}

Only for x = 2, equation (i) Satisfy

Hence, number of integral values of m = 7

So, only one solution (x = 2)

x

8.

x

Þ

(b)

x

æ 3ö æ 4ö f (x) = ç ÷ + ç ÷ - 1 è 5ø è 5ø

Þ

(a) Q (a – b)2 + (b – c)2 + (c – a)2 > 0

Þ 2(a2 + b2 + c2 – ab – bc – ca) > 0 [Q a2 + b2 + c2 = 1]

Put f (x) = 0 x

9.

Þ 2 > 2(ab + bc + ca) Þ ab + bc + ca < 1 x

æ 3ö æ 4ö 0 = ç ÷ + ç ÷ -1 è 5ø è 5ø

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Permutations and Combinations

7

Permutations and Combinations TOPIC Ć

1.

2.

3.

4.

5.

Fundamental Principle of Counting, Factorials, Permutations, Counting Formula for Permutations, Permutations in Which Things may be Repeated, Permutations in Which all Things are Different, Number of Permutations Under Certain Restricted Conditions, Circular Permutations

Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated? [Sep. 06, 2020 (I)] (a) 2! 3! 4! (b) (3!)3 × (4!) (c) (3!)2 × (4!) (d) 3! (4!)3 The value of (2 × 1P0 - 3 × 2 P1 + 4 × 3 P2 - ... up to 51th term) + (1! – 2! + 3! – ... up to 51th term) is equal to : [Sep. 03, 2020 (I)] (a) 1 – 51(51)! (b) 1 + (51)! (c) 1 + (52)! (d) 1 If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without meaning) be listed as in a dictionary, then the position of the word 'MOTHER' is ________. [NA Sep. 02, 2020 (I)] If the number of five digit numbers with distinct digits and 2 at the 10th place is 336 k, then k is equal to: [Jan. 9, 2020 (I)] (a) 4 (b) 6 (c) 7 (d) 8 Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is: [Jan. 7, 2020 (I)] 1 (6!) (a) 2

(b) 6!

(c) 56

(d)

5 (6!) 2

6.

The number of 6 digit numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no digit is repeated, is: [April 10, 2019 (I)] (a) 72 (b) 60 (c) 48 (d) 36

7.

The number of four-digit numbers strictly greater than 4321 that can be formed using the digits 0, 1, 2, 3, 4, 5 (repetition of digits is allowed) is: [April 08, 2019 (II)] (a) 288 (b) 360 (c) 306 (d) 310 Consider three boxes, each containing 10 balls labelled 1, 2,..., 10. Suppose one ball is randomly drawn from each of the boxes. Denote by n i, the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is : [Jan. 12, 2019 (I)] (a) 120 (b) 82 (c) 240 (d) 164 The number of natural numbers less than 7,000 which can be formed by using the digits 0,1,3,7,9 (repetition of digits allowed) is equal to: [Jan. 09, 2019 (II)] (a) 374 (b) 372 (c) 375 (d) 250 Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is:[Jan. 09, 2019 (II)] (a) 9 (b) 18 (c) 36 (d) 32 The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4, (repetition of digits is not allowed) and are multiple of 3 is? [Online April 16, 2018] (a) 30 (b) 48

8.

9.

10.

11.

(c) 24

(d) 36

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12.

Mathematics

n – digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is [Online April 15, 2018] (a) 6 (b) 8 (c) 9

13.

14.

15.

21.

(d) 7

The number of ways in which 5 boys and 3 girls can be seated on a round table if a particular boy B1 and a particular girl G1 never sit adacent to each other, is : [Online April 9, 2017] (a) 5 × 6! (b) 6 × 6! (c) 7! (d) 5 × 7! If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is : [Online April 8, 2017] (a) 44th (b) 45th (c) 46th (d) 47th If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is : [2016] (a) 52nd (b) 58th (c) 46th (d) 59th

22.

23.

24.

10

16.

17.

The sum

å (r 2 + 1) × (r!) is equal to : r =1

[Online April 10, 2016] (a) 11 × (11!) (b) 10 × (11!) (c) (11!) (d) 101 × (10!) If the four letter words (need not be meaningful) are to be formed using the letters from the word "MEDITERRANEAN" such that the first letter is R and the fourth letter is E, then the total number of all such words is : [Online April 9, 2016] (a) 110 (b) 59

25.

26.

11!

(c) 18.

19.

(2!)3

The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0) is : [2015] (a) 820 (b) 780 (c) 901 (d) 861 The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is : [2015] (a) 120 (b) 72 (c) 216

20.

(d) 56

(d) 192

The number of ways of selecting 15 teams from 15 men and 15 women, such that each team consists of a man and a woman, is: [Online April 10, 2015]

27.

28.

(a) 1120 (b) 1880 (c) 1960 (d) 1240 Two women and some men participated in a chess tournament in which every participant played two games with each of the other participants. If the number of games that the men played between themselves exceeds the number of games that the men played with the women by 66, then the number of men who participated in the tournament lies in the interval: [Online April 19, 2014] (a) [8, 9] (b) [10, 12) (c) (11, 13] (d) (14, 17) 8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such numbers in which the odd digits do no occupy odd places, is: [Online April 12, 2014] (a) 160 (b) 120 (c) 60 (d) 48 An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The number of ways in which this can be done is: [Online April 11, 2014] (a) 72 (7!) (b) 18 (7!) (c) 40 (7!) (d) 36 (7!) The sum of the digits in the unit’s place of all the 4-digit numbers formed by using the numbers 3, 4, 5 and 6, without repetition, is: [Online April 9, 2014] (a) 432 (b) 108 (c) 36 (d) 18 5 - digit numbers are to be formed using 2, 3, 5, 7, 9 without repeating the digits. If p be the number of such numbers that exceed 20000 and q be the number of those that lie between 30000 and 90000, then p : q is : [Online April 25, 2013] (a) 6 : 5 (b) 3 : 2 (c) 4 : 3 (d) 5 : 3 Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is: [2012] (a) 880 (b) 629 (c) 630 (d) 879 If seven women and seven men are to be seated around a circular table such that there is a man on either side of every woman, then the number of seating arrangements is [Online May 26, 2012] (a) 6! 7! (b) (6!)2 (c) (7!)2 (d) 7! If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number [2005] (a) 601

(b) 600

(c) 603

(d) 602

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Permutations and Combinations

29.

30.

31.

32.

33.

34.

How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order[2004] (a) 480 (b) 240 (c) 360 (d) 120 The range of the function f ( x) =7 - x Px -3 is [2004] (a) {1, 2, 3, 4, 5} (b) {1, 2, 3, 4, 5, 6} (c) {1, 2, 3, 4,} (d) {1, 2, 3,} The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by [2003] (a) 6! × 5! (b) 6 × 5 (c) 30 (d) 5 × 4 The sum of integers from 1 to 100 that are divisible by 2 or 5 is [2002] (a) 3000 (b) 3050 (c) 3600 (d) 3250 Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is [2002] (a) 125 (b) 105 (c) 374 (d) 625 Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are [2002] (a) 216 (b) 375 (c) 400 (d) 720

TOPIC n

35.

36.

37.

38.

Combinations, Counting Formula for Combinations, Division and Distribution of Objects, Dearrangement Theorem, Sum of Numbers, Important Result About Point

The number of words (with or without meaning) that can be formed from all the letters of the word “LETTER” in which vowels never come together is ______. [NA Sep. 06, 2020 (II)] The number of words, with or without meaning, that can be formed by taking 4 letters at a time from the letters of the word ‘SYLLABUS’ such that two letters are distinct and two letters are alike, is ______. [NA Sep. 05, 2020 (I)] There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which the candidate can choose the questions, is : [Sep. 05, 2020 (II)] (a) 3000 (b) 1500 (c) 2255 (d) 2250 A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is __________. [NA Sep. 04, 2020 (II)]

39. 40.

41. 42.

43.

The total number of 3-digit numbers, whose sum of digits is 10, is _________. [NA Sep. 03, 2020 (II)] Let n > 2 be an integer. Suppose that there are n Metro stations in a city located along a circular path. Each pair of stations is connected by a straight track only. Further, each pair of nearest stations is connected by blue line, whereas all remaining pairs of stations are connected by red line. If the number of red lines is 99 times the number of blue lines, then the value of n is : [Sep. 02, 2020 (II)] (a) 201 (b) 200 (c) 101 (d) 199 If Cr º 25Cr and C0 + 5×C1 + 9×C2 + ... + (101)×C25 = 225×k, then [NA Jan. 9, 2020 (II)] k is equal to ________. An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is _________. [NA Jan. 8, 2020 (I)] If a, b and c are the greatest values of 19Cp, 20Cq and 21Cr respectively, then: [Jan. 8, 2020 (I)] (a)

a b c = = 11 22 21

(b)

a b c = = 10 11 21

a b c a b c = = = = (d) 11 22 42 10 11 42 The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of the word ‘EXAMINATION’ is _________. [NA Jan. 8, 2020 (II)] The number of ordered pairs (r, k) for which 6. 35Cr = (k2 – 3).36Cr + 1, where k is an integer, is: [Jan. 7, 2020 (II)] (a) 3 (b) 2 (c) 6 (d) 4

(c)

44. 45.

46.

The number of ways of choosing 10 obects out of 31 obects of which 10 are identical and the remaining 21 are distinct is: [April 12, 2019 (I)] (a) 220 – 1 (b) 221 (c) 220 (d) 220+1

47.

A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to : [April 12, 2019 (II)] (a) 28 (b) 27 (c) 25 (d) 24

48.

Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the top of each pillar has been connected by beams with the top of all its non-adacent pillars, then the total number of beams is : [April 10, 2019 (II)] (a) 170 (b) 180 (c) 210 (d) 190

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49.

50.

51.

Mathematics

A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then: [April 9, 2019 (I)] (a) m + n = 68 (b) m = n = 78 (c) n = m – 8 (d) m = n = 68 All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is : [April 8, 2019 (I)] (a) 180 (b) 175 (c) 160 (d) 162 There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is [Jan. 12, 2019 (II)] (a) 12 (b) 11 (c) 9 (d) 7 20

52.

53.

54.

If

å

i =1

56.

n+2

57.

58.

59.

60.

61.

62.

63.

(d) 270

From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is : [2018] (a) less than 500 (b) at least 500 but less than 750 (c) at least 750 but less than 1000 (d) at least 1000 A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and

n-2

C6 P2

= 11, then n satisfies the equation : [Online April 10, 2016] (b) n2 + 2n – 80 = 0 (d) n2 + 5n – 84 = 0

æ 15 C r ö r2 ç ÷ ç 15 C ÷ is equal to : r =1 r -1 ø è [Online April 9, 2016] (a) 1240 (b) 560 (c) 1085 (d) 680 Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A × B, each having at least three elements is : [2015] (a) 275 (b) 510 (c) 219 (d) 256 If in a regular polygon the number of diagonals is 54, then the number of sides of this polygon is [Online April 11, 2015] (a) 12 (b) 6 (c) 10 (d) 9 Let A and B two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B having 3 or more elements is [2013] (a) 256 (b) 220 (c) 219 (d) 211 Let Tn be the number of all possible triangles formed by oining vertices of an n-sided regular polygon. If [2013] Tn+1 – Tn = 10, then the value of n is : (a) 7 (b) 5 (c) 10 (d) 8 On the sides AB, BC, CA of a DABC, 3, 4, 5 distinct points (excluding vertices A, B, C) are respectively chosen. The number of triangles that can be constructed using these chosen points as vertices are : [Online April 23, 2013] (a) 210 (b) 205 (c) 215 (d) 220 The number of ways in which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question, is : [Online April 22, 2013] 15

3

[Jan. 10, 2019 (I)] (a) 400 (b) 50 (c) 200 (d) 100 Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is: [Jan. 9, 2019 (I)] (a) 500 (b) 200 (c) 300 (d) 350 The number of four letter words that can be formed using the letters of the word BARRACK is [Online April 15, 2018] (a) 144 (b) 120

If

(a) n2 + n – 110 = 0 (c) n2 + 3n – 108 = 0

20 æ ö Ci - 1 k ç 20 ÷ = , then k equals: 20 ç Ci + Ci - 1 ÷ 21 è ø

(c) 264 55.

Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is : [2017] (a) 484 (b) 485 (c) 468 (d) 469

64.

The value of å

(a)

C7

(b)

21

C8

(d) 30 C8 C7 A committee of 4 per sons is to be formed from 2 ladies, 2 old men and 4 young men such that it includes at least 1 lady, at least 1 old man and at most 2 young men. Then the total number of ways in which this committee can be formed is : [Online April 9, 2013] (a) 40 (b) 41 (c) 16 (d) 32 (c)

65.

30 21

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Permutations and Combinations

66.

67.

68.

69.

The number of arrangements that can be formed from the letters a, b, c, d, e, f taken 3 at a time without repetition and each arrangement containing at least one vowel, is [Online May 19, 2012] (a) 96 (b) 128 (c) 24 (d) 72 If n = mC2, then the value of nC2 is given by [Online May 19, 2012] (b) m – 1 C4 (a) 3(m + 1C4) (c) m + 1C4 (d) 2(m + 2C4) Statement 1: If A and B be two sets having p and q elements respectively, where q > p. Then the total number of functions from set A to set B is qp. [Online May 12, 2012] Statement 2: The total number of selections of p different obects out of q obects is qCp. (a) Statement 1 is true, Statement 2 is false. (b) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1. (c) Statement 1 is false, Statement 2 is true (d) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1. If the number of 5-elemen t subsets of the set A= {a1, a2, ...., a20} of 20 distinct elements is k times the number of 5-element subsets containing a4, then k is [Online May 7, 2012] (a) 5

70.

71.

72.

73.

(b)

74.

75.

Thus A È B È C = S, A Ç B = B Ç C = A Ç C = f. The number of ways to partition S is [2007] 12!

(a)

(4!)

76.

77.

78.

80.

(4!) 4 12!

(d)

3

The value of

50

C4 +

6

å 56 - r C3

is

[2005]

r =1

(a)

55

C4

(b)

(c)

56

C3

(d)

55

C3

56

C4

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is [2004] 8

C3

(b) 21

(c) 38 (d) 5 A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is [2003] (a) 346 (b) 140 (c) 196 (d) 280 If nCr denotes the number of combination of n things taken r at a time, then the expression n

Cr +1 + nC r -1 + 2´n Cr equals

(a)

n +1

Cr +1

n+2

81.

12!

3!(4!) 4 3!(4!) At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is [2006] (a) 5040 (b) 6210 (c) 385 (d) 1110

(a)

79.

(b)

3

12!

(c)

20 7

10 (c) 4 (d) 3 There are 10 points in a plane, out of these 6 are collinear. If N is the number of triangles formed by oining these points. Then : [2011RS] (a) N £ 100 (b) 100 < N £ 140 (c) 140 < N £ 190 (d) N > 190 Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9C3 . Statement-2: The number of ways of choosing any 3 places from 9 different places is 9C3. [2011] (a) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is [2010] (a) 36 (b) 66 (c) 108 (d) 3 From 6 different novels and 3 different dictionaries,4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is: [2009]

(a) at least 500 but less than 750 (b) at least 750 but less than 1000 (c) at least 1000 (d) less than 500 How many different words can be formed by umbling the letters in the word MISSISSIPPI in which no two S are adacent? [2008] (a) 8. 6C4. 7C4 (b) 6.7. 8C4 (c) 6. 8. 7C4. (d) 7. 6C4. 8C4 The set S = {1, 2, 3, ......., 12} is to be partitioned into three sets A, B, C of equal si e.

(b)

[2003] n+ 2 n +1

Cr

(d) (c) Cr Cr +1 Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4, 6 and 7 without repetition. Total number of such numbers are [2002] (a) 312 (b) 3125 (c) 120 (d) 216

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1.

Mathematics

(b) Number of arrangement

Number of four-digit numbers starting with 44 is,

= (3! ´ 3! ´ 4!) ´ 3! = (3!)3 4! 2.

4

r! (c) We know, (r + 1) × r Pr -1 = ( r + 1) × = ( r + 1)! 1!

= 6 × 6 = 36 6 6 Number of four-digit numbers starting with 43 and greater than 4321 is,

So, (2 × 1P0 - 3 × 2 P1 + .....51 terms) +

3.

(1! – 2! + 3! – ... upto 51 terms) = [2! – 3! + 4! – ... + 52!] + [1! – 2! + 3! – ... + 51!] = 52! + 1! = 52! + 1 (309)

4

Þ 2 ´ 5! + 2 ´ 4! + 3 ´ 3! + 2! + 1 4.

5.

6.

(d) Five digits numbers be 1, 3, 5, 7, 9 For selection of one digit, we have 5C1 choice. 6! And six digits can be arrange in ways. 2! 5.6! 5 = .6! Hence, total such numbers = 2! 2 (b) Given digit 0, 1, 2, 5, 7, 9 a1 a 2

7.

a3 a 4

a5

5

= 6 × 6 × 6 = 216 6

6

6

Number of four-digit numbers starting with 45 is, 4

5

= 6 × 6 = 36 6

4

3

2 =4

8.

9.

a6

(a1 + a3 + a5) – (a2 + a4 + a6 ) = 11 K Therefore, (1, 2, 9) (0, 5, 7) Number of ways to arranging them = 3! × 3! + 3! × 2 × 2 = 6 × 6 + 6 × 4 = 6 × 10 = 60 (d) 0, 1, 2, 3, 4, 5 Number of four-digit number starting with 5 is,

3

= 3 × 6 = 18 (5/4/3) 3 6 Number of four-digit numbers starting with 432 and greater than 4321 is,

M O T H E R 3 4 6 2 1 5

= 240 + 48 + 18 + 2 + 1 = 309 (d) Number of five digit numbers with 2 at 10th place = 8 × 8 × 7 × 6 = 2688 Q It is given that, number of five digit number with 2 at 10th place = 336k \ 336 k = 2688 Þ k = 8

4

10.

4 Hence, required numbers = 216 + 36 + 36 + 18 + 4 = 310. (a) Collecting different labels of balls drawn = 10 × 9 × 8 Q arrangement is not required. \ the number of ways in which the balls can be chosen is, 10 ´ 9 ´ 8 = 120 3! (a) Number of numbers with one digit = 4 = 4 Number of numbers with two digits = 4 ´ 5 = 20 Number of numbers with three digits = 4 ´ 5 ´ 5 = 100 Number of numbers with four digits = 2 ´ 5 ´ 5 ´ 5 = 250 \ Total number of numbers = 4 + 20 + 100 + 250 = 374 (c) One of the possible DOAB is A(a, 0) and B(0, b).

Area of DOAB =

1 |ab| . 2

\

|ab| = 100 |a| |b| = 100 But 100 = 1 ´ 100, 2 ´ 50, 4 ´ 25, 5 ´ 20 or 10 ´ 10 \ For 1 ´ 100, a = 1 or –1 and b = 100 or – 100 \ Total possible pairs are 8. Total possible pairs for 1 ´ 100, 2 ´ 50, 4 ´ 25 or 5 ´ 20 are 4 ´ 8. And for 10 ´ 10 total possible pairs are 4. \ Total number of possible triangles with integral coordinates are 4 ´ 8 + 4 = 36.

6

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Permutations and Combinations

11.

(a) The thousands place can only be filled with 2, 3 or 4, since the number is greater than 2000. For the remaining 3 places, we have pick out digits such that the resultant number is divisible by 3. It the sum of digits of the number is divisible by 3, then the number itself is divisible by 3. Case 1: If we take 2 at thousands place.

15.

(b) ALLMS No. of words starting with

_____ A: A

4! = 12 2!

_ _ _ _ _ 4! = 24 L : L _____ M: M

The remaining digits can be filled as:

4! = 12 2!

0, 1 and 3 as 2 + 1 + 0 + 3 = 6 is divisible by 3. _A _____ S : S

0, 3 and 4 as 2+ 3 + 0 + 4 = 9 is divisible by 3. In both the above combinations the remaining three digits can be arranged in 3! ways.

: S_ L_ _ _ _ 3! = 6

\ Total number of numbers in this case = 2 × 3! = 12. Case 2: If we take 3 at thousands place. The remaining digits can be filled as: 0, 1 and 2 as 3 + 1 + 0 + 2 = 6 is divisible by 3.

SMALL ® 58 th word 10

16.

0, 2 and 4 as 3 + 2 + 0 + 4 = 9 is divisible by 3.

R -1

T1 = r r + r – (r – 1) r

Case 3: If we take 4 at thousands place.

T2 = 2 3 – 1 2

T1 = 1 2 – 0

T3 = 3 4 – 2 3 T10 = 10 11– 9 10

In the above combination, the remaining three digits can be arranged in 3! ways.

10

å (r 2 + 1) r = 10 11

\ Total number of numbers in this case = 3! = 6. \ Total number of numbers between 2000 and 5000 divisible by 3 are 12 + 12 + 6 = 30.

R -1

17.

(d) Required n digit numbers is 3n as each place can be filled by 2, 5, 7. So smallest value of n such that 3n > 900. Therefore n = 7. (a) 4 boys and 2 girls in circle Þ 5! ´

14.

å (r 2 + 1) r

T1 = (r2 + 1 + r – r) r = (r2 + r) r – (r – r) r

0, 2 and 3 as 4 + 2 + 0 + 3 = 9 is divisible by 3.

13.

(b)

In both the above combinations, the remaining three digits can be arranged in 3! ways. Total number of numbers in this case = 2 × 3! = 12. The remaining digits can be filled as:

12.

3! =3 2!

6! ´ 2! 4!2!

18.

Þ 5 ´ 6! (c) E, E, N, Q, U (i) E ................. = 4! = 24

(b) M, EEE, D. I, T, RR, AA, NN R––E Two empty places can be filled with identical letters [EE, AA, NN] Þ 3 ways Two empty places, can be filled with distinct letters [M, E, D, I, T, R, A, N] Þ 8P2 \ Number of words 3 + 8P2 = 59 (b) Total number of integral points inside the square OABC = 40 × 40 = 1600 No. of integral points on AC

(0, 41) C

(41, 41) B

4! = 12 2 (iii) Q E ............... = 3! = 6 (ii) N ................. =

(iv) Q N ............... =

3! =3 2!

(v) Q U E E N = 1 \ Required rank = (24) + (12) + (6) + (c) + (a) = 46th

O (0, 0)

A (41, 0)

= No. of integral points on OB

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Mathematics

= 40 [namely (1, 1), (2, 2) ... (40, 40)] \ =

23.

No. of integral points inside the DOAC 1600 – 40 = 780 2

(d) Four digits number can be arranged in 3 × 4! ways. Five digits number can be arranged in 5! ways. Number of integers = 3 × 4! + 5! = 192. 20. (d) Number of ways of selecting a man and a woman for a team from 15 men and 15 women = 15 × 15 = (15)2 Number of ways of selecting a man and a woman for next team out of the remaining 14 men and 14 women. = 14 × 14 = (14)2 Similarly for other teams Hence required number of ways 19.

= (15)2 + (14)2 + .... + (1)2 = 21.

22.

Digits which are not used Number of 8 digits to form 8 digits number numbers which are divisible by 9 divisible by 9 0 and 9 8 ´ 7!

15 ´16 ´ 31 = 1240 6

(b) Let no. of men = n No. of women = 2 Total participants = n + 2 No. of games that M1 plays with all other men = 2(n – 1) These games are played by all men M2, M3, ......, Mn. So, total no. of games among men = n.2(n – 1). However, we must divide it by ‘2’, since each game is counted twice (for both players). So, total no. of games among all men = n(n – 1) ....... (i) Now, no. of games M1 plays with W1 and W2 = 4 (2 games with each) Total no. of games that M1, M2, ....., Mn play with W1 and W2 = 4n ....... (ii) Given : n(n – 1) – 4n = 66 Þ n = 11, –6 As the number of men can't be negative. So, n = 11 (b) In 8 digits numbers, 4 places are odd places. Also, in the given 8 digits, there are three odd digits 1, 1 and 3. No. of ways three odd digits arranged at four even

24.

25.

=

4! 5! ´ = 120 2! 3!2!

7 ´ 7!

2 and 7 3 and 6

7 ´ 7! 7 ´ 7!

4 and 5

7 ´ 7!

Hence total number of 8 digits numbers which are divisible by 9 = 8 × (7!) + 7 × (7!) + 7 × (7!) + 7 × (7!) + 7 × (7!) = 36 × (7!) (b) With 3 at unit place, total possible four digit number (without repetition) will be 3 ! = 6 With 4 at unit place, total possible four digit numbers will be 3! = 6 With 5 at unit place, total possible four digit numbers will be 3! = 6 With 6 at unit place, total possible four digit numbers will be 3! = 6 Sum of unit digits of all possible numbers =6×3+6×4+6×5+6×6 = 6 [3 + 4 + 5 + 6] = 6 [18]= 108 (d)

q:

4 P3 4! = 2! 2! No. of ways the remaining five digits 2, 2, 2, 4 and 4 5! 3!2! Hence, required number of 8 digits number

1 and 8

p : 0 0 0 0 0 place 5 4 3 2 1 ways

Total no. of ways = 5 ! = 120 Since all numbers are > 20,000 \ all numbers 2, 3, 5, 7, 9 can come at first place.

places =

arranged at remaining five places =

(d) We know that any number is divisible by 9 if sum of the digits of the number is divisible by 9. Now sum of the digits from 0 to 9 = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 Hence to form 8 digits numbers which are divisible by 9, a pair of digits either 0 and 9, 1 and 8, 2 and 7, 3 and 6 or 4 and 5 are not used.

0 0 0 0 0 place 3 4 3 2 1 ways

Total no. of ways = 3 × 4 ! = 72 (Q 2 and 9 can not be put at first place) So, p : q = 120 : 72 = 5 : 3 26.

(d) Given that number of white balls = 10 Number of green balls = 9 and Number of black balls = 7 \ Required probability

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Permutations and Combinations

= (10 + 1) (9 + 1) (7 + 1) – 1 = 11.10.8 –1 = 879 [Q The total number of ways of selecting one or more items from p identical items of one kind, q identical items of second kind; r identical items of third kind is (p + 1) (q + 1) (r + 1) –1 ] 27. (a) 7 women can be arranged around a circular table in 6! ways. Among these 7 men can sit in 7! ways. Hence, number of seating arrangement = 7! × 6! 28. (a) Alphabetical order is A, C, H, I, N, S No. of words starting with A = 5! = 120 No. of words starting with C = 5! = 120 No. of words starting with H = 5! = 120 No. of words starting with I = 5! = 120 No. of words starting with N = 5! = 120 SACHIN – 1 \ Sachin appears at serial no. 601 29. (c) Total number of arrangements of letters in the word GARDEN = 6 ! = 720 there are two vowels A and E, in half of the arrangements A preceeds E and other half A follows E. So, numbers of word with vowels in alphabetical order in

32.

33. 34. 35.

7- x

4! 2!

Then put both E in 5 gaps formed in 5 C2 ways.

\ No. of ways = 36.

4! 5 × C2 = 120 2!

(240)

S ® 2, L ® 2, A, B, Y ,U . 2 5 \ Required number of ways = C1 ´ C2 ´

37.

4! = 240. 2!

(d) Since, each section has 5 questions. \ Total number of selection of 5 questions

= 3 ´ 5C1 ´ 5C1 ´ 5C3 + 3 ´ 5C1 ´ 5C2 ´ 5C2

\ x = 3, 4, 5

= 3 × 5 × 5 × 10 + 3 × 5 × 10 × 10 = 750 + 1500 = 2250. (135) Select any 4 correct questions in 6C4 ways. Number of ways of answering wrong question = 3

\ f (3) =7 -3 P3-3 =4 P0 = 1

\ Required number of ways = 6C4 (1)4 ´ 32 = 135.

(d)

Px -3 is defined if

7 - x ³ 0, x - 3 ³ 0 and 7 - x ³ x - 3 Þ 3 £ x £ 5 and x ÎI

38.

39.

\ f (4) =7 - 4 P4 -3 =3 P1 = 3

31.

(120.00) For vowels not together Number of ways to arrange L, T, T, R =

1 ´ 720 = 360 2

30.

(b) Required sum = (2 + 4 + 6 + ... + 100) + (5 + 10 + 15 + ... + 100) – (10 + 20 + ... + 100) = 2(1 + 2 + 3 ...+ 50) + 5(1 + 2 + 3 + ...+50) –10(1 + 2 + 3 + ....+10) = 2550 + 1050 – 530 = 3050. (c) Total number of numbers = 3 ´ 5 ´ 5 ´ 5 –1 = 374 (d) Total number of numbers formed using 0, 1, 2, 3, 5, 7 = 5 ´ 6 ´ 6 ´ 4 = 36 ´ 20 = 720.

(54) Let xyz be the three digit number

\ f (5) =7 -5 P5 -3 =2 P2 = 2

x + y + z = 10, x £ 1, y ³ 0, z ³ 0

Hence range = {1, 2, 3}

x -1 = t Þ x = 1+ t

x - 1 ³ 0, t ³ 0

(a) No. of ways in which 6 men can be arranged at a round table = (6 – 1)! = 5!

t + y + z = 10 - 1 = 9

0 £ t , z, z £ 9

\ Total number of non-negative integral solution

M M

M

M

M M

40. 6

Now women can be arranged in P5 = 6! ways. Total Number of ways = 6! × 5!

11×10 = 55 2 But for t = 9, x = 10, so required number of integers = 55 – 1 = 54. (a) Number of two consecutive stations (Blue lines) = n Number of two non-consecutive stations (Red lines) = 9 + 3-1C3-1 = 11C2 =

= n C2 - n

Now, according to the question, n C2 - n = 99n

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Mathematics

Þ

n(n - 1) - 100n = 0 Þ n(n - 1 - 200) = 0 2

46.

Þ n - 1 - 200 = 0 Þ n = 201 25

25

25

r =0

r =0

r =0

å (4r + 1) 25Cr = 4 å r. 25Cr + å 25 Cr

41. (51)

25

= 4å r ´ r =1

25 25 24 Cr -1 + 225 = 100 å 24 Cr -1 + 2 25 r r =1

= 100.224 + 225 = 225(50 + 1) = 51.225 Hence, by comparison k = 51 42. (490) 0 Red, 1 Red, 2 Red, 3 Red Number of ways of selecting atmost three red balls = 7C4 + 5C1 · 7C3 + 5C2 · 7C2 + 5C3 · 7C1 = 35 + 175 + 210 + 70 = 490 43. (c) We know nCr is greatest at middle term. So, a = (19Cp)max = 19C10 = 19C9 b = (20Cq)max = 20C10 c = (21C6)max = 21C10 = 21C11 a

Now,

19

=

C9

221 = 220 2 (c) Number of ways of selecting three persons such that there is atleast one boy and atleast one girl in the selected persons = n + 5C3 – nC3 – 5C3 = 1750

= 1 + 21C1 + 21C2 + ... + 21C10 = 47.

Þ

(n + 5)! 5! n! 3!(n + 2)! 3!(n - 3)! 3! 2! = 1750

(n + 5) (n + 4)(n + 3) n( n - 1)( n - 2) = 1760 6 6 n = 25 [n = – 28 reected] Þ n2 + 3n – 700 = 0 Þ 20 (a) Total number of beams = C2 – 20 = 190 – 20 = 170 (b) Since, m = number of ways the committee is formed with at least 6 males = 8C6 . 5C5 + 8C7 . 5C4 + 8C8 . 5C3 = 78 and n = number of ways the committee is formed with at least 3 females = 5C3 . 8C8 + 5C4 . 8C7 + 5C5 . 8C6 = 78 Hence, m = n = 78 (a) Q There are total 9 digits and out of which only 3 digits are odd.

Þ

48. 49.

b c = 20 19 21 20 19 C9 . C9 . 10 11 10

a b c a b c = = = = \ 1 2 42 / 11 11 22 42 44. (2454) EXAMINATION 2N, 2A, 2I, E, X, M, T, O Case I : If all are different, then

(c) Number of ways of selecting 10 obects = (10I, 0D) or (9I, 1D) or (8I, 1D) or ... (0I, 10D) Here, D signifies distinct obect and I indicates identical obect

Þ

50.

8! = 8.7.6.5 = 1680 4! Case II : If two are same and two are different, then 8

p4 =

\

4! = 3.21.12 = 756 2! Case III : If two are same and other two are same, then

Hence, total number of 9 digit numbers

3

C1 . 7C2 .

4! = 3.6 = 18 2!2! Total cases = 1680 + 756 + 18 = 2454

3! ö 6! æ4 = ç C3 . 2! ÷ . 2! 4! = 180 è ø

3

C2 .

\ 45. (1) Þ

r +1 6

r +1 6 r can be 5, 35 for k Î I r = 5, k = ± 2 r = 35, k = ± 3 Hence, number of ordered pairs = 4.

Þ

51.

36 × 35Cr(k2 – 3) = 35Cr.6 r +1

k2 - 3 =

4 Number of ways to arrange odd digits first = C3.

k2 = 3 +

52.

(a) mC2 ´ 2 = mC1 × 2C1 ´ 2 + 84 m(m – 1) = 4m + 84 m2 – 5m – 84 = 0 m2 – 12m – 7m – 84 = 0 m(m – 12) + 7 (m – 12) = 0 m = 12, m = – 7 Q m>0 m = 12 (d) Consider the expression, 20 20

Ci -1 20

Ci + Ci -1

20

=

Ci -1

21

C1

3! 2!

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Permutations and Combinations

=

(2) 2 ladies, 1 man (3) 1 lady, 2 men (4) 0 ladies, 3 men Possible cases for Y are

20! i !(21 - i )! i ´ = (i - 1)!(21 - i )! 21! 21 3

20 æ ö 20 i 3 C (1) 20 å çç 20C + 20i -C1 ÷÷ = å æç 21 ö÷ = (21)3 å i3 ø i =1 è i =1 è i=1 i i -1 ø 20

\

2

=

1 æ 20 ´ 21 ö 100 ´ç ÷ = 21 (21)3 è 2 ø æ å çç i =1 è 20

Q

(3) 2 ladies, 1 man (4) 3 ladies, 0 man No. of ways = 4C3 . 4C3 + (4C2 . 3C1)2 + (4C1 . 3C2)2

3

Ci-1 ö k ÷ = 20 Ci + 20 Ci-1 ÷ø 21 20

+ (3C3)2 = 16 + 324 + 144 + 1 = 485

\ k = 100 53. (c) Since, the number of ways to select 2 girls is 5C2. Now, 3 boys can be selected in 3 ways. (a) Selection of A and selection of any 2 other boys (except B) in 5C2 ways (b) Selection of B and selection of any 2 two other boys (except A) in 5C2 ways (c) Selection of 3 boys (except A and B) in 5C3 ways Hence, required number of different teams = 5C2 (5C2 + 5C2 + 5C3) = 300 54. (d) If all four letters are different then the number of words 5C × 4! = 120 4 If two letters are R and other two different letters are chosen from B, A, C, K then the number of words = 4 C2 ´

n +2

57.

15

58.

15Cr 15Cr -1

4! =6 2! 2!

15 1 15 - r r 15 - r r - 1 15 1 = = r - 1| 15 - r .(16 - r ) r - 1 16 - r

=

16 - r r

=

å r 2 çè

15

Therefore, the number of four-letter words that can be formed = 120 + 72 + 72 + 6 = 270

æ 15C ö r ÷ è Cr -1 ø

å r 2 ç 15

(d)

r =1

4! = 72 2!

r =1

æ 16 - r ö ÷ å r (16 - r ) r ø = r =1 15

15

15

r =1

r =1

= 16å r - å r

(d) \ Required number of ways = C4 ´ C1 ´ 4! 6

= 11

P2

(n + 2)(n + 1) n (n - 1)(n - 2)(n - 3) 6.5.4.3.2.1 = 11 (n - 2)(n - 3) 2.1

4! = 72 2!

of words =

C6

Þ (n + 2) (n + 1) n (n – 1) = 11 . 10. 9. 4 Þ n=9 n2 + 3n – 108 = (9)2 + 3(9) – 108 = 81 + 27 – 108 = 108 – 108 = 0

If word is formed using two R’s and two A’s then the number

55.

n-2

Þ

If two letters are A and other two different letters are chosen from B, R, C, K then the number of words = 4 C2 ´

(c)

3

= 15 × 3 × 24 = 1080

2

16 ´ 15 ´ 16 15 ´ 31 ´16 2 6 = 8 × 15 × 16 – 5 × 8 × 31 = 1920 – 1240 = 680 (c) Given n(A) = 4, n(B) = 2, n(A × B) = 8

=

(b)

X 3 men

Possible cases for X are (1) 3 ladies, 0 man

59.

4 men

Required number of subsets = 8C3 + 8C4 +.... + 8C8 = 28 – 8C0 – 8C1 – 8C2 = 256 – 1 – 8 – 28 = 219

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60.

61.

62.

63. 64.

Mathematics

(a) Number of diagonal = 54 n(n - 3) Þ = 54 2 2 Þ n – 3n – 108 = 0 Þ n2 – 12 n + 9n – 108 = 0 Þ n (n – 12)+ 9 (n – 12) = 0 Þ n = 12, –9 Þ n = 12 (Q n ¹ –9) (c) Given n(A) = 2, n(B) = 4, n(A × B) = 8 Required number of subsets = 8C + 8C +.... + 8C = 28 – 8C – 8C – 8C 3 4 8 0 1 2 = 256 – 1 – 8 – 28 = 219 (b) We know, Tn = nC3, Tn+1 = n+1C3 ATQ, Tn+1 – Tn = n+1C3 – nC3 = 10 Þ nC2 = 10 Þ n = 5. (b) Required number of triangles = 12C3 – (3C3 + 4C3 + 5C3) = 205 (c) 30 marks to be alloted to 8 questions. Each question has to be given ³ 2 marks Let questions be a, b, c, d, e, f, g, h and a + b + c + d + e + f + g + h = 30 Let a = a1 + 2 so, a1 ³ 0 b = a2 + 2 so, a2 ³ 0,......, a8 ³ 0

= 2 C2 ´ 4C1 ´ 3! = 1´ 4 ´ 6 = 24

Hence, total number of arrangements = 72 + 24 = 96 67.

m(m - 1) is also a natural number.. 2

Now

Þ a1 + a2 + ...... + a8 = 30 – 16 = 14 So, this is a problem of distributing 14 articles in 8 groups. Number of ways = 14+8–1C8–1 = 21C7

65.

(b)

L 2 ³1

O 2 ³1

Y 4 2£

Þ

O 1 2 1 2

Y 2 1 1 0

68.

Required number of ways = 2C1 × 2C1 × 2C2 + 2C1 × 2C2 × 4C1 + 2C2 × 2C1 × 4C1 + 2C2 × 2C2 × 4C0 4´3 +2×1×4+1×2×4+1×1×1 2 = 24 + 8 + 8 + 1 = 41 (a) There are 2 vowels and 4 consonants in the letters a, b, c, d, e, f. If we select one vowel, then number of arrangements

=2×2× 66.

4´3 = 2C1 ´ 4 C2 ´ 3! = 2 ´ ´ 3 ´ 2 = 72 2 If we select two vowels, then number of arrangements

m2 - m m(m - 1) = 2 2

æ m 2 - m öæ m2 - m ö - 1÷ ç ÷ç ç 2 ÷ç 2 ÷ m(m - 1) è øè ø C2 = \ 2 2

So, a1 + a2 + ...... + a8 ü = 30 + 2 + 2 +...... + 2 ýþ

L 1 1 2 2

m(m - 1) 2 Since m and (m – 1) are two consecutive natural numbers, therefore their product is an even natural number. So

(a) n = mC2 =

69.

=

m(m - 1) (m2 - m - 2) 8

=

m(m - 1)[m2 - 2m + m - 2] 8

=

m(m - 1)[m(m - 2) + 1(m - 2)] 8

=

m(m - 1)(m - 2)(m + 1) 8

=

3 ´ ( m + 1) m (m - 1)(m - 2) =3 4 ´ 3 ´ 2 ´1

(

m +1

C4

)

(d) Statement - 1 : n(A) = p, n(B) = q, q > p Total number of functions from A ® B = qp It is a true statement. Statement - 2 : The total number of selections of p different obects out of q obects is qCp. It is also a true statement and it is a correct explanation for statement - 1 also. (c) Set A = {a1, a2, ....., a20} has 20 distinct elements. We have to select 5-element subset. \ Number of 5-element subsets = 20C5 According to question

(

)

20 C = 19 C .k 5 4

Þ

æ 19! ö 20! = k. ç 5! 15! è 4! 15!÷ø

Þ

20 = k Þk=4 5

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Permutations and Combinations

70. (a) Number of required triangles =

10

= (50 C4 +50 C3 ) +51 C3 +52 C3 + 53 C3 +54 C3 +55 C3

C3 -6 C3

10 ´ 9 ´ 8 6 ´ 5 ´ 4 = 120 - 20 = 100 6 6 (a) The number of ways of distributing 10 identical balls in 4 distinct boxes

= (51 C4 + 51C3 ) + 52 C3 + 53 C3 + 54 C3 + 55 C3

=

71.

= 10 -1 C4 -1 = 9 C3 72.

Proceeding in the same way, we get = 78.

(c) Two balls are taken from each urn 3

74.

Cr -1

9´8 = 3 ´ 36 = 108 2 (c) 4 novels, out of 6 novels and 1 dictionary out of 3 can

be selected in 6 C4 ´3C1 ways

7! 7´ 6 = = 21 2!5! 2 ´1 (c) According to given question two cases are possible. (i) Selecting 4 out of first five question and 6 out of remaining question

Then 4 novels with one dictionary in the middle can be arranged in 4! ways. \ Total ways of arrangement

= 5C4 ´8 C6 = 140 ways (ii) Selecting 5 out of first five question and 5 out of remaining

= 6 C4 ´3C1 ´ 4! = 1080 (d) First let us arrange M, I, I, I, I, P, P

8 questions = 5C5 ´8 C5 = 56 ways Therefore, total number of choices =140 + 56 = 196.

7! 8 7! 8 C4 = C4 = 7 ´ 6C4 ´ 8C4 4!2! 4!2! (a) Set S = {1, 2, 3, ...... 12} AÈ B È C = S, AÇ B = B ÇC = A ÇC = f \ Each sets contain 4 elements. \ The number of ways to partition = 12C4 × 8C4 × 4C4

=

=

12! 12! 8! 4! = ´ ´ 4!8! 4!4! 4!0! (4!)3

(c) The number of ways can vote = 10 C1 + 10 C2 + 10 C3 + 10 C4 = 10 + 45 + 120 + 210 = 385

77.

n -1

\ The required number of ways

7! ways 4!2! *M*I*I*I*I*P*P* Now 4 S can be kept at any of the * places in 8C4 ways so that no two S are adacent. Total required ways

76.

(b) We know that the number of ways of distributing n identical items among r persons, when each one of them

Total number of ways = C2 ´ C2

Which can be done in

75.

C4 + 55 C3 = 56 C4 .

receives at least one item is

9

=3×

73.

55

50

(d)

=

50

C4 +

6

å 56- r C3

r =1

é 55 C3 + 54 C3 + 53C3 + 52 C3 ù ú C4 + ê êë + 51C3 + 50 C3 úû

=

79.

80.

8 -1

(c)

C3-1 = 7C2 =

n

Cr +1 + n Cr -1 + 2 n Cr

= n Cr -1 + n Cr + n Cr + n Cr +1

éëQ nCr + nCr -1 = n+1Cr ùû 81.

= n +1Cr + n +1Cr +1 = n + 2 Cr +1 (d) We know that a number is divisible by 3 only when the sum of the digits is divisible by 3. The given digits are 0, 1, 2, 3, 4, 5. Herethe possible number of combinations of 5 digits out of 6 are 5C4 = 5, which are as follows– 1 + 2 + 3 + 4 + 5 = 15 = 3 × 5 (divisible by 3) 0 + 2 + 3 + 4 + 5 = 14 (not divisible by 3) 0 + 1 + 3 + 4 + 5 = 13 (not divisible by 3) 0 + 1 + 2 + 4 + 5 = 12 = 3 × 4 (divisible by 3) 0 + 1 + 2 + 3 + 5 = 11 (not divisible by 3) 0 + 1 + 2 + 3 + 4 = 10 ( not divisible by 3) Thus the number should contain the digits 1, 2, 3, 4, 5 or the digits 0, 1, 2, 4, 5. Taking 1, 2, 3, 4, 5, the 5 digit numbers are = 5! = 120 Taking 0, 1, 2, 4, 5, the 5 digit numbers are = 5! – 4! = 96 \ Total number of numbers = 120 + 96 = 216

n n n +1 Cr ù We know éë Cr + Cr -1 = û

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Mathematics

8

Binomial Theorem 7.

TOPIC Ć 1.

Binomial Theorem for a Positive Integral Index ‘x’, Expansion of Binomial, General Term, Coefficient of any Power of ‘x’

2.

1

[Sep. 06, 2020 (I)]

(a)

5 8

(b)

7 8

(c)

3 8

(d)

1 8

The natural number m, for which the coefficient of x in the æ 1ö binomial expansion of ç x m + 2 ÷ è x ø

3.

6

5.

10.

æ in the expansion of ç x + è 1) is equal to : (a) 29 (c) 26 If the fourth term in

11.

1 ö æ 1 ç 1 + log x + x 12 ÷ is equal to 200, and x > 1, then the 10 ç x ÷ è ø value of x is: [April 08, 2019 (II)] (a) 100 (b) 10 (c) 103 (d) 104 Let (x + 10)50 + (x – 10)50 = a0 + a1x + a2x2 + .... + a50x50,

8.

is 1540, is ______.

[NA Sep. 05, 2020 (I)] The coefficient of x4 in the expansion of (1 + x + x2 + x3)6 in powers of x, is ____________.[NA Sep. 05, 2020 (II)] a7 is equal to a13 r =0 ___________. [NA Sep. 04, 2020 (I)] If a and b be the coefficients of x4 and x2 respectively in the expansion of

Let (2 x 2 + 3 x + 4)10 = å ar x r . Then

(x + 6.

9.

[April 9, 2019 (I)] (a) 83 (b) 82 (c) 8 (d) 8–2 If some three consecutive coefficients in the binomial expansion of (x + 1)n in powers of x are in the ratio 2:15:70, then the average of these three coefficients is: [April 09, 2019 (II)] (a) 964 (b) 232 (c) 227 (d) 625 The sum of the co-efficients of all even degree terms in x

22

20

4.

æ2 log x ö ç + x 8 ÷ (x > 0) is 20 × 87, then a value of x is: èx ø

If {p} denotes the fractional part of the number p, then 200 ïì 3 ïü í ý , is equal to : îï 8 ïþ

) ( 6

If the fourth term in the Binomial expansion of

)

6

for all x Î R; then

n

æ 2 1 ö of x in the expansion of ç x + 3 ÷ is nC23 , is : x ø è [April 10, 2019 (II)] (a) 38 (b) 58 (c) 23 (d) 35

6

6

x 2 - 1 + x - x 2 - 1 , then: [Jan. 8, 2020 (II)]

(a) a + b = 60 (b) a + b = –30 (c) a – b = 60 (d) a – b = –132 The smallest natural number n, such that the coefficient

6

x 3 - 1 ö÷ + æç x - x3 - 1 ö÷ , (x > ø è ø [April 8, 2019 (I)] (b) 32 (d) 24 the binomial expansion of

(a) 12.50 (c) 12.25 12.

a2 is equal to : [Jan. 11, 2019 (II)] a0 (b) 12.00 (d) 12.75

(

If the third term in the binomial expansion of 1 + x log 2 x

)

5

equals 2560, then a possible value of x is: [Jan. 10, 2019 (I)]

www.jeebooks.in M-73

Binomial Theorem

21.

1 (a) 4

(b) 4 2

18

æ 1 ö ç 3 1 ÷ m x + ç 1 ÷ , (x > 0), are m and n respectively, then n ç ÷ 2x 3 ø è is equal to : [Online April 10, 2016] (a) 27 (b) 182

1 (d) 2 2 8 The positive value of l for which the co-efficient of x2 (c)

13.

10

l ö æ in the expression x 2 ç x + 2 ÷ x ø è

is 720, is:

(c)

22.

(b) 2 2 (d) 3

5

k 2403 is , then k 15 15 [Jan. 9, 2019 (I)]

14.

If the fractional part of the number

15.

is equal to: (a) 6 (b) 8 (c) 4 (d) 14 The coefficient of x10 in the expansion of (1 + x)2 (1 + x2)3 (1 + x3)4 is equal to [Online April 15, 2018] (a) 52 (b) 44 (c) 50 (d) 56 If n is the degree of the polynomial,

16.

8

17.

18.

23.

(1 + ax + bx ) (1 - 2 x ) (a, b) is equal to:

24.

19.

æ x +1 x -1 ç 1 1 ç 23 è x - x3 +1 x - x 2

20.

æ 272 ö (b) ç16, ÷ 3 ø è

æ 251 ö (c) ç 16, ÷ 3 ø è

æ 251 ö (d) ç 14, ÷ 3 ø è

{

}

If X = 4n - 3n - 1 : n Î N and

5

25.

If 1 + x4 + x5 =

å a i (1 + x )

i =0

i

[2014]

, for all x in R, then a2 is: [Online April 12, 2014]

(a) – 4 (c) – 8 26.

where x ¹ 0, 1, is :

[Online April 9, 2017] (a) 1 (b) 4 (c) – 4 (d) – 1 If (27)999 is divided by 7, then the remainder is : [Online April 8, 2017] (a) 1 (b) 2 (c) 3 (d) 6

æ 272 ö (a) ç14, ÷ 3 ø è

numbers, then X È Y is equal to: (a) X (b) Y (c) N (d) Y – X

10

ö ÷ ÷ ø

in powers of x are both ero, then [2014]

Y = {9 ( n - 1) : n Î N } , where N is the set of natural

(x + x3 - 1)5 + (x - x3 - 1)5 ,(x > 1) is : (a) 0 (b) 1 (c) 2 (d) – 1 The coefficient of x –5 in the binomial expansion of

18

2

8

é ù é ù 1 1 ê ú +ê ú and ê 5 x 3 + 1 – 5 x3 – 1 ú ê 5 x 3 + 1 + 5 x3 – 1 ú ë û ë û m is the coefficient of xn in it, then the ordered pair (n, m) is equal to [Online April 15, 2018] (a) (12 , (20)4) (b) (8, 5 (10)4) (c) (24 , (10)8) (d) (12, 8 (10)4) 2 The coefficient of x in the expansion of the product (2 – x2). ((1 + 2x + 3x2)6 + (1 – 4x2)6) is [Online April 16, 2018] (a) 106 (b) 107 (c) 155 (d) 108 The sum of the co-efficients of all odd degree terms in the expansion of [2018]

5 4 (d) 4 5 If the coefficients of the three successive terms in the binomial expansion of (1 + x)n are in the ratio 1 : 7 : 42, then the first of these terms in the expansion is: [Online April 10, 2015] (a) 8th (b) 6th (c) 7th (d) 9th If the coefficents of x 3 and x 4 in the expansion of

(c)

[Jan. 10, 2019 (II)] (a) 4

If the coefficients of x–2 and x–4 in the expansion of

27.

xö æ If ç 2 + ÷ è 3ø

(b) 6 (d) 10 55

is expanded in the ascending powers of x and

the coefficients of powers of x in two consecutive terms of the expansion are equal, then these terms are: [Online April 12, 2014] (a) 7th and 8th (b) 8th and 9th (c) 28th and 29th (d) 27th and 28th The number of terms in the expansion of (1 + x)101 (1 + x2 – x)100 in powers of x is: [Online April 9, 2014] (a) 302 (b) 301 (c) 202 (d) 101

EBD_8344

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Mathematics

29.

If for positive integers r > 1, n > 2, the coefficients of the (3r)th and (r + 2)th powers of x in the expansion of (1 + x)2n are equal, then n is equal to : [Online April 25, 2013] (a) 2r + 1 (b) 2r –1 (c) 3r (d) r + 1 The sum of the rational terms in the binomial expansion of

30.

1ö æ 1 ç 2 2 + 35 ÷ is : [Online April 23, 2013] ç ÷ è ø (a) 25 (b) 32 (c) 9 (d) 41 If the 7th term in the binomial expansion of

28.

35.

36.

10

31.

(

3 +1

)

2n

-

(

3 -1

)

(d) 144

The remainder left out when 82n – (62)2n+1 is divided by 9 is: [2009]

37.

2n

38.

(b) an odd positive integer (c) an even positive integer

32.

33.

34.

(

The number of terms in the expansion of y

+x

)

39.

(b) Statement-1 is true, Statement-2 is true; Statement-2 is NOT a correct explanation for Statement-1 (c) Statement-1 is true, Statement-2 is false (d) Statement-1 is false, Statement-2 is true

r =0

= (1 + x)n + nx(1 + x )n –1.

n -5 6

(b)

n-4 5

(c)

5 n-4

(d)

6 . n -5

For natural numbers m, n if (1 - y ) m (1 + y ) n = 1 + a1 y + a2 y 2 + ....... and a1 = a2 = 10, then (m, n) is (a) (20, 45) (c) (45, 35)

(b) (35, 20) (d) (35, 45)

[2006] 11

40.

7 If the coefficient of x in éê ax 2 + æç 1 ö÷ ùú è bx ø û ë

equals the

11

é -7 1 öù coefficient of x in ê ax - æç ÷ ú , then a and b satisfy è bx 2 ø û ë the relation [2005] (a) a – b = 1 (b) a + b = 1

Statement - 2 : For each natural number n, n 7 - n is divisible by 7. [2011 RS] (a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

å (r + 1) nCr x r

(a)

,

in which powers of x and y are free from radical signs are [Online May 12, 2012] (a) six (b) twelve (c) seven (d) five If f(y) = 1 – (y – 1) + (y – 1)2 – (y – 1)3 + ... – (y – 1)17, then the coefficient of y2 in it is [Online May 7, 2012] (a) 17 C2 (b) 17 C3 (c) 18 C2 (d) 18 C3 Statement - 1 : For each natural number n, (n + 1)7–1 is divisible by 7.

r =0

[2008] (a) Statement -1 is false, Statement-2 is true (b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1 (c) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1 (d) Statement -1 is true, Statement-2 is false In the binomial expansion of (a – b)n, n ³ 5, the sum of 5th and 6th terms is ero, then a/b equals [2007]

(d) a rational number other than positive integers 1/10 55

å (r + 1) nCr = (n + 2)2n –1.

n

(a) an irrational number

1/5

(d) 0

Statement -1 :

Statement-2:

is : [2012]

(b) 7 n

(d) 2e

If n is a positive integer, then

(b) –144

(c) 132

(c) 8

9

e 2

(a) –132

(a) 2

æ 3 ö + 3 ln x ÷ , x > 0, is equal to 729, then x can be : ç3 è 84 ø [Online April 22, 2013] (a) e 2 (b) e

(c)

The coefficient of x7 in the expansion of (1– x – x2 + x3 )6 is [2011]

(c) 41.

a =1 b

(d) ab = 1

The coefficient of xn in expansion of (1 + x )(1 - x )n is [2004] (a) ( -1)n -1 n

(b) ( -1)n (1 - n)

(c) ( -1)n -1 (n - 1)2

(d) (n - 1)

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Binomial Theorem

42.

43.

44.

The number of integral terms in the expansion of ( 3 + 8 5)256 is [2003] (a) 35 (b) 32 (c) 33 (d) 34 r and n are positive integers r > 1, n > 2 and coefficient of (r+2)th term and 3rth term in the expansion of (1 + x)2n are equal, then n equals [2002] (a) 3r (b) 3r + 1 (c) 2r (d) 2r + 1 The coefficients of xp and xq in the expansion of (1+ x )p+q are [2002] (a) equal (b) equal with opposite signs (c) reciprocals of each other (d) none of these

Middle Term, Greatest Term, Independent Term, Particular Term from end in Binomial Expansion, Greatest Binomial Coefficients

TOPIC n

n

50.

1 For a positive integer n, æç1 + ö÷ is expanded in increasing è xø

powers of x. If three consecutive coefficients in this expansion are in the ratio, 2 : 5 : 12, then n is equal to __________. [NA Sep. 02, 2020 (II)] 16

51.

1 ö æ x + In the expansion of ç ÷ , if l1 is the least è cos q x sin q ø p p £ q £ and 8 4 l2 is the least value of the term independent of x when

value of the term independent of x when

p p £ q £ , then the ratio l2 : l1 is equal to : 16 8 [Jan. 9, 2020 (II)] (a) 1 : 8 (b) 16 : 1

52.

(c) 8 : 1 (d) 1 : 16 The total number is irrational terms in the binomial

53.

1 ö æ 1 ç 7 5 - 310 ÷ expansion of ç [Jan. 12, 2019 (II)] ÷ is : è ø (a) 55 (b) 49 (c) 48 (d) 54 A ratio of the 5th term from the begining to the 5th term

60

45.

If the constant term in the binomial expansion of 10

æ k ö ç x 2÷ x ø è

46.

47.

is 405, then |k| equals:

[Sep. 06, 2020 (II)]

(a) 9 (b) 1 (c) 3 (d) 2 If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1 + x)n +5 are in the ratio 5 : 10 : 14, then the largest coefficient in this expansion is : [Sep. 04, 2020 (II)] (a) 462 (b) 330 (c) 792 (d) 252 If the number of integral terms in the expansion of

10

æ 1 ö ç ÷ 1 23 + 1 ÷ from the end in the binomial expansion of ç çç ÷ 2(3) 3 ÷ø è is: (a)

(31 2 + 51 8 )n is exactly 33, then the least value of n is :

48.

49.

Let a > 0, b > 0 be such that a + b = 4. If the maximum value of the term independent of x in the binomial 3

expansion of (a) 336 (c) 84

1 (a x 9

1 + b x 6 )10

2

is 10k, then k is equal to : [Sep. 02, 2020 (I)] (b) 352 (d) 176

1

(c) 4(36) 3 :1 54.

(d) 2(36) 3 :1

The term independent of x in the binomial expansion of 8

æ 1 5 öæ 2 1 ö ç 1 - + 3 x ÷ç 2 x - ÷ is : [Online April 11, 2015] xø è x øè

9

æ3 2 1 ö ç x - ÷ is k, then 18k is equal to : 3x ø è2 [Sep. 03, 2020 (II)] (a) 5 (b) 9 (c) 7 (d) 11

1

(b) 1: 4(16) 3

1

[Sep. 03, 2020 (I)] (a) 264 (b) 128 (c) 256 (d) 248 If the term independent of x in the expansion of

[Jan. 12, 2019 (I)] 1 1 : 2(6) 3

55.

(a) 496 (b) –496 (c) 400 (d) –400 The term independent of x in expansion of x +1 x -1 ö æ çè 2/ 3 1/ 3 ÷ x - x + 1 x - x1/ 2 ø

(a) 4 (c) 210

10

is

(b) 120 (d) 310

[2013]

EBD_8344

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56.

Mathematics

The ratio of the coefficient of x15 to the term independent 2ö æ of x in the expansion of ç x 2 + ÷ è xø

57.

67.

68.

æ x3 3 ö the binomial expansion of çç + ÷÷ equals 5670 is : è 3 xø [Jan. 11, 2019 (I)] (a) 0 (b) 6 (c) 4 (d) 8 The value of r for which

is :

æ 1ö The middle term in the expansion of ç1 - ÷ è xø

powers of x is (a) – 2nCn–1 (c) 2nCn – 1 58.

15

[Online April 9, 2013] (b) 7 : 64 (d) 1 : 32

(a) 7 : 16 (c) 1 : 4

If 20C1 + (22) 20C2 +(32) 20C3+ ………. + (202) 20C20 = A(2b), then the ordered pair (A, b) is equal to : [April 12, 2019 (II)] (a) (420, 19) (b) (420, 18) (c) (380, 18) (d) (380, 19) 18 The coefficient of x in the product (1+x)(1–x) 10 (1+x+x2)9 is : [April 12, 2019 (I)] (a) 84 (b) –126 (c) –84 (d) 126 If the coefficients of x2 and x3 are both ero, in the expansion of the expression (1 + ax + bx2) (1–3x)15 in powers of x, then the ordered pair (a, b) is equal to: [April 10, 2019 (I)] (a) (28, 861) (b) (–54, 315) (c) (28, 315) (d) (–21, 714) The sum of the series 2·20C0 + 5·20C1 + 8·20C2 + 11·20C3 + … + 62·20C20 �矄 ⼄盐 ᨨ  [April 8, 2019 (I)] (a) 226 (b) 225 (c) 223 (d) 224 The sum of the real values of x for which the middle term in

63.

n

(1 - x )n

in

64.

65.

[Online May 26, 2012] (b) – 2nCn (d) 2n Cn

The coefficient of the middle term in the binomial expansion 4

66.

6

in powers of x of (1 + ax ) and of (1 - ax ) is the same if a equals (a)

[2004]

3 5

(b)

10 3

8

(c)

-3 10

TOPIC Đ

(d)

-5 3

Properties of Binomial Coefficients, Number of Terms in the Expansion of (x+y+z)n, Binomial theorem for any Index, Multinomial theorem, Infinite Series

20

Cr 20C0 + 20 Cr -120C1 + 20 Cr - 220C2 + ... + 20 C020 Cr is maximum, is : [Jan. 11, 2019 (I)] (a) 15 (b) 20 (c) 11 (d) 10

20

59.

The value of

å 50-r C6

is equal to : [Sep. 04, 2020 (I)]

r=0

(a) (c) 60.

51C 7 50C 6

– –

30C 7 30C 6

(b) (d)

– +

30C 7 30C 7

If

to: (a) (25)2 (c) 224

[NA Jan. 9, 2020 (I)]

[Jan. 10, 2019 (II)] (b) 225 – 1 (d) 225 3

If the sum of the coefficients of all even powers of x in the

æ 1 - t6 ö ÷÷ è 1- t ø [Jan. 09, 2019 (II)]

70.

The coefficient of t4 in the expansion of ç ç

71.

(a) 14 (b) 15 (c) 10 (d) 12 The value of (21C1 – 10C1) + (21C2 – 10C2) + (21C3 – 10C3) + (21C4 – 10C4)

product (1 + x + x2 + ... + x2n) (1 – x + x2 – x3 + ... + x2n) is 61, then n is equal to ¾¾¾. 62.

[NA Jan. 7, 2020 (I)]

The term independent of x in the expansion of æ 1 x8 ö æ 2 3 ö 6 çç - ÷÷ . ç 2 x - 2 ÷ is equal to : x ø è 60 81 ø è [NA April 12, 2019 (II)] (a) –72 (b) 36 (c) –36 (d) –108

then K is equal

r=0

The coefficient of x4 in the expansion of (1 + x + x2)10 is _______.

61.

50C 7 51C 7

å { 50 Cr × 50 - r C25 - r } = K ( 50 C25 ), 25

69.

+ .... + (21C10 – 10C10) is :

[2017]

(a) 220 – 210

(b) 221 – 211

(c) 221 – 210

(d) 220 – 29

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Binomial Theorem

(a) Statement -1 is false, Statement-2 is true

n

72.

73.

æ 2 4ö If the number of terms in the expansion of ç1 - + 2 ÷ , è x x ø

x ¹ 0, is 28, then the sum of the coefficients of all the terms in this expansion, is : [2016] (a) 243 (b) 729 (c) 64 (d) 2187 The sum of coefficients of integral power of x in the

(

binomial expansion 1 - 2 x

74.

75.

)

is :

(b)

1 50 2 +1 2

(

)

(d)

1 50 3 2

(c)

1 50 3 +1 2

(d) Statement -1 is true, Statement-2 is false 77.

(

)

( )

The coefficient of x 1012 in the expansion of (1 + xn + x253)10, (where n £ 22 is any positive integer), is [Online April 19, 2014] (a) 1 (b) 10 C4 (c) 4n (d) 253 C 4 10

10

j =1

j =1

78.

C0 -

20

C1 +

20

and S3 = å j j =1

2 10

Cj.

C2 -

[2007] 20

C3 + ..... -..... +

(a) 0

(b)

20

C10

(c)

(d)

1 2

20

- 20 C10

20

(1 +

neglected,

C10

then

3 x) 2

æ 1 ö - ç 1 + x÷ è 2 ø

3

1 (1 - x ) 2

approximated as [2010]

3 (a) 1 - x 2 8

Statement -1 : S3 = 55 × 29. 79.

(c) Statement -1 is false, Statement -2 is true . 80.

In a shop there are five types of ice-creams available. A child buys six ice-creams. Statement-1 : The number of different ways the child can buy the six ice-creams is 10C5. Statement -2 : The number of different ways the child can buy the six ice-creams is equal to the number of different ways of arranging 6 A’s and 4 B’s in a row. [2008]

be

[2005] (b)

3 3 x + x2 8

x 3 2 3 (d) - x - x2 2 8 8 If x is positive, the first negative term in the expansion of (1 + x)27 5 is

(b) Statement -1 is true, Statement -2 is false.

(d) Statement - 1 is true, Statement 2 is true ; Statement -2 is a correct explanation for Statement -1.

may

(c)

Statement - 2: S1 = 90 × 28 and S2 = 10 × 28 . (a) Statement -1 is true, Statement -2 is true ; Statement 2 is not a correct explanation for Statement -1.

C10 is

3 If x is so small that x and higher powers of x may be

Let S1 = å j ( j - 1)10C J , S2 = å j10C j 10

76.

The sum of the series 20

[2015]

)

1 50 3 -1 2

(c) Statement -1 is true, Statement-2 is true; Statement 2 is not a correct explanation for Statement-1

50

(

(a)

(b) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1

81.

[2003]

(a) 6th term (b) 7th term (c) 5th term (d) 8th term The positive integer ust greater than (1 + 0.0001)10000 is [2002] (a) 4 (b) 5 (c) 2 (d) 3 If the sum of the coefficients in the expansion of (a + b)n is 4096, then the greatest coefficient in the expansion is [2002] (a) 1594 (b) 792 (c) 924 (d) 2924

EBD_8344

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1.

Mathematics

(d)

3200 1 100 = (9 ) 8 8

r1 r2

r3

n(n + 1) 2 1 1é ù = (1 + 8)100 = ê1 + n × 8 + × 8 + ....ú 8 8ë 2 û

0

7

3

1

5

4

2

3

5

1 = + Integer 8

3

1

6

ìï 3200 üï ì 1 ü 1 \í ý = í + integer ý = þ 8 ïî 8 ïþ î 8

2.

a7 =

10!37 43 10!(2)(3)5 (4) 4 + 7!3! 5!4!

10!(2) 2 (3)3 (4)5 10!(2) 3 (3)(4) 6 + 2!3!5! 3!6! a13 = Coeff. of x13 2r1 + r2 = 13 and r1 + r2 + r3 = 10 Possibilities are

(13)

+

Tr +1 =

22

Tr +1 =

22

æ 1ö Cr × ( xm )22- r × ç 2 ÷ èx ø

r

Cr × x 22 m - mr - 2r

Q 22m - mr - 2r = 1

3.

r1 r2

22 m - 1 3×3×5 Þr= Þ r = 22 m+ 2 m+2 So, possible value of m = 1, 3, 7, 13, 43 But 22Cr = 1540 \ Only possible value of m = 13. (120.00)

æ 1- x Coefficient of x 4 in çç è 1- x

4

a13 =

6

ö ÷÷ = coefficient of x 4 in ø

(1 - 6 x4 )(1 - x)-6

= 9C4 - 6 ×1 = 126 - 6 = 120. 4.

\

5.

(8.00)

r =0

General term =

10! (2 x 2 )r1 (3 x) r2 (4) r3 r1 !r2 !r3 !

Since, a7 = Coeff. of x7 2r1 + r2 = 7 and r1 + r2 + r3 = 10 Possibilities are

0

4

5

1

5

3

2

6

1

3

10!(25 )(33 )(42 ) 10!(26 )(3)(43 ) + 5! 3! 2! 6!1! 3!

a7 = 23 = 8 a 13

(d) Using Binomial expansion (x + a)n + (x – a)n = 2(T1 + T3 + T5 + T7...)

20

The given expression is (2 x 2 + 3x + 4)10 = å ar x r

7

10!(23 )(37 ) 10!(24 )(35 )(4) + 3!7! 4!5!

+

4 6 7 2 = coefficient of x4 in (1 - 6 x ) éë1 + C1 x + C2 x + ....ùû

r3

3

6

6

\ æç x + x 2 - 1 ö÷ + æç x - x 2 - 1 ö÷ = 2 (T1 + T3 + T5 + T7 ) è ø è ø

2[6C0 x5 + 6C2 x4 (x2 – 1) + 6C4 x2 (x2 – 1)2 + 6C6 (x2 – 1)3] = 2[x6 + 15(x6 – x4) + 15x2(x4 – 2x2 + 1) + (–1 + 3x2 – 3x4 + x6)] =2(32x6 – 48x4 + 18x2 – 1) a = – 96 and b = 36 \

a – b = – 132

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Binomial Theorem

6.

(a)

æ 2 1 ö çx + 2 ÷ x ø è

n

Þ 1 n

( )

General term Tr +1 = Cr x

7.

2 n-r æ

According to the question x > 1, \ x = 10.

3

æ xlog8 x ö log8 x 7 Þ 8 ´ 20 ´ ç ÷ = 20 ´ 8 Þ x = 64 ç x ÷ x è ø

11.

= a0 + a1x + a2x2 + ¼ + a50x50 \ \

1 or x = 82 8 (b) Given nCr–1 : nCr : nCr+1 = 2 : 15 : 70

\

Cr -1

=

Cr

=

2 r r +1 3 = and = n - r + 1 15 n - r 14 Þ 17r = 2n + 2 and 17r = 3n – 14 i.e., 2n + 2 = 3n – 14 Þ n = 16 & r = 2 \ Average =

=

16

C1 + 16C2 + 3

16

C3

50

C2 ´1048

50

C50 1050

50 ´ 49 49 = = 12.25 2 ´100 4

(a) Third term of (1 + xlog2 x )5 = 5C2 ( xlog2 x )5-3

Given, 5C2 ( xlog2 x )2 = 2560 Þ

( xlog2 x )2 = 256 = (±16)2

Þ

x log 2 x = 16 or x log 2 x = –16 (reected)

Þ

x log 2 x = 16 Þ log2 xlog2 x = log2 16 = 4 Þ log2 x = ±2 Þ x = 22 or 2–2

( x + x 3 - 1)6 + ( x - x 3 - 1)6

+ 6C6(x3 – 1)3] 6 7 4 8 5 2 = 2[x + 15x – 15x + 15x – 30x + 15x + x9 – 3x6 + 3x3 – 1] Hence, the sum of coefficients of even powers of x = 2[1 – 15 + 15 + 15 – 3 – 1] = 24 (b) \ fourth term is equal to 200. æ æ 1 öö ÷÷ ç ç T4 = C ç x è 1+ log10 x ø ÷ 3 ç ÷ ç ÷ è ø 6

a2 a0 =

=5C2 ( xlog2 x )2

16 + 120 + 560 = 3

= 2[6C0x6 + 6C2x4 (x3 – 1) + 6C4x2 (x3 – 1)2

10.

12.

696 = 232 3

(d)

a0 = 2.50C501050 a2 = 2.50C2.1048

n 2 15 Cr = and n 15 Cr +1 70

Þ

a0 + a1x + a2x2 + ¼ + a50x50

= 2(50C0x50 + 50C2x48 . 102 + 50C4x46 . 104 + ¼)

Þx=

9.

(c) (x + 10)50 + (x – 10)50

Now, take log8 on both sides, then (log8x)2 – (log8x) = 2 Þ log8x = –1 or log8x =2

n

3

Þ t2 + 4t – t – 4 = 0 Þ t (t + 4) – 1 (t + 4) = 0 Þ t = 1 or t = – 4 log10 x = 1 Þ x = 10 or log10 x = – 4 Þ x = 10–4

3

n

+

æ1 3 ö ç + ÷ t = 1 Þ t2 + 3t – 4 = 0 4 2(1 + t) ø è

To find coefficient of x, 2n – 5r = 1 Given nCr = nC23 Þ r = 23 or n – r = 23 \ n = 58 or n = 38 Minimum value is n = 38 (b) Q T4 = 20 ´ 87

Þ

1

x 4 2(1+ log10 x ) = 10 Taking log10 on both sides and putting log10 x = t

r

1 ö n 2n – 5r ç 3 ÷ = Cr × x èx ø

æ 2ö Þ 6C3 ç ÷ ´ ( xlog8 x )3 = 20 ´ 87 è xø

8.

3

20 x 2(1+ log10 x ) .x 4 = 200

3

3

æ 1 ö ç x12 ÷ = 200 ç ÷ è ø

1 4 (a) Since, coefficient of x 2 in the expression x 2

Þ x = 4 or 13.

lö æ çè x + 2 ÷ø is a constant term, then x

lö æ Coefficient of x2 in x2 ç x + 2 ÷ è x ø

10

lö æ = co-efficient of constant term in ç x + 2 ÷ è x ø lö æ General term in ç x + 2 ÷ è x ø

10

=

10

Cr

( x)

10

10- r

æ lö çè 2 ÷ø x

r

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Mathematics

= 10C ( x ) r

10 - r - 2r 2

× l2

720 = 16 Þ l = 4d Þ 5´ 9 Hence, required value of l is 4. (b) 2403 = 2400 × 23 = 24 ×100 × 23 = (24)100 × 8 = 8(24)100 = 8(16)100 = 8 (1 + 15)100 = 8 + 15 m When 2403 is divided by 15, then remainder is 8.

l2 =

Hence, fractional part of the number is

15.

é ù é ù 1 1 ê ú +ê ú ê 5 x 3 + 1 – 5 x3 – 1 ú ê 5 x 3 + 1 + 5 x3 – 1 ú ë û ë û

17.

=

1 é ê 28 ë

(

5 x3 + 1 + 5 x3 – 1

) +(

)

28 24 = 54 × 24 × 2 2

(a) Let a = ((1 + 2x + 3x2)6 + (1 – 4x2)6) \

Coefficient of x2 in the expansion of the product

(2 – x2) ((1 + 2x + 3x2)6 + (1 – 4x2)6) = 2 (Coefficient of x2 in a) – 1 (Constant of expansion) In the expansion of ((1 + 2x + 3x2)6 + (1 – 4x2)6). Constant = 1 + 1 = 2 Coefficient of x2 = [Coefficient of x2 in (6C0 (1 + 2x)6 (3x2)0)] + [Cofficient of x2 in (6C1 (1 + 2x)5 (3x2)1)]

8

– [6C1 (4x2)] = 60 + 6 × 3 – 24 = 54 \

8

The coefficient of x2 in (2 – x2) ((1 + 2x + 3x2)6 +

(1 – 4x2)6) = 2 × 54 – 1 (2) = 108 – 2 = 106 18.

8

é 5 x 3 + 1 + 5 x3 – 1 ù é 5 x3 + 1 – 5 x 3 – 1 ù ú +ê ú =ê ê (5 x 3 + 1) – (5 x3 – 1) ú ê (5 x3 + 1) – (5 x 3 – 1) ú ë û ë û 8

(

= 104 × 23 = 8 (10)4

After rationalise the polynomial we get

8

) )

ù ú ú ú ú ú 8 ú ú û

+ 8C854 ] = 54 ×

é 5 x 3 + 1 – 5 x3 – 1 ù 1 ú +ê ´ ê 5 x3 + 1 + 5 x3 – 1 5 x 3 + 1 – 5 x 3 – 1 úû ë

)

Now, coefficient of x12 = [8C054 + 8C254 + 8C454 + 8C654

8 15

é 5 x3 + 1 + 5 x 3 – 1 ù 1 ê ú ´ 3 3 ê 5 x3 + 1 – 5x3 – 1 ú + + 5 x 1 5 x – 1 ë û

)(

So, the degree of polynomial is 12,

Therefore value of k is 8 (a) Q (1 + x)2 = 1 + 2x + x2, (1 + x2)3 = 1 + 3x2 + 3x4 + x6, and (1 + x3)4 = 1 + 4x3 + 6x6 + 4x9 + x12 So, the possible combinations for x10 are: x . x9, x . x6 . x3, x2 . x2 . x6, x4 . x6 Corresponding coefficients are 2 × 4, 2 × 1 × 4, 1 × 3 × 6, 3 × 6 or 8, 8, 18, 18. \ Sum of the coefficient is 8 + 8 + 18 + 18 = 52 Therefore, the coefficient of x 10 in the expansion of (1 + x)2 (1 + x2)3 (1 + x3)4 is 52. (d)

(

é 8C0 (5 x 3 + 1)4 + 8C2 (5 x 3 + 1)3 (5 x 3 – 1) + 8C ù 4 ú 1 ê 3 2 3 2 ê ú = 8 (5 x + 1) (5 x – 1) 2 ê 8 ú 3 3 3 3 4 ê + C6 (5 x + 1) (5x – 1) + 8C8 (5 x – 1) ú ë û

8

16.

)( )(

( (

10 - r - 2r = 0 Þ r = 2 2 Co-efficient is x2 in expression = 10C2l2 = 720

14.

)

(

8 6 2 é8 3 8 3 5 x3 – 1 ê C0 5 x + 1 + C2 5 x + 1 ê 4 4 1 ê = 8 ê + 8 C4 5 x 3 + 1 5 x3 – 1 2 ê 2 6 ê 8 3 5 x3 – 1 + 8C8 5 x3 - 1 ê + C6 5 x + 1 ë

Then, for constant term,

(c) Since we know that, (x + a)5 + (x – a)5 = 2[5C0 x5 + 5C2 x3× a2 + 5C4x×a4] \

8

(x +

x3 - 1

) + (x 5

x3 - 1

)

5

= 2[5C0 x5 + 5C2 x3(x3 – 1) + 5C4x(x3 – 1)2]

)

5 x3 + 1 - 5 x3 – 1 ú û

Þ 2[x5 + 10x6 – 10x3 + 5x7 – 10x4 + 5x] \

Sum of coefficients of odd degree terms = 2.

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Binomial Theorem

19.

(a)

(

)(

) (

é x1 3 + 1 x 2 3 - x1 3 + 1 ê 23 13 ê + x x 1 ëê

(

)

(

= x1 3 + 1 - 1 - 1 x1 2 Tr + 1 = 10Cr x for r = 10 T11 =

10

C10 x

) = (x 10

13

)(

)

10

x -1 x +1 ù ú x x -1 ú ûú

(

- 1 x1 2

)

24.

(b) 4n – 3n – 1 = (1 + 3)n – 3n –1 = [nC0 + nC1.3 + nC2.32 +......+ nCn3n] – 3n – 1 = 9 [nC2 +nC3.3+....+nCn.3n–2] \ 4n – 3n – 1 is a multiple of 9 for all n. \ X = {x : x is a multiple of 9} Also, Y = {9 (n – 1) : n ÎN} = {All multiples of 9} Clearly X Ì Y. \ X È Y = Y

25.

(a)

)

10

20-5r 6

5

-5

4

5

1+ x + x =

Coefficient of x–5 = 10C10 (1) (–1)10 = 1 20.

(d)

( 28 - 1)

Þ \

21.

7

(b) Tr + 1 =

18

Cr

çè

÷ø

Þ 1 + x 4 + x5 = a0 + a1 (1 + x) + a2 (1 + 2 x + x 2 ) + a3 (1 + 3x + 3x 2 + x3 ) + a4 (1 + 4 x + 6 x 2 + 4 x3 + x 4 ) + a5 (1 + 5x + 10 x 2 + 10 x3 + 5x 4 + x5 )

r

æ ö ç 1 ÷ = ç 1÷ çè 3 ÷ø 2x

18

Cr x

6-

2r 3

coefficient of x -2 coefficient of x -4 n

22. (c)

18

=

18

C12 C15

Þ 1 + x 4 + x5

1 2

r

2 = a0 + a1 + a1 x + a2 + 2a2 x + a2 x + a3 + 3a3 x

+3a3 x 2 + a3 x3 + a4 + 4a4 x + 6a4 x 2 + 4a4 x3 + a4 x 4 + a5

2r ì ü ïï 6 - 3 = -2 Þ r = 12 ïï í ý ï & 6 - 2r = -4 Þ r = 15 ï îï þï 3

Þ

+5a5 x + 10a5 x 2 + 10a5 x3 + 5a5 x 4 + a5 x5

Þ 1 + x 4 + x5 1

= (a0 + a1 + a2 + a3 + a4 + a5 ) + x(a1 + 2a2 + 3a3 + 4a4 + 5a5 ) + x 2 (a2 + 3a3 + 6a4 + 10a5 ) + x3 (a3 + 4 a4 + 10a5 )

212 = 182 1

+ x 4 (a4 + 5a5 ) + x5 (a5 ) On comparing the like coefficients, we get

15

2

a5 = 1

C r n Cr +1 n C r + 2 = = 1 7 42

18 ´ 17 ´ 16 4 a + 18 ´ 17 .8 + - 36b = 0 6 2

= –51 × 16 × 8 + a × 36 × 17 – 36b = 0 = –34 × 16 + 51a – 3b = 0 = 51a – 3b = 34 × 16 = 544 = 51a – 3b = 544 ....(i) Only option number (b) satisfies the equation number (i)

...(i)

;

a4 + 5a= 5 1 ...(ii);

a3 + 4a4 + 10 a= 5 0 ...(iii)

By solving we get r = 6 so, it is 7th term. 23. (b) Consider (1 + ax + bx2) (1 – 2x)18 = (1 + ax + bx2) [18C0 – 18C1 (2x) + 18C2(2x)2– 18C3(2x)3 + 18C4(2x)4 –.......] 3 18 Coeff. of x = C3 (–2)3 + a. (–2)2.18C2 + b (–2).18C1 = 0 Coeff. of x3 = – 18C3.8 + a × 4. 18C2 – 2b × 18 = 0 =-

ai (1 + x)i

+ a4 (1 + x )4 + a5 (1 + x )5

28l - 7 + 7 - 1 7 ( 4l - 1) + 6 = 7 7 Remainder = 6 18 - r æ 1ö ç x3 ÷

i= 0

1 2 3 = a0 + a1 (1 + x) + a2 (1 + x ) + a3 (1 + x )

28l - 1 = 7

999

å

and a2 + 3a3 + 6 a4 + 10a5 = 0 ...(iv) from (i) & (ii), we get a4 = -4 ...(v) from (i), (iii) & (v), we get a3 = +6 ...(vi)

26.

Now, from (i), (v) and (vi), we get a2 = – 4 (a) Let rth and (r + 1)th term has equal coefficient

xö æ çè 2 + ÷ø 3

55

xö 55 æ = 2 ç1 + ÷ è 6ø

æ xö rth term = 255 55Cr ç ÷ è 6ø

55

r

EBD_8344

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Mathematics

1 Coefficient of xr is 255 55Cr r 6 æ xö (r +1)th term = 255 55Cr +1 ç ÷ è 6ø

30.

9

r +1

æ 3 ö + 3 ln x ÷ , x > 0 ç3 84 è ø

We have Tr+1 = nCr (x)n–r ar for (x + a)n. \ According to the question

1 55 55 Cr +1. r +1 Coefficient of xr+1 is 2 6 Both coefficients are equal 255 55Cr

1 6

r

= 255 55Cr +1

1 6

729 =

r +1

1 1 1 . = r 55 - r r + 1 54 - r 6

27.

6 (r + 1) = 55 – r 6r + 6 = 55 – r 7r = 49 r=7 (r + 1) = 8 Coefficient of 7th and 8th terms are equal. (c) Given expansion is (1 + x)101 (1 – x + x2)100 = (1 + x) (1 + x)100 (1 – x + x2)100 = (1 + x) [(1 – Expansion (1 –

x3)100

As given 2nCr+2 = 2nC3r (2n)! (2n)! = Þ (r + 2)! (2n - r - 2)! (3r )! (2 n - 3r )!

Þ (3r) ! (2n – 3r) ! = (r + 2) ! (2n – r – 2) ! ...(1) Now, put value of n from the given choices.

= 2 é 2n C1 êë

(

( 3)

3 +1 2 n -1

)

2n

-

+ 2 n C3

(

( 3)

)

2n

2 n -3

( 3)

2n-5

+ ....ù úû

Q (a + b)n – (a – b)n = 2[n C1a n -1b + nC3 a n -3b3 ...] = which is an irrational number. 55

32.

1 ö æ 1 5 10 (a) Given expansion is çç y + x ÷÷ è ø The general term is

Tr +1 =

55

æ 1ö Cr ç y 5 ÷ ç ÷ è ø

55- r

æ 1 ö .ç x10 ÷ ç ÷ è ø

r

Tr + 1 would free from radical sign if powers of y and x are integers.

r 55 - r and are integer. 5 10

LHS : (3r) ! (4r + 2 – 3r) ! = (3r) ! (r + 2) !

i.e.

RHS : (r + 2) ! (3r) !

Þ r is multiple of 10.

Þ LHS = RHS

Hence, r = 0, 10, 20, 30, 40, 50

(d) (21/2 + 31/5)10 = 10C0(21/2)10

It is an A.P.

+ 10C1(21/2)9 (31/5) + ...... + 10C10(31/5)10

3 -1

+ 2 n C5

Choice (a) put n = 2r + 1 in (1)

29.

33 3 ´ 3 ´ (6 ln x) 84

31. (a) Consider

will have 100 + 1 = 101 terms

(a) Expansion of (1 + x)2n is 1 + 2nC1x + 2nC2 x2 + ...... + 2nC xr + 2nC r+1 + ...... + 2nC x2n r r+1 x 2n

3

æ 3 ö 6 C6 ç ÷ .( 3 ln x ) 3 è 84 ø

Þ (ln x)6 = 1 Þ (ln x)6 = (ln e)6 Þ x=e

So, (1 + x) (1 – x3)100 will have 2 × 101 = 202 terms 28.

9

Þ 36 = 84 ´

= (1 + x) [(1 + x) (1 – x + x2)]100 x3)100]

(b) Let r + 1 = 7 Þ r = 6 Given expansion is

Thus, 50 = 0 + (k – 1) 10

There are only two rational terms – first term and last term.

50 = 10k – 10 Þ k = 6

Now sum of two rational terms

Thus, the six terms of the given expansion in which x and y are free from radical signs.

= (2)5 + (3)2 = 32 + 9 = 41

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Binomial Theorem

33.

(d) Given function is

36.

f(y) = 1 – (y – 1) + (y – 1)2 – (y – 1)3 + ........ – (y – 1)17 In the expansion of (y – 1)n

(a) (8)2n – (62) 2n + 1 = (64) n – (62)2n + 1 = (63 + 1)n – (63 – 1)2n + 1 = é nC0 (63) n + n C1 (63) n -1 + nC2 (63)n - 2 ë

Tr + 1 = nCr yn – r (– 1)r coeff of y2 in (y – 1)2 = 2C0 coeff of y2 in (y – 1)3 = – 3C1 coeff of y2 in (y – 1)4 = 4C2 So, coeff of termwise is

+ ........+

Cn-1 (63) + nCn ù û

- é 2n+1C0 (63)2n+1 - 2n+1C1 (63)2n ë + 2 n +1C2 (63)2 n -1 – ........+ (–1)2n+1 2n+1C2n+1 ù û

2C + 3C + 4C + 5C + .......... + 17C 0 1 2 3 15 = 1 + 3C1 + 4C2 + 5C3 + .......... + 17C15 = (3C0 + 3C1) + 4C2+ 5C3 + .......... + 17C15 = 4C1 + 4C2 + 5C3 + .......... + 17C15 = 5C2 + 5C3 + .......... + 17C15 = 18C15 = 18C3

n n-1 n n- 2 n + C2 (63) n-3 = 63 × éë C0 (63) + C1 (63)

+........ + nCn -1 ùû + 1 – 63 × é 2n+1C0 (63) 2n - 2n+1C1 (63)2 n-1 + ....... + 2 n+1 C2n ù + 1 ë û

34. (a) Statement 2 : P ( n ) : n 7 - n is divisible by 7

37.

Put n = 1, 1 – 1 = 0 is divisible by 7, which is true Let n = k, P (k) : k7 – k is divisible by 7, true Put n = k + 1 \

n

( n + 1) 7 - n7 - 1 is divisible by 7

Hence both Statements 1 and 2 are correct and Statement 2 is the correct explanation of Statement -1. 35. (b) (1 – x – x2 + x3)6 = [(1– x) – x2 (1 – x)]6 = (1– x)6 (1 – x2)6 = (1 – 6x + 15x2 – 20x3 + 15x4 – 6x5 + x6) × (1 – 6x2 + 15x4 – 20x6 + 15x8 – 6x10 + x12) 7 Coefficient of x = (– 6) (– 20) + (– 20)(15) + (– 6) (–6) = – 144

r =0

r =1

n –1

Cr –1 x r + (1 + x)n

n –1

Cr –1 x r –1 + (1 + x )n

= nx (1+ x) n–1 + (1+ x) n = RHS \ Statement 2 is correct. Putting x = 1, we get n

å (r + 1)n Cr = n × 2n –1 + 2n = (n + 2) × 2n –1.

r =0

Statement 1 : n - n is divisible by 7

Þ

r =0

n

7

is divisible by 7

r =0 n

= nx å

\ P(k + 1) is divisible by 7 Hence P(n) : n7 – n is divisible by 7

( n + 1) 7 - n7 - 1 + ( n7 - n)

n

r =1

So 7 C1 , 7 C 2 , ......7 C 6 are all divisible by 7

Þ

n

n =år× r

P(k + 1) : k7 + 7C1k6 + 7C2k2 +......+ 7C6k + 1 – k – 1, is div. by 7. P(k + 1) : (k7 – k) + (7C1k6 + 7C2k5 +........+ 7C6k) is div. by 7. Since 7 is coprime with 1, 2, 3, 4, 5, 6.

( n + 1) 7 - ( n + 1) is divisible by 7

n

n r n r å (r + 1) nCr x r = å r. Cr x + å Cr x

7 P (k + 1) : (k + 1) – ( k + 1) is div. by 7

Þ

= 63 × some integral value + 2 Hence, when divided by 9 leaves 2 as the remainder. (b) From statement 2:

38.

\ Statement 1 is also true and statement 2 is a correct explanation for statement 1. (b) Tr + 1 = (–1)r. nCr (a)n – r. (b)r is an expansion of (a – b)n \ 5th term = t5 = t4+1 = (–1)4. nC4 (a)n–4.(b)4 = nC4 . an–4 . b4 6th term = t6 = t5+1 = (–1)5 nC5 (a)n–5 (b)5 Given t5 + t6 = 0 \ nC4 . an–4 . b4 + (– nC5 . an–5 . b5) = 0 Þ

n! an n! a n b5 . .b4 . =0 4!(n - 4)! a 4 5!(n - 5)! a5

Þ

é 1 b ù ê ú = 0 [Q a ¹ 0, b ¹ 0] 4!( n - 5)!.a 4 ë (n - 4) 5.a û

Þ

b 1 a n-4 =0 Þ = n - 4 5a 5 b

n !.a n b4

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39.

Mathematics

(d)

(1 - y ) m (1 + y ) n

42.

= [1 - m C1 y + m C2 y 2 - ......] [1 + n C1 y + n C 2 y 2 + .....]

=

ì m(m - 1) n(n - 1) ü + - mn ý y2 + ..... = 1 + (n - m) y + í 2 2 î þ \ a1 = n - m = 10

and a2 =

...(i)

é 2 1 ù ê ax + bx ú ë û

43.

(c) tr + 2 = 2nCr + 1 xr + 1;t3r = 2nC3r – 1 x3r – 1 Given that, 2nCr + 1 = 2nC3r – 1 ; Þ r + 1 + 3r – 1 = 2n Þ 2n = 4r Þ n = 2r

[from (i)] 44.

(a) We know that tp + 1 = p + qCp xp and tq + 1 =

2 11 - r

11

= Cr (ax )

æ 1ö çè ÷ø bx

256 - r 2 (5) r / 8

It is possible if r is an integral multiple of 8 and 0 £ r £ 256

Tr +1 in the expansion 11

Cr (3)

\ r = 33

m2 + n2 - m - n - 2mn = 10 2

...(ii) Þ m + n = 80 Solving (i) and (ii), we get \ m = 35, n = 45 (d)

256

Cr ( 3)256 - r (8 5)r

256

Terms will be integral if 256 - r & r both are +ve integer.. 2 8

(m - n) 2 - (m + n) = 20

40.

Tr +1 =

(c)

Q

r

45.

p+qC xq q

p + qC = p + qC .[ Remember nC = nC p q r n–r

(c) General term = Tr +1 =

= 11 Cr (a)11 - r (b) - r ( x)22 - 2r - r

æ kö Cr ( x )10- r × ç - 2 ÷ è x ø

For the Coefficient of x7, we have 22 – 3r = 7 Þ r = 5

=

10

Cr ( - k ) r × x

\ Coefficient of x 7

=

10

Cr ( - k ) r × x

= 11C5 (a)6 (b)- 5

11

1 ù é ê ax - 2 ú bx û ë

-r

r

r

11

Þ k2 =

- 2 r +11- r

46.

-7

...(ii)

2

nn

Þ

= (-1) n + ( -1) n -1.n

= ( -1)n (1 - n)

Cr

Cr +1

=

n+ 5

Cr ,

n+ 5

Cr +1 ,

n+ 5

Cr + 2

1 2

(Given)

r +1 1 = Þ 3r = n + 3 n+ 5- r 2 n +5

n

(1 + x )(1 - C1 x + C2 x - .... + (-1) Cn x ) = (-1) nn Cn + ( -1) n -1n Cn -1

n +5

Q n+ 5

\ 11 C5 (a)6 (b) - 5 = 11C6 a5 ´ (b) - 6 Þ ab = 1. (b) Coeff. of xn in ( 1+x) (1 – x)n = coeff of xn in n

(a) Consider the three consecutive coefficients of

(1 + x) n+ 5 be

= C6 a ´ 1 ´ (b) Q Coefficient of x7 = Coefficient of x–7 From (i) and (ii),

n

405 ´ 2 81 = =9 10 ´ 9 9

\| k | = 3

-6

5

10 - 5 r 2

\10C2 (- k )2 = 405

(-1) ´ (b) ( x) = Cr (a) For the Coefficient of x–7, we have Now 11 – 3r = – 7 Þ 3r = 18 Þ r = 6 \ Coefficient of x

10 - r - 2r 2

10 - 5r =0Þr =2 2

æ 1 ö = 11Cr (ax 2 )11 - r ç è bx 2 ÷ø

11- r

11

r

Since, it is constant term, then

...(i)

Again T r + 1 in the expansion

41.

10

]

and

n+ 5

Cr +1 Cr + 2

=

...(i)

5 7

r+2 5 = Þ 12r = 5n + 6 ...(ii) n+4-r 7 Solving (i) and (ii) we get r = 4 and n = 6 \ Largest coefficient in the expansion is 11C6 = 462. Þ

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Binomial Theorem

47.

n- r (3) 2

Tr +1 = Cr n

50.

n

1ö æ 1 (c) Here, ç 3 2 + 5 8 ÷ è ø

n

r (5) 8

n-r r and are integer 2 8 So, r must be 0, 8, 16, 24 ......

n

n

Cr

æ 3x2 ö 9 (c) General term = Tr +1 = Cr ç ÷ è 2 ø

9- r

æ 1ö çè - ÷ø 3x

r

Cr

=

12 n - r 12 Þ = 5 r +1 5

16- r

r

T9 = 16C8

18 - 3r = 0 Þ r = 6 6

æ 3 ö æ 1ö æ 1ö \ T7 = C6 ç ÷ ç - ÷ = 9C3 ç ÷ è 2 ø è 3ø è 6ø

T9 = 16C8

3

9

7 = 7. 18 (a) General term of \18k = 18 ´

1

-1

=

Cr (ax 9 )10 - r (bx 6 ) r

10

Cr

10

Then, by AM-GM inequality 1

a 3 + b2 ³ ( a 3b 2 ) 2 2

Þ (2)2 ³ a3b2 Þ a 6b4 £ 16 Q The maximum value of the term independent of x = 10k 10

C4 ×16 Þ k = 336.

l

2 Now, l = 1

C4 a 6b4

Since a 3 + b2 = 4

\10k =

(sin 2q)8

28 8

æ 1 ö ç ÷ è 2ø

= 16C8.28.24 [Q min. value of l2 at q = p/8]

10 - r r - = 0 Þ r = 4. 9 6

\ Term independent of x =

28

l2 = 16C8 =

10 - r r a10 - r b r ( x ) 9 6

Term independent of x if

1 sin8 q cos8 q

é p pù When qÎ ê , ú , th en least value of the term ë16 8 û independent of x,

-1

1

10

r

p p When qÎ éê , ùú , then least value of the term ë8 4û independent of x, l 1 = 16C8 28 [Q min. value of l1 at q = p/4]

3

9 ´ 8 ´ 7 æ 1ö æ 7ö = ç ÷ = çè ÷ø . 3 ´ 2 ´ 1 è 6ø 18

(ax 9 + bx 6 )10 =

...(i)

æ x ö æ 1 ö Tr +1 =16Cr ç ÷ ç x cos q ÷ è sin q ø è ø For r = 8 term is free from ‘x’

The term is independent of x, then

49.

5 n - r +1 5 Þ = 2 r 2

...(ii) Þ 5n - 17r - 12 = 0 Solving eqns. (i) and (ii), n = 118, r = 34 51. (b) General term of the given expansion

æ 1 ö 18-3r çè - ÷ø x 3

3

=

Cr -1

Cr +1 n

Þ n = 256

9-r

n

Þ 2n - 7r + 2 = 0

Now n = t33 = a + (n - 1)d = 0 + 32 ´ 8 = 256

æ 3ö = 9Cr ç ÷ è 2ø

Cr -1 : nCr : n Cr +1 = 2 : 5 :12

Þ

Q

48.

(118) According to the question,

52.

16

C8 .28.24

16

C8.28

= 16

(d) Let the general term of the expansion Tr+1 = =

60

60

æ 1ö Cr ç 7 5 ÷ è ø 12-

Cr × (7)

60- r

r 5 ( -1) r

æ 1ö ç -310 ÷ è ø

r

r

× (3)10

Then, for getting rational terms, r should be multiple of L.C.M. of (5, 10) Then, r can be 0, 10, 20, 30, 40, 50, 60. Since, total number of terms = 61 Hence, total irrational terms = 61 – 7 = 54

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55.

10

53.

æ 1 ö 0 æ 1 ö æ 1 ö 10 ç 23 + 1 ÷ 10 3 C 2 + ÷ ç 1÷ = 0ç ç è 2(3)1/ 3 ÷ø è ø çè ÷ 3 2(3) ø

(c)

(c) Given expression can be written as 10

é 1/3 3 3 ù ( x )2 - 12 ú ê ( x ) +1 ê 2/3 1/3 x ( x - 1) úû ë x - x +1

10

æ 1ö æ 1 ö0 L + C10 ç 2 3 ÷ ç 1/ 3 ÷ è ø è 2(3) ø

10

æ æ x +1 ö ö = ç ( x1/3 + 1) - çç ÷÷ ç x ÷ø ÷ø è è

10

6

æ 1ö 1 ç 23 ÷ 1 4 5th term from beginning T5 =10C4 è ø æ 3 ö ç 2.3 ÷ è ø

and 5th term from end T11–5+1 = æ 1ö 3 T5 : T7 = C4 ç 2 ÷ è ø 10

\

2 æ 1ö æ 1 ö 3 ç ÷ 2 = ç ÷ : ç 1÷ è ø è 3ø 2.3 2

2

10

4 æ 1ö æ 23 ç

C6 ç è

ö 1 ÷ ÷ 1 ø çè 3 ÷ø 2.3

4

= (x1/3 – x–1/2)10 General term = Tr+1 = 10Cr (x1/3)10–r(–x–1/2)r =

6

4 ö ö æ 1ö æ ç 1 ÷ : 10C ç 2 3 ÷ ç 1 ÷ 6 1 ç 1÷ è ø çè 3 ø÷ è 2.33 ø 2.3

= 10 Cr ( -1) r · x

6

56. 1

8

æ 2 1ö (c) General term of ç 2x - ÷ is è xø

8

15

Required ratio =

r

8

-r

2

10 - r r - =0 3 2

C10 ´ (2)10

15 ! 10 ! 5 ! = 15 ! ´ (2)5 5 !10 !

= 1 : 32 57. (d) Given expansion can be written as

r

æ 1ö +3x 5 . 8 Cr (2x 2 )8- r ç - ÷ è xø =

15

C5 ´ (2)5

r

æ 1ö 1 æ 1ö Cr (2x 2 )8- r ç - ÷ - 8Cr (2x 2 )8-r ç - ÷ è xø x è xø

· ( -1) r · x

(d) Tr+1 = 15Cr(x2)15–r . (2x–1)r = 15Cr × (2)r × x30–3r For independent term, 30 – 3r = 0 Þ r = 10 Hence the term independent of x, T11 = 15C10 × (2)10 For term involve x15, 30 – 3r = 15 Þ r = 5 Hence coefficient of x15 = 15C5 × (2)5

r

æ -1 ö Cr (2x 2 )8- r ç ÷ è x ø \ Given expression is equal to

=

10 - r 3

10- r - r 3 2

8

æ 1 5ö 8 2 8- r æ 1 ö çè1 - + 3x ÷ø Cr (2x ) çè - ÷ø x x

Cr x

Þ r=4 So, required term = T5 = 10C4 = 210

2 3 3 = 2 × 2 × 3 = 4(6) 3 :1 = 4.(36) 3 :1 1

54.

10

Term will be independent of x when

2

2

10

æ 1 ö = ç x1/ 3 + 1 - 1 ÷ xø è

n

r

Cr 28- r (-1)r x16 -3r – 8 C r 28- r ( -1)r x15-3r r

æ 1ö +3. 8Cr 2(8- r) ç - ÷ (-1)r x 21-3r è xø For the term independent of x, we should have 16 – 3r = 0, 15 – 3r = 0, 21 – 3r = 0 From the simplification we get r = 5 and r = 7 \ – 8C5 (23) (–1)5 – 3. 8C7.2 é 8! ù é 8! ù +ê ´ 8ú - 3 ´ ê ´ 2ú ë 5!3! û ë 7!1! û

æ x -1 ö n n –n 2n ç x ÷ .(1 - x ) = (–1) x (1 – x) è ø Total number of terms will be 2n + 1 which is odd (Q 2n is always even)

\ Middle term =

2n + 1 + 1 = (n + 1) th 2

Now, Tr +1 = nCr (1)r x n - r 2n

So,

Cn × x 2 n - n n n

x .( -1)

= 2 n Cn .( -1)n

Middle term is an odd term. So, n + 1 will be odd. So, n will be even. \ Required answer is 2nCn.

= (56 × 8) – 48 = 448 – 6 × 8 = 448 – 48 = 400

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Binomial Theorem

58.

(c) The middle term in the expansion of

6

1 æ 2 3 ö 3 ö x8 æ 2x - 2 ÷ - ç 2x2 - 2 ÷ ç 60 è x ø 81 è x ø Term independent of x,

(1 + ax) 4 = T3 =4 C2 (ax )2 = 6a 2 x 2 The middle term in the expansion of (1 - ax)6 = T4 =6 C3 ( -ax )3 = -20a 3 x3 According to the question 6a 2 = -20a 3 Þ a = -

6

1 æ 2 3 ö 1 2x - 2 ÷ - . = Coefficient of x in 60 çè 81 x ø

3 10

æ 2 3 ö coefficient of x–8 in ç 2x - 2 ÷ x ø è

20

59.

(a) The given series,

å 50- r C6

=

= 50C6 + 49C6 + 48C6 + 47 C6 + ... + 32C6 + 31C6 + 30C6

= (30 C7 + 30C6 ) + 31C6 + 32C6 + .... + 49C6 + 50C6 - 30C7

63.

= (31 C7 + 31C6 ) + 32C6 + .... + 49C6 + 50C6 - 30C7 = (32 C7 + 32C6 ) + .... + 49C6 + 50C6 - 30C7

=

10! b+ 2 g x a !b!g !

Now, compare it with R.H.S., A = 420 and b = 18 64.

Case-2 : When g = 1, b = 2, a = 7

10! = 360 7! 2!1!

(a) Given expression, (1 – x)10 (1 + x + x2)9 (1 + x) = (1 – x3)9 (1 – x2) = (1 – x3)9 – x2(1 – x3)9 Þ Coefficient of x18 in (1 – x3)9 – coeff. of x16 in (1 – x3)9 9! 7 ´ 8 ´ 9 = = 84 6!3! 6 (c) Given expression is (1 + ax + bx2) (1 – 3x)15 Co-efficient of x2 = 0 Þ 15C2 (– 3)2 + a . 15C1(– 3) + b . 15C0 = 0

9 = C6 - 0 =

Case-3 : When g = 2, b = 0, a = 8

10! = 45 8!0! 2!

æ 1 x8 ö æ 2 3 ö6 ç - ÷ ç 2x - 2 ÷ ç 60 81 ÷ è x ø è ø

r =1

= 420 × 218

10! b+ 2 g x a !b! g !

Hence, total = 615 61. (30) Let (1– x + x2 ..... x2n) (1 + x + x2 ..... x2n) = a0 + a1x + a2x2 + ..... put x = 1 1(2n + 1) = a0 + a1 + a2 + ..... a2n put x = – 1 (2n + 1) × 1 = a0 – a1 + a2 + ...... a2n Adding (i) and (ii), we get, 4n + 2 = 2(a0 + a2 + .....) = 2 × 61 Þ 2n + 1 = 61 Þ n = 30. 62. (d) Given expression is,

r =1

20

19 = 20 å r. Cr -1

é 20 ù 18 19 = 20 êê19 å Cr - 2 + 2 úú = 20[19 . 218 + 219] ë r=2 û

For coefficient of x4; b + 2g = 4 Here, three cases arise Case-1 : When g = 0, b = 4, a = 6

Þ

å r 2 . 20 Cr

20 é 20 ù 19 19 ê ú + 20 ( r 1) C C å å 1 1 r r = ê úû r =1 ër = 1

= 51C7 - 30C7

Þ

(b) Given, 20C1+ 22.20C2 + 32.20C3 + ... + 202.20C20 = A(2b) Taking L.H.S., 20

................................................................. ................................................................. .................................................................

Þ

6

-1 6 1 6 C3 (2)3 (3)3 + C (2) (3)5 60 81 5 = – 72 + 36 = – 36

r=0

60. (615) General term of the expansion =

6

=

65.

15 ´14 ´ 9 - 15 ´ 3a + b = 0 2 Þ 945 – 45a + b = 0 ...(i) Now, co-efficient of x3 = 0 Þ 15C3 (– 3)3 + a . 15C2(– 3)2 + b .15C1 (– 3) = 0

Þ ...(i) ...(ii)

15 ´14 ´13 15 ´14 ´ 9 ´ (-3 ´ 3 ´ 3) + a ´ – b × 3 × 15 = 0 3´ 2 2 Þ 15 × 3 [– 3 × 7 × 13 + a × 7 × 3 – b] = 0 Þ 21a – b = 273 ...(ii) From (i) and (ii), we get, a = 28, b = 315 Þ (a, b) º (28, 31,5)

Þ

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66.

Mathematics

(b) 2.20C0 + 5.20C1 + 8.20C2 + ...... + 62.20C20 20

=

å (3r + 2) 20Cr

20

20

r =0

r =0

+

20 20 = 3 å r. Cr + 2 å Cr

r =0

20

20

r =1

r=0

Hence, the coefficient of t4 = 1 ×

19 20 = 60å Cn -1 + 2 å Cr

71.

3´ 4´ 5´ 6 = 15 4 ´ 3 ´ 2 ´1 21 (a) We have ( C1 + 21C2 ...... + 21C10)

– (10C1 + 10C2 ..... 10C10)

th

æn ö (a) Middle Term, ç + 1÷ term in the binomial è2 ø

= 8

æ x3 3 ö expansion of ç + ÷ is, è 3 xø 3ö4

=

69.

4

(d)

25

r =0

r =0

æ

å ( 50Cr ×50-r C25- r ) = å çè 50 - r 50

æ 50 1 æ 25 ö ö = å ç 25 ´ 25 ´ çè 25 - r r ÷ø ÷ ø r =0 è 25

= C25 å Cr = 25

50

25

C25 (2 )

r =0

70.

.... + 10C10 = 210 – 1)

1 21 [2 - 2] - (210 - 1) 2

72.

(b) Total number of terms = n+2C2 = 28 (n + 2) (n + 1) = 56; n = 6 6

æ 2 4 ö \ Put x = 1 in expansion ç 1 - + 2 ÷ , è x x ø

we get sum of coefficient = (1 – 2 + 4)6 = 36 = 729. 73.

(c) We know that (a + b)n + (a – b)n = 2[n C0 a n b0 + n C2 a n - 2b 2 + n C4 a n -4b 4 ...]

(1 - 2

Then, by comparison, K = 225 (b) Consider the expression 3

æ 1 - t6 ö ç 1 - t ÷ = (1 – t6)3(1 – t)–3 è ø

3× 4 2 æ t = (1 – 3t6 + 3t12 – t18) çè 1 + 3t + 2!

x)

50

+ (1 + 2 x )

50

2 éë 50 C0 + 50C2 (2 x )2 + 50C4 (2 x ) 4 ...ùû = 2 éë 50 C0 + 50C2 22 x +

50 - r ö r 25 25 - r ÷ø

50

C4 24 x 2 + ...ùû

Putting x = 1, we get, 50

25

50

10C + 10C + 1 2

= (220 – 1) – (210 – 1) = 220 – 210

8 ´ 7 ´ 6 ´ 5 12- 4 ´x Þ = 5670 4´3´2 Þ x8 = 81 Þ x8 – 81 = 0 \ sum of all values of x = sum of roots of equation (x8 – 81 = 0). (b) Consider the expression 20C 20C + 20C 20 20 20 20 20 r 0 r – 1 C1 + Cr – 2 C2 + ... + C0 × Cr For maximum value of above expression r should be equal to 20. as 20C20 × 20C0 + 20C19 × 20C1 + ... + 20C20 × 20C0 = (20C0)2 + (20C1)2 + ... + (20C20)2 = 40C20. Which is the maximum value of the expression, So, r = 20. 25

1 21 [( C1 + .... + 21C10 ) + (21C11 + .... 21C20)] – (210 – 1) 2

(Q

æx æ 3ö 8 T4 + 1 = C4 ç ÷ çè ÷ø = 5670 3 è ø x

68.

3× 4 × 5 × 6 4!

=

= 60 × 219 + 2 × 220 = 221 [15 + 1] = 225 67.

3× 4×5 3 3× 4×5×6 4 ö t + t + K ฀÷ ø 3! 4!

74.

(b)

C0 + 50C2 22 + 50C4 24... =

350 + 1 2

Given expansion (1 + x n + x 253 )10

Let x1012 = (1)a (xn)b. (x253)c Here a, b, c, n are all +ve integers and a £ 10, b £ 10, c £ 4, n £ 22, a + b + c = 10 Now bn + 253c = 1012 Þ bn = 253 (4 – c) For c < 4 and n £ 22; b > 10, which is not possible. \ c = 4, b = 0, a =6 \ x1012 = (1)6. (xn)0 . (x253)4 Hence the coefficient of x1012 =

10! = 6!0!4!

10C 4

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Binomial Theorem 10

75.

S2 = å j

(b)

10

j =1

10

C j = å 10 9C j -1

= (1 - x )

j =1

n n -1 é n ù Cr -1 ú êëQ Cr = r û

76.

= 10 é 9 C0 + 9C1 + 9 C2 + .... + 9 C9 ù = 10.29 ë û (a) The given situation in statement 1 is equivalent to find the non negative integral solutions of the equation x1 + x2 + x3 + x4 + x5 = 6 which is coeff. of x6 in the expansion of (1 + x + x2 + x3 + .....¥)5= coeff. of x6 in (1– x)–5

= coeff. of x6 in 1 + 5x +

é ê x = ê1 + + ë 2

79.

5.6 2 x .......... 2!

5 × 6 × 7 × 8 × 9 ×10 10! 10 = = C6 6! 6!4! \ Statement 1 is wrong. Number of ways of arranging 6A’s and 4B’s in a row

1 3 ù . 2 2 x 2 ú é -3 x 2 ù = -3 x 2 ú ú 8 2! û êë 8 û

Tr +1 =

27 ö æ 32 \ r = 7 . çèQ n = 5 ÷ø 5

Therefore, first negative term is T8 . n

80.

10! 10 = C 6 which is same as the number of ways the 6!4! child can buy six icecreams. \ Statement 2 is true. (d) We know that, (1 + x)20 = 20C0 + 20C1x + 20C2 x2

(d)

æ 1ö (1 + 0.0001)10000 = ç1 + ÷ , n = 10000 è nø

1 n ( n - 1) 1 n ( n - 1)( n - 2) 1 1 = 1 + n. + + + ... + n 2! n 2 3! n n3 n

=

20C x10 + ..... 20C x20 10 20 x = –1, (0) = 20C0 – 20C1 + 20C2 – 20C3 + ...... + 20C10 – 20C11 .... + 20C20 0 = 2[20C0 – 20C1 + 20C2 – 20C3 + ..... – 20C9] + 20C10

ù 3 1 ö . 2 2 2 x 2 ÷ - æç1 + 3 x + 3.2 x ö÷ ú ú ÷ ç 2! 2 2! 4 ÷ø ûú ø è

n(n - 1)(n - 2).........( n - r + 1) r ( x) r! For first negative term, n - r +1 < 0 Þ r > n +1

(d)

Þr >

=

77.

éæ êç 3 êç 1 + x + ëêè 2

-1 2

=1+1 +

1 æ 1 ö 1 æ 1 öæ 2 ö 1 1- + 11 - ÷ + ... + n 2! çè n ÷ø 3! çè n ÷ç øè n ø n

+ ......

Put Þ

Þ 20C

10

= 2[20C0 – 20C1 + 20C2 – 20C3 + ...... – 20C9 + 20C10]

Þ 20C – 20C + 20C – 20C + .... + 20C 0 1 2 3 10

= 78.

1 20 C10 2

< 1+

1 1 1 1 + + + ........+ 1! 2! 3! (9999)!

1 1 + + .......¥ = e < 3 1! 2! (c) Take a = 1 and b = 1 in (a + b)n × =1+

81.

2n = 4096 = 212 Þ n = 12; The greatest coeff = coeff of middle term. So middle term = t7. Þ t7 = t6 + 1 = 12C6a6b6

Þ Coeff of t7 = 12C6 =

12! = 924. 6!6!

(c) Q x3 and higher powers of x may be neglected 3

\

(1 + x) 2 - æçè1 +

(

1-

1 x2

)

xö ÷ 2ø

3

EBD_8344

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Mathematics

9

Sequences and Series 7.

TOPIC Ć Arithmetic Progression 1.

2.

3.

The common difference of the A.P. b1, b2, ..., bm is 2 more than the common difference of A.P. a 1, a 2, ..., a n. If a40 = – 159, a100 = – 399 and b100 = a70, then b1 is equal to: [Sep. 06, 2020 (II)] (a) 81 (b) – 127 (c) – 81 (d) 127 2sin2a–1 4–2sin2a If 3 , 14 and 3 are the first three terms of an A.P. for some a, then the sixth term of this A.P is: [Sep. 05, 2020 (I)] (a) 66 (b) 81 (c) 65 (d) 78 If the sum of the first 20 terms of the series

(a1 ¹ 0), then the sum of the A.P., a1 , a3 , a5 , ..., a23 is ka1, where k is equal to : [Sep. 02, 2020 (II)] (a) 8.

9.

log (71/ 2 ) x + log (71/ 3 ) x + log (71/ 4 ) x + ... is 460, then x is

equal to : (a) 72 4.

(b) 71/2

(c) e 2

[Sep. 05, 2020 (II)] (d) 746/21

10.

difference is an integer and Sn = a1 + a2 + .... + an . If ( Sn - 4 , an - 4 ) is equal to :

5.

(a) 6.

1 6

(b)

1 5

(c)

1 4

(d)

(d) nth term is -4

2 5

121 10

(c)

72 5

(d) -

72 5

The number of terms common to the two A.P.’s 3, 7, 11, ..., 407 and 2, 9, 16, ..., 709 is _____. [NA Jan. 9, 2020 (II)] 1 1 and its 20th term is , 20 10 then the sum of its first 200 terms is: [Jan. 8, 2020 (II)]

If the 10th term of an A.P. is

(b) 50

1 4

(c) 100

(d) 100

1 2

Let f : R ® R be such that for all x Î R, (21+x + 21–x), f (x) and (3x + 3–x) are in A.P., then the minimum value of f (x) is: [Jan. 8, 2020 (I)] (a) 2 (b) 3 (c) 0 (d) 4 Five numbers are in A.P., whose sum is 25 and product is

(a) 27 12.

(b) 7

(c)

21 2

(d) 16

Let Sn denote the sum of the first n terms of an A.P. If S4 = 16 and S6 = –48, then S10 is equal to : [April 12, 2019 (I)] (a) –260

13.

3 1 4 20 + 19 + 19 + 18 + ... upto nth term is 488 and then 5 5 5 nth term is negative, then : [Sep. 03, 2020 (II)] (a) n = 60 (b) nth term is –4

(b)

1 2520. If one of these five numbers is - , then the greatest 2 number amongst them is: [Jan. 7, 2020 (I)]

1 7

In the sum of the series

(c) n = 41

11.

[Sep. 04, 2020 (II)]

(a) (2490, 249) (b) (2480, 249) (c) (2480, 248) (d) (2490, 248) If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is : [Sep. 03, 2020 (I)]

121 10

(a) 50

Let a1 , a2 ,....., an be a given A.P. whose common a1 = 1, an = 300 and 15 £ n £ 50, then the ordered pair

If the sum of first 11 terms of an A.P., a1 , a2 , a3 , ... is 0

(b) –410

(c) –320

(d) –380

If a1, a2, a3, …… are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P. is : [April 12, 2019 (II)] (a) 200

14.

(b) 280

(c) 120

(d) 150

If a1, a2, a3, ..... an are in A.P. and a1 + a4 + a7 + ….+ a16 = 114, then a1 + a6 + a11 + a16 is equal to : [April 10, 2019 (I)] (a) 98

(b) 76

(c) 38

(d) 64

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Sequences and Series

15.

Let the sum of the first n terms of a non-constant A.P., n(n - 7) a1, a2, a3, ………….. be 50n + A, where A is a 2 constant. If d is the common difference of this A.P., then the ordered pair (d, a50) is equal to: [April 09, 2019 (I)] (a) (50, 50 + 46A) (b) (50, 50 + 45A) (c) (A, 50 + 45A) (d) (A, 50 + 46A)

24.

Let

25.

12

k =0

k =1

17.

18.

19.

20.

21.

22.

Let a 1 , a 2 , ....., a 30 be an A.P., S =

T=

Let a1,a 2 ,a 3 ,..., a 49 be in A.P. such that

å a 4k +1 = 416 and a 9 + a 43 = 66. If

å f (a + k) = 16(210 - 1) , where the function f satisfies

f(x + y) = f(x) f(y) for all natural numbers x, y and f(a) = 2. Then the natural number ‘a’ is: [ April 09, 2019 (I)] (a) 2 (b) 16 (c) 4 (d) 3 If the sum and product of the first three terms in an A.P. are 33 and 1155, respectively, then a value of its 11th term is: [April 09, 2019 (II)] (a) –35 (b) 25 (c) –36 (d) –25 The sum of all natural numbers ‘n’ such that 100 < n < 200 and H.C.F. (91, n) > 1 is : [April 08, 2019 (I)] (a) 3203 (b) 3303 (c) 3221 (d) 3121 If nC4, nC5 and nC6 are in A.P., then n can be : [Jan. 12, 2019 (II)] (a) 9 (b) 14 (c) 11 (d) 12 If 19th term of a non- ero A.P. is ero, then its (49th term) : (29th term) is : [Jan. 11, 2019 (II)] (a) 4 : 1 (b) 1 : 3 (c) 3 : 1 (d) 2 : 1 The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is: [Jan. 10, 2019 (I)] (a) 1256 (b) 1465 (c) 1365 (d) 1356

1 1 1 , 2 ,..... are two A.P's such that h1 h hn

x3 = h2 = 8 and x8 = h7 = 20, then x5. h10 equals. [Online April 15, 2018] (a) 2560 (b) 2650 (c) 3200 (d) 1600

10

16.

If x1 , x2, ....., xn and

2 a12 + a 22 + ... + a17 = 140m , then m is equal to :

26.

27.

(a) 68 (b) 34 (c) 33 (d) 66 For any three positive real numbers a, b and c, 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c). Then : [2017] (a) a, b and c are in G.P. (b) b, c and a are in G.P. (c) b, c and a are in A.P. (d) a, b and c are in A.P. If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is : [Online April 9, 2017] 1

(a) 2 28.

29.

30

is:

i=1

(a)

15

å a (2i - 1) .

i= 1

30.

If a5 = 27 and S – 2T = 75, then a10 is equal to: [Jan. 09, 2019 (I)] (a) 52 (b) 57 (c) 47 (d) 42

1 1 1 , , ,....., (xi ¹ 0 for i = 1, 2, ...., n) be in A.P.. 23. Let x1 x2 x3

2

(b) 4 3 (c) 4 3 (d) 4 Let a1, a2, a3, ...., an, be in A.P. If a3 + a7 + a11 + a15 = 72, then the sum of its first 17 terms is equal to : [Online April 10, 2016] (a) 306 (b) 204 (c) 153 (d) 612 Let a and b be the roots of equation px2 + qx + r = 0, p ¹ 0. If p, q, r are in A.P and

å a i and

34 9

(b)

2 13 9

1 1 + = 4, then the value of | a – b| a b [2014]

(c)

61 9

æ1ö ÷ is equal to. i =1 è i ø

[Online April 16, 2018] (a) 3

(b)

13 8

(c)

13 4

(d)

1 8

2 17 9

[Online April 11, 2014] (a) 4000 31.

(b) 4020

(c) 4200

(d) 4220

Given an A.P. whose terms are all positive integers. The sum of its first nine terms is greater than 200 and less than 220. If the second term in it is 12, then its 4th term is: [Online April 9, 2014]

n

åç x

(d)

The sum of the first 20 terms common between the series 3 + 7 + 11 + 15 + ......... and 1 + 6 + 11 + 16 + ......, is

such that x1 = 4 and x21 = 20. If n is the least positive integer for which xn > 50, then

[2018]

(a) 8 32.

(b) 16

(c) 20

(d) 24

If a1, a2, a3,...., an, .... are in A.P. such that a4 – a7 + a10 = m, then the sum of first 13 terms of this A.P., is : [Online April 23, 2013] (a) 10 m

(b) 12 m

(c) 13 m

(d) 15 m

EBD_8344

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33.

34.

Mathematics

Given sum of the first n terms of an A.P. is 2n + 3n2. Another A.P. is formed with the same first term and double of the common difference, the sum of n terms of the new A.P. is : [Online April 22, 2013] (a) n + 4n2 (b) 6n2 – n (c) n2 + 4n (d) 3n + 2n2 Let a1, a2, a3,... be an A.P, such that a1 + a2 + ... + a p a1 + a2 + a3 + ... + aq

=

p3 q3

(a) 34 minutes (c) 135 minutes 41.

35.

36.

37.

41 11

(b)

31 121

a1 + a2 + ........... + aq

11 41

(d)

121 1861

If 100 times the 100th term of an AP with non ero common difference equals the 50 times its 50th term, then the 150th term of this AP is : [2012] th (a) – 150 (b) 150 times its 50 term (c) 150 (d) Zero If the A.M. between pth and qth terms of an A.P. is equal to the A.M. between rth and sth terms of the same A.P., then p + q is equal to [Online May 26, 2012] (a) r + s – 1 (b) r + s – 2 (c) r + s + 1 (d) r + s Suppose q and f (¹ 0) are such that sec (q + f), sec q and

(a) ± 2

42.

(c) ±

1 2

(d) ± 2

Let an be the n th term of an A.P. If

41 11

(b)

2

(c) m2 – m (4r + 1) + 4 r 2 – 2 = 0 43.

2 2 (d) m – m (4r – 1) + 4 r + 2 = 0 Let Tr be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers

m, n, m ¹ n, Tm =

1 1 and Tn = , then a – d equals n m [2004]

1 1 1 + (b) 1 (c) (d) 0 m n mn If 1, log9 (31–x + 2), log3 (4.3x – 1) are in A.P. then x equals [2002] (b) 1 – log3 4 (a) log3 4 (c) 1 – log4 3 (d) log4 3

(a)

å a2r = a and r=1

å a2r –1 = β, then the common difference of the A.P. is

a6 equals a21

(b) m 2 – m (4r + 1) + 4 r + 2 = 0

44.

100

, p ¹ q , then

2 (a) m 2 – m (4r – 1) + 4 r – 2 = 0

100

38.

q2

the binomial expansion of (1 + y )m are in A.P., then m and r satisfy the equation [2005]

[Online May 19, 2012] (b) ± 1

p2

2 7 11 (c) (d) 7 2 41 If the coefficients of rth, (r + 1)th, and (r + 2)th terms in the

(a)

æ fö sec (q – f) are in A.P. If cos q = k cos ç ÷ for some k, then è 2ø k is equal to

=

[2006]

; p ¹ q . Then a6 is equal to: a21

(c)

Let a1 , a2 , a3 ............ be terms on A.P. If a1 + a2 + ...........a p

[Online April 9, 2013] (a)

(b) 125 minutes (d) 24 minutes

TOPIC n Geometric Progression

r=1

[2011] (a) a - b 39.

40.

(b)

a -b 100

(c) b - a

(d)

α–β 200

A man saves ` 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ` 40 more than the saving of immediately previous month. His total saving from the start of service [2011] will be ` 11040 after (a) 19 months (b) 20 months (c) 21 months (d) 18 months A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a1 = a2 = ... = a10 = 150 and a10, a11, ... are in an AP with common difference –2, then the time taken by him to count all notes is [2010]

¥

45.

If f (x + y) = f (x) f (y) and

å f ( x) = 2 , x, y Î N, where N is x =1

the set of all natural numbers, then the value of

f (4) is : f (2)

[Sep. 06, 2020 (I)] (a) 46.

2 3

(b)

1 9

(c)

1 3

(d)

4 9

Let a, b, c, d and p be any non ero distinct real numbers such that (a2 + b2 + c2)p2 – 2 (ab + bc + cd)p + (b2 + c2 + d2) = 0. Then : [Sep. 06, 2020 (I)] (a) a, c, p are in A.P.

(b) a, c, p are in G.P.

(c) a, b, c, d are in G.P.

(d) a, b, c, d are in A.P.

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Sequences and Series

47.

Suppose that a function f : R®R satisfies f (x + y) = f(x)f(y) for all x, y Î R and f (a) = 3. If

S=

n

å f (i) = 363 , then n is

x10 - x + 45a( x - 1) , then k is equal to : x -1

i =1

equal to ______. 48.

(a) – 5

[NA Sep. 06, 2020 (II)]

If 210 + 29×31 + 28×32 + . . . . + 2×39 + 310 = S – 211 then S is equal to: [Sep. 05, 2020 (I)] (a) 311 – 212

1 49 (3 - 1) 26

(b)

2 50 (3 - 1) (c) 13

50.

52.

53.

56.

100

(a) (-¥, - 9] È [3, ¥)

(b) [-3, ¥)

(c) (-¥, - 3] È [9, ¥)

(d) (-¥, 9]

to : (a) 300

58.

59.

x + y - xy (1 + x)(1 + y )

(b)

2

2

3

x + y + xy (1 + x)(1 + y )

x + y + xy (d) (1 - x )(1 - y )

{x + ka} + {x 2 + (k + 2)a} + {x 3 + (k + 4)a} +{x4 + (k + 6)a} + ... where

a¹0

and

x ¹ 1. If

n =1

n =1

(c) 175

¥

¥

n= 0

n=0

is equal

[Jan. 9, 2020 (II)] (d) 150 p , 4

then : [Jan. 9, 2020 (II)] (a) x(1 + y) = 1 (b) y(1 – x) = 1 (c) y(1 + x) = 1 (d) x(1 – y) = 1 The greatest positive integer k, for which 49k + 1 is a factor of the sum 49125 + 49124 + ... + 492 + 49 + l, is: [Jan. 7, 2020 (I)] (a) 32 (b) 63 (c) 60 (d) 65 Let a1, a2, a3, ... be a G. P. such that a1 < 0, a1 + a2 = 4 and

å

i =1

ai = 4l, then l is equal to: [Jan. 7, 2020 (II)]

(a) –513

(b) –171

(c) 171

(d)

511 3

The coefficient of x7 in the expression (1 + x)10 + x(l + x)9 + x2(l + x)8 + ... + x10 is: [Jan. 7, 2020 (II)] (a) 210

(b) 330

(c) 120

(d) 420

61. If a, b and g are three consecutive terms of a nonconstant G.P. such that the equations ax 2 + 2bx + g = 0 and x 2 + x – 1 = 0 have a common root, then a (b+ g) is equal to : [April 12, 2019 (II)] 62.

Let S be the sum of the first 9 terms of the series :

200

å a2n = 100, then å an

2n n If x = å (-1) tan2n q and y = å cos q, for 0 < q
0, such that 3 is areal number. Then the sum 1 + + 2 + .... + 11 is equal to: (Online April 10, 2016)

(a) 1

The sum of an infinite geometric series with positive terms

n

3

æ3ö æ3ö æ 3ö æ 3ö Let A n = ç ÷ - ç ÷ + ç ÷ - ... + (-1)n - 1 ç ÷ and è4ø è4ø è 4ø è 4ø Bn = 1 – An. Then, the least odd natural number p, so that Bn > An, for all n ³ p is [Online April 15, 2018] (a) 5 (b) 7 (c) 11 (d) 9 If a, b, c are in A.P. and a2, b2, c2 are in G.P. such that

1 1 + 1 + cos q 1 - sin q

a3 a9 Let a1, a2, ..., a10 be a G.P. If a = 25, then a equals : 5 1

(a) 54

(b)

2

72.

( -1)n ö÷

is equal to : bn ÷ø

a is equal to: c [Jan. 09, 2019 (II)]

(b)

(a) 1365 3i

(b) -1365 3i

(c) -1250 3i

(d) 1250 3i

If m is the A.M. of two distinct real numbers l and n(l, n > 1) and G1, G2 and G3 are three geometric means between l and n, then G14 + 2G 42 + G 34 equals. (a) 4 lmn2 (b) 4 l2m2n2 (c) 4 l2 mn

77.

[2015] (d) 4 lm2n

The sum of the 3rd and the 4th terms of a G.P. is 60 and the product of its first three terms is 1000. If the first term of this G.P. is positive, then its 7th term is : [Online April 11, 2015] (a) 7290

(b) 640

(c) 2430

(d) 320

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Sequences and Series

78.

Three positive numbers form an increasing G. P. If the middle term in this G.P. is doubled, the new numbers are in A.P. then the common ratio of the G.P. is: [2014]

(a)

(b) 2 + 3

(a) 2 - 3

(c)

(d) 3 + 2 2+ 3 The least positive integer n such that

5

(

1 1- 5 2

(c) 79.

10

87.

2 2 2 1 1 - - 2 - .... - n -1 < , is: [Online April 12, 2014] 3 3 100 3 (a) 4 80.

81.

(b) 5

(c) 6

(1000 )!

(c) 82.

84.

85.

86.

(b)

(1001)! ( 51)!( 950 )!

(d)

( 49 )!( 951)!

(b) 8

(c) 4

(a2 + b2 + c2) p2 – 2p (ab + bc + cd) + (b2 + c2 + d2) £ 0, then [Online May 12, 2012] (a) a, b, c, d are in A.P. (b) ab = cd (c) ac = bd (d) a, b, c, d are in G.P. The difference between the fourth term and the first term of a Geometrical Progresssion is 52. If the sum of its first three terms is 26, then the sum of the first six terms of the progression is [Online May 7, 2012] (a) 63 (b) 189 (c) 728 (d) 364 The first two terms of a geometric progression add up to 12. the sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is [2008] (a) –4 (b) –12 (c) 12 (d) 4 In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals [2007]

1 2

(d)

1 5 2

)

5 -1

2k p 2k p ö + i cos ÷ is 11 11 ø

[2006]

(c) – 1 (d) – i (a) i (b) 1 If the expan sion in powers of x of the fun ction a0 + a1 x + a2 x 2 + a3 x 3 ......

then an is [2006]

89.

n n (a) b - a b-a

n n (b) a - b b-a

n +1 - bn +1 (c) a b-a

(d)

bn +1 - a n +1 b-a

Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation [2004] (a) x 2 - 18 x - 16 = 0

(b) x 2 - 18 x + 16 = 0

(c) x 2 + 18 x - 16 = 0 90.

(d) x 2 + 18 x + 16 = 0 Sum of infinite number of terms of GP is 20 and sum of their square is 100. The common ratio of GP is [2002] (a) 5

(d) 2

If a, b, c, d and p are distinct real numbers such that

æ

å çè sin

1 is (1 - ax)(1 - bx)

(1001)! (50 )!( 951)!

Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal, then the last term is : [Online April 25, 2013] (a) 16

83.

88.

(d) 7

(1000 )!

(50 )!( 950 )!

The value of

k =1

In a geometric progression, if the ratio of the sum of first 5 terms to the sum of their reciprocals is 49, and the sum of the first and the third term is 35. Then the first term of this geometric progression is: [Online April 11, 2014] (a) 7 (b) 21 (c) 28 (d) 42 50 The coefficient of x in the binomial expansion of (1 + x) 1000 + x (1 + x) 999 + x 2 (1 + x) 99 8 + .... + x1000 is: [Online April 11, 2014] (a)

)

(

(b)

91.

92.

(b) 3/5

(c) 8/5

(d) 1/5

Fifth term of a GP is 2, then the product of its 9 terms is [2002] (a) 256 (b) 512 (c) 1024 (d) none of these

æ pö The sum of all values of qÎ ç 0, ÷ satisfying è 2ø sin22q + cos4 2q = (a) p

(b)

3 is: 4

5p 4

[Jan. 10, 2019 (I)] (c)

p 2

(d)

3p 8

Harmonic Progression, Relation

TOPIC Đ Between A. M., G. M. and H.M. of two Positive Numbers 93.

If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4th A.M. is equal to 2nd G.M., then m is equal to ___________. [Sep. 03, 2020 (II)]

EBD_8344

www.jeebooks.in M-96

94.

Mathematics

If the arithmetic mean of two numbers a and b, a > b > 0, is a+b five times their geometric mean, then is equal to : a-b

[Online April 8, 2017] (a)

95.

6 2

(b)

3 2 4

If A > 0, B > 0 and A + B = tanA + tanB is :

96.

(a)

3- 2

(c)

2 3

(c)

7 3 12

(d)

5 6 12

p , then the minimum value of 6 [Online April 10, 2016]

(b) 4 - 2 3

97.

f :R ® R

be a function which satisfies

f ( x + y ) = f ( x) + f ( y ), " x, y ÎR. If f (a) = 2 and

å f (k ), n Î N , then the value of n, for which

g(n) = 20, is : (a) 5 (b) 20

(c) 4

1 1 1 :G and . If b a M [Online April 12, 2014]

(a) 1 : 4

(b) 1 : 2

(c) 2 : 3

(d) 3 : 4

If a1, a2, .........., an are in H.P., then the expression a1a2 + a2a3 + .......... + an–1an is equal to [2006] (a) n(a1 - an )

(b) (n - 1)(a1 - an )

(c) na1an

(d) (n - 1)a1an

¥

If x =

¥

n(n + 1)(2 n + 1) is equal to __________. 4 n =1

å

[Jan. 8, 2020 (II)]

n =0

n=0

n=0

where a, b, c are in

A.P and |a | < 1, | b | < 1, | c | < 1 then x, y, z are in [2005] (a) G. P. (b) A.P. (c) Arithmetic - Geometric Progression (d) H.P. 100. If the sum of the roots of the quadratic equation ax 2 + bx + c = 0 is equal to the sum of the squares of

a b c their reciprocals, then , and are in c a b (a) Arithmetic - Geometric Progression (b) Arithmetic Progression (c) Geometric Progression (d) Harmonic Progression.

20

104. The sum

å (1 + 2 + 3 + ... + k )

is _______.

k =1

[Jan. 8, 2020 (I)] 105. If the sum of the first 40 terms of the series, 3 + 4 + 8 + 9 + 13 + 14 + 18 +19 + ... is (102)m, then m is equal to: [Jan. 7, 2020 (II)] (a) 20 (b) 25 (c) 5 (d) 10 106. For x Î R, let [x] denote the greatest integer < x, then the sum of the series

¥

å a n , y = å bn , z = å c n

[Sep. 02, 2020 (II)] (d) 9

7

103. The sum,

Let G be the geometric mean of two positive numbers a

is 4 : 5, then a : b can be:

99.

102. Let

(d) 270

and b, and M be the arithmetic mean of

98.

[Sep. 04, 2020 (I)] (b) (11, 103) (d) (11, 97)

(a) (10, 97) (c) (10, 103)

k =1

[Online April 9, 2016] (c) 258

19) = a - 220b, then an ordered pair (a, b) is equal to :

( n -1)

(d) 2 - 3

(b) 216

101. If 1 + (1 – 22 × 1) + (1 – 42 × 3) + (1 – 62 × 5) + . . . . . . . + (1 – 202 ×

g (n) =

Let x, y, be positive real numbers such that x + y + = 12 and x3y4 5 = (0.1) (600)3. Then x3 + y3 + 3 is equal to : (a) 342

Arithmetic-Geometric Sequence

TOPIC Ė (A.G.S.), Some Special Sequences

[2003]

é 1ù é 1 1 ù é 1 2 ù é 1 99 ù êë - 3 úû + êë - 3 - 100 úû + êë- 3 - 100 úû + L + êë - 3 - 100 úû is [April 12, 2019 (I)] (a) –153

(b) –133

(c) –131

(d) –135

3 ´ 13 5 ´ (13 + 23 ) 7 ´ (13 + 23 + 33 ) + + ) + …... 12 12 + 22 12 + 22 + 32 upto 10th term, is : [April 10, 2019 (I)]

107. The sum

(a) 680 108. The sum 1 +

(b) 600

(c) 660

(d) 620

13 + 23 13 + 23 + 33 + + ¼¼ + 1+ 2 1+ 2 + 3

13 + 23 + 33 + ... + 153 1 - (1 + 2 + 3 + ... + 15 is equal to : 1 + 2 + 3 + ... + 15 2 (a) 620

(b) 1240

(c) 1860

[April 10, 2019 (II)] (d) 660

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Sequences and Series

109. The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 + ….. upto 11th term is: [April 09, 2019 (II)] (a) 915 (b) 946 (c) 945 (d) 916 110. Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is: [April 09, 2019 (II)] (a) 157 2) 262 (c) 225 (d) 190 20

111. The sum

1

å k 2k

is equal to :

k =1

3 11 (a) 2 – 17 (b) 1 – 20 2 2

112. Let Sk =

[April 08, 2019 (II)]

11 (c) 2 – 19 2

21 (d) 2 – 20 2

1 + 2 + 3 + ..... + k . 2

2 If S12 + S22 + ..... + S10 =

5 A. Then A is equal to 12

[Jan. 12, 2019 (I)] (a) 283 (b) 301 (c) 303 (d) 156 113. If the sum of the first 15 terms of the series 3

3

3

3

æ3ö æ 1ö æ 1ö 3 æ 3ö ç ÷ + ç1 ÷ + ç 2 ÷ + 3 + ç 3 ÷ + ....... is equal to è4ø è 2ø è 4ø è 4ø 225 k then k is equal to : [Jan. 12, 2019 (II)] (a) 108 (b) 27 (c) 54 (d) 9 114. The sum of the following series [Jan. 09, 2019 (II)]

1+ 6 +

9 (1 + 2 + 3 ) 12 (1 + 2 + 3 + 4 ) + 7 9 2

2

2

2

2

2

15 (12 + 22 + .... + 52 ) +... up to 15 terms, is: 11 (a) 7520 (b) 7510 (c) 7830 (d) 7820 115. The sum of the first 20 terms of the series

(a) 38 +

1 220

[Online April 16, 2018]

(b) 39 +

If B - 2A = 100 l , then l is equal to : [2018] (a) 248 (b) 464 (c) 496 (d) 232 117. Let a, b, c Î R. If f(x) = ax2 + bx + c is such that a + b + c = 3 and f(x + y) = f(x) + f(y) + xy, " x, y Î R, then

1 219

1 1 (c) 39 + 20 (d) 38 + 19 2 2 116. Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series

10

å f ( n ) is equal

n =1

to : (a) 255

[2017] (b) 330

(c) 165

(d) 190

1 1+ 2 1+ 2 + 3 118. Let Sn = 3 + 3 3 + 3 3 3 + ..... 1 1 +2 1 +2 +3 +

1 + 2 + ...... + n 1 + 23 + ....... + n3

3

, If 100 Sn = n, then n is equal to :

[Online April 9, 2017] (a) 199 (b) 99 (c) 200 (d) 19 119. If the sum of the first n terms of the series 3 + 75 + 243 + 507 + ...... is 435 3 , then n equals : [Online April 8, 2017] (a) 18 (b) 15 (c) 13 (d) 29 120. If the sum of the first ten terms of the series 2

2

2

2

16 æ 3ö æ 2ö æ 1ö 2 æ 4ö ç 1 5 ÷ + ç 2 5 ÷ + ç 3 5 ÷ + 4 + ç 4 5 ÷ + ......., is 5 m, è ø è ø è ø è ø

then m is equal to : (a) 100 (b) 99

[2016] (c) 102

(d) 101

121. For x Î R, x ¹ -1 , if (1 + x)2016 + x (1 + x)2015 + x2 2016

(1 + x)2014 + .... + x2016 =

å ai xi , then a17 is equal to :

i =0

[Online April 9, 2016]

2

+

3 7 15 31 1 + + + + + ... is? 2 4 8 16

12 + 2 × 22 + 32 + 2.4 2 + 52 + 2.62 + ...

(a)

2016! (b) 17! 1999!

2017! 17!2000!

2016! 2017! (d) 16! 2000! 122. The sum of first 9 terms of the series.

(c)

13 13 + 23 13 + 23 + 33 + + + .... 1 1+ 3 1+ 3 + 5 (a) 142 (b) 192 (c) 71 5

123. If

1

å n(n + 1)(n + 2)( n + 3) =

n =1

[2015]

(d) 96

k , then k is equal to 3

[Online April 11, 2015] (a)

1 6

(b)

17 105

(c)

55 336

(d)

19 112

EBD_8344

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Mathematics 30

124. The value of

å ( r + 2) ( r - 3)

r =16

(a) 7770

[Online April 10, 2015] (c) 7775 (d) 7780

(b) 7785

( ) + 3(11)

125. If (10 ) + 2 (11) 10 9

1

8

2

(10 )

7

9

9

(a) 100

(b) 110

121 10

(c)

k =1

[2014] (d)

441 100

126. The number of terms in an A.P. is even; the sum of the odd terms in it is 24 and that the even terms is 30. If the last term exceeds the first term by 10 the A.P. is: (a) 4 (b) 8 127. If the sum

3 12

+

5 12 + 22

+

7 12 + 22 + 32

1 , then the number of terms in 2 [Online April 19, 2014] (c) 12 (d) 16

+ ...... + up to 20 terms is equal

k to , then k is equal to: [Online April 9, 2014] 21 (a) 120 (b) 180 (c) 240 (d) 60 128. The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,....., is [2013]

number n. [2012] (a) Statement-1 is false, Statement-2 is true. (b) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1. (c) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1. (d) Statement-1 is true, statement-2 is false. 134. If the sum of the series 12 + 2.22 + 32 + 2.42 + 52 + ... 2.62 +... 2

of the series, when n is odd, is (a) n2(n + 1)

135. The sum of the series 1 +

(a)

(c)

7 (179 + 10 -20 ) 81

(d)

7 (99 + 10 -20 ) 9

(c) n +

1

+

5 2

1 +2

2

is: 7 11 (b) 2 4 132. The sum of the series :

(a)

+

7

+ .... upto 11-terms 1 + 22 + 32 [Online April 22, 2013] 2

(c)

11 60 (d) 2 11 [Online April 9, 2013]

1 1 1+ + + ....... upto 10 terms, is : 1+ 2 1+ 2 + 3

4 10 28 + + + ... upto n terms is 3 9 27

[Online May 19, 2012]

7 (99 - 10-20 ) 9

2

n 2 ( n - 1) 2

(d) n2(n – 1)

2

(b)

3

(b)

n2 ( n + 1)

(c)

7 (179 - 10-20 ) 81

129. The value of l2 + 32 + 52 + .......................+ 252 is : [Online April 25, 2013] (a) 2925 (b) 1469 (c) 1728 (d) 1456 130. The sum of the series : (b)2 + 2(d)2 + 3(6)2 + ... upto 10 terms is : [Online April 23, 2013] (a) 11300 (b) 11200 (c) 12100 (d) 12300

n ( n + 1) , then the sum 2 [Online May 26, 2012]

upto n terms, when n is even, is

(a)

131. The sum

å ( k 3 - ( k - 1)3 ) = n3, for any natural n

Statement-2:

+ .....

+10 (11) = k (10 ) , then k is equal to:

16 20 18 22 (b) (c) (d) 9 11 11 13 133. Statement-1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + .... + (361 + 380 + 400) is 8000.

(a)

is equal to :

7 1 2 n + - n -1 6 6 3.2

(b)

1 1 2 2.3 n

5 7 1 n - + n -1 3 6 2.3

1 1 (d) n - - n -1 3 3.2

136. The sum of the series

1 1+ 2

+

1 2+ 3

+

1 3+ 4

+ ...

upto 15 terms is [Online May 12, 2012] (a) 1 (b) 2 (c) 3 (d) 4 137. The sum of the series 12 + 2.22 + 32 + 2.42 + 52 + 2.62 + .... + 2(2m)2 is [Online May 7, 2012] (a) m(2m + 1)2 (b) m2(m + 2) (c) m2(2m + 1) (d) m(m + 2)2 138. The sum to infinite term of the series 1+

2 6 10 14 ... is + + + + 3 32 33 34

(a) 3

(b) 4

[2009] (c) 6

(d) 2

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Sequences and Series

1 1 1 - + - ....... upto infinity is 2! 3! 4! [2007]

139. The sum of series

(a)

-

1 2

+

(b)

1 2

e e 140. The sum of the series

(c) e –2

(d) e –1

1 1 1 1+ + + + ....................ad inf. is 4.2! 16.4! 64.6 !

é n(n + 1) ù (a) ê ë 2 úû (c)

[2005]

2

n(n + 1)2 4

143. If Sn =

n

1

å nC

r=0

and t n =

r

(b)

n2 (n + 1) 2

(d)

3n( n + 1) 2

n

r

å nC

r =0

r

t , then n is equal to Sn

[2004] (a)

e -1 e

(b)

e +1 e

141. The sum of series

(a)

(e2 - 2) e 2

(c)

e -1 2 e

1 1 1 + + + ..... is 2! 4! 6!

(b)

(d)

e +1 2 e

(c) n – 1

(d)

1 n 2

[2003]

[2004]

(e - 1)2 2e 2

(e - 1) (e - 1) (d) 2e 2 142. The sum of the first n terms of the series

(c)

12 + 2.2 2 + 3 2 + 2.4 2 + 5 2 + 2.6 2 + ...

is

2n – 1 1 n -1 (b) 2 2 144. The sum of the series

(a)

n(n + 1)2 when n is even. When n is odd the sum is 2 [2004]

1 1 1 + .......... .. up to ¥ is equal to 1.2 2.3 3.4 æ 4ö èeø

(a) log e ç ÷

(b) 2 log e 2

(d) log e 2 (c) log e 2 - 1 3 3 3 3 3 145. 1 – 2 + 3 – 4 +...+9 = (a) 425 (b) –425 (c) 475 146. The value of 21/4. 41/8. 81/16 ... ¥ is (a) 1 (b) 2 (c) 3/2

[2002] (d) –475 [2002] (d) 4

EBD_8344

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1.

Mathematics

(c) Let common difference of series

4.

\ 300 = 1 + (n - 1)d

a1 , a2 , a3 ,......., an be d. Q a40 = a1 + 39d = -159

...(i)

and a100 = a1 + 99d = -399

...(ii)

Þd =

From equations (i) and (ii), d = – 4 and a1 = – 3 Since, the common difference of b1, b2, ......, bn is 2 more than common difference of a1, a2, ......, an. \ Common difference of b1, b2, b3, ..... is (– 2).

\ n - 1 = 13 or 23

Þ n = 24 and d = 13

a20 = 1 + 19 ´ 13 = 248 s20 =

Þ b1 = 198 - 279 Þ b1 = -81 5.

32sin 2a 81 + 2sin 2 a = 28 3 3

(a) Given a = 3 and S25 = S40 – S25

Þ 2´

Þ 5(2 + 8d ) = 4(2 + 13d )

x 81 + = 28 3 x

Þ 10 + 40 d = 8 + 52d

Þ x - 84 x + 243 = 0 Þ x = 81, x = 3 2

Þd =

When x = 81 Þ sin 2a = 2 (Not possible) When x = 3 Þ a =

p 12

6.

\ a = 30 = 1, d = 14 - 1 = 13 a6 = a + 5d = 1 + 65 = 66. 3.

Q S = 460

Þ log 7 ( x × x × x × ......x ) = 460 3

4

21

Þ log7 x(2+3+ 4......21) = 460 Þ (2 + 3 + 4 + ..... + 21) log 7 x = 460 20 (2 + 21) log 7 x = 460 Þ 2 460 Þ log 7 x = = 2 Þ x = 7 2 = 49 230

1 6

(b) Sn = 20 + 19 3 + 19 1 + 18 4 + .... 5 5 5 Q Sn = 488 488 =

n é æ 100 ö æ 2ö ù 2ç ÷ + ( n - 1) çè - ÷ø ú 2 êë è 5 ø 5 û

488 =

n (101 - n) Þ n2 - 101n + 2440 = 0 2

(a) S = log 7 x 2 + log 7 x 3 + log 7 x 4 + ...20 terms 2

25 40 [6 + 24 d ] = [6 + 39 d ] 2 2

Þ 25[6 + 24d ] = 20[6 + 39d ]

Let 32sin 2 a = x Þ

20 (2 + 19 ´ 13) = 2490. 2

Þ 2S25 = S40

So, 32sin 2a-1 + 34 - 2sin 2a = 28 Þ

(Q15 £ n £ 50)

Þ n = 14 or 24

Þ b1 + 99(-2) = (-3) + 69( -4)

(a) Given that 32sin 2a -1 , 14, 34 - 2sin 2a are in A.P..

299 23 ´ 13 , = (n - 1) (n - 1)

Q d is an integer

Q b100 = a70

2.

(d) Given that a1 = 1 and an = 300 and d ÎZ

Þ n = 61 or 40

For n = 40 Þ Tn > 0 For n = 61 Þ Tn < 0 nth term = T61 =

100 æ 2ö + (61 - 1) ç - ÷ = -4 è 5ø 5

www.jeebooks.in M-101

Sequences and Series

7.

(d) Let common difference be d. Q S11 = 0 \

11.

11 {2a1 + 10 × d } = 0 2

Þ a1 + 5d = 0 Þ d = -

a1 5

...(i)

Now, S = a1 + a3 + a5 + ... + a23 = a1 + (a1 + 2d ) + (a1 + 4d ) + .... + (a1 + 22d ) = 12 a1 + 2d

11 ´ 12 2

é æ a öù = 12 êa1 + 11 × ç - 1 ÷ ú è 5 øû ë

(From (i))

72 æ 6ö = 12 ´ ç - ÷ a1 = - a1 è 5ø 5 8.

(14)First common term of both the series is 23 and common difference is 7 × 4 = 28 Q Last term £ 407 Þ 23 + (n – 1) × 28 £ 407 Þ (n – 1) × 28 £ 384 384 +1 28 Þ n £ 14.71 Hence, n = 14

12.

Þ n£

9.

(d) T10 =

T20 =

(d) Let 5 terms of A.P. be a – 2d, a – d, a, a + d, a + 2d. Sum = 25 Þ 5a = 25 Þ a = 5 Product = 2520 (5 – 2d) (5 – d) 5(5 + d) (5 + 2d) = 2520 Þ (25 – 4d 2) (25 – d 2) = 504 Þ 625 – 100d 2 – 25d 2 + 4d 4 = 504 Þ 4d 4 – 125d 2 + 625 – 504 = 0 Þ 4d 4 – 125d 2 + 121 = 0 Þ 4d 4 – 121d 2 – 4d 2 + 121 = 0 Þ (d 2 – 1) (4d 2 – 121) = 0 11 Þ d = ± 1, d = ± 2 -1 11 as a term d = ± 1 and d = – , does not give 2 2 11 \ d= 2 \ Largest term = 5 + 2d = 5 + 11 = 16 (c) Given, S4 = 16 and S6 = – 48 Þ 2(2a + 3d) = 16 Þ 2a + 3d = 8 ...(i) And 3[2a + 5d] = – 48 Þ 2a + 5d = – 16 Þ 2d = – 24 [using equation (i)] Þ d = – 12 and a = 22

10 = (44 + 9(– 12)) = – 320 2 (a) Let the common difference of the A.P. is ‘d’. Given, a1 + a7 + a16 = 40 Þ a1 + a1 + 6d + a1 + 15d = 40 Þ 3a1 + 21d = 40

\ 13.

1 = a + 9d 20

...(i)

1 = a + 19d 10

40 ...(i) 3 Now, sum of first 15 terms of this A.P. is,

Þ a1 + 7d =

...(ii)

Solving equations (i) and (ii), we get S15 =

1 1 a= ,d = 200 200

Þ S200 10.

(b) If 2

200 é 2 199 ù 201 1 = + = = 100 2 êë 200 200 úû 2 2

1– x

+2

1+ x

x

14.

15.

Using AM ³ GM f(x) ³ 3

(b) a1 + a4 + a7 + .... + a16 = 114 Þ a1 + a16 = 38 Now, a1 + a6 + a11 + a16 = 2(a1 + a16 ) = 2 × 38 = 76

ö ÷ ÷ ø

ö æ x 1 ö ÷ +ç3 + x ÷ 3 ø ø è

[Using (i)]

Þ 3(a1 + a 16 ) = 114

, f(x), 3 + 3 are in A.P., then

1 æ 2 f ( x) = 2 ç 2 x + x 2 è

15 [2a1 + 14d] = 15 (a1 + 7d) 2

æ 40 ö = 15 ç ÷ = 200 è 3 ø

–x

æ 21+ x + 21- x + 3x + 3- x f ( x) = ç ç 2 è

S10

A 7Aö æ n + n2 ´ (d) Q Sn = ç 50 Þ a1 = 50 – 3S 2 ÷ø 2 è \ d = a2 – a1 = Sn 2 - Sn1 - Sn1 Þ d = Now, a50 = a1 + 49 × d = (50 – 3A) + 49 A = 50 + 46 A So, (d, a50) = (A, 50 + 46 A)

A ´2 = A 2

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16.

Mathematics

(d) Q f(x + y) = f(x) . f(y) Þ Let f(x) = tx Q f(1) = 2 \ t' = 2 Þ f(x) = 2x

tn = a + (n – 1)d \ t19 = a + 18d = 0 \ a = –18d t49 a + 48d t29 = a + 28d

\

10

Since,

å f (a + k ) = 16(210 - 1)

k =1

=

10

a +k = 16(210 - 1) Then, å 2 k =1

21.

10

Þ 2a å 2k = 16(210 - 1) k =1

a Þ2 ´

17.

18.

19.

((210 ) - 1) ´ 2 = 16 ´ (210 - 1)2.2a = 16 (2 - 1)

Þa=3 (d) Let three terms of A.P. are a – d, a, a + d Sum of terms is, a – d + a + a + d = 33 Þ a = 11 Product of terms is, (a – d) a (a + d) = 11(121 – d2) = 1155 Þ 121 – d2 = 105 Þ d = ± 4 if d = 4 T11 = T1 + 10d = 7 + 10(4) = 47 if d = – 4 T11 = T1 + 10d = 15 + 10 (– 4) = – 25 (d) Q 91 = 13 × 7 Then, the required numbers are either divisible by 7 or 13. \ Sum of such numbers = Sum of no. divisible by 7 + sum of the no. divisible by 13 – Sum of the numbers divisible by 91 = (105 + 112 + ... + 196) + (104 + 117 + ... +195) – 182 = 2107 + 1196 – 182 = 3121 (b) Since nC4, nC5 and nC6 are in A.P. 2nC5 = nC4 + nC6 n

2=

n

C4 C5

+

n

C6

n

C5

5 n-5 + 5 n-4 Þ 12(n – 4) = 30 + n2 – 9n + 20 Þ n2 – 21n + 98 = 0 (n – 7) (n – 14) = 0 (n – 7) (n – 14) = 0 n = 7, n = 14

20.

–18d + 48d 30d = =3 –18d + 28d 10d

t49 : t29 = 3 : 1 (d) Two digit positive numbers which when divided by 7 yield 2 as remainder are 12 terms i.e,16, 23, 30, ..., 93 Two digit positive numbers which when divided by 7 yield 5 as remainder are 13 terms i.e,12, 19, 26, ..., 96 By using AP sum of 16, 23, ..., 93, we get S1 = 16 + 23 + 30 + ... + 93 = 654 By using AP sum of 12, 19, 26, ..., 96, we get S1 = 12 + 19 + 26 + ... + 96 = 702 \ required Sum = S1 + S2 = 654 + 702 = 1356 30

22. (a) S = å ai = i =1

(c) Let first term and common difference of AP be a and d respectively, then

30 [2a1 + 29d] 2

15 15 T = å a(2i -1) = 2 [2a1 + 28d ] i =1

Since, S – 2T = 75 Þ 30 a1 + 435d – 30a1 – 420d = 75 Þd=5 Also, a5 = 27 Þ a1 + 4d = 27 Þ a1 = 7, Hence, a10 = a1 + 9d = 7 + 9 ´ 5 = 52 23.

(c) Q

1 1 1 1 , , ,....., are in A.P x1 x2 x3 xn

x1 = 4 and x21 = 20 Let 'd ' be the common difference of this A.P \ its 21st term = Þd=

2=

¼(i)

1 1 = + [(21 – 1) ´ d ] x21 x1

1 æ 1 1ö ´ç – ÷ 20 è 20 4 ø

Þ d=–

Also xn > 50 (given). \

1 1 = + [(n – 1) ´ d ] xn x1

Þ xn = \

x1 1 + (n – 1) ´ d ´ x1

x1 > 50 1 + ( n – 1) ´ d ´ x1

1 100

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Sequences and Series

Þ

26.

4 > 50 æ 1 ö 1 + (n – 1) ´ ç – ÷´4 è 100 ø

4 æ 1 ö Þ 1 + ( n – 1) ´ ç – ÷´4< 50 è 100 ø

1 [(15 a – 3b)2 + (3b – 5c)2 + (5c – 15a)2 ] = 0 2 It is possible when 15a – 3b = 0, 3b – 5 c = 0 and 5c – 15a = 0 Þ 15a = 3b Þ b = 5a

1 23 (n – 1) < – 100 100 Þ n – 1 > 23 Þ n > 24 Therefore, n = 25.

Þ –

Þ

25

1

åx

i =1 i

24.

=

Þ b=

25 éæ 1ö æ 1 ö ù 13 ç 2 ´ ÷ + (25 – 1) ´ ç – ÷ú = ê 2 ëè 4ø è 100 ø û 4

Þ x5 = x3 + 2d1 = 8 + 2 ×

12 = 2.4 5

27.

12 = 12.8 5

Suppose d 2 is the common difference of the A.P

1 1 1 , ,..... h1 h2 hn then 5d2 =

28.

1 1 –3 –3 – = Þ d2 = 20 8 40 200

29.

Þ x5 . h10 = 12.8 × 200 = 2560 12

Þ a1 + 24d = 32 Now, a9 + a43 = 66 Þ 2a1 + 50d = 66 From eq. (i) & (ii) we get; d = 1 and a1 = 8 Also,

17

17

r =1

r =1

å a 2r = å [8 + (r - 1)1]2 = 140 m

17

Þ

å (r + 7)

Þ

å (r 2 + 14r + 49) = 140 m

2

= 140 m

r =1 17

r =1

æ 17 ´18 ´ 35 ö æ 17 ´18 ö ÷ + 14 ç ÷ + (49 ´17) = 140 Þ ç è ø è 2 ø 6 Þ m = 34

17 [a + a ] = 17 × 18 = 306 2 1 17

(b) Let p, q, r are in AP Þ 2q = p + r

...(i) ...(ii)

Þ

...(i)

1 1 + =4 a b

Given

13 (b) Q å a 4k +1 = 416 Þ [2a1 + 48d] = 416 2 k =0

c 5c 6c + = 3 3 3

Þ a + b = 2c Þ b, c, a are in A.P. (a) By Arithmetic Mean: a + c = 2b Consider a = b = c = 2 Þ abc = 8 Þ a + b = 2b \ minimum possible value of b = 2 (a) a3 + a7 + a11 + a15 = 72 (a3 + a15) + (a7 + a11) = 72 a3 + a15 + a7 + a11 = 2 (a1+ a17) a1 + a17 = 36 S17 =

1 1 1 Q = + 3d 2 = Þ h10 = 200 h10 h7 200

25.

5c c ,a = 3 3

Þ a+b=

(a) Suppose d1 is the common difference of the A.P. x1, x2, .... xn then Q x8 – x3 = 5d1 = 12 Þ d1 =

(c) We have 9(25a2 + b2) + 25 (c2 – 3ac) = 15b (3 a + c) Þ 225a2 + 9b2 + 25c2 – 75ac = 45ab + 15bc Þ (15a)2 + (3b)2 + (5c)2 – 75ac – 45ab– 15 bc = 0

a +b =4 ab

We have a + b = – q/p and ab = q p = 4 Þ q = - 4r r p

r p

-

Þ

....(ii)

From (i), we have 2( – 4r) = p + r p = –9r

....(iii)

Now, | a - b | = (a + b)2 - 4ab 2

æ -q ö 4r = ç ÷ = è pø p

q 2 - 4 pr | p|

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Mathematics

From (ii) and (iii) 16r 2 + 36r 2 2 13 = | -9 r | 9 (b) Given n = 20; S20 = ? Series (1) ® 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59... Series (2) ® 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71. The common terms between both the series are 11, 31, 51, 71... Above series forms an Arithmetic progression (A.P). Therefore, first term (a) = 11 and common difference (d) = 20 n Now, Sn = [2a + (n - 1)d ] 2 20 S20 = [2 × 11 + (20 – 1) 20] 2 S20 = 10 [22 + 19 × 20] S20 = 10 × 402 = 4020 \ S20 = 4020 (c) Let a be the first term and d be the common difference of given A.P. Second term, a+ d = 12 ...(i) Sum of first nine terms,

33.

(b) Given Sn = 2n + 3n2 Now, first term = 2 + 3 = 5 second term = 2(2) + 3(4) = 16 third term = 2(3) + 3 (9) = 33 Now, sum given in option (b) only has the same first term and difference between 2nd and 1st term is double also.

34.

(b)

a1 + a2 + a3 + ...... + a p p3 = a1 + a2 + a3 + ...... + aq q 3

Þ

a1 + a2 8 = Þ a1 + (a1 + d) = 8a1 1 a1

=

30.

31.

9 (2a + 8d ) = 9( a + 4d ) 2 Given that S9 is more than 200 and less than 220 Þ 200 < S9 < 220 Þ 200 < 9 (a + 4d) < 220 Þ 200 < 9 (a + d + 3d) < 220 Putting value of (a + d) from equation (i) 200 < 9 (12 + 3d) < 220 Þ 200 < 108 + 27d < 220 Þ 200 – 108 < 108 + 27d – 108 < 220 – 108 Þ 92 < 27d < 112 Possible value of d is 4 27 × 4 = 108 Thus, 92 < 108 < 112 Putting value of d in equation (i) a + d = 12 a = 12 – 4 = 8 4th term = a + 3d = 8 + 3 × 4 = 20 (c) If d be the common difference, then

Þ d = 6a1

=

m = a4 – a7 + a10 = a4 – a7 + a7 + 3d = a7 S13 =

13 13 [a1 + a13 ] = [ a1 + a7 + 6d ] 2 2

13 [2a7 ] = 13a7 = 13 m = 2

a1 + 5 ´ 6a1 1 + 30 31 = = a1 + 20 ´ 6a1 1 + 120 121

35.

(d) Let 'a' is the first term and 'd ' is the common difference of an A.P. Now, According to the question 100a100 = 50a50 100 (a + 99d) = 50 (a + 49d) Þ 2a + 198 d = a + 49d Þ a + 149 d = 0 Hence, T150 = a + 149 d = 0

36.

(d) Given :

37.

(a) Since, sec (q – f), secq and sec (q + f) are in A.P.,

S9 =

32.

a + 5d a6 = 1 a21 a1 + 20d

Now

a p + aq

ar + as 2 2 Þ a + (p – 1) d +a + (q – 1)d = a + (r – 1)d + a + (s – 1)d Þ 2a + (p + q)d – 2d = 2a + (r + s) d – 2d Þ (p + q)d = (r + s)d Þ p + q = r + s. =

\ 2 secq = sec (q – f) + sec (q + f) Þ

cos ( q + f) + cos ( q - f ) 2 = cos q cos ( q - f ) cos ( q + f )

(

)

Þ 2 cos 2 q - sin 2 f = cos q [ 2 cos q cos f] Þ cos 2 q (1 - cos f ) = sin 2 f = 1 - cos 2 f 2 2 Þ cos q = 1 + cos f = 2 cos

f 2 f k But given cosq = cos 2 \ k= 2

f 2

\ cos q = 2 cos

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Sequences and Series

38.

(b) Let A.P. be a, a + d , a + 2d ,......... a2 + a4 + ........... + a200 = a 100 é2 ( a + d ) + (100 - 1) 2d ùû = a ...(i) Þ 2 ë and a1 + a3 + a5 + ......... + a199 =b 100 Þ [ 2a + (100 – 1) 2d] = β 2 Subtracting (ii) from (i), we get

39.

40.

43.

(m - n)d =

...(ii)

44.

Þ 31– x + 2 = 3 (4.3x – 1) Þ 3.3–x + 2 = 12.3x – 3. Put 3x = t 3 + 2 = 12t - 3 Þ 12t2 – 5t – 3 = 0; Þ t 1 3 Hence t = - , 3 4 3 Þ 3x = (as 3 x ¹ - ve ) 4 æ 3ö è 4ø

Þ x = log3 ç ÷ or x = log3 3 – log3 4

45.

Cr ,

m

¥

å f ( x) = 2 Þ k + k 2 + k 3 + .....¥ = 2

m

Cr

m

Cr +1

m

Cr 2

4 f (4) k 4 = = k2 = . 9 f (2) k 2 (c) Rearrange given equation, we get

Now,

46.

(a 2 p 2 - 2abp + b2 ) + (b2 p 2 - 2bcp + c 2 ) +(c2 p 2 - 2cdp + d 2 ) = 0 Þ (ap - b)2 + (bp - c)2 + (cp - d )2 = 0

=

r m-r + m - r +1 r +1

Þ m - m(4r + 1) + 4r - 2 = 0 . 2

2 k =2Þk = 1- k 3

Cr +1 are in A.P..

2m Cr =m Cr -1 + m Cr +1

+

Þ x = 1 – log3 4 (d) Let f (1) = k, then f (2) = f (1 + 1) = k2 f (3) = f (2 + 1) = k3

Þ

(c) Coeficient of rth, (r + 1)th and (r + 2)th terms is mCr–1, mC and mC r r+1 resp.

Cr -1

p log b a q

x =1

a 11 a1 + 5d 11 Þ 6 = = a21 41 a1 + 20d 41

m

ap =

Þ log3 (31– x + 2) = log3 [3(4 × 3x – 1)]

Put p = 11 and q = 41

Þ 2=

bq

Þ log3 (31 – x + 2) = log33 + log3 (4.3x – 1)

2a1 + ( p - 1)d p = 2a1 + (q - 1)d q

m

1 Þa-d =0 mn (b) 1, log9 (31 – x + 2), log3 (4.3x – 1) are in A.P. Q a, b, c are in A.P then b = a + c

Q log

But n = 125 is not possible \ Total time = 24 + 10 = 34 minutes. (d) Given that

Given that m Cr -1 ,

1 1 1 - Þd = n m mn

Þ 2 log9 (31– x+2) = 1 + log3 (4.3x – 1)

Þ n - 3 [ 2 ´ 240 + (n - 4) ´ 40] = 11040 - 600 2 Þ (n - 3)[240 + 20n - 80] = 10440 Þ (n - 3)(20n + 160) = 10440 Þ (n - 3)(n + 8) = 522 Þ n2 + 5n - 546 = 0 Þ (n + 26) (n – 21) = 0 \ n = 21 (a) Till 10th minute number of counted notes = 1500 Remaining notes = 4500 – 1500 = 3000 n 3000 = [ 2 ´ 148 + ( n - 1)( -2) ] = n [148 - n + 1] 2

p S p 2 [2a1 + ( p - 1)d ] p 2 = = 2 q Sq [2a1 + (q - 1)d ] q 2

.....(ii)

From (i) a =

Þ n = 125, 24

42.

.....(i)

1 m Subtracting (ii) from (i), we get

a -b d= 100 (c) Let number of months = n \ 200 × 3 + (240 + 280 + 320 + ... + (n – 3)th term) = 11040

Þ

1 n

Tn = a + (n - 1) d =

n 2 - 149 n + 3000 = 0

41.

(d) Tm = a+ (m – 1) d =

\ ap - b = bp - c = cp - d = 0 Þ

b c d = = a b c

\ a, b, c, d are in G.P.

EBD_8344

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47.

Mathematics

(5.00)

51.

Q f ( x + y ) = f ( x) × f ( y )

(4)

"x Î R and f(1) = 3

(0.16)

Þ f ( x) = 3x Þ f (i ) = 3i

æ1 1 1 ö log 2.5 ç + 2 + 3 + .....¥÷ è3 3 3 ø

æ 1 ö ç ÷ log 2.5 ç 3 ÷ 1 ç 1- ÷ è 3ø

n

Þ å f (i) = 363 i =1

= 0.16

Þ 3 + 3 + 3 + .... + 3 = 363 2

Þ

3

n

= 0.16

é a (r - 1) ù êQ S n = ú (r - 1) û ë n

3(3n - 1) = 363 3 -1

52.

Þ 3 = 243 = 3 Þ n = 5 (b) Given sequence are in G.P. and common ratio æ æ 3ö 210 ç ç ÷ è è 2ø

ö - 1÷ ø

11

(c) Let terms of G.P. be

æ3 ö çè - 1÷ø 2

-2

= 4.

a , a, ar r

. . . (i)

and a3 = 27 Þa=3 Put a = 3 in eqn. (1), we get

= S - 211

If f ( x) = x +

1 , then f ( x) Î ( -¥, - 2] È [2, ¥) x

Þ 311 - 211 = S - 211 Þ S = 311 (b) Let the first term be 'a' and common ratio be 'r'.

Þ 3 f ( x) Î ( -¥, - 6] È [6, ¥)

Q ar (1 + r + r ) = 3

...(i)

Then, it concludes that

and ar 5 (1 + r + r 2 ) = 243

...(ii)

\ S50 =

Þ 3 + 3 f ( x) Î ( -¥, - 3] È [9, ¥) S Î (-¥, - 3] È [9, ¥) 53.

From (i) and (ii), r 4 = 81 Þ r = 3 and a =

1 13

a(r 50 - 1) 350 - 1 = r -1 26

. . . (ii)

1ö æ S = 3 + 3ç r + ÷ è rø

2

50.

æ 1ö =ç ÷ è 2ø

æ1 ö \ a ç + 1 + r÷ = S èr ø

3 2

æ 311 - 211 ö ç ÷ è 211 ø Þ 210 = S - 211 1 2

49.

æ 1ö -2log 2.5 ç ÷ è 2ø

5

n

\

æ 1ö log 2.5 ç ÷ è 2ø

= (2.5)

363 ´ 2 Þ 3n - 1 = = 242 3

48.

a ù é êëQ S¥ = 1 - r úû

é a(r n - 1) ù êQ Sn = ú (r - 1) û ë

2 2 3 2 2 3 (c) S = ( x + y) + ( x + y + xy) + ( x + x y + xy + y ) + ....¥

=

1 é( x 2 - y 2 ) + ( x 3 - y 3 ) + ( x 4 - y 4 + ....¥ù û x- y ë

=

1 é x2 y 2 ù ( x - y )( x + y - xy ) ê ú= x - y ë1 - x 1 - y û ( x - y )(1 - x )(1 - y )

(b) Let a, b, g , d be in G.P., then ad = bg Þ

Þ

a ù é êëQ S¥ = 1 - r úû

a g a -b g -d = Þ = b d a+b g +d

9- 4p 36 - 4q = 3 6

= 54.

x + y - xy (1 - x)(1 - y)

(c) S = ( x + x2 + x 3 + ....9 terms)

Þ 36 - 16 p = 36 - 4q Þ q = 4 p \

2q + p 8 p + p 9 p 9 = = = 2q - p 8 p - p 7 p 7

+ a[k + (k + 2) + + (k + 4) + ...9 terms]

ÞS=

x ( x 9 - 1) 9 + [2ak + 8 ´ (2a)] x -1 2

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Sequences and Series

x10 - x 9a(k + 8) x10 - x + 45a( x - 1) (Given) + = x -1 x -1 1

ÞS = Þ

æ 4 ö ç ÷ è 1+ r ø

(a)

1

24

+

2 3 + +.....¥ 16 48

a

a1 (r q - 1) (-4)((-2)9 - 1) = r -1 -2 - 1

å ai =

100

4 = (-513) = 4l Þ l = – 171 3 (b) The given series is in G.P. then

Þ

i =1

60. ar 2 ( r 200 - 1) 2

r -1

= 200

...(i)

Sn =

100

å a2n = a2 + a4 + ..... + a200 = 100 ar (r 200 - 1)

= 100 r2 -1 From equations (i) and (ii), r = 2 and a2 + a3 + ..... + a200 + a201 = 300 Þ r(a1 + ..... + a200) = 300

Þ

Þ

200

å an =

n =1

...(ii)

(1 + x)10 é(1 + x )11 - x11 ù ë û Þ = (1 + x)11 – x11 1 11 (1 + x) ´ (1 + x )

300 = 150 r

(b) y = 1 + cos q + cos q + ..... 2

61.

4

1 1 Þ = sin 2 q y 1 - cos 2 q x = 1 – tan 2q + tan4q + .....

Þ y=

x=

1 2

1 - (- tan q) y=

1

sin 2 q \ y(1 – x) = 1

=

1 2

sec q

Þ y=

1 1- x

n é ù (49)126 - 1 ((49)63 + 1)(4963 - 1) êQS = a(r -1) ú n = (b) r -1 úû êë 48 48

\ K = 63

From eqn. (i), a1 =

62. ...(i) ...(ii)

4 and substituting the value of a1, 1+ r

\ Coefficient of x7 is 11C7 = 11C11–7 = 11C4 = 330 (d) Q a, b, g are three consecutive terms of a nonconstant G.P. \ b2 = ag So roots of the equation ax2 + 2bx + g = 0 are

-2b ± 2 b2 - ag b = 2a a 2 Q ax + 2bx + g = 0 and x2 + x – 1 = 0 have a common root. \ this root satisfy the equation x2 + x – 1 = 0 b2 – ab – a2 = 0 Þ ag – ab – a2 = 0 Þ a + b = g Now, a(b + g) = ab + ag

Þ x = cos2q

(b) Since, a1 + a2 = 4 Þ a1 + a1r = 4 Þ a1r2 + a1r3 = 16 a3 + a4 = 16

a(1 - r n ) 1- r

é æ x ö11 ù (1 + x)10 ê1 - ç ÷ ú êë è 1 + x ø úû x ö æ ç1 ÷ è 1+ x ø

n =1

59.

4 3

= = 2 (d) Let G.P. be a, ar, ar2 .....

n =1

58.

= 16

r = – 2, a1(1 – 2) = 4 Þ a1 = – 4

å a2n+1 = a3 + a5 + ..... + a201 = 200

57.

r3

r = 2, a1(1 + 2) = 4 Þ a1 =

1 1 1 + + +.....¥ 2 4 8 16

56.

æ 4 ö +ç ÷ è 1+ r ø

Þ 4r2 (1 + r) = 16(1 + r) Þ r2 = 4 \ r=±2

x10 - x + 9a(k + 8)( x - 1) x10 - x + 45a( x - 1) = x -1 x -1

Þ 9a(k + 8) = 45a Þ k + 8 = 5 Þ k = -3. 55.

r2

= ab + b2 = (a + b)b = bg (d) Q a, b, c are in G.P. Þ b = ar, c = ar2 Q 3a, 7b, 15c are in A.P. Þ 3a, 7ar, 15ar2 are in A.P. \ 14ar = 3a + 15ar2 1 3 Þ 15r2 – 14r + 3 = 0 Þ r = or 3 5

in eqn (ii),

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Mathematics

\ x = cos q, cosec q, qÎ (0, 45°)

1 3 \ r = reected 2 5 Fourth term = 15ar2 + 7ar – 3a Qr
An

Þ An =

3 7

Þ 1 – An > An

Þ 1 > 2 × An Þ An
0 è xø è xø îï

(3, 2)

C (x2, y2)

B (x1, y1)

ì æ 1ö æ 1ö 3 2 ï (20 x - x )sin çè x ÷ø - 8 x cos çè x ÷ø + 10, ïï 0, f ''( x ) = í ï 1 1 ï(20 x 3 - x) cos çæ ÷ö + 8 x 2 sin çæ ÷ö + 2l, è ø è x xø îï

1 + x1 1 + y1 =-1, =2 2 2 Þ B (–3, 3) 1 + x2 1 + y2 = 3, =2 2 2 Þ C (–5, 3) \ Centroid is 1 - 3 + 5 , 1 + 3 + 3 3 3 æ 7ö Þ ç1, ÷ è 3ø

17.

18.

2

2

2

(3 x - 1) + (3 y ) = a + b

19.

(a) AB = BC =

2

21.

CA = ( 4 - 3) 2 + ( 0 - 5) 2 = 26 ; Q AB = CA \ Isosceles triangle Q ( 26 )2 + ( 26 )2 = 52

x=0 x>0

(a) Coordinates of centroides

æ x + x + x y + y + y3 ö C =ç 1 2 3, 1 2 ÷ 3 3 è ø æ 3 + 1 + 2 -1 + 3 + 4 ö =ç , ÷ = (2, 2) 3 è 3 ø The given equation of lines are x + 3y – 1 = 0 3x – y + 1 = 0 Then, from (i) and (ii)

22.

...(i) ...(ii)

æ 1 2ö point of intersection P ç - , ÷ è 5 5ø equation of line DP 8x – 11y + 6 = 0 (b) Y

4

?

2 3

( 4 + 1) 2 + (0 + 1) 2 = 26 ;

(3 + 1) 2 + (5 + 1) 2 = 52

x C1C 2 , r + 3 > 5 Þ r > 2 From (1) and (2), 2 < r < 8.

a 2 + b2 - 2ax - 2by + 4 = 0 2

Þ x2 + y 2 - x - y = 0 (b) Q Given two circles intersect at two points

\ r1 - r2 < C1C2

\ 2( g ´ 0 + f ´ 0) = c - 4 Þ c = 4

83.

(d) Solving y = x and the circle

we should have

X

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90.

Mathematics

OA £ OP £ OB

[From (ii)]

Þ (5 - 3) £ x 2 + y 2 £ 5 + 3

Þ 2m12 + ym1 + ( x + 2) = 0

Þ 4 £ x 2 + y 2 £ 64 (b) Let the required circle be x2 + y2 + 2gx + 2fy + c = 0 Since it passes through (0, 0) and (1, 0) On putting these values, we get

From (i) and (iii),

1 x +1 - y = = Þ x+3=0 y x+2 2 93.

1 2 Points (0, 0) and (1, 0) lie inside the circle x2 + y2 = 9, so two circles touch internally Þ c1c2 = r1 – r2 3 \ g 2 + f 2 = 3 - g 2 + f 2 Þ g2 + f 2 = 2 Squaring both side, we get

Þ c = 0 and g = -

5 2 = - ( x - 1) Þ 4 x + 6 y = 19 ...(i) 2 3 Equation of normal to the parabola at (2, 4) is,

1 y - 4 = - ( x - 2) Þ x + 4 y = 18 4

\ From (i) and (ii), x = -

ö ö æ1 æ1 ç , 2 ÷ or ç ,- 2 ÷ ø ø è2 è2 (c) Let ABC be an equilateral triangle, whose median is AD.

In equilateral triangle median is also altitude So, AD ^ BC A Given AD = 3a. Let AB = BC = AC = x. O In DABD, AB2 = AD2 + BD2 ; Þ x2 = 9a2 + (x2/4) B

D

94.

1 ...(i) m Q Equation of tangent to x2 = 4y with slope m be :

1 = - m 2 Þ m = -1 m \ Equation tangent : x + y + 1 = 0 It is tangent to circle x2 + y2 = c2

C

Þc= 95.

2 m2

[Tangent to y2 = 8(x + 2)]

m12 ( x + 1) - ym1 + 1 = 0

...(i)

m22 ( x + 2) - ym2 + 2 = 0

...(ii)

1 m1

(Q L1 ^ L2 )

(c)

1 2 P

R

M

Q

1 [Tangent to y2 = 4(x + 1)] m1

Q m2 = -

...(ii)

From eq. (i) and (ii),

L1 : y = m1 ( x + 1) +

L2 : y = m2 ( x + 2) +

(b) Equation tangent to parabola y2 = 4x with slope m be: y = mx +

Þ r2 = (3a – r)2 +

(a)

16 53 , y= 5 10

y = mx - am 2

x2 4 Þ r2 = 9a2 – 6ar + r2 + 3a2 Þ 6ar = 12a2 Þ r = 2a So equation of circle is x2 + y2 = 4a2

...(ii)

æ 16 53ö \ Centre of the circle is ç - , ÷ è 5 10 ø

3 2 x = 9a2 Þ x2 = 12a2. 4 In D OBD, OB2 = OD2 + BD2

92.

(d) Circle passes through A(0, 1) and B(2, 4). So its centre is the point of intersection of perpendicular bisector of AB and normal to the parabola at (2, 4). Perpendicular bisector of AB; y-

9 1 Þ f2 = - =2 \f =± 2. 4 4 Hence, the centres of required circle are

91.

...(iii)

N

Q y 2 = 12 x \a = 3

Let P(at 2 , 2at )

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Conic Sections

Þ N (at 2 , 0) Þ M (at 2 , at ) Q Equation of QM is y = at So, y 2 = 4ax Þ x =

tan 30° =

at 2 4

Area =

æ at 2 ö Þ Qç , at ÷ è 4 ø

98.

Þ Equation of QN is y =

-4 ( x - at 2 ) 3t

96.

3 2 1 at = and PN = 2at = 2 4 4 2

(b)

y = 4x

L

1 3

=

4t 2t 2

Þt = 2 3

1 × 8(2 3) × 2 × 24 = 192 3. 2

-1 2

Parameter of other end of focal chord is 2 So, coordinates of B is (8, 8) Þ Equation of tangent at B is 8y – 4(x + 8) = 0 Þ 2y – x = 8 Þ x – 2y + 8 = 0

4 4 1 = - ( - at 2 ) Þ at = 1 Þ t = 3 3t 3

Now, MQ =

Þ

æ1 ö (b) Let parabola y2 = 8x at point ç , -2 ÷ is (2t 2, 4t) è2 ø Þ t=

æ 4ö Q QN passes through ç 0, ÷ , then è 3ø

4t 2t 2

99.

(a) Let point P be (2t, t2) and Q be (h, k) Using section formula,

2 5

h=

C1 S(1,0)

-2 + t 2 2t ,k = 3 3

C2 L'

æ 3h ö Hence, locus is 3k + 2 = ç ÷ è 2 ø

2

Þ 9x2 = 12y + 8 100. (0.5) Let the coordinates of P = P(t2, t) Distance between the centres

= C1C2 = 2C1S = 2 20 - 4 = 8. 97.

2 2 (c) Let A = (2t , 4t ) and B = (2t , - 4t )

2

A (2 t , 4t)

O (0, 0)

30 30

M

2

B (2 t , –4t) For equilateral triangle (ÐAOM = 30°)

Tangent at P(t2, t) is ty =

x + t2 2

Þ 2ty = x + t2 Q(–t2, 0), O(0, 0)

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Mathematics

0 1 2 t \ Area of DOPQ = 2 -t 2

0 1

0 1

Þ |t|3 = 8 t = ± 2 (t > 0) \ 4y = x + 4 is a tangent \ P is (4, 2)

101. (c) y = mx + 4 Tangent of y2 = 4x is Þ y = mx +

1 m

B

ᔴ (h k) Þ h=

x–y㦨3

5 ,k=–1 2

103. (d) Equation of tangent on y 2 = 4 2 x is yt = x + 2t 2

1 \m= 2

Now, y = mx

A

t 1 =4

This is also tangent on circle ...(i)

...(ii)

2t 2 1+ t2

= 1 Þ 2t4 = 1 + t2 Þ t2 = 1

Hence, equation is ± y = x + 2 Þ | c | =

2

104. (d) The circle and parabola will have common tangent at P (1, 2). P( 1

2)

a [Q Equation of tangent of y2 = 4 ax is y = mx + ] m

\

From (i) and (ii) 4=

1 1 Þ m= m 4

So, line y =

1 x + 4 is also tangent to parabola 4

x2 = 2by, so solve both equations.

æ x + 16 ö x 2 = 2b ç ÷ è 4 ø Þ 2x2 – bx – 16b = 0 Þ D=0 [For tangent] Þ b2 – 4 × 2 × (–16b) = 0 Þ b2 + 32 × 4b = 0 b = – 128, b = 0 (not possible) 102. (c) Tangent to the curve y = (x – 2)2 – 1 at any point (h, k) is, Þ

1 ( y + k ) = ( x - 2)(h - 2) - 1 2

y+k = xh - 2 x - 2h + 3 Þ 2 Þ (2h – 4) x – y – 4h + 6 – k = 0 Given line, x – y – 3 = 0

Þ

2h - 4 4h - 6 + k = =1 1 3

So, equation of tangent to parabola is, 4 ( x + 1) Þ 2 y = 2x + 2 Þ y = x + 1 2 Let equation of circle (by family of circles) is (x – x1)2 + (y – y1)2 + lT = 0 Þ c º (x – 1)2 + (y – 2)2 + l(x – y + 1) = 0 Q circles touches x-axis. \ y-coordinate of centre = radius Þ c = x2 + y2 + (l – 2)x + (– l – 4)y + (l + 5) = 0 y ´ ( 2) =

2

2

l+4 æ l - 2 ö æ -l - 4 ö = ç ÷ +ç ÷ - ( l + 5) 2 è 2 ø è 2 ø

Þ

l 2 - 4l + 4 = l + 5 Þ l2 – 4l + 4 = 4l + 20 4

Þ l2 – 8l – 16 = 0 Þ l = 4 ± 4 2

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Conic Sections

Þ l = 4 - 4 2 (Q l = 4 + 4 2 forms bigger circle)

(

)

Hence, centre of circle 2 2 - 2, 4 - 2 2 and radius =4-2 2

(

\ area = p 4 - 2 2

)

2

(

= 8p 3 - 2 2

)

-4a 8 + . b b Since, tangents are perpendicular to each other.

105. (a) Q Þa=4 One end of focal chord of the parabola is at (1, 4) y – coordinate of focal chord is 2at \ 2 at = 4 y2 = 16x

4ax + by = 8 Þ y =

-4a -1 = Þ b = 8a b 2 from (1) & (2) we get:

Þ

1 2 Hence, the required length of focal chord Þt =

2

2

106. (c) The shortest distance between line y = x and parabola = the distance LM between line y = x and tangent of parabola having slope 1. Y y㦨x

2 2 Þ a2 = 34 17 108. (c) To find intersection point of x2 + y2 = 5 and y2 = 4x, substitute y2 = 4x in x2 + y2 = 5, we get x2 + 4x – 5 = 0 Þ x2 + 5x – x – 5 = 0 Þ x (x + 5) – 1 (x + 5) = 0 \ x = 1, – 5 Intersection point in 1st quadrant be (1, 2). Now, equation of tangent to y2 = 4x at (1, 2) is y × 2 = 2 (x + 1) Þ y = x + 1 Þx–y+1=0 ...(i)

æ3 7ö Hence, ç , ÷ lies on (i) è4 4ø

L y 2= x – 2 X

(2 0)

...(ii)

Þa=±

1ö æ 1ö æ = a ç t + ÷ = 4 ´ ç 2 + ÷ = 25 2ø è tø è

M

107. (d) Since (a, b) touches the given ellipse 4x 2 + y 2 = 8 \ 4a2 + b2 = 8 ...(i) Equation of tangent on the ellipse at the point A (1, 2) is: 4x + 2y = 8 Þ 2x + y = 4 Þ y = –2x + 4 But, also equation of tangent at P (a, b) is:

109. (c) x2 = 8y

P(2at, at2)

q

Let equation of tangent of parabola having slope 1 is, y = m (x – 2) +

a m

Here m = 1 and a =

1 4

\ equation of tangent is: y = x –

7 4

Distance between the line y – x = 0 and y – x + 7 -0 4

=

2

2

1 +1

=

7 4 2

7 =0 4

Then, equation of tangent at P tx = y + at2 Þ y = tx – at2 Then, slope t = tan q Now, y = tan qx – 2 tan2 q Þ cot qy = x – 2 tan q x = y cot q + 2tan q 110. (c) Equation of a tangent to parabola y2 = 4x is: y = mx +

1 m

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Mathematics

This line is a tangent to xy = 2 1ö æ 1 2 \ x çè mx + ÷ø = 2 Þ mx + x - 2 = 0 m m Q Tangent is common for parabola and hyperbola. 2

x- 2

x2 +4 2 =0 4

2 x 2 - 4 x - 16 2 = 0

æ 1ö \ D = ç ÷ - 4 × m × (-2) = 0 è mø

x1 + x2 = 2 2, x1x2 = –16, (x1 – x2)2 = 8 + 64 = 72 Since, points P and Q both satisfy the equations (ii), then

1 + 8m = 0 m2 1 + 8 m3 = 0

x1 - 2 y1 + 4 2 = 0

m3 = -

x1 - 2 y2 + 4 2 = 0 (x2 – x1) =

1 1 Þm= 8 2

1 \ Equation of common tangent: y = - x - 2 2 Þ 2y = –x – 4 Þ x + 2y + 4 = 0 111. (d) y2 = –4(x – a2)

Þ PQ = ( x2 - x1 )2 + ( y2 - y1 )2 = ( x2 - x1 )2 + = | x2 - x1 | ×

Q

(0, 2a) P (a2, 0)

R

2( y2 - y1 ) Þ (x2 – x1)2 = 2(y2 – y1)2

( x2 - y1) 2 2

3 3 =6 2´ =6 3 2 2

Hence, length of chord = 6 3. 114. (b) Since, vertex and focus of given parabola is (2, 0) and (4, 0) respectively y

(0, –2a)

Area =

1 (4a)(a2) = 2a3 2

Since 2a3 = 250 Þ a = 5 112. (a, b, c, d) Normal to y2 = 8ax is y = mx – 4am – 2am3 ...(i) and normal to y2 = 4b (x – c) with slope m is y = m(x – c) – 2bm – bm3 ...(ii) Since, both parabolas have a common normal. \ 4am + 2am3 = cm + 2bm + bm3 Þ 4a + 2am2 = c + 2b + bm2 or m = 0 Þ (4a – c – 2b) = (b – 2a) m2 or (X-axis is common normal always) Since, x-axis is a common normal. Hence all the options are correct for m = 0. 113. (d) Let intersection points be P(x1, y1) and Q(x2, y2) The given equations x2 = 4y ...(i)

x - 2y + 4 2 = 0 Use eqn (i) in eqn (ii)

...(ii)

O

(4, 0)

(2, 0)

x

y¢ Then, equation of parabola is (y – 0)2 = 4 ´ 2(x – 2) Þ y2 = 8x – 16 Hence, the point (8, 6) does not lie on given parabola. 115. (b) Since, the equation of tangent to parabola y2 = 4x is y = mx +

1 m

...(i)

The line (i) is also the tangent to circle x2 + y2 – 6x = 0 Then centre of circle = (3, 0) radius of circle = 3 The perpendicular distance from centre to tangent is equal to the radius of circle

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Conic Sections

\

1 m =3Þ 1 + m2

3m +

, 16 P(

1 3

nt ge n Ta A (–16, 0)

X'

1 x± 3 Then, from equation (i): y = ± 3 Hence,

) 16

al rm No

Þ m= ±

Y

2

1ö æ 2 ç 3m + ÷ = 9 (1 + m ) è mø

X

B(24, 0)

3 y = x + 3 is one of the required common

tangent.

Y'

116. (a)

4 3 Slope of PB (m2) = –2

Slope of PC (m1) =

4 +2 m1 - m 2 3 Hence, tan q = = 1 + m1 × m 2 1 - 4 × 2 3

Let the coordinates of C is (t2, 2t). Since, area of DACB t2 1 9 =2 4

2t 1 6 1 -4 1

1 = |t2(6 + 4) – 2t(9 – 4) + 1(–36 – 24)| 2

Þ tan q = 2 118. (a) Equation of the chord of contact PQ is given by: T=0 or T º yy1 – 4(x + x1), where (x1, y1) º (– 8, 0) \ Equation becomes: x = 8 & Chord of contact is x = 8 \ Coordinates of point P and Q are (8, 8) and (8, – 8) and focus of the parabola is F (2, 0) 1 \ Area of triangle PQF = × (8 – 2) × (8 + 8) = 48 sq. units 2 119. (c) c = –29m – 9m3 a= 2 Given (at2 – a)2 + 4a2t2 = 64 Þ (a(t2 + 1)) = 8 Þ t2 + 1 = 4 Þ t2 = 3 Þ t=

1 = |10t2 – 10t – 60| 2 = 5|t2 – t – 6|

\

3

c = 2at (2 + t2) = 2 3 (5) c = 10 3

2

25 æ 1ö = 5 èç t - 2 ÷ø - 4

[Here, t Î (0, 3)]

For maximum area, t =

1 2

125 1 = 31 sq. units Hence, maximum area = 4 4 117. (a) Equation of tnagent at P(16, 16) is given as: x – 2y + 16 = 0

120. (c)

y

x’

x P

y’

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Mathematics

Tangent to x2 + y2 = 4 is

123. (b) Let P(h, k) divides OQ in the ratio 1 : 3 Let any point Q on x2 = 8y is (4t, 2t2).

y = mx + 2 1 + m2 Also, x2 = 4y x2 = 4mx + 8 1 + m 2 or x2 = 4mx – 8 1 + m 2 For D = 0

1

2

we have; 16 m2 + 4.8 1 + m = 0

P3

Q (4t, 2t2)

O

Þ m2 + 2 1 + m2 = 0 2 Þ m2 = -2 1 + m 4 2 Þ m = 4 + 4m Þ m4 – 4m2 – 4 = 0

Then by section formula

Þ m2 =

4 ± 16 + 16 2

Þ m2 =

4± 4 2 2

Þ

2

P (2t , 4t) C Centre of new circle = P(2t2, 4t) = P(2, – 4) Radius = PC = (2 – 0)2 + (–4 + 6)2 = 2 2 \ Equation of circle is :

( )

2

(x –2)2 + (y + 4) = 2 2 Þ x2 + y2 – 4x + 8y + 12 = 0

t12 = t 2 +

t2 +

4 t2

4 t2

2 t

124. (d) Let P ( -at12 , 2at1 ), Q( -at12 , -2at1 ) and R (h, k) By using section formula, we have h = -at12 , k =

-2at1 3

2at1 3 Þ 3k = –2at1 Þ 9k2 = 4a2 t12 = 4a (–h) Þ 9k2 = –4ah Þ 9k2 = –4h Þ 9y2 = –4x 125. (c) Given parabolas are y2 = 4x x2 = –32y Let m be slope of common tangent Equation of tangent of parabola (1) 1 y = mx + m Equation of tangent of parabola (2) y = mx + 8m2 (i) and (ii) are identical

k= -

1 1 1 = 8m2 Þ m3 = Þ m = 2 m 8 ALTERNATIVE METHOD:

Þ

1 m Since this is also tangent to x2 = – 32y

Let tangent to y2 = 4x be y = mx +

+4

³ 2 t 2.

t2 and h = t 2

Þ 2k = h 2 Required locus of P is x2 = 2y

Þ m2 = 2 + 2 2 121. (c) Minimum distance Þ perpendicular distance Eqn of normal at p(2t2, 4t) y = –tx + 4t + 2t3 It passes through C(0, –6) t3 + 2t + 3 = 0 Þ t = – 1

122. (a) t1 = – t –

k=

\

4 t2

=4

Minimum value of t12 = 8

1ö æ x 2 = -32 ç mx + ÷ è mø

Þ x2 + 32mx +

32 =0 m

Now, D = 0

...(1) ...(2)

...(i) ...(ii)

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Conic Sections

æ 32 ö (32) 2 - 4 ç ÷ = 0 è mø 4 1 3 Þm = Þ m= 32 2 126. (a) Equation of parabola, y2 = 6x 3 Þ y2 = 4 ´ x 2

128. (c)

A x2 + y 2 = 32

B

æ3 ö \ Focus = çè , 0÷ø 2 Let equation of chord passing through focus be ax + by + c = 0 ...(1) æ3 ö Since chord is passing through ç , 0÷ è2 ø 3 \ Put x = , y = 0 in eqn (1), we get 2 3 a+c = 0 2 3 Þ c = - a ...(2) 2 distance of chord from origin is a (0) + b(0) + c

=

2

2

=

5 5 2 2

c

a +b a + b2 Squaring both sides c2 5 = 2 4 a + b2

2

4 2 c 5 Putting value of c from (2), we get 4 9 2 a2 + b2 = ´ a 5 4 9 4 2 2 2 b2 = a - a = a 5 5 a2 5 a 5 2 = 4,b = ± 2 b

Þ a2 + b2 =

æ ± 5ö dy 5 a = - = -ç ÷=m 2 dx b è 2 ø 127. (d) The locus of the point of intersection of tangents to the parabola y2 = 4 ax inclined at an angle a to each other is tan2a (x + a)2 = y2 – 4ax Given equation of Parabola y2 = 4x {a = 1} Point of intersection (–2, –1) tan2a (–2 + 1)2 = (–1)2 – 4 × 1 × (–2) Þ tan2a = 9 Þ tan a = ± 3 Þ |tan a| = 3

Slope of chord,

y2 = 8x

We have x2 + (8x) = 9 x2 + 9x – x – 9 = 0 x (x + 9) – 1 (x + 9) = 0 (x + 9) (x – 1) = 0 x = –9, 1 for x = 1, y = ± 2 2 x = ± 2 2 (2 2 + 2 2) 2 + (1 - 1) 2 = 4 2 L2 = Length of latus rectum = 4a = 4 × 2 = 8 L1 < L2 129. (b) Let common tangent be L1 = Length of AB =

5 m Since, perpendicular distance from centre of the circle to the common tangent is equal to radius of the circle, therefore

y = mx +

5 m

=

5 2

1+ m On squaring both the side, we get m2 (1 + m2) = 2 Þ m4 + m2 – 2 = 0 Þ (m2 + 2)(m2 – 1) = 0 Þ m = ± 1 (Q m2 ¹ –2) 2

(

)

y = ± x + 5 , both statements are correct as m = ±1

satisfies the given equation of statement-2. 130. (b) We know that point of intersection of the normal to the parabola y 2 = 4ax at the ends of its latus rectum is (3a, 0) Hence required point of intersection = (3, 0) 131. (b) Both statements are true and statement-2 is the correct explanation of statement-1 a \ The straight line y = mx + is always a tangent to the m 2 parabola y = 4ax for any value of m. æ a 2a ö The co-ordinates of point of contact ç 2 , ÷ èm m ø

Now, required radius = OB =

9 + 16 = 25 = 5

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Mathematics

x2 y 2 + =1 16 3 Now, equation of normal at (2, 3/2) is

135. (a)

132. (a) Ellipse is

y

(t2, t)

16 x 3 y = 16 - 3 2 3/ 2 Þ 8x – 2y = 13

Þ

y = 4x -

Let y = 4 x -

13 2

13 touches a parabola 2

y2 = 4ax. We know, a straight line y = mx + c touches a parabola y2 = 4ax if a – mc = 0 \

x

O

æ 13ö a - ( 4) ç - ÷ = 0 Þ a = – 26 è 2ø

\ Distance = =

Hence, required equation of parabola is y2 = 4 (– 26)x = – 104 x 133. (a) Point P is (4, –2) and PQ ^ x-axis So, Q = (4, 2)

Y

Let (t2, t) be point on parabola from that line have shortest distance.

t 2 - t +1 2

2 1 éæ 1 ö 3ù êç t - ÷ + ú 4ú 2 êëè 2 ø û

Distance is minimum when t -

1 =0 2

1 é 3ù 3 2 0+ = 8 2 êë 4 úû 136. (b) We know that the locus of perpendicular tangents is directrix i.e., x = - a; x = -1 137. (b) We know that vertex of a parabola is the mid point of focus and the point \ Shortest distance =

nt ange T Q

Normal X P (4, – 2)

Y

X

OA

B

x= 2

Equation of tangent at (4, 2) is yy1 = Þ 2y =

1 (x + x1) 2 1 (x + 2) Þ 4y = x + 2 2

x 1 Þy= + 4 2

where directrix meets the axis of the parabola. Given that focus is O(0, 0) and directrix meets the axis at B(2, 0)

æ 0+ 2 ö \ Vertex of the parabola is ç , 0÷ = (1, 0) è 2 ø 138. (b) Given that parabola y2 = 8x

1 4 \ Slope of normal = – 4 134. (d) Both the given statements are true. Statement - 2 is not the correct explanation for statement - 1. So, slope of tangent =

y 2 = 8x

Y

(2,0) X' x+2=0

F

Y'

X

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Conic Sections

We know that the locus of point of intersection of two perpendicular tangents to a parabola is its directrix. Point must be on the directrix of parabola Q Equation of directrix x + 2 = 0 Þ x = –2 Hence the point is (–2, 0) 139. (a) Given that family of parabolas is y=

a3 x 2 a 2 x + - 2a 3 2

9 ö 3a a3 æ 2 3 x+ - 2a Þ y = çè x + ÷2a 3 16a2 ø 16

æ -3 -35a ö ÷ \ Vertex of parabola is çè , 4a 16 ø To find locus of this vertex,

Þ 64xy = 105

105 which is the required equation of locus. 64 140. (a) Given P = (1, 0), let Q = (h, k) Since Q lies on y2 = 8x

Þ xy =

\ K = 8h 2

...(i)

Let (a, b) be the midpoint of PQ k+0 h +1 \a= , b= 2 2

2 a -1 = h 2 b = k. Putting value of h and k in (i) (2b) 2 = 8(2a - 1) Þ b 2 = 4a - 2

Þ y2 - 4x + 2 = 0 . 141. (d) Equation of circle with centre (0, 3) and radius 2 is x 2 + ( y - 3)2 = 4

Let locus of the centre of the variable circle is (a , b ) Q It touches x - axis.

\

Circle touch externally Þ c1c2 = r1 + r2 \ a 2 + (b - 3) 2 = 2 + b

Þ

a 2 = 10(b - 1 / 2)

\

1ö æ Locus is x 2 = 10 ç y - ÷ è 2ø

y 2 = 4ax and x 2 = 4ay , we get (0, 0) and ( 4a, 4a) Putting in the given equation of line 2bx + 3cy + 4d = 0, we get d = 0 and 2b +3c = 0

-3 16 y and a = 4x 35

-3 -16 y = 4x 35

r2 c2 (a, b)

Which is equation of parabola. 142. (d) Solving equations of parabolas

-3 -35a x= and y = 4a 16

Þ

r1

a 2 + (b - 3) 2 = b 2 + 4 + 4b a2 + b2 – 6b + 9 = b2 + 4 + 4b

2 35a a3 æ 3ö = çx+ ÷ Þy+ 16 3 è 4a ø

Þ a=

c1

It¢s equation is ( x - a )2 + ( y – b)2 = b2

Þ d 2 + (2b + 3c )2 = 0 143. (b) Equation of the normal to a parabola y2 = 4bx at point

(bt

2 1 , 2bt1

) is

3

y = – t1 x + 2bt1 + bt1

(

)

2 Given that, it also passes through bt 2 , 2bt 2 then

2bt2 = – t1 bt22 + 2 bt1 + bt13

(

2 2 Þ 2t2 – 2t1 = – t1 t2 – t1

)

Þ 2(t2 – t1) = –t1(t2 + t1) (t2 – t1) 2 Þ 2 = – t1(t2 + t1) Þ t2 + t1 = – t 1 2 t1 144. (b) The equation of any tangent to the parabola y2 = 8ax is

Þ t2 = – t1 –

2a ...(i) m 2 If (i) is also a tangent to the circle, x + y2 = 2a2 then,

y = mx+

2a = ±

2a

m m2 + 1 Þ ( m2 + 2)(m2 – 1) = 0 Þ m = ± 1. Þ Putting the value of m in eqn (i), we get y = ± (x + 2a).

m2(1 + m2) = 2

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Mathematics

145. (c) We know that the locus of the feet of the perpendicular draw from foci to any tangent of the ellipse the auxiliary circle x + y = a 2

2

2

x

2

a2

+

y

2

b2

= 1 is

\ Auxiliary circle : x2 + y 2 = 4

149. (c) The given ellipse is x2 a

2

+

y2 b2

= 1, (a > b)

Length of latus rectum =

\ (-1, 3) satisfies the given equation. æ b2 ö 146. (c) Normal to the ellipse 2 + 2 = 1 at ç ae, ÷ is aø è a b

x2

y2

Þ x - ey =

2

...(i)

Q (0, - b) lies on equation (i), then

be =

e( a - b ) a

1 2

f ''(t ) = -2 < 0 Þ maximum Þ f (t ) max =

2

5 1 1 8 2 + - = = 12 2 4 12 3

Since, f(t )max. = eccentricity

Þ ab = a 2 e 2 Þ b = ae 2 Þ

b2 a2

= e4

\1 - e 2 = e 4 Þ e 4 + e 2 - 1 = 0

x2 y 2 147. (b) Ellipse : + = 1, 16 9

a = 4, b = 3, c = 16 - 9 = 7

Þe=

on it; PA + PB = 2a Þ PA + PB = 2(4) = 8.

x2 y 2 + =1 5 4

2 3

Now, b 2 = a 2 (1 - e2 ) 5a 2 æ 4ö 5a = a 2 ç 1 - ÷ Þ 5a = Þ a 2 - 9a = 0 è 9ø 9 2 Þ a = 9 Þ a 2 = 81 and b = 45

\ a 2 + b2 = 81 + 45 = 126

\ (± 7, 0) are the foci of given ellipse. So for any point P

148. (a) Ellipse º

...(i)

5 + t - t2 12

f '(t ) = 1 - 2t = 0 Þ t =

e( a - b ) a

2

2b 2 = 10 Þ b2 = 5a a

Now f(t ) =

a 2 x b2 y = a 2 - b2 ae b2 / a 2

Þ

2b 2 a

150. (d)

1 a = 4Þ a =4´ =2 e 2

Now, b 2 = a 2 (1 - e2 )

3 æ 1ö Þ b2 = 4 ç1 - ÷ = 4 ´ = 3 è 4ø 4

Let a point on ellipse be ( 5 cos q, 2sin q) So, equation

\ PQ 2 = ( 5 cos q)2 + (-4 - 2sin q)2 = 5cos 2 q + 4sin 2 q + 16 + 16sin q

x2 y 2 + =1 4 3

Þ 3x2 + 4 y 2 = 12 Now, P (1, b) lies on it

= 21 + 16 sin q - sin 2 q

= 21 + 64 - (sin q - 8) 2 = 85 - (sin q - 8) 2 PQ2 to be maximum when sin q = 1 2 \ PQmax = 85 - 49 = 36.

...(i)

Þ 3 + 4b 2 = 12 Þ b =

3 2

æ 3ö So, equation of normal at P ç1, ÷ è 2ø

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Conic Sections

a 2 x b2 y = a 2 - b2 Þ 4 x - 2 y = 1 1 3/ 2 151. (d) The given ellipse : Þ

x2 y 2 + =1 4 3 2

Given : 2a = 2 Þ a =

Q c 2 = a 2 + b2 Þ 1 =

154. (a) Let

1 2

2b =

1 1 + b2 Þ b = 2 2

x2

y2 + = 1; a > b a 2 b2

4 2 Þ b= 3 3

Equation of tangent º y = mx ±

So, equation of hyperbola is Comparing with º y =

x2 y 2 =1 1 1 2 2

m=

Þ 2x2 - 2 y2 = 1 So, option (d) does not satisfy it.

(0, 3) Q

152. (a)

(–2, 0)

|x| | y| + = 1 2 3

O P

x2 y 2 + = 1 4 9 (2, 0)

Q Area of ellipse = pab = p ´ 2 ´ 3 = 6p \ Required area = Area of ellipse – 4 (Area of triangle OPQ) æ1 ö = 6p - 4 ç ´ 2 ´ 3÷ è2 ø = 6p - 12 = 6(p - 2) sq. units 153. (a) Eccentricity of ellipse

a 2 m2 + b2

-x 4 + 6 3

-1 16 and a 2m2 + b 2 = 6 9

a 2 4 16 a 2 16 4 4 + = = - = Þ 36 3 9 36 9 3 9 Þ a2 = 16 Þ a = ± 4 Þ

Now, eccentricity of ellipse (e) = 1 -

Þ e = 1-

b2 a2

4 11 1 11 = = 3 ´16 12 2 3

155. (d) Let P be (x1, y1). So, equation of normal at P is 1 x y - =2 x1 y1 2

(0, –3)

4 7 7 e1 = 1 - = = 18 9 3 Eccentricity of hyperbola

Since, the point (e1, e2) is

7 13 Þ k = 15 æç ö÷ + 3 æç ö÷ è9ø è 9 ø Þ k = 16

Qc = a - b = 4 - 3 = 1 \ Foci = (±1, 0) Now for hyperbola : 2

4 13 13 = = 9 9 3 on the ellipse 15x2 + 3y2 = k. Then, 15e12 + 3e22 = k e2 = 1 +

1 æ ö , 0÷ It passes through ç è 3 2 ø

Þ

-1 6 2x1

So, y1 =

=-

1 1 x = 2 Þ 1 3 2

2 2 (as P lies in Ist quadrant) 3

y 2 So, b = 1 = 2 3 156. (b) 2ae = 6 and

2a = 12 e

Þ ae = 3

...(i)

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Mathematics

a a ...(ii) =6 Þ e= 6 e Þ a2 = 18 [From (i) and (ii)] Þ b2 = a2 – a2e2 = 18 – 9 = 9

Case-2: q =

and

\ Latus rectum =

5p , then tangent does not pass through 3

Q (4, 4). 159. (a) Let the equation of ellipse :

2b 2 2 ´ 9 = =3 2 a 3 2

x2

y2 + =1 a 2 b2

157. (a) 3 x + 4 y = 12 2

Given that length of minor axis is 4 i.e. a = 4. Also given be = 2

Þ 4 y = -3x + 12 2

Q a2 = b2 (1 – e2) Þ 4 = b2 – 4 Þ b = 2 2

3 Þ y =- x+3 2 4

Hence, equation of ellipse will be

Now, condition of tangency, c2 = a2m2 + b2 2 \ 18 = a .

Q ( 2 , 2) satisfies this equation.

9 9 + 9 Þ a2. = 9 16 16

\ ellipse passes through ( 2 , 2).

Þ a2 = 16 Þ a = 4 160. (a) Equation of tangent to Eccentricity e = 1 -

b2 a

2

= 1-

9 7 = 16 4

3x a

\ ae =

-

2

9y 2b 2

\

\ Focus are (± 7,0)

3 a

\ Distance between foci of ellipse = 2 7 158. (a) Slope of tangent on the line 2x + y = 4 at point P is Given ellipse is,

=1

2

-9

=

2

2b .(-2)

=

1 12

Þ a2 = 3 × 12 and b2 = 1 . 2

(On comparing) 9 ´12 4

Þ a = 6 and b = 3 3 Therefore, latus rectum =

x2 22

+

9ö æ y2 + = 1 at ç 3, - ÷ is, 2 2 2ø è a b x2

But given equation of tangent is, x – 2y = 12

7 .4 = 7 4

3x2 +4y2 = 12 Þ

y2 ( 3) 2

=1

161. (d) 3x2 + 5y2 = 32 Þ

3 sin q) \ equation of tangent on the ellipse, at P is,

Let point P(2cos q,

2b2 2 ´ 27 = =9 6 a

3x 2 5 y 2 + =1 32 32

P(2 2)

x y cos q + sin q = 1 2 3

Þ mT = –

R

3 cot q 2

Q both the tangents are parallel Þ Þ tan q = – Case-1: q =

x2 y 2 + =1 4 8

3 1 cot q = 2 2

p p 3 Þ q = p – 3 or q = 2p – 3

2p , then point P 3

3ö æ 5 5 ç -1, 2 ÷ and PQ = è ø 2

Tangent on the ellipse at P is 3(2) x 5(2) y 3x 5 y + =1Þ + =1 32 32 16 16

æ 16 ö \ co-ordinates of Q will be ç ,0 ÷ è 3 ø

Q

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Conic Sections

Now, normal at P is

32 32 y 32 32 = 3(2) 5(2) 3 5

æ4 ö \ co-ordinates of R will be ç , 0 ÷ è5 ø

(-ae, 0) S(ae, 0)

1 Hence, area of DPQR = ( PQ)( PR) 2 =

68 1 136 136 = . 15 2 9 25

1 ö æ 1 162. (d) Let tangent to parabola at point ç 2 , - 2m ÷ is è 4m ø

2 and tangent to ellipse is, y = mx ± m +

1 × 2ae × b = 8 Þ b2 = 8 2 From eqn (i) a2e2 = 8

1 2

Also, e2 = 1 -

Now, condition for common tangency,

2 Þ 16m4 + 8m2 – 1 = 0 Þ m =

4m

2

=

b2 a2

Hence, required length of latus rectum =

-8 ± 64 + 64 2 (16 )

x2 + y2 = 1 ( 2)2

1 = 2 +1 2 -1 4 4

( 2 cos q, sin q)

163. (b) Given that focus is (0,5 3) Þ| b | > | a |

( 2, 0)

Let b > a > 0 and foci is (0, ± be) Q a2 = b2 – b2e2 Þ b2e2 = b2 – a2 be =

2 2 b2 - a 2 Þ b – a = 75

...(i)

Q 2b – 2a = 10 Þ b – a = 5 From (i) and (ii) b + a = 15 On solving (ii) and (iii), we get Þ b = 10, a = 5

...(ii) ...(iii)

2 cos q x + y sin q = 1 2

æ 2 ö 1 ö æ Pç , 0÷ and Q ç 0, ÷ è sin q ø è cos q ø

Let mid point be (h, k) 2

Now, length of latus rectum =

2b 2 2(8) = 4 a

= 4 units 165. (c) Given the equation of ellipse,

-8 ± 8 2 2 -1 = 2 (16 ) 4

1

...(ii)

Þ a2e2 = a2 – b2 Þ 8 = a2 – 8 Þ a2 = 16

1 1 Þ 1 = m2 + 1 = ± m2 + 2 16m 2 4m 2

a=

...(i)

Þ

1 y = mx + 4m

=

b b ´ = –1 Þ b2 = a2e2 - ae ae Since, area of DS¢BS = 8

2a 50 = =5 b 10

164. (a) Q DS¢BS is right angled triangle, then (Slope of BS) ´ (Slope of BS¢) = –1

1 1 ,k= 2 sin q 2 cos q As cos2q + sin 2q = 1

Þ h=

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\

Mathematics

1 1 + 2 =1 2 2h 4k

Locus is

Þ a2 +

9 4 \ from eq. (iii) we get;

1 1 + =1 2 x2 4 y 2

Þ a=

166. (a) A ={(a,b) Î R ´ R :| a - 5 |< 1,| b - 5 |< 1}

e2 = 1 -

Let a – 5, = x, b – 5 = y Set A contains all points inside | x |< 1,| y |< 1 2

1 3 168. (c) Centre at (0, 0) x2

y2 + 2 =1 a b at point (4, –1)

(x - 1)2 y 2 + =1 9 4

2

Y

16

1 + =1 a 2 b2 Þ 16b2 + a2 = a2b2 at point (–2, 2)

(0, 1) (0, 1)

(–1, 0)

(1, 0) (1, –1) (–1, –1) (0, –1)

X

4 + 2 =1 a b Þ 4b2 + 4a2 = a2b2 2

\ (±1, ±1) lies inside the ellipse. Hence, A Ì B . 167. (d) Let for ellipse coordinates of focus and vertex are (ae, 0) and (a, 0) respectively.

3 Þ a - = ae 2 9 - 3a = a 2e 2 4

Length of latus rectum = Þ b2 = 2a e2 = 1 -

b2 a2

Þ e2 = 1 -

2a a2

3 2 (given)

2

Þ 3a 2 = 12b 2 b2 = a2 (1– e2)

2

Þ

2

a 2 = 4b2

3 3 Þ e= 2 4 169. (d) e = 3/5 & 2ae = 6 Þ a = 5 Þ e2 =

2 2 2 Q b = a (1 – e ) 2 Þ b = 25 (1 – 9/25)

...(i)

2b 2 =4 a ...(ii)

(from (ii))

2 ... (iii) a Substituting the value of e2 in eq. (i) we get;

Þ e2 = 1 -

...(ii)

Þ 16b + a = 4a + 4b From equations (i) and (ii) 2

\ Distance between focus and vertex = a(1 – e) =

...(i)

4

Þ a2 +

2 8 1 =1 - = a 9 9

Þ e=

2

B = {(a,b) ÎR ´ R: 4(a - 6) + 9(B- 5) £ 36} Set B contains all points inside or on

(–1, 1)

9 2ö æ - 3a = a 2 ç1 - ÷ 4 a è ø

Þ b= 4 \ area of required quadrilateral = 4 (1/2 ab) = 2ab = 40

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Conic Sections

Þ 2a2e = b2 = a2 (1 – e2)

170. (c) Equation of tangent to ellipse

Þ 2e = 1 – e2 Þ (e + 1)2 = 2 Þ e =

2 -1 173. (a) Given equation of ellipse can be written as

x y cos q + sin q = 1 27 3

Area bounded by line and co-ordinate axis D=

1 27 3 9 . . = 2 cos q sin q sin 2q

Þ a2 = 6, b2 = 2 Now, equation of any variable tangent is

D = will be minimum when sin 2q = 1 Dmin = 9

y = mx ± a 2 m2 + b2

171. (b) The end point of latus rectum of ellipse æ b2 ö x y + = 1 in first quadrant is ç ae, ÷ and the tangent aø è a 2 b2 2

2

æa ö at this point intersects x-axis at ç , 0÷ and y-axis at (0, a). èe ø x2 y 2 The given ellipse is + =1 9 5

Then

...(i)

where m is slope of the tangent So, equation of perpendicular line drawn from centre to tangent is y=

-x m

...(ii)

Eliminating m, we get ( x4 + y 4 + 2 x2 y 2 ) = a 2 x2 + b2 y 2

a2 = 9, b2 = 5

Þ ( x 2 + y 2 )2 = a 2 x 2 + b2 y 2

5 2 Þ e = 1– = 9 3

\ End point of latus rectum in first quadrant is L (2, 5/3)

2x y Equation of tangent at L is + =1 9 3 [Q It meets x-axis at A (9/2, 0) and y-axis at B (0, 3)]

\ Area of DOAB =

x2 y 2 + =1 6 2

Þ ( x2 + y 2 )2 = 6 x2 + 2 y 2 174. (b) Let point A (a, 0) is on x-axis and B (0, b) is on y-axis.

Y

1 9 27 ´ ´3= 2 2 4

B (0, b) P (h, k)

Y

l

B (0, 3) L(2,5/3) C

O

S

A X (9/2, 0)

D

By symmetry area of quadrilateral = 4 × (Area DOAB) = 4 ´

27 = 27 sq. units. 4

172. (b) Focus of an ellipse is given as (± ae, 0) Distance between them = 2ae According to the question, 2ae =

b2 a

A(a, 0)

X

Let P (h, k) divides AB in the ratio 1 : 2. So, by section formula h=

2(0) + 1( a ) a = 1+ 2 3

k=

2(b) + 1(0) 2b = 3 3

Þ a = 3h and b =

3k 2

Now, a 2 + b 2 = l2 Þ 9 h2 +

9k 2 = l2 4

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Þ

Mathematics

h2 æ lö çè ÷ø 3

+

2

k2 æ 2l ö çè ÷ø 3

2

=1

æ l2 9 ö 1- ç ´ 2 ÷ = è 9 4l ø

Now e =

1-

1 3 = 4 2

Thus, required locus of P is an ellipse with eccentricity

3 . 2 x2

Þ Focii = (± 7, 0)

y2

+ = 1 be the equation of ellipse. a 2 b2 Given that F1B and F2B are perpendicular to each other. Slope of F1B × slope of F2B = – 1

175. (a) Let

7 4 Now, radius of this circle = a2 = 16 Þ e=

æ 0-b ö æ 0-b ö çè ÷ ´ç ÷ =–1 - ae - 0 ø è ae - 0 ø

x2 y 2 + =1 9 4 Normal at the point is parallel to the line

Þ

B (0, b)

F1 (–a, 0) (–ae, 0) F2(ae, 0)

Now equation of circle is (x – 0)2 + (y – 3)2 = 16 x2 + y2 – 6y – 7 = 0 177. (c) Given ellipse is 4x2 + 9y2 =36

4x – 2y – 5 = 0 Slope of normal = 2 (a, 0)

Slope of tangent =

-1 2

Point of contact to ellipse (0, – b)

æ b ö æ -b ö çè ÷ø ´ çè ÷ø = – 1 ae ae

e = 1-

b

b2 ïü ïì 2 íQ e = 1 - 2 ý a ïþ ïî

b2 a

2

2

a2

=

b2 a2

b2 b2 1 1= 2 2 Þ 2 = 2 a a 1 1 b2 e2 = 1 - 2 = 1 - = 2 2 a

y2 b2

=1

æ a2m b and line is ç , ç 2 2 2 2 2 a m + b2 è a m +b

ö ÷ ÷ ø

1 2 No common tangents for these two circles. 176. (a) From the given equation of ellipse, we have

æ -9 8 ö \ Point = ç , ÷ è 5 5ø

178. (c) Given equations of ellipses x2 y 2 + =1 3 2 2 1 Þ e1 = 1 - = 3 3

E1 :

and E2 :

e2 =

9 a = 4, b = 3, e = 1 16

a2

+

Now, a2 = 9, b2 = 4

b2 = a2e2 2

x2

Þ e2 =

x2 y 2 + =1 16 b 2

1 - b2 16 - b2 = 16 4

Also, given e1 × e2 = Þ

1 3

´

1 2

16 - b 2 1 = Þ 16 - b 2 = 12 4 2

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Conic Sections

Þ b2 = 4 \ Length of minor axis of E2 = 2b = 2 × 2 = 4

Þ x = 2 or – 2

179. (d) x2 = 8y

...(ii)

2

When y = – 3, then x2 = – 24, which is not possible.

2

æ 1ö c = 4 ç – ÷ + 12 = 2 è 2ø

We put y = – 2

x2 a2

x a

2

+

y

2

or x 2 – 2 2 x + 2 = 0

y2 b2

=1

b

2

= 1 is given by

...(ii)

32 2 32 ,b = 3 5 So, the equation of the ellipse is

Solving (i) and (ii) we get a2 =

3x 2 + 5 y 2 = 32 182. (a) The given equation of ellipse is x2 y2 + =1 4 1 So, A = (2, 0) and B= (0, 1) If PQRS is the rectangle in which it is inscribed, then P = (2, 1).

x2

Q

B (0,1)

O

2

Þ x2 + 2 2 x + 2 = 0

+

Þ 5b2 = 3a 2 2

1 x+ 2 2

x æ x ö + ç – + 2÷ = 1 ø 4 è 2 ö x 2 æ x2 æ xö +ç – 2 ç ÷ 2 + 2÷ = 1 è ø 4 è 4 2 ø

\

2

= 2 + 8 = 10

y2 + = 1 be the ellipse a 2 b2 the rectangle PQRS.

x2 y 2 + =1 4 1

)}

b 2 = a 2 (1 - e 2 ) = a 2 (1 - 2 / 5)

Let

1 So, y = – x ± 2 2

For ellipse :

2

(

2– – 2

Given it passes through (–3, 1) so 9 1 + 2 = 1 ...(i) 2 a b Also, we know that

1 = 0 Þ 3y – 1 = 0 3

1– 0 1 =– 0–2 2

)

{

181. (d) Let the equation of ellipse be

æ 2 6 1ö æ 2 6 1ö , ÷ and ç , ÷ ç 3 3 3 3ø è ø è Required equation of the line,

y = mx ± a 2 m2 + b2 Here a = 2, b = 1

(

æ 2 ö = ç + 2 2 è 2 ÷ø

2 6 1 When y = , then x = ± 3 3

180. (d) Any tangent on an ellipse

2

1 ö æ 1 ö æ \ Points are ç 2, , ç – 2, – ÷ ÷ è 2ø è 2ø 2

Point of intersection are

1

and x = – 2, y = –

ì 1 æ 1 öü \ P1 P2 = í –ç– ÷ý + 2øþ î 2 è

8y 1 + y 2 = 1 Þ y = – 3, 3 3

m=

2

...(i)

x2 + y2 = 1 3 From (i) and (ii),

y-

1

If x = 2, y =

R

Then it passed through P (2,1 ) 4 1 \ + = 1 ....(i) 2 a b2

circumscribing

P (2, 1)

A (2,0) (4,0) S

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Mathematics

Also, given that, it passes through (4, 0) \ 16 + 0 = 1 Þ a 2 = 16 a2 Þ b2 = 4/3 [putting a2 = 16 in eqn (i)] \ The required equation of ellipse is

[from (i)]

Þ 2e2 = 1, e = x2 y2 + =1 16 4 / 3

or x2 + 12y2 =16 a 183. (a) Perpendicular distance of directrix x = ± from e

focus ( ± ae, 0)

X

O S (ae, 0)

a x= a/e Y´ – ae = 4 e 1ö æ Þaç2 – ÷ = 4 è 2ø 8 Þa = 3 \ Semi maor axis = 8/3 184. (a) Given that distance between foci is 2ae = 6 Þ ae = 3 and length of minor axis is 2b = 8 Þ =

b=4

we know that b 2 = a 2 (1 - e 2 ) Þ 16 = a 2 - a 2 e 2 Þ a 2 = 16 + 9 = 25 Þ a = 5

1 a 186. (b) Given that e = . Directrix , x = = 4 2 e 1 1 = 2 \b = 2 1- = 3 2 4 Equation of ellipse is \a = 4´

y = mx ± 6 1 + m2 For common tangent,

36(1 + m 2 ) = 100m 2 - 64 100 64 ´ 36 369 100 164 æ ö \ c 2 = 36 ç1 + = ÷= 64 ø 64 4 è Þ 100 = 64m 2 Þ m 2 =

Þ 4c 2 = 369 188. (a) Q The equation of hyperbola is x2

y2

=1 a 2 b2 Q Equation of hyperbola passes through (3, 3) 1 2

-

-

1 2

=

1 9

Þ FB 2 + F ' B 2 = FF ' 2

a 2e2 + b2 2

) ( 2

+

a 2e2 + b2 2 2

)

Þ 2(a e + b ) = 4a e Þ e = 2

2

= (2ae) 2

b2 a2

...(i)

B (0, b)

Q It passes through (9, 0)

6 -3 = 1 1 - 2 a2 b 1 1 = b 2 2a 2 From equations (i) and (ii), \

F' ( -ae, 0)

O

...(i)

x -3 y -3 = 1 1 ×3 - 2 ×3 a2 b

185. (a) Given that ÐFBF ' = 90°

2 2

.

a b Equation of normal at point (3, 3) is :

3 3 \ e= = a 5

(

2

y = mx ± 100m2 - 64 and the general tangent to the circle in slope form is

æa ö çè - ae÷ø e

\

1

x2 y 2 + = 1 Þ 3x 2 + 4 y 2 = 12 4 3 187. (c) General tangent to hyperbola in slope form is

Y

We know that e 2 = 1 - b 2 / a 2 = 1 - e2

F (ae, 0)

9 a2 = , b2 = 9 2

...(ii)

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Conic Sections

Q Eccentricity = e, then e 2 = 1 +

b2 a

2

=3

æ9 ö \ (a 2 , e 2 ) = ç , 3÷ è2 ø x2 y 2 + =1 25 b2

\ x12 + 5 y12 =

b2 25 x2 y 2 + =1 16 b2

x2 5cos q 2

+

y2 =1 5

According to the question, e1 = 5e2 2 3 Now length of latus rectum of ellipse 2 Þ 1 + cos 2 q = 5sin 2 q Þ cos q =

æ b öæ b ö Þ (e1e2 ) 2 = 1 Þ çç 1 - ÷ç ÷ç 1 + ÷÷ = 1 è 25 øè 16 ø 2

2

=

b2 b 2 b4 Þ 1+ - =1 16 25 25 ´ 16 9 b4 b2 Þ = 0 Þ b2 = 9 16 × 25 25 ×16

2a 2 10 cos 2 q 20 4 5 = = = 3 b 5 3 5

192. (b) Let the hyperbola is

x2

y2 =1 a 2 b2

If a hyperbola passes through vertices at (± 6, 0), then \ a=6 As hyperbola passes through the point P(10, 16)

9 4 \ e1 = 1 = 25 5

\

9 5 = 16 4

Distance between focii of ellipse

100 256 = 1 Þ b2 = 144 36 b 2

\ Required hyperbola is

= a = 2ae1 = 2(5)(e1 ) = 8 Distance between focii of hyperbola

x2 y 2 =1 36 144

36 x 144 y + = 36 + 144 10 16 \ At P(10, 16) normal is

Equation of normal is

= b = 2ae2 = 2(4)(e2 ) = 10 \ (a, b) = (8, 10) 190. (a) The tangent to the hyperbola at the point (x1, y1) is, xx1 - 2 yy1 - 4 = 0 The given equation of tangent is 2x – y = 0 Þ

and Ellipse :

x2 y2 2 = 1 Þ e1 = 1 + cos q 10 10cos 2 q

Þ e2 = 1 - cos 2 q = sin q

b2 16

e1e2 = 1

And, e2 = 1 +

32 2 +5´ = 6 7 7

191. (d) Hyperbola :

The equation of hyperbola,

Then, e2 = 1 +

...(ii)

2 32 y12 = , x12 = 7 7

189. (d) Equation of ellipse is

Then, e1 = 1 -

x12 y12 -1 = 0 4 2 On solving eqs. (i) and (ii) \

36 x 144 y + = 36 + 144 10 16 \ 2x + 5y = 100.

193. (c) Equation of tangent to y2 = 12x is y = mx +

x1 =2 2 y1

3 m

Equation of tangent to

Þ x1 = 4 y1

...(i)

Since, point (x1, y1) lie on hyperbola.

x2 y 2 = 1 is y = mx ± m2 - 8 1 8 Q parabola and hyperbola have common tangent.

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Mathematics

9 3 = ± m2 - 8 Þ 2 = m2 – 8 m m

\

Put m2 = u u2 – 8u – 9 = 0 Þ u2 – 9u + u – 9 = 0 Þ (u + 1) (u – 9) = 0 Q u = m2 ³ 0 Þ u = m2 = 9 Þ m = ± 3 \ equation of tangent is y = 3x + 1 or y = – 3x – 1

æ 1 ö \ intersection point is P ç - ,0 ÷ . è 3 ø e = 1+

b2 a2

Þ e = 1+

8 Þe=3 1

\ foci (± 3, 0)

\

16 a2

S æ 1 ö ç - 3 ,0÷ P è ø

12 - 2 2 =1 a e - a2

Þ

æ a2 ö 4 é4 3 ù Þ 4e 2 - 4 - 3 = (e2 - 1) ç ÷ = 1 ê ú ç 4 ÷ a 2 ë 1 e2 - 1û è ø

æ 4e ö Þ 4(4e2 - 7) = (e2 - 1) ç ÷ è 5ø

(3, 0)

x2 y 2 =1 9 16 Then focus is S’ (– ae, 0)

S' (–3, 0)

4 m Q This is common tangent.

y = mx +

So, put y = mx +

x=

4 in xy = – 4. m

= 16m Þ m3 = 1Þ m = 1

\ equation of common tangent is y = x + 4 195. (c) Q directrix of a hyperbola is, 5x = 4 5 Þ x =

4 a 4 Þ = e 5 5

(3, 0)

–9 5

2 a = 3, b = 4 Þ e = 1 +

4ö æ 4 x ç mx + ÷ + 4 = 0 Þ mx 2 + x + 4 = 0 mø m è 16

2

196. (d) 16x2 – 9y2 = 144 Þ

194. (a) Given curves, y2 = 16x and xy = – 4 Equation of tangent to the given parabola;

m2

y2 - 2 = 1 passes throug (4, -2 3) a b 2

é 2 ù b2 2 2 2 2 êQ e = 1 + 2 Þ a e - a = b ú a êë úû

1 3+ SP 3 = 10 = 5 = 1 ¢ SP 3 8 4 3

D=0Þ

x2

Þ 4e 4 - 24e 2 + 35 = 0

S' (–3, 0)

Now, hyperbola

é 2 ù 16 25 êQ e = 1 + b ú = a2 úû 9 9 êë

5 ö æ \ the focus S ¢ º ç 3 - ´ ,0 ÷ º (-5,0) 3 ø è 197. (c) Since, lx + my + n = 0 is a normal to

then

x2

y2 = 1, a 2 b2

a2

b 2 (a 2 + b2 )2 = l 2 m2 n2

but it is given that mx - y + 7 3 is normal to hyperbola

x2 y 2 =1 24 18 24

æa ö ç e 0÷ è ø

(4 -2 3)

then

m

2

-

18 (-1)

2

=

(24 + 18) 2 (7 3) 2

Þm=

2 5

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Conic Sections

Þ a = 8, b2 = 32

198. (c) Let equation of hyperbola be

Then, the equation of the ellipse

x2

y2 =1 a2 b2

Qe = 1+

b2 a2

...(i)

16 2

-

36

=1 a b2 On solving (i) and (ii), we get a2 = 4, b2 = 12

\

(

Þ b2 = a2 (e2 – 1)

e = 2 Þ b2 = 3a2 Equation (i) passes through (4, 6),

...(ii)

...(iii)

x2 y 2 =1 4 12 Now equation of tangent to the hyperbola at (4, 6) is

y 4x 6 y = 1 Þ x - = 1 Þ 2x – y = 2 4 12 2 199. (d) Let the points are, A(2, 0), A¢(–2, 0) and S(–3, 0) Þ Centre of hyperbola is O(0, 0) A A¢ = 2a Þ 4 = 2a Þ a = 2 Q Distance between the centre and foci is ae.

200. (a) \ Conugate axis = 5 \ 2b = 5 Distance between foci =13

x2 y 2 = 1 Þ a2 = 5, b2 = 4 5 4 The equation of tangent to the hyperbola is, y = mx ± a 2 m 2 - b2 = x ± 5-4 Þy = x ±1 203. (b) Since, r ¹ ±1, then there are two cases, when r > 1

Þ e2 = 1 Þ e= ...(i)

(1 – r) = (1 + r) (e2 – 1) Þ e2 = 1 +

13 13 Þe= 2 12

201. (b) Let the ellipse be Then,

(r - 1) 2r = (r + 1) ( r + 1)

2r r +1

204. (a) Q a2 = cos2 q, b2 = sin 2 q and e > 2 Þ e2 > 4 Þ 1 + b2/a2 > 4 Þ 1 + tan2 q > 4

\ a=6 ae =

2 (r + 1)

x2 y2 = –1 (Hyperbola) 1- r 1+ r Then,

Þ e=

Þ a2 = 36

(r - 1) 2 = (r + 1) (r + 1)

When 0 < r < 1, then

2ae = 13 Then, b2 = a2 (e2 – 1)

( r - 1) ( r + 1)

(r – 1) = (r + 1) (1 – e2) Þ 1 – e2 =

3 2 Þ b2 = a2 (e2 – 1) = a2e2 – a2 = 9 – 4 = 5

\ (6, 5 2) does not lie on this hyperbola.

202. (a) Given, the equation of line, x–y=2Þy=x–2 \ its slope = m = 1 Equation of hyperbola is:

x2 y2 + = 1 (Ellipse) r -1 r +1 Then,

\ OS = ae Þ 3 = 2e Þ e =

x2 y 2 =1 4 5 Q (6, 552) does not satisfy eq (i).

)

Hence, the point 4 3,2 2 lies on the ellipse.

Now equation of hyperbola is

Þ Equation of hyperbola is

x2 y2 + =1 64 32

x2 y 2 + =1 a2 b2

2b 2 = 8, 2ae = b2 and b2 = a2(1 – e2) a

æ p pö Þ sec2 q > 4 Þ q Î çè , ÷ø 3 2 Latus rectum,

EBD_8344

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Mathematics

2b2 2sin 2 q = = 2(sec q – cos q) a cos q æ p pö d ( LR ) Þ = 2(sec q tan q + sin q) > 0 "q Î çè , ÷ø 3 2 dq p p 1 æ ö æ ö \ min (LR) = 2 çè sec - cos ÷ø = 2 çè 2 - ÷ø = 3 3 3 2 LR =

p 2 Hence, length of latus rectum lies in the interval (3, ¥)

max (LR) tends to infinity as q ®

1 \ Area of DPQT = ´ TR ´ PQ 2

Q P º (3 5, -12) \ TR = 3 + 12 = 15,

1 \ Area of DPQT = ´15 ´ 6 5 = 45 5 sq. units 2 207. (a) Here, lines are:

2 x – y + 4 2k = 0 2 x + 4 2k = y

Þ

2kx + ky - 4 2 = 0

and

205. (d)

...(i) ...(ii)

Put the value of y from (i) in (ii) we get;

(

)

Þ 2 2 kx + 4 2 k 2 – 1 = 0 Þ x=

2 (1 – k 2 ) 2 2(1 + k 2 ) ,y= k k

Consider equation of hyperbola 2

2

æ y ö æxö \ç ÷ – ç ÷ =1 è4ø è4 2ø

x2 y 2 =1 22 b 2 Q (4, 2) lies on hyperbola

\ length of transverse axis

16 4 =1 4 b2 4 \ b2 = 3

2a = 2 × 4 2 = 8 2 Hence, the locus is a hyperbola with length of its transverse

\

axis equal to 8 2

b2 Since, eccentricity = 1 + 2 a 4 Hence, eccentricity = 1 + 3 = 1 + 1 = 2 4 3 3

206. (d) Here equation of hyperbola is

x 2 y2 =1 9 36

Now, PQ is the chord of contant \ Equation of PQ is : Þ y = –12

x(0) y(3) =1 9 36

\

2 a2 2

4 -b

x 2 y2 =1 9 36 (0, 3) X

X'

Q R Y'

-

y2

...(i)

Hyperbola passes through ( 2, 3 )

2

Y

x2

=1 a 2 b2 foci is (±2, 0) Þ ae = ±2 Þ a2e2 = 4 Since b2 = a2 (e2 – 1) b2 = a2 e2 – a2 \ a2 + b2 = 4

208. (c) Equation of hyperbola is

-

3 b2 3 b2

=1

...(ii)

=1

Þ b4 + b2 – 12 = 0 Þ (b2 – 3) (b2 + 4) = 0 Þ b2 = 3 b2 = – 4 For b2 = 3

[from (i)]

(Not possible)

2 2 Þ a2 = 1 \ x - y = 1 1 3

P

Equation of tangent is

2x 3y =1 1 3

Clearly (2 2,3 3) satisfies it.

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Conic Sections

209. (d) Here, tx – 2y – 3t = 0 & x – 2ty + 3 = 0 On solving, we get; 6t

3t 2 + 3

3t

= & x= 2t 2 - 2 t 2 - 1 t2 -1 Put t = tan q \ x = – 3 sec 2q & 2y = 3 (–tan 2q) 2 2 Q sec 2q – tan 2q = 1

y=

(

2b 2 =8 a

13 2 If e, be the eccentricity of the ellipse, then

94 1 5 =1+ Þ e= 9 4 2

x2 a2

+

2

Now

x2 y 2 + =1 211. (c) 12 16

æ9 ö Þ a2 = 4 ç - 1÷ è4 ø Þ a2 = 5

a 2 - b 2 =ae1

The point (5, 2 3 ) does not satisfy the above equation. 212. (a) S(5, 0) is focus Þ ae = 5 (focus)

Þ b2 = 12 Hence, equation of ellipse is x 2 y2 + =1 13 12 æ 13 3 ö Now putting the coordinate of the point ç 2 , 2 ÷ in è ø the equation of the ellipse, we get

13 3 + =1 4 ´ 13 4 ´ 12

x2 y 2 \ It's equation is =–1 5 4

5 (a) Þ (e) = 3 b2 = a2 (e2 – 1) Þ b2 = 16

=1

\ 13 – b2 = 1

1 12 e = 1= 2 16 Foci (0, 2) & (0, – 2) So, transverse axis of hyperbola = 2b = 4 Þ b = 2 & a2 = 12 (e2 – 1)

(a) & (b) Þ a2 = 9

b2

a2 = 13

3

a a 9 Þ = (directrix) 5 e 5

y2

Since ellipse passes through the foci (+ 13 , 0) of the hyperbola, therefore

Þ 4b2 = a 2 e2 Þ 4a 2 (e 2 - 1) = a 2 e 2

x=

1 13 1 = Þ e1 = 2 2 13 Equation of ellipse is e1 ´

1 and 2b = (2ae) 2

Þ 3e2 = 4 Þ e =

)

e=

x2 y 2 =1 9 94 which represents a hyperbola \ a2 = 9 & b2 = 9/4

210. (a)

x 2 y2 =1 4 9

Its Foci = ± 13, 0

Þ

l(T.A.) = 6; e2 = 1 +

213. (c) Equation of hyperbola is

––––– (a)

––––– (b)

Þ

1 1 + = 1 , which is not true, 4 16

æ 13 3 ö Hence the point ç 2 , 2 ÷ does not lie on the ellipse. è ø 214. (a)

æ b ç - ae, ÷ aø è

a2 – b2 = 9 – 16 = – 7

æ b2 ö ç ae, ÷ aø è L

(ae, 0)

EBD_8344

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Mathematics

æp ö æp ö 3 x cos ç - q÷ + 2 y cot ç - q÷ = 32 + 22 è2 ø è2 ø

x2 y 2 =1 4 5 Þ a2 = 4, b2 = 5

Given

a 2 + b2

e=

a2

=

Þ

4+5 3 = 4 2

3h sin q + 2k tan q = 32 + 2 2

...(3)

and 3h cos q + 2k cot q = 32 + 22

...(4)

3 5ö æ æ 5ö L = çè 2 ´ , ÷ø = çè 3, ÷ø 2 2 2

Comparing equation (3) & (4), we get

Equation of tangent at (x1, y1) is x x1 y y1 - 2 =1 a2 b 5 Here x1 = 3, y1 = 2

3h cos q - 3h sin q = 2k tan q - 2k cot q

Þ

3h cos q + 2 k cot q = 3h sin q + 2 k tan q

3h(cos q - sin q) = 2k (tan q - cot q) 3h(cos q - sin q) = 2k

3x y x y + =1 =1Þ 4 -2 4 2 3

or, 3h =

20 16 -4 = 9 9 215. (d) Let the coordinate at point of intersection of normals at P and Q be (h, k) Since, equation of normals to the hyperbola

a

-

y2 b

2

= 1 At point (x1, y1) is

a 2 x b2 y + = a2 + b2 x1 y1

therefore equation of normal to the hyperbola

2

x

2

-

3

y

2

22

= 1 at point P (3 secq , 2 tanq) is

Þ 3 x cos q + 2 y cot q = 3 + 2

–2k = 13 Þ k =

-13 2

Hence, ordinate of point of intersection of normals at P and Q is

-13 2

216. (a) x2 – 6y = 0

...(i)

2

2x – 4y = 9

...(ii)

Consider the line, 2

...(1)

at point Q (3 sec f, 2 tanf) is

x2 32

-

y2 22

32 x 22 y + = 32 + 22 3sec f 2 tan f Þ 3 x cos f + 2 y cot f = 32 + 2 2

3 2 On solving (i) and (iii), we get only x- y =

Similarly, Equation of normal to the hyperbola

Given q + f =

Þ 2k tan q - 2 k + 2k tan q = 13

2

32 x 22 y + = 32 + 22 3sec q 2 tan q 2

...(5)

-2k (sin q + cos q)sin q + 2k tan q = 32 + 22 sin q cos q

OA2 – OB2 =

2

-2 k (sin q + cos q) sin q cos q

Now, putting the value of equation (5) in eq. (3)

4 3 y-intercept of the tangent, OB = –2

x-intercept of the tangent, OA =

x2

(sin q - cos q)(sin q + cos q) sin q cos q

x = 3, y =

3 2

æ Hence ç 3, è

...(2)

p p Þ f = - q and these passes through 2 2

...(iii)

3ö ÷ is the point of contact of conic (i), and 2ø

line (iii) On solving (ii) and (iii), we get only x = 3, y =

(h, k) \ From eq. (2)

3 2

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Conic Sections

æ 3ö Hence ç 3, ÷ is also the point of contact of conic (ii) and è 2ø

b2 16

eccentricity = e = 1 -

line (iii). Hence line (iii) is the common tangent to both the given conics.

foci: ± ae = ± 4 1 -

b2 16

217. (d) Equation of the tangent at the point ‘q’ is Equation of hyperbola is

x sec q y tan q =1 a b

Þ

Þ P = (a cos q, 0) and Q = (0, – b cot q) Let R be (h, k) Þ h = a cos q, k = –b cot q Þ

-b -bh h k = Þ sin q = and cos q = ak a h a sin q

x2 y2 =1 144 81 25 25

2 2

a k

+

h2 a2

=

=1

81 25 81 ´ = 1+ 25 144 144

eccentricity = e = 1 +

By squaring and adding, b2 h2

x2 y 2 1 = 144 81 25

foci: ± ae = ±

225 15 = 144 12

12 15 ´ = ±3 5 12

Since, foci of ellipse and hyperbola coincide Y \ ± 4 1R

Q O

P

b2 = ± 3 Þ b2 = 7 16

K2 4 - =1 9 4

X

(Qb = ± 2)

Þ K2 = 18 219. (a) Given hyperbola is

Þ

b2 k2

+1 =

a2

Þ

h2

a2 h2

-

b2 k2

Now, given eqn of hyperbola is

x2 y 2 =1 9 b2 =1

Since this passes through (K, 2), therefore

x2 y 2 =1 4 2

K2 4 =1 9 b2

...(1)

Þ a 2 = 4, b 2 = 2

\ R lies on

a

2

x2

-

b

2

y2

= 1 i.e.,

4 x2

218. (c) Given equation of ellipse is x2 y 2 + =1 16 b 2

-

2 y2

=1

Also, given e = 1 +

Þ

1+

b2 a

2

=

13 3

13 b2 = Þ 9 + b2 = 13 9 3

Þ b=±2

EBD_8344

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Mathematics

Now, from eqn (1), we have K2 4 - =1 9 4

Co-ordinates of foci are (± ae, 0)

(Qb = ± 2)

Þ K2 = 18 220. (b) Given that ae = 2 and e = 2 \ a =1

(

222. (d) We know that tangent to the hyperbola

y = mx ±

b 2 = 1( 4 - 1)

\ Equation of hyperbola,

x2 a2

-

y2 b2

=1

cos 2 a

2 2 2 2 Þ m = a and a m - b = b

\

a 2a 2 - b 2 = b 2

Locus is a 2 x 2 - y 2 = b2 which is hyperbola.

3 x2 - y 2 = 3 221. (b) Given, equation of hyperbola is -

y2 sin 2 a

=1

223. (d)

a=

1 x2 y2 = 144 81 25

144 12 = ,b = 25 5

81 9 = , 25 5

Compare with equation of hyperbola x2 a2

-

y2 b2

e = 1+

= 1, we get a 2 = cos 2 a and

81 15 5 = = 144 12 4

\ Foci = ( ± ae, 0) = ( ±3 , 0) \ foci of ellipse = foci of hyperbola \ for ellipse ae = 3 but a = 4,

b 2 = sin 2 a

We know that, b 2 = a 2 (e2 - 1) Þ sin 2 a = cos 2 a (e 2 - 1)

\ e=

3 4

Þ sin 2 a + cos2 a = cos 2 a.e2

Then, b 2 = a 2 (1 - e 2 )

Þ e 2 = sec2 a

Þ b 2 = 16æç1 - 9 ö÷ = 7 è 16 ø

Þ e = sec a

\ ae = cos a .

y2 b2

a 2 m2 - b2

2

x y =1 1 3

x2

a2

-

Given that y = a x + b is the tangent of hyperbola.

b2 = 3

Þ

x2

is

)

We know, b 2 = a 2 e 2 - 1

2

i.e. ( ± 1, 0) Hence, abscissae of foci remain constant when a varies.

1 =1 cos a

=1

www.jeebooks.in

12 Limits and Derivatives TOPIC Ć

1.

x ®a +

(a)

3.

lim

3 2

( x - 1)( x 2 - 5 x + 6) is equal to : x ®-ab x2 - 6 x + 8

then lim

x (e (

x®0

1 - cos(p( x )) is equal to: [Sep. 05, 2020 (I)] x+a-4

3

(b)

(c)

1 2

2 1+ x 2 + x 4 -1) / x

- 1)

(d)

(a) 1/2 7.

1 2

[Sep. 05, 2020 (II)]

1 + x2 + x4 - 1

x ®0

1 - x+ | x | = L, then L is equal to : l - x + [ x]

8.

9.

4.

5.

(b) 2

(c)

1 2

(d) 0

ìï 1 æ x2 x2 x2 x 2 ö üï If lim í 8 çç1 - cos - cos + cos cos ÷÷ ý = 2 - k , x ®0 ï x 2 4 2 4 ø ïþ î è then the value of k is __________. [NA Sep. 03, 2020 (I)] 3x + 33- x - 12 lim - x /2 1- x is equal to ¾¾¾. x ®2 3 -3 [NA Jan. 7, 2020 (I)]

x + 2sin x + 1 - sin 2 x - x + 1 2

(b) 2

(c) 3

is [April 12, 2019 (II)] (d) 1

x4 - 1 x3 - k 3 = lim 2 , then k is:[April 10, 2019 (I)] x ®1 x - 1 x ®k x - k 2

8 3

(b)

3 8

(c)

3 2

(d)

4 3

x 2 - ax + b = 5 , then a + b is equal to : x ®1 x -1

If lim

(b) 5

(c) –7

sin 2 x equals : x ®0 2 - 1 + cos x lim

(a) 4 2

11.

[April 12, 2019 (II)] (d) 3/2

If lim

(a) –4 10.

(c) –1/2

x + 2 sin x

lim

x ®0

(a)

[Sep. 03, 2020 (I)] (a) 1

(b) –3/2

(a) 6

(a) is equal to e (b) is equal to 1 (c) is equal to 0 (d) does not exist Let [t] denote the greatest integer £ t. If for some l Î R - {0, 1}, lim

Let f(x) = 5 - x - 2 and g ( x) = x + 1 , xÎ R. If f(x) attains maximum value at a and g(x) attains minimum value at b,

If a is the positive root of the equation, p(x) = x2 – x – 2 = 0, then lim

2.

6.

Limit of a Function, Left Hand & Right Hand limits, Existance of Limits, Sandwitch Theorem, Evaluation of Limits when X® ¥, Limits by Factorisation, Substitution & Rationalisation

(b)

2

(c) 2 2

cot3 x - tanx is: p pö æ x ® cos x + ç 4 4 ÷ø è (b) 4 2

[April 8, 2019 (I)] (d) 4

[Jan. 12, 2019 (I)]

lim

(a) 4

[April 10, 2019 (II)] (d) 1

(c) 8 2

(d) 8

EBD_8344

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12.

Mathematics

x ®1-

lim

(

) (

tan p sin x + | x | - sin ( x [ x ]) 2

x2

)

x cot ( 4 x )

lim

x ®0 sin 2

x cot 2 ( 2 x )

is equal to :

21.

3x - 3

lim

2x - 4 - 2

(a)

3

lim x ®0

(b) exists and equals 2 2

23.

(b) e x - cos x

lim

sin 2 x

x® 0

25.

2 2

(d) does not exist For each xÎR, let [x] be greatest integer less than or equal to x. Then [Jan. 09, 2019 (II)]

x ([ x ] + | x |) sin [ x] lim is equal to: x ®0 x (c) sin 1 (a) – sin 1 (b) 1 lim

x ®0

(

lim

x®0

x

(b) -

1 2

(c)

1 4

(d)

2

1 8

1 2 2 [2015]

(d) 3

(c)

lim

x®0

27.

[Online April 10, 2015] 3 2

(d)

5 4

) is equal to: (c)

[2014] p 2

(d) 1

{

2

x - 4x + 4

x ®2

} = 5,

[Online April 11, 2014] (c) 2 (d) 3

(1 - cos 2 x)(3 + cos x) is equal to x tan 4 x 1 4

lim

x ®0

(a) – p

(d) 0

1 2

(d)

(c) 4

is equal to :

(b) p

(b)

(

sin p cos 2 x

(d)

3 2

(c)

tan ( x - 2 ) x 2 + ( k - 2 ) x - 2k

If lim

(a) -

26.

x tan 2 x - 2 x tan x equals. [Online April 15, 2018] (1 - cos 2 x) 2

(a) 1

1 2

then k is equal to: (a) 0 (b) 1

)

2 +1

1

(c) exists and equals

2

(b) 3

(a) -p

24.

(

1 16

is equal to : [Online April 8, 2017]

1

(b)

sin p cos 2 x

4 2

1

(c)

(1 - cos 2x)(3 + cosx) is equal to : x tan 4x

1

(a) exists and equals

18.

[2017]

2

22.

[Jan. 9, 2019 (I)]

y4

1 24

x ®3

(a) 2

(b) equals 0 (d) does not exist

1 + 1 + y4 - 2

y®0

(b)

[Jan. 11, 2019 (II)]

(a) 0 (b) 2 (c) 4 (d) 1 For each t Î R, let [t] be the greatest integer less than or equal to t. Then, [Jan. 10, 2019 (I)]

lim

1 4

(a) 2

(a) equals 1 (c) equals – 1

17.

equals :

( p - 2x )3

p x® 2

:

æp ö (1 - | x | + sin |1 - x |) sin ç [1 - x] ÷ 2 è ø lim x ®1+ |1 - x | [1 - x]

16.

cot x - cos x

lim

(a)

20.

2

[Jan. 11, 2019 (I)] (b) equals p (d) equals 0

(a) does not exist (c) equals p + 1

15.

19.

2 p (b) (c) (d) p 2p p 2 Let [x] denote the greatest integer less than or equal to x. Then :

x ®0

14.

[Jan. 12, 2019 (II)]

1

(a)

13.

p - 2 sin -1 x is equal to: 1- x

lim

x

2

1 2

) equals

(b) 1

(c) 1

(d) 2

[Online May 26, 2012] (c) – 1

æ x - sin x ö æ 1ö lim ç ÷ sin çè ÷ø x ø x

(d) p [Online May 7, 2012]

x®0 è

(a) equals 1 (c) does not exist

[2013]

(b) equals 0 (d) equals – 1

www.jeebooks.in M-209

Limits and Derivatives

28.

Let f : R ® [0, ¥) be such that lim f(x) exists and x® 5

( f ( x ))

lim

2

-9

x -5

x® 5

35.

=0

lim x ®0

1 - cos 2 x

29.

(c) equals 30.

1

1¥, Limits by Expansion Method

[2011] (b) equals –

2

TOPIC n Evaluation of Limits of the form

(d) 3

1

36.

2

(a)

2 3

4

1

[2010]

æ 2 öæ 2 ö 3 (b) ç ÷ç ÷ è 3 øè 9 ø

(c) 3

1 - cos(ax 2 + bx + c) ( x - a) 2

is equal to

1

æ 2 ö3 (c) ç ÷ è9ø

(d) 1

[2005]

37.

æ 2 öæ 2 ö 3 (d) ç ÷ç ÷ è 9 øè 3 ø

x + x 2 + x3 + ... + x n - n = 820, (n Î N) then the x ®1 x -1

If lim

value of n is equal to ________.

2

32.

(a)

a (a - b ) 2 2

(b) 0

(c)

-a 2 (a - b ) 2 2

(d)

33.

34.

38.

1 (a - b) 2 2

(b)

1 8

39.

(a)

(c) 0

(d)

1 32

40.

log x n - [ x] , n Î N , ([x] denotes greatest integer less [ x] x ®0 than or equal to x) [2002] (a) has value -1 (b) has value 0 (c) has value 1 (d) does not exist

(a)

e4

(b) e2

1/ x

is equal to :

(b) 2

æ 3 x2 + 2 ö lim ç 2 ÷ x ®0 è 7 x + 2 ø

[NA Sep. 02, 2020 (I)]

(d) e 2

(c) 1

1/ x 2

is equal to:

[Jan. 8, 2020 (I)]

1 e

lim ò

(b)

x t sin(10t ) dt

x® 0 0

x

(c) e2

is equal to:

[Jan. 8, 2020 (II)]

(b)

n

n

r =1

r =1

å a r + nlim å br is equal to : then nlim ®¥ ®¥ [April 12, 2019 (I)]

[2002] e3

(a) (d) 1

(d) e

1 1 1 (c) (d) 10 5 10 If a and b are the roots of the equation 375x2–25x–2=0,

(a) 0 41.

1 e2

x

(c)

[Sep. 02, 2020 (II)]

[2003]

lim

æ x2 + 5 x + 3 ö lim ç 2 ÷ x ®¥ çè x + x + 2 ÷ø

æ æp öö lim ç tan ç + x÷ ÷ è4 øø x®0 è

(a) e

é æ xö ù ê1 - tan çè 2 ÷ø ú [1 - sin x] ë û lim is p é x ® 1 + tan æ x ö ù [ p - 2 x ]3 2 ê èç 2 ø÷ úû ë

(a) ¥

( a ¹ 0) is equal to :

æ 2 ö3 (a) ç ÷ è3ø

Let a and b be the distinct roots of ax 2 + bx + c = 0 , x®a

1 - (4 x ) 3

4

3 2

(b)

then lim

(3a +

1 x) 3

[Sep. 03, 2020 (II)]

f (3 x) f (2 x) = 1 then lim = f ( x) x ®¥ f ( x)

lim

lim

x ®a

1

(a + 2 x) 3 - (3 x ) 3

(d) does not exist

2

Let f : R ® R be a positive increasing function with x ®¥

31.

(c) 2

æ 1 - cos{2( x - 2)} ö lim ç ÷ x®2 è x-2 ø (a) equals

(b) –1 (d) does not exist Limits Using L-hospital's Rule,

x®5

(b) 1

[2002]

(a) 1 (c) ero

[2011RS]

Then lim f(x) equals : (a) 0

is

2x

21 346

(b)

29 358

(c)

1 12

(d)

7 116

EBD_8344

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42.

Mathematics

Let f : R ® R be a differentiable function satisfying 1 öx

43.

æ 1 + f ( 3 + x ) – f ( 3) f ¢(3) + f ¢(2) = 0. Then lim ç ÷ is equal x ®0 è 1 + f ( 2 – x ) – f ( 2 ) ø to : [April 08, 2019 (II)] (a) 1 (b) e–1 (c) e (d) e2 For each t Î R , let [t] be the greatest integer less than or equal to t. Then [2018]

48.

æ a b ö If lim çç1+ + ÷÷ x ®¥ è x x 2 ø

49.

(a) a = 1 and b = 2

(b) a = 1, b Î R

(c) a Î R, b = 2

(d) a Î R, b Î R x ®2

is given by (a) 2 (b) –2

1

lim

x®0

2

45.

[Online April 16, 2018]

9 - (27 + x ) 3

(a) -

1 3

1 6

(b)

(c) -

(

Let p = lim 1 + tan 2 x x ®0+

)

1 2x

1 6

(d)

50.

1 3

Let f (x) be a polynomial of degree 4 having extreme values

æ f ( x) ö at x = 1 and x = 2. If lim ç 2 + 1÷ = 3 then f (– 1) is equal x ®0 è x ø to [Online April 15, 2018]

then log p is equal to : (a) [2016]

(a)

46.

1 2

51.

1 4

(c) 2

(1 - cos 2x) 2 is : x ®0 2x tan x - x tan 2x lim

(a) 2

47.

(b)

(b) -

1 2

æ a 4ö If lim ç1 + - 2 ÷ x x ø x ®¥ è

(c) –2

(d) 1

[Online April 10, 2016]

(d)

1 2

52.

2x

= e3, then 'a' is equal to : 53.

(a) 2

1 (c) 2

2 (d) 3

1 2

(b)

3 2

(c)

5 2

(d)

9 2

Let f(x) be a polynomial of degree four having extreme

é f (x) ù 1 + 2 ú = 3, then f(2) is values at x = 1 and x = 2. If xlim ®0 ê x û ë equal to : [2015] (a) 0 (b) 4 (c) – 8 (d) – 4 Let f (1) = –2 and f ¢ (x) ³ 4.2 for 1 £ x £ 6 . The possible value of f (6) lies in the interval : [April 25, 2013] (a) [15, 19) (b) (– ¥ , 12) (c) [12, 15) (d) [19, ¥ ) If f (x) = 3x10 – 7x8 + 5x6 – 21x3 + 3x2 – 7, then

[Online April 9, 2016] 3 (b) 2

(c) – 4

Derivatives of Polynomial & Trigonometric Functions, Derivative of Sum, Difference, Product & Quotient of two functions

TOPIC Đ equals.

xf (2) - 2 f ( x) x-2 [2002] (d) 3

Let f (x) = 4 and f ¢ (x) = 4. Then lim

(a) is equal to 15. (b) is equal to 120. (c) does not exist (in R). (d) is equal to 0.

44.

= e 2 , then the values of a and b, are

[2004]

æé 1 ù é 2 ù é15 ù ö lim x ç ê ú + ê ú + ... + ê ú ÷ è x x ë x ûø x ®0+ ë û ë û

(27 + x ) 3 - 3

2x

lim

f (1 - a ) - f (1)

a®0

(a) -

a3 + 3a 53 3

(b)

53 3

is

[Online May 19, 2012] (c) -

55 3

(d)

55 3

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Limits and Derivatives

1.

(b) x2 - x - 2 = 0 Þ ( x - 2)( x + 1) = 0

R.H.L. = lim

h® 0

Þ x = 2, - 1 Þ a = 2

1- h + h 1 = l+h+0 l

Given that limit exists. Hence L.H.L. = R.H.L. 1 - cos( x - x - 2) x-2 2

\ lim

x ® 2+

Þ | l - 1| = | l |

1 1 =2 and L = l 2

Þl=

æ x - x - 2ö 2 sin ç ÷ 2 è ø 2

= lim

x ® 2+

( x 2 - x - 2) ( x 2 - x - 2) 2 ´ 2( x - 2) æ x2 - x - 2ö ç ÷ 2 è ø

lim

2 sin

2.

1 2

Put

x4

Þ

1+ x 2 + x 4 -1 x

e

5.

1 + x2 + x4 - 1 x

(36)Let 3x = t2 t2 + lim

t ®3

-1

1 + x2 + x4 - 1 = t when x ® 0 Þ t ® 0 x

27 t2

- 12

1 3 t t2

t 4 - 12t 2 + 27 t -3 t ®3

= lim

(t 2 - 3)(t + 3)(t - 3) t -3 t ®3 2 = (3 – 3) (3 + 3) = 36. (a) f (x) = 5 – | x – 2 | Graph of y = f (x) y = lim

6.

et - 1 =1 t ®0 t

(b) Given lim

x®0

1- x + | x | =L l - x + [ x]

Here, L.H.L. = lim

h® 0

sin q ù é lim = 1ú êëQ q® 0 q û

4 = 2-8 = 2- k 16 ´ 64

\ L = lim

3.

= 2-k

\k = 8

æ 1+ x2 + x 4 -1 ö x xçe - 1÷ ç ÷ ç ÷ ø (b) Let L = lim è 2 4 x®0 1 + x + x -1

x®0

x4

x2 x2 2sin 2 8 = 2- k Þ lim 4 4 ´ 4 x®0 x x ´ 16 ´ 64 16 64

2

= lim

æ x2 ö 1 cos ç 4 ÷ø è

2sin 2

3

´ 1´ 3 =

æ x2 ö 1 cos ç 2 ÷ø è

x®0

æ æ x2 - x - 2ö ö ç sin ç ÷÷ 2 è ø÷ 1 ( x - 2)( x + 1) = ´ lim lim ç ( x - 2) x 2 - x - 2 ÷ x ® 2+ 2 x ® 2+ ç ç ÷ 2 è ø

=

(8)

x-2

x ® 2+

= lim

4.

1+ h + h 1 = l + h -1 l -1

(2 5)

O

2

By the graph f (x) is maximum at x = 2 \ a = 2 g (x) = | x + 1 |

x

EBD_8344

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Mathematics

Graph of y = g (x) y

9.

x 2 - ax + b =5 x -1 x ®1

(c) lim

Q limit is finite. \ 1 – a + b = 0

Þ lim

x ®1

–1

O

2

x

By the graph g (x) is minimum at x = – 1 \ b=–1 Now, xlim ®2 = lim

x ®2

7.

x® 0

10.

( x - 1)( x - 2)( x - 3) ( x - 2)( x - 4)

= lim

x ®0

x + 2sin x x 2 + 2sin x + 1 - sin 2 x - x + 1

8.

11.

4 x3 =4 x ®1 1

x® K

x3 - k 3 x2 - k 2

tan x ö æ cot 3 x ç1 è cos3 x ÷ø cos ( x + p /4 )

(1 - tan 4 x) 3 p x ® tan x cos( x + p / 4) 4

(1 + tan 2 x)(1 - tan x)(1 + tan x) p æ cos x - sin x ö x® tan 3 x ç 4 ÷ø è 2

= lim

æ0 ö ç 0 form ÷ è ø

(1 + tan 2 x )(1 + tan x )(cos x - sin x) sin 3 x æ cos x - sin x ö x® 4 ç ÷ø cos 2 x è 2

[Using L Hospital’s Rule]

=4

3x 2 =4 x ® K 2x

Þ lim

Þ

cot 3 x - tan x = lim (d) lim p p x ® cos æ x + p ö x® 4 çè ÷ø 4 4

= limp

Lt

\ lim

sin 2 x sin 2 x = lim x ù x®0 2 2 sin 2 x é 2 ê1 - cos ú 4 2û ë

= lim

æ x3 - k 3 ö x4 - 1 = lim ç 2 ÷ (a) Given, xlim x ® K çè x - k 2 ÷ø ®1 x - 1 x4 -1 x ®1 x - 1

é0ù êë 0 úû

2

é sin ù = 1ú êQ xlim x ® 0 ë û

Taking L.H.S. lim

x 2 - 2cos2 2

æ sin x ö ç ÷ .16 16 è x ø =4 2 = lim = 2 2 2 x ®0 xö æ ç sin 4 ÷ 2 2ç ÷ ç x ÷ è 4 ø

( x + 2sin x) ëé x 2 + 2sin x + 1 + sin 2 x - x + 1 ûù ( x 2 - sin 2 x ) + ( x + 2sin x)

3´ 2 =2 3

sin 2 x

x ®0

( x - 1)( x - 3) 1 = 2 x-4

é æ sin x ö ù é 2 2 ù ê1 + 2 ç x ÷ ú ë x + 2sin x + 1 + sin x - x + 1 û è øû ë lim = x® 0 æ sin 2 x ö æ æ sin x ö ö çx÷ + 1+ 2ç ÷÷ x ø çè è è x øø

=

sin 2 x x ®0 2 - 1 + cos x

(a) lim

= lim

On rationalising, = lim x ®0

æ0 ö ç 0 form ÷ (By L Hospital’s rule) è ø

Þ 2 – a = 5 Þ a = – 3 and b = – 4 Then a + b = – 3 – 4 = – 7

(b) Given limit is,

lim

2x - a =5 1

=8

( 2 )( 2 )

[Using L Hospital’s Rule] 12.

3 8 k =4 Þk= 3 2

(2)(2) 1

=

(b)

lim-

x ®1

= lim

h®0

p - 2sin -1 x = lim f (1 - h ) h®0 1- x

p - 2sin -1 (1 - h) 1 - (1 - h)

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Limits and Derivatives

p - 2sin -1 (1 - h)

= lim

h® 0

-

= lim

16.

h 1 -1

2 2sin (1 - h)

h®0

1 -1

2sin (1 - h ) 1 2 h

= lim

h® 0

=2´

1

´2´

1 - (1 - h)2

2 1 1 ´ = p p 2

y ®0 4 æ

y ç è

= lim

y® 0

æ tan(p sin x ö + 1÷ = 1 + p = lim+ ç x ®0 è x2 ø

x ®0

tan(p sin x) + ( - x + sin x) x2

=

2

17.

tan(p sin x ) + x + sin x - 2 x sin x x ®0 x2 = p+1+1–2=p Since, LHL ¹ RHL Hence, limit does not exist. 2

2

(a)

2

(d) lim x ®0

lim

x ® 0-

2

15.

2

æ 4x ö 4 .ç . =1 è tan 4 x ø÷ 22

æp ö (1 - | x | + sin (|1 - x |))sin ç [1 - x]÷ è2 ø (b) lim x ®1+ |1 - x | [1 - x]

æp ö (1 - |1 + h | + sin (|1 - 1 - h |))sin ç [1 - 1 - h]÷ è2 ø = lim h®0 |1 - 1 - h | [1 - 1 - h] æp ö (1 - 1 - h + sinh)sin ç ( -1)÷ è2 ø = lim h®0 h([0 - h])

= lim

(- h)( -1 + h)sin( -1) h

= lim (1 - h)sin(-1) = –sin1 h®0

18.

(d) Let, L = lim

x ®0

( x tan 2 x - 2 x tan x) = lim K (say) x ®0 (1 - cos 2 x)2

é 2 tan x ù xê ú - 2 x tan x 1 - (tan x )2 û ë ÞK= (1 - (1 - 2sin 2 x ))2

=

2 x tan x - [2 x tan x - 2 x tan 3 x ] 4 sin 4 x ´ (1 - tan 2 x )

=

2 x tan 3 x = 4 sin x ´ (1 - tan 2 x) 4

2 x tan 3 x æ cos 2 x - sin 2 x ö 4 sin 4 x ´ çç ÷÷ cos 2 x è ø

sin 3 x cos3 x = æ cos 2 x - sin 2 x ö 4 sin 4 x ´ çç ÷÷ cos 2 x è ø 2x

æ pö (- h + sin h) sin ç - ÷ è 2ø = lim =0 h® 0 h (-1)

)

x ([ x ] + | x |) × sin[ x] | x|

(0 - h)([0 - h]+ |0 - h |) × sin[0 - h] |0 - h |

h®0

2

æ x ö æ tan 2 x ö .ç = lim ç ÷ x ®0 è sin x ÷ ø è 2x ø

(

= lim

h®0

x cot 4 x x .tan 2 2 x = lim 2 x®0 sin 2 x .tan 4 x sin x .cot 2 x

æ y4 ç è

æ 1 + y4 –1öæ 1 + y4 + 1 ö ç ÷ç ÷ è øè ø ö 1 + 1 + y 4 + 2 ÷ æç 1 + y4 + 1ö÷ ø øè 1+ y4 -1 ö 1+ 1+ y4 + 2 ÷ 1+ y4 +1 ø

1 1 = 2 2´2 4 2

= lim-

14.

æ ö y4 ç 1+ 1+ y4 + 2÷ è ø

= lim 2

2

And LHL is, lim-

1+ 1+ y4 - 2

y® 0

2

y4

y® 0

= lim

tan(p sin x ) + ( x - 0) (a) RHL is, lim+ x ®0 x2

1+ 1+ y4 - 2

æ öæ ö 4 4 çè 1 + 1 + y - 2 ÷ø çè 1 + 1 + y + 2 ÷ø = lim y® 0 æ ö y4 ç 1+ 1+ y4 + 2÷ è ø

1 2 h 1 h(2 - h )

2

13.

(-1)

(a) L = lim

EBD_8344

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Mathematics

x 2 sin x ´ (cos x - sin 2 x) cos x

ÞK=

1 x ´ lim 2 sin x x ® 0 cos x (cos2 x - sin 2 x)

\ L = lim

x ®0

= lim

x®0

19.

(c)

23.

2

cot x(1 - sin x)

p x® 2

pö æ -8 ç x - ÷ 2ø è

3

= limp x®

æp ö 8ç - x ÷ è2 ø

x2

= lim sin = lim sin x ®0

tan t(1 - cos t)

t®0

8t 1 1 1 = .1. = 8 2 16 20.

3

= lim

t ®0

24.

tan t 1 - cos t . 8t t2

3 ( x - 3)

x ®3 2

21.

( x - 3)

´

2x - 4 + 2

(

3x + 3

)

or k + 2 = 5 25.

3 2 2 1 ´ = 2 6 2

=

x 2sin x 3 + cos x · · = lim x®0 1 tan 4 x x2

= 2 lim

= 2.4

x

· lim 3 + cos x · lim x®0

x ®0

= 2.4. x tan 4 x

1 4x 1 lim = 2.4. = 2 4 x®0 tan 4 x 4

(c) lim

x ®0

sin 2 x

x ®0

x2

22.

Þ k=3 (d) Multiply and divide by x in the given expression, we get x (1 - cos 2 x ) (3 + cos x) lim · x®0 1 tan 4 x x2 é 2 xù êëQ1 - cos 2 x = 2sin 2 úû x 2sin 2 x 3 + cos x · · 2 x®0 1 tan 4x x

2

2

=5

tan h ü ì = 1ý íQ lim î h®0 h þ

(k + x ) = 5 Þ 1 ´ xlim ®2

= lim

x®0

( x - 2) 2

tan( x - 2){x ( x - 2) + k ( x - 2)} =5 ( x - 2) ´ ( x - 2) x® 2

x (1 - cos 2 x ) (3 + cos x) lim · x®0 1 tan 4 x x2

sin 2 x

tan( x - 2){x 2 + kx - 2 x - 2 k}

=5

Þ lim

(a) Multiply and divide by x in the given expression, we get

= 2 lim

x2 - 4 x + 4

x ®2

æ (k + x )( x - 2 ) ö æ tan( x - 2) ö ´ lim ç =5 Þ xlim ç ÷ è ( x 2) ø ®2 x® 2 è ( x - 2 ) ÷ø

( 3x - 9 ) ´ ( 2x - 4 + 2 ) A = lim x ®3 {( 2x - 4 ) - 2} ´ ( 3x + 3 )

= lim

tan( x - 2){x 2 + (k - 2) x - 2k}

x®2

Rationalise Þ

(d) lim Þ lim

2x - 4 - 2

x ®3

(p sin 2 x) p sin 2 x ´ x2 p sin 2 x 2

3x - 3

(b) Let A = lim

[Q sin (p – q) = sin q]

x2

æ sin x ö = lim 1´ p ç ÷ =p x ®0 è x ø

Put

= lim

( p - p sin 2 x)

x ®0

3

p p - x = t Þ as x ® Þ t ® 0 2 2 æp öæ æ p öö cot ç - t ÷ ç 1 - sin ç - t÷ ÷ è2 øè è 2 øø = lim 3 t® 0 8t

sin éëp(1 - sin 2 x) ùû

x ®0

cot x(1 - sin x)

2

x2

x®0

= lim

1 x 1 ´ lim = 2 2 x 0 ® 2 sin x cos 0 (cos 0 - sin 0) 2

lim

sin( p cos 2 x)

(b) lim

2xe + sin x 2sin x cos x

1 3 æ x x2 1 ö 1 = 1+ = lim ç e + ÷ 2 ø cos x 2 2 x ®0 è sin x

x2

· lim ( 3 + cos x ) · lim x® 0

x® 0

1 =2 4

(

4x 1 ´ tan 4 x 4

sin p cos 2 x

26.

(d) Consider, lim

x®0

(

sin p - p sin 2 x

= lim

x®0

x

2

(

sin p sin 2 x

= lim

x®0

p sin 2 x

x

)

)

2

éëQ sin ( p - q ) = sin q ùû

) ´ ( p sin x) = p 2

x2

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Limits and Derivatives

27.

æ ( x - a )( x - b) ö 2sin 2 ç a ÷ø è 2 = lim x® a ( x - a)2

æ x - sin x ö æ 1ö (b) Consider lim çè ÷ sin çè ÷ø x ø x x ®0 é æ sin x ö ù ê x ç1 - x ÷ø ú 1 = lim ê è ú ´ lim sin æç ö÷ x x ®0 ë û x® 0 è x ø

2

= lim

x ® a ( x - a )2

´

æ ( x - a ) ( x - b) ö sin 2 ç a ÷ø è 2

é sin x ù æ 1ö ´ lim sin ç ÷ = lim ê1 è xø x úû x®0 x® 0 ë

28.

sin x ù é æ 1ö ´ lim sin ç ÷ = ê1 - lim ú è xø ë x® 0 x û x ® 0

=

æ 1ö = 0 ´ lim sin ç ÷ = 0 è xø x ®0

æp xö tan ç - ÷ .(1 - sin x ) è4 2ø (d) lim p ( p - 2 x )3 x®

(d) Given that lim Þ lim

x®5

32.

( f ( x) ) 2 - 9

x®5

x-5

2

Let x =

=0

[(f (x) )2 – 9] = 0 = lim

2

Þ é lim f ( x) ù = 9 Þ lim f (x) = 3 ë x®5 û x® 5 29.

a 2 (a - b)2 . 2

(d) lim

x®2

1 - cos{2( x - 2)} x-2

é 2 vù êëQ1 - cos q = 2sin 2 úû

x-2

x®2

L.H.L = - lim

x® 2

(at x = 2)

R.H.L = lim

x®2

(at x = 2)

2 sin( x - 2) =- 2 ( x - 2) 2 sin( x - 2) = 2 ( x - 2)

Thus L.H.L ¹ R.H.L (at x = 2)

(at x = 2)

1 - cos{2( x - 2)} does not exist. x-2 (d) Given that f(x) is a positive increasing function. \ 0 < f ( x ) < f (2 x ) < f (3 x ) Divided by f (x)

33.

( -2 y )3 y y - tan 2sin 2 2 2 é = lim 2 qù y ®0 y3 êëQ1 - cos q = 2sin 2 úû ( -8). .8 8 y 2 tan é 1 2 . sin y / 2 ù = 1 = lim ê ú 32 y ®0 32 æ y ö ë y / 2 û ç ÷ è2ø sin q tan q ù é = lim = 1ú êëQ lim q® 0 q® 0 q q û (d) Since, lim [ x ] = -1 ¹ lim [ x] = 0. So lim [ x] does x ®0-

x ®¥

x ®¥

f (2 x ) f (3 x ) £ lim f ( x) x ®¥ f ( x)

By Sandwich Theorem. f (2 x) =1 f ( x) (a) Given that ax2 + bx + c = a(x – a)(x – b) 1 - cos a ( x - a )( x - b ) lim x ®a ( x - a )2

Þ lim

34.

(a)

x ®0

æ x 2 + 5 x + 3ö æ 4x +1 ö lim ç 2 çè1 + 2 ÷ ÷ = xlim ®¥ x + x + 2ø x ®¥ è x + x + 2 ø

( 4 x +1) x

é x2 + x + 2 ù x2 + x + 2 êæ 4 x + 1 ö 4 x+1 ú ú = lim êç1 + 2 ÷ x ®¥ ê è x + x + 2ø ú ëê ûú lim

4x2 + x

= e x®¥ x

x ®¥

31.

x ®0 +

x

x® 2

Þ lim 1 £ lim

æ yö tan ç - ÷ .(1 - cos y ) è 2ø

not exist, hence the required limit does not exist.

Hence, lim

f (2 x ) f (3 x) < Þ 0 0 Also a + b + c > 0 \ – (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2] < 0

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Determinants

b2 + c 2

ab

ab

c2 + a 2

40. (c) Let D =

ac

0 2 ( a + b) a - c D = 2 ( a + b) 0 b-c

ac bc

a-c

2

a + b2

bc

Multiply C1 by a, C2 by b and C3 by c and hence divide by abc.

(

a b2 + c 2 =

1 abc

)

ab2

ac 2

(

a 2b

b c2 + a 2

a2c

b 2c

)

bc2

(

c a 2 + b2

)

Take out a, b, c common from R1, R2 and R3 respectively.

\

D=

b2 + c 2

b2

c2

a2

c2 + a 2

c2

a2

b2

a 2 + b2

abc abc

Apply C1® C1 – C2 – C3 b2

c2

c2 + a 2

c2

b2

a2 + b2

0

D = -2c 2 -2b2

b2

c2

c2 + a 2

c2

b2

a2 + b2

0

= -2 c 2 b2

1 x y 1 1 -y x \ Area of triangle = 2 x- y x+ y 1 1 = [x (x – x – y) – y (– y – x + y) + 1 (– yx – y2 – x2 + xy)] 2

1 1 [– xy + xy – y2 – x2] = (x2 + y2) 2 2 (Q Area can not be negative)

(

1 2 2 z Q z = x + iy , z = x 2 + y 2 2 43. (b) Given parabola is x2 = 8y Þ 4a = 8 Þ a = 2 To find: Area of DABC A = (– 2a, a) = (– 4, 2) B = (2a, a) = (4, 2) C = (0, 0)

=

0

b2

c2

2 = -2 c

a2

0 = – 2 [– b2 (c2a2) + c2 (– a2b2)]

b2

0

a2

= 2a2b2c2 + 2a2b2c2 = 4a2b2c2 But D = ka2b2c2 \ k = 4

-2c

On expanding, we get D = – 2 (a + b) {– 2c [2(a + b)] – (a – c) (b – c)} + (a – c) [2(a + b) (b – c)] D = 8c (a + b) (a + b) + 4 (a + b) (a – c) (b – c) = 4 (a + b) [2ac + 2bc + ab – bc – ac + c2] = 4 (a + b) [ac + bc + ab + c2] = 4(a + b) [c(a + c) + b (a + c)] = 4 (a + b) (b + c) (c + a) = a (a + b) (b + c) (c + a) Hence, a = 4 42. (b) Vertices of triangle in complex form is z, iz, z + iz In cartesian form vertices are (x, y), (– y, x) and (x – y, x + y)

=

Apply C2 – C1 and C3 – C1

-2a

b-c

A

B

(– 2a, a)

a+b a+c

)

(2a, a)

41. (c) Let D = b + a -2b b + c c + a b + c -2c C (0, 0)

Applying C1 + C3 and C2 + C3 -a + c

D=

2a + b + c a + c

2b + a + c

-b + c

b+c

a-c

b-c

-2c

Now, applying R1 + R3 and R2 + R3

-4 2 1 1 1 4 2 1 = [– 4 (2) – 2(4) + 1(0)] \ Area = 2 2 0 0 1

=

-16 = -8 » 8 sq. unit (Q area cannot be negative ) 2

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44. (b)

46. (b) Let r be the common ratio of an G.P., then a +1 a -1

a a +1 a -1 -b b + 1 b - 1 + c

c -1 c +1

a

b +1 b -1

c -1 c +1 = 0

(-1) n+ 2 a ( -1)n +1 b (-1)n c

a + 1 a -1

a +1 a -1

Þ -b b + 1 b - 1 + b + 1 b - 1 c c -1 c + 1 c -1 c + 1

( -1)

n +2

( -1)

n+1

b =0

Þ

( -1)n + 2 a

-b b + 1 b - 1 c

c -1 c +1

( -1)

n+ 2

(-1)

a -1 a + 1

( -b) b - 1 b + 1 = 0

n+2

c

c + 1 c -1

Þ

a + 1 a -1

a

Þ

a + 1 a -1

-b b + 1 b - 1 + (-1)n+ 2 -b b + 1 b - 1 = 0 c c -1 c +1 c c -1 c + 1 a

a + 1 a -1

é1 + (-1)n+ 2 ù -b b + 1 b - 1 = 0 ë û c c -1 c + 1

C2 – C1, C3 – C1 a Þ

Þ Þ Þ Þ Þ

log an +5

log an+ 6

log an+ 7

log an+8

log a1r n -1

log a1r n

log a1r n +1

= log a1r n+ 2

log a1r n +3

log a1r n + 4

log a1r n+ 5

log a1r n+ 6

log a1r n+ 7

log a1 + ( n - 1) log r

log a1 + n log r

log a1 + (n + 1) log r

= log a1 + ( n + 2) log r log a1 + (n + 3) log r log a1 + (n + 4) log r log a1 + ( n + 5) log r log a1 + (n + 6)log r log a2 + (n + 7) log r

Applying C3 ® C3 + C1, we get log a1 + ( n - 1) log r log a1 + n log r 2 [log a1 + n log r ] = log a1 + ( n + 2)log r log a1 + (n + 3) log r 2 [log a1 + ( n + 3) log r ] log a1 + ( n + 5) log r log a1 + (n + 6) log r 2 [log a1 + (n + 6) log r ]

=0

C2 Û C3 a

log an+ 2

( -1) n c

C1 Û C3 a + 1 a -1

log an+1 log an+ 4

a

(Taking transpose of second determinant)

a

log an log an+ 3

-1

1

47. (d) Applying, C1 ® C1 + C2 + C3 , we get 1 + (a 2 + b 2 + c 2 + 2) x (1 + b 2 ) x (1 + c 2 ) x 2 2 2 f (x) = 1 + (a + b + c + 2) x

1 + b2 x

(1 + c 2 ) x

1 + (a 2 + b 2 + c 2 + 2) x (1 + b 2 ) x

1 + c2 x

[Q a2 + b2 + c2 = –2] 1 (1 + b2 ) x (1 + c 2 ) x

= 1

1 + b2 x

(1 + c 2 ) x

1 (1 + b2 ) x

1 + c2 x

é1 + (-1)n+ 2 ù -b 2b + 1 2b - 1 = 0 R + R 1 3 ë û c -1 1

Applying, R2 ® R2 - R1 , R3 ® R3 - R1

0 0 a+c é1 + ( -1) n+ 2 ù -b 2b + 1 2b - 1 = 0 ë û -1 c 1

\ f (x) = 0 0

n+2

[1+ (– 1) ](a + c) (2b + 1+ 2b – 1) = 0 4b (a + c) [1 + (–1)n + 2] = 0 1 + (–1)n + 2 = 0 as b (a + c) ¹ 0 n should be an odd integer. 1

1

1

1 45. (d) Given that, D = 1 1 + x 1 1 1+ y

Applying R2 ® R2 – R1 and R3 ® R3 – R1 1 1 1 \ D = 0 x 0 = xy 0 0 y

Hence, D is divisible by both x and y

1 (1 + b2 ) x (1 + c 2 ) x 1- x

0

0

1- x

f (x) = ( x - 1)2 Hence degree = 2. 48. (d) Let r be the common ratio of an G.P., then log an

log an+1

log an+ 2

log an+ 3

log an+ 4

log an +5

log an+ 6

log an+ 7

log an+8

log a1r n-1

log a1r n

log a1r n+1

= log a1r n + 2

log a1r n+ 3

log a1r n + 4

log a1r n +5

log a1r n+ 6

log a1r n + 7

log a1 + ( n - 1) log r log a1 + n log r log a1 + (n + 1) log r = log a1 + ( n + 2) log r log a1 + (n + 3) log r log a1 + (n + 4) log r log a1 + ( n + 5) log r log a1 + (n + 6)log r log a2 + (n + 7) log r

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Determinants

Applying C3 ® C3 + C1, we get

5

Now, | B | =

log a1 + n log r 2 [log a1 + n log r ] = log a1 + ( n + 2)log r log a1 + (n + 3) log r 2 [log a1 + ( n + 3) log r ] log a1 + ( n + 5) log r log a1 + (n + 6) log r 2 [log a1 + (n + 6) log r ] log a1 + ( n - 1) log r

49. (c) Given that

b

ax + b

b

c

bx + c

ax + b bx + c

Þ Þ

0

Applying R3 ® R3 – (xR1 + R2); bx + c

= b c

0 0 -(ax + 2bx + c)

= (ax2 + 2bx + c)(b2 – ac) = (+)(–) = –ve. [Given that discriminant of ax2 + 2bx + c is –ve \ 4b2 – 4ac < 0 Þ b2 – ac < 0] 50. (d) l = ARp –1 Þ log l = log A + (p – 1) log R m = ARq –1 Þ log m = log A + (q – 1) log R n = ARr –1 Þ log n = log A + (r – 1) log R

[Q | ad A | = | A |

= 2a2 – 2a – 25

1 2a 2 - 2a - 25

2a 2 - 2a - 24 2a 2 - 2a - 25

+1 = 0

=0

(n - 1)n = 78 Þ n2 – n – 15 = 0 2 Þ n = 13 Þ

p 1 q 1 =0 r

é1 n ù é1 13ù Now, the matrix ê ú=ê ú ë 0 1 û ë0 1 û é1 13ù é1 -13ù Then, the required inverse of ê ú =ê ú ë 0 1 û ë0 1 û

55. (c) Let |A| = a, |B| = b

1

Þ |AT| = a |A–1| =

| ad A | = | A |2 = 9 n -1

1 -1

é1 1 + 2 + 3 + ... + ( n - 1) ù é1 78 ù Þê ú = ê0 1 ú 1 ë0 û ë û

log l p 1 log A + ( p - 1) log R p 1 Now, log m q 1 = log A + (q - 1) log R q 1 log n r 1 log A + (r - 1) log R r 1 Operating

51. (a)

2 3

é1 1ù é1 2 ù é1 3ù é1 4 ù é1 n - 1ù é1 78 ù = ê 0 1ú ê0 1 ú ê0 1 ú ê0 1 ú ... ê 0 1 úû êë 0 1 úû ë ûë ûë ûë û ë

2

0 C1 – (log R)C2 + (log R – log A) C3 = 0 0

0 a

Þ a = 4, – 3 Þ Sum of values = 1 54. (b)

ax + b

a b

1

Given, det. (A) + 1 = 0

=0 a

2a

Q

1 1 , |BT| = b, |B–1| = a b

|ABAT| = 8 Þ |A| |B| |AT| = 8¼(1)

Þ a.b.a = 8 Þ a2b = 8

]

Þ | A | = ±3 = l Þ | l | = 3

Q

Þ | B | = | ad A |2 = 81

|AB–1| = 8 Þ |A| |B–1| = 8 Þ a .

1 =8 b

¼(2)

From (1) & (2) -1 T

m = | (B ) | = | B

52. (a)

1 A =1

1 3

-1

1 1 = | = |B| = | B | 81 -1

a = 4, b =

2 4 = ((9 + 4) - 1(3 - 4) + 2(-1 - 3))

Then, |BA–1BT| = |B| |A–1| |BT| = b .

1 -1 3

= 13 + 1 – 8 = 6 2 |adjB| = |adj(adjA)| = |A|(n – 1) = |A|4 = (36)2 3 |C| = |3A| = 3 × 6 Hence, 53.

| adjB | 36 ´ 36 = 3 =8 |C | 3 ´6

(c) Q B = A = Þ | B | = –1

1 | A|

1 2

56.

1 1 b2 .b= = a 16 a

écos q - sin q ù (c) A = ê sin q cos q ú Þ | A| = 1 ë û T é cos q sin q ù é+ cos q - sin q ù ad(A) = ê ú = ê - sin q cos q ú sin cos + q + q ë û ë û

é cos q sin q ù Þ A–1 = ê - sin q cos q ú = B ë û

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é cos q sin q ù é cos q sin q ù B2 = ê - sin q cos q ú ê- sin q cos q ú ë ûë û

é cos 2q sin 2q ù = ê- sin 2q cos 2q ú ë û

é36 – 32 ù Þ A2 = ê 4 úû ë0

é cos3q sin 3q ù Þ B3 = ê - sin3q cos3q ú ë û

For k = – 6

é cos(50q) sin(50q) ù Þ A–50 = B50 = ê- sin(50q) cos(50q) ú ë û

é 3 ê 2 ( A-50 ) p ê q= = ê 1 12 êë 2

1 ù ú 2 ú 3ú ú 2 û

é 3ù pö p æ 50 p ö æ ÷ = cos ç 4 p + ÷ = cos = êQ cos ç ú 12 6 6 2 è ø è ø ë û

57.

é6 – 4ù A= ê ú ... From (1) ë0 2 û

é1 2ù (d) Since A . ê ú is a scalar matrix and |3A| = 108 ë 0 3û ék 0 ù suppose the scalar matrix is ê ú ë0 k û é1 2ù é k 0 ù \A. ê ú ú =ê ë 0 3û ë 0 k û é k 0 ù é1 2 ù ÞA= ê ú ê ú ë 0 k û ë0 3 û

–1

1 é k 0 ù é 3 – 2ù ê úê ú 3 ë0 k û ë0 1 û

2ù é 1 – ú ék 0 ù ê 3 Þ A=ê ú úê k 0 1 ú ë û ê0 3 úû ëê

58. (a) We have (A – 3I) (A – 5I) = O Þ A2 – 8A + 15I = O Multiplying both sides by A – 1, we get; A – 1 A . A – 8A – 1 A + 15A – 1 I = A – 1 O Þ A – 8I + 15A – 1 = O A + 15A – 1 = 8I

A 15 A -1 + = 4I 2 2

1 15 16 + = =8 2 2 2 é 2 -3ù 59. (c) We have A = ê ú ë -4 1 û \

a+b=

é 48 -27 ù ê -36 39 ú ë û

é 48 -27 ù é 24 -36 ù \ 3A2 + 12A = ê -36 39 ú + ê -48 12 ú ë û ë û é 72 -63ù =ê ú ë -84 51 û é 51 63ù ad (3A2 + 12A) = ê ú ë84 72 û

60. (b) 61. (d) Given that A(ad A) = A AT

2 ù é êk – 3 k ú Þ A=ê ú k ú ... (1) ê0 êë 3 úû

–1

Pre-multiply by A

both side, we get

Þ A A (ad A) = A A A –1

–1

T

T

Q |3 A| = 108 Þ 108 =

é36 – 32 ù Þ A2 = ê 4 úû ë0

é 16 -9 ù Þ A2 = ê -12 13 ú Þ 3A2 = ë û é 24 -36 ù Also 12A = ê -48 12 ú ë û

[ \ AB = C Þ ABB–1 = CB–1 Þ A = CB–1] Þ A=

é– 6 4 ù ÞA=ê ú .... From (1) ë 0 – 2û

3k

– 2k

0

k

Þ 3k2 = 108 Þ k2 = 36 For k = 6

é 2 b ù é 5a 3 ù Þê ú=ê ú ë -3 5a û ë- b 2 û

Þk =±6

2 Þ a = and b = 3 5

Þ 5a + b = 5

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Determinants

62. (a) A2 – 5A = – 7I AAA–1 – 5AA–1 = –7I A–1 AI – 5I = –7A–1 A – 5I = –7A–1 1 A–1 = (5I – A) 7 A3 – 2A2 – 3A + I = A (5A – 7I) – 2A2 – 3A + I = 5A2 – 7A – 2A2 – 3A + I = 3A2 – 10A + I = 3 (5A – 7I) – 10A + I = 5A – 20I = 5(A – 4I) 63. (a) |5. ad A | = 5 Þ 53. |A|3–1 = 5 Þ 125 |A|2 = 5 Þ |A| = ±

1 5

64. (d) BB' = B ( A-1 A ') ' = B ( A ') '( A-1 ) ' = BA (A–1)' = ( A -1 A ')( A( A -1) ') = A–1A .A'.(A–1)' {as AA' = A'A} –1 2 = I(A A)' = I × I = I = I é0 0 1ù é1 2 3ù 65. (a) Given A ê0 2 3ú = êê1 0 0úú ê ú êë0 1 0úû ëê0 1 1ûú Applying C1 « C3 é3 A ê3 ê ëê1 Again

é1 2 1ù ê ú 2 0 = ê0 ú êë0 1 0ûú Applying C2

0 0ù 0 1ú ú 1 0úû « C3

é3 1 2ù é1 ê3 0 2ú ê0 ú = ê A ê êë1 0 1úû êë0 pre-multiplying both

0 0ù 1 0ú ú 0 1úû

sides by A–1

é3 1 2ù A–1 A êê3 0 2úú = A–1 ëê1 0 1úû

é 1 0 0ù ê 0 1 0ú ê ú êë0 0 1úû

é3 1 2ù ê3 0 2ú ú = A–1 I = A–1 I ê ëê1 0 1úû (Q A–1A = I and I = Identity matrix) é3 1 2ù ê3 0 2ú ú Hence, A = ê ëê1 0 1úû 66. (b) | P | = 1(12 – 12) – a(4 – 6) + 3(4 – 6) = 2a – 6 Now, ad A = P Þ | ad A | = | P | Þ | A |2 = | P | Þ | P | = 16 Þ 2a – 6 = 16 Þ a = 11 –1

67. (c) Given that P3 = Q3 and P2Q = Q2P Subtracting (1) and (2), we get P3 – P2Q = Q3 – Q2P Þ P2 (P–Q) + Q2 (P – Q) = 0 Þ (P2 + Q2) (P–Q) = 0 2 2 Q P ¹ Q, \ P + Q = 0 Hence |P2 + Q2| = 0

...(1) ...(2)

æ1ö æ0ö ç ÷ ç ÷ 68. (d) Let Au1 = ç 0 ÷ and Au2 = ç 1 ÷ ç0÷ ç0÷ è ø è ø æ 1ö æ0ö ç ÷ ç ÷ Then, Au1 + Au2 = ç 0 ÷ + ç 1 ÷ ç0÷ ç0÷ è ø è ø

Þ

æ 1ö ç ÷ A ( u1 + u2 ) = ç 1 ÷ ç0÷ è ø

...(1)

æ1 0 0ö ç ÷ A = Given that ç2 1 0÷ ç 3 2 1÷ è ø Þ |A| = 1(1) – 0 (2) + 0 (4 – 3) = 1 C11 = 1 C21 = 0 C31 = 0 C12 = –2 C22 = 1 C32 = 0 C13 = 1 C23 = –2 C33 = 1 é 1 0 0ù \ adjA = êê -2 0 0 úú êë 1 -2 1 úû We know,

A-1 =

Þ A -1 = adj ( A )

(Q

A = 1)

Now, from equation (1), we have u1 + u2 = A

æ 1ö ÷ ç 1÷ ç0÷ è ø

-1 ç

é 1 0 0ù æ 1 ö é 1 ù ç ÷ = êê -2 1 0 úú ç 1 ÷ = êê -1úú êë 1 -2 1úû èç 0 ø÷ êë -1úû é0 0 ê 69. (c) A = ê 0 b êë d e

aù ú c ú , |A| = – abd ¹ 0 f úû

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c11 = + (bf – ce), c12 = – (– cd) = cd, c13 = + (– bd) = –bd c21 = –(–ea) = ae, c22 = + (–ad) = –ad, c23 = –(0) = 0 c31 = + (–ab) = – ab, c32 = – (0) = 0, c33 = 0 é (bf - ce) ae ê cd -ad Ad A = ê êë -bd 0

A– 1 =

1 1 (ad A) = | A| abd

-ab ù ú 0 ú 0 úû ébf - ce ae ê -ad ê cd êë -bd 0

-ab ù ú 0 ú 0 úû

Now A–1 = AT ébf - ce ae 1 ê cd -ad -abd ê êë -bd 0

ébf - ce ae ê - ad Þ ê cd 0 ëê -bd

é0 0 d ù -ab ù ê ú ú 0 ú = ê0 b e ú êë a c f úû 0 úû

é 0 - ab ù ê ú 0 ú = ê 0 ê 0 ûú ê -a 2bd ë

0 -ab 2d - abcd

- abd 2 ù ú -abde ú ú - abdf ú û

\ bf – ce = ae = cd = 0 ...(i) 2 2 2 abd = ab, ab d = ad, a bd = bd ...(ii) abde = abcd = abdf = 0 ...(iii) From (ii), (abd 2 ). (ab2d). (a2bd) = ab. ad. bd Þ (abd)4 – (abd)2 = 0 Þ (abd)2 [(abd)2 – 1] = 0 ...(iv) Q abd ¹ 0 , \ abd = ±1 From (iii) and (iv), e=c=f=0 ...(v) From (i) and (v), bf = ae = cd = 0 ...(vi) From (iv), (v) and (vi), it is clear that a, b, d can be any non- ero integer such that abd = ± 1 But it is only possible, if a = b = d = ± 1 Hence, there are 2 choices for each a, b and d. there fore, there are 2×2×2 choices for a, b and d. Hence number of required matrices = 2×2×2=(2)3 éa 0ù 70. (a) Let A and B be real matrices such that A = ê ú ë 0 bû é0 g ù and B = ê ú ëd 0û

é 0 gb ù and BA = ê ú ë da 0 û Statement - 1 :

é 0 g ( a - b)ù AB - BA = ê ú d b a 0 û ( ) ë

AB - BA = ( a - b) gd ¹ 0 2

\ AB – BA is always an invertible matrix. Hence, statement - 1 is true. But AB – BA can be identity matrix if g = – d or d = – g So, statement - 2 is false. 71. (b) For reflexive

é0 0 d ù ê ú T A = ê0 b e ú êë a c f úû

Þ

é 0 ag ù Now, AB = ê ú ëbd 0 û

A = P -1 AP is true, For P = I, which is an invertible matrix.

( A, A) Î R \ R is reflexive. For symmetry As ( A, B) ÎR for matrix P

A = P-1 BP Þ PAP-1 = B Þ B = PAP -1 Þ

( )

B = P –1

-1

A (P–1)

\ (B, A) Î R for matrix P -1 \ R is symmetric. For transitivity

A = P-1 BP and B = P–1CP Þ Þ

Þ

(

)

A = P –1 P -1CP P

( ) CP A = (P ) C (P ) A = P -1

2 -1

2

2

2

2 \ (A, C) ÎR for matrix P R is transitive. \ So R is equivalence. So, statement-1 is true. We know that if A and B are two invertible matrices of order n, then (AB)–1 = B–1 A–1 So, statement-2 is true. 72. (a) We know that if A is square matrix of order n then adj (adj A) = | A | n–2 A. = | A |0 A = A

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Determinants

Also | adj A | = | A | n–1 = | A | 2–1 = | A | \ Both the statements are true but statement-2 is not a correct explanation for statement-1 . 73. (c) Given that all entries of square matrix A are integers, therefore all cofactors should also be integers. If det A = ± 1 then A–1 exists. Also all entries of A–1 are integers. 74. (d) Given that A2 – A + I = 0 Pre-multiply by A–1 both side, we get A-1 A2 - A-1 A + A-1.I = A-1.0 -1 Þ A - I + A-1 = 0 or A = I - A .

é 4 2 2ù 75. (a) Given that 10 B = ê -5 0 a ú ê ú êë 1 -2 3 úû é 4 2 2ù 1 Þ B = ê -5 0 a ú ú 10 ê ëê 1 -2 3 úû

Given that B = A-1 Þ AB = I

Þ

é 1 -1 1 ù é 4 2 2 ù é 1 0 0 ù 1 ê úê ú ê ú 2 1 -3ú ê -5 0 a ú = ê0 1 0 ú 10 ê êë 1 1 1 úû êë 1 -2 3 úû êë0 0 1 úû

é10 0 5 - 2 ù é1 0 0ù 1 ê 0 10 -5 + a úú = êê0 1 0úú Þ 10 ê êë 0 0 5 + a úû êë0 0 1 úû

Þ

5- a =0Þa=5 10

é 0 0 -1ù ê ú 76. (a) Given that A = ê 0 -1 0 ú êë -1 0 0 úû

clearly A ¹ 0. Also |A| = -1 ¹ 0 é -1 0 0 ù ê 0 -1 0 ú ¹ A -1 I = ( 1) \ A exists, further ê ú êë 0 0 -1ûú é 0 0 -1ù é 0 0 -1ù Also A = ê 0 -1 0 ú ê 0 -1 0 ú ê úê ú êë -1 0 0 úû êë -1 0 0 úû 2

é1 0 0ù = ê0 1 0ú = I ê ú êë0 0 1úû

1 1 1

77. (c)

D= 1 2 3 =0Þl =5 1 3 l 2 1 1

Dx = 5 2 3 = 0 Þ m = 8 m 3 5 78. (3.00) For non- ero solution, D = 0 l - 1 3l + 1 Þ l - 1 4l - 2 2

2l l+3

=0

3l + 1 3(l - 1)

Þ 6l 3 - 36l 2 + 54l = 0

Þ 6l[l 2 - 6l + 9] = 0 Þ l = 0, l = 3 [Distinct values] Then, the sum of distinct values of l = 0 + 3 = 3. 2 -4 l 79. (a) Q 1 -6 1 = 0 Þ 3l 2 - 7l - 12 = 0 l -10 4 2 Þ l = 3 or 3 1 -4 l D1 = 2 -6 1 = 2(3 - l ) 3 -10 4 2 \ When l = - , D1 ¹ 0. 3 2 Hence, equations will be inconsistent when l = - . 3 80. (a) Since, system of linear equations has non- ero solution \D = 0 1 1 3

Þ 1 3 k2 = 0 3 1

3

Þ 1(9 - k 2 ) - 1(3 - 3k 2 ) + 3(1 - 9) = 0 Þ 9 - k 2 - 3 + 3k 2 - 24 = 0 Þ 2k 2 = 18 Þ k 2 = 9, k = ±3 So, equations are x + y + 3z = 0 x + 3y + 9z = 0 3x + y + 3z = 0 Now, from equation (i) – (ii), y -2 y - 6 z = 0 Þ y = -3z Þ = -3 z Now, from equation (i) – (iii),

...(i) ...(ii) ...(iii) ...(iv)

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-2 x = 0 Þ x = 0

Put z = 0 in equation (i), we get x = 2y

Q15 £ x2 + y 2 + z 2 £ 150

y = 0 - 3 = -3 z

So, x +

Þ 15 £ 4 y 2 + y 2 £ 150

81. (5.00) For infinitely many solutions,

[Q x = 2 y, z = 0]

D = D1 = D 2 = D 3 = 0

Þ 3 £ y 2 £ 30

1 -2 3 D= 2

Þ y = ±2, ± 3, ± 4, ± 5 Þ 8 solutions.

1 =0

1

1 -7 a

2 -1 2

Þ (a + 7) - 2(1 - 2a) + 3(-15) = 0

85. (d)

D = 1 -2 l = -(l - 1)(2l + 1)

Þ a =8

1

1 -2 D3 = 2

9

D1 = -4 -2 l = -2(l 2 + 6l - 4)

1 -7 24

Þ (24 + 7b) - 2(b - 48) + 9(-15) = 0 Þ b =3

82.

\ a - b = 5. (b) Given that Ax = b has solutions x1, x2, x3 and b is equal to b1, b2 and b3

\ x1 + y1 + z1 = 1 Þ 2 y1 + z1 = 2 Þ z1 = 2 Determinant of coefficient matrix

1

2 4 -1 = 0

[Q Equation has many solutions]

l

But given that x2 + y 2 + z 2 = 1

7

\k = ±

9 Þ -15 + 6 + 2l = 0 Þ l = 2

1 1 \ DZ = 2 4

86. (d) Q | P | = 1( -3 + 36) - 2(2 + 4) + 1( -18 - 3) = 0 Given that PX = 0 \ System of equations

Let z = k ÎR and solve above equations, we get 11k 2k x= , y= ,z=k 7 7

0 0 1

3 2

1 and D1 ¹ 0 2 1ü ì Hence, S = í1, - ý 2þ î \ D = 0 Þ l = 1, -

and x + 9 y - z = 0 has infinitely many solution.

| A| = 0 2 1 = 2

1 1

4 l 1 For no solution D = 0 and at least one of D1, D2 and D3 is non- ero.

x + 2 y + z = 0 ; 2x - 3 y + 4z = 0

1 1 1

83. (d)

1

-1 2

2

b =0

1

l

2 6 =0 Þ m = 5

3 2 2m

\ 2l + m = 14. 84. (8) The given system of equations

x - 2 y + 5z = 0

...(i)

-2 x + 4 y + z = 0

...(ii)

-7 x + 14 y + 9 z = 0

...(iii)

From equation, 2 × (i) + (ii) Þ z = 0

174 \ Two solutions only. 87. (c) The given system of linear equations 7x + 6y – 2z = 0 3x + 4y + 2z = 0 x – 2y – 6z = 0 Now, determinant of coefficient matrix 7

6

-2

D=3

4

2

1 -2 -6

= 7(– 20) – 6(–20) – 2(–10) = – 140 + 120 + 20 = 0 So, there are infinite non-trivial solutions.

...(i) ...(ii) ...(iii)

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Determinants

From eqn. (i) + 3 × (iii); we get 10x – 20z = 0 Þ x = 2z Hence, there are infinitely many solutions (x, y, z) satisfying x = 2z. 88. (a) From the given linear equation, we get 1 2 3 D = 3 4 5 ( R3 ® R3 - 2 R2 + 3R3 ) 4 4 4 1 2 3 = 3 4 5 =0 0 0 0

Now, let P3 = 4x + 4y + 4z – d = 0. If the system has solutions it will have infinite solution. So, P3 º aP1 + bP2 Hence, 3a + b = 4 and 4a + 2b = 4 Þ a = 2 and b = – 2 So, for infinite solution 2m – 2 = d Þ For 2m ¹ d + 2 system is inconsistent l

89. (c)

2 2

D = 2l 3 5 l 6

4

D = l2 + 6l – 16 D = (l + 8) (2 – l) For no solutions, D = 0 Þ l = – 8, 2 when l = 2 5

2 2

D1 = 8

3 5

Þ

1 1 1 , , in A.P. a b c 91. (13) x + y + z = 6 ...(i) x + 2y + 3z = 10 ...(ii) 3x + 2y + lz = µ ...(iii) From (i) and (ii), If z = 0 Þ x + y = 6 and x + 2y = 10 Þ y = 4, x = 2 (2, 4, 0) If y = 0 Þ x + z = 6 and x + 3z = 10 Þ z = 2 and x = 4 (4, 0, 2) So, 3x + 2y + lz = m, must pass through (2, 4, 0) and (4, 0, 2) So, 6 + 8 = m Þ m = 14 and 12 + 2l = m 12 + 2l = 14 Þ l = 1 So, m – l2 = 14 –1 = 13 92. (d) Given system of linear equations: x + y + z = 5;

Þ

x + 2y + 2z = 6 and x + 3y + lz = m have infinite solution. \ D = 0, Dx = Dy = Dz = 0 1 1 1

Now, D =

Dy =

There exist no solutions for l = 2 90. (a) For non- ero solution

1 6 2 1 m 3

=0Þ

1 1

b =0

2 4c

c

1

5

1

0

1

1

0 m-5 2

Here,

1

D = 4 l -l = 0 -4

3 2

C1 ® C1 – C2 & C2 ® C2 – C3

1 2a a 1 3b b = 0 c

Þ (3bc – 4bc) – (2ac – 4ac) + (2ab – 3ab) = 0 Þ – bc + 2ac – ab = 0 Þ ab + bc + 2ac

=0

Þ 1 (2 – m + 5) = 0 Þ m = 7 \ l + m = 10 93. (d) Q system of equations has infinitely many solutions. \ D = D1 = D2 = D3 = 0

2 2a a 2 3b

=0

1 5 1

= 5[18 – 10] – 2 [48 – 50] + 2 (16 – 30] = 40 + 4 – 28 ¹ 0

1 4c

1 2 2 1 3 l

Þ 1 (2l – 6) – 1 (l – 2) + 1 (3 – 2) = 0 Þ 2l – 6 – l + 2 + 1 = 0 Þ l = 3

10 2 6

Þ

2 1 1 = + b a c

0

0

1

Þ 4 - l 2l -l = 0 1

6

-4

Þl=3

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Mathematics

6

Now, for l = 3,

1

1

D1 = l - 2 l -l = 0 -5 1

6

2

= 1 (– k + 1) + 2 (– 2k – 3) + k (– 2 – 3) = – k + 1 – 4k – 6 – 5k = – 10k – 5 = – 5(2k + 1)

-4

D1 =

1

3

-5

1 1

-4

D2 = 1 1 6 For l = 3, D3 = 4 l l - 2 = 0 3 2 -5

2

3

-1

Þ D =0Þ 1

k

-2 = 0 Þ k =

2 -1

\ equations are

1

9 2

2x + 3y – z = 0 2x – y + z = 0 2x + 9y – 4z = 0

...(i) ...(ii) ...(iii)

By (i) – (ii), 2y = z \ z = – 4x and 2x + y = 0 -1 1 9 1 x y z + + +k = + -4+ = y z x 2 2 2 2

95. (b) If the system of equations has non-trivial solutions, then the determinant of coefficient matrix is ero

c

-1

Þ Þ Þ Þ

c

3 3 -k

-2 k 1

3 -1 -k

-2 1

2 1

= 0, D3 =

2

3 -1 3

=0

10 - 3l 2l and y = – 10 5 \ (x, y) must lie on line 4x – 3y – 4 = 0 97. (a) Q The system of linear equations has a unique solution. \ D¹0 Let z = l ¹ 0 then x =

1+ a

D= a a

b

1

1+ b 1 ¹ 0 b

2

1+ a + b +1

Þ

b

1

a +1+ b +1 b + 1 1 ¹ 0 a+b+2

b

2

b

1

[C1 ® C1 + C2 + C3]

(a + b + 2) 1 b + 1 1 ¹ 0 1

b

2

1 b 1

Þ

(a + b + 2) 0 1 0 ¹ 0 0 0 1

R2 ® R2 - R1 R3 ® R3 - R1

Þ (a + b + 2) 1(1) ¹ 0 Þ a+b+2¹0

1 Hence, the greatest value of c is for which the system 2 of linear equations has non-trivial solution. 96. (b) Given system of linear equations, x – 2y + kz = 1, 2x + y + z = 2, 3x – y – kz = 3

D=

1

1 2 \ System of equation has infinite many solutions.

Þ

=0

1 Þ c = – 1 or 2

2 1

k

2 2 1

1

(1 – c2) + c (– c – c2) – c (c2 + c) = 0 (1 + c) (1 – c) – 2c2 (1 + c) = 0 (1 + c) (1 – c – 2c2) = 0 (1 + c)2 (1 – 2c) = 0

1

= – 5 (2k + 1)

Þ –5 (2k + 1) = 0 Þ k = -

1 -c -c c

1

Qz ¹ 0 Þ D = 0

\ for l = 3, system of equations has infinitely many solutions. 94. (b) Given system of equations has a non-trivial solution.

c -1

2 1

3 -1 -k

4 l - 2 -l = 0

For l = 3, D2 =

\

-2 k

1

Q

Ordered pair (2, 4) satisfies this condition

\

a = 2 and b = 4.

98. (a) Consider the given system of linear equations x(1 –l) – 2y – 2z = 0 x + (2 – l)y + z = 0 –x – y – lz= 0 Now, for a non-trivial solution, the determinant of coefficient matrix is ero.

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Determinants

1- l

-2

-2

1 5

1

1

2- l

1

1 9

3

-1

-1

-l

D2 =

=0

Þ (1 – l)3 = 0 l=1 99. (b) Q System of equations has more than one solution \ D = D1 = D2 = D3 = 0 for infinite solution 2

a

D1 =

3

b -1 5 c -3 2

= a(13) + 2(5c – 2b) + 3(–3b + c)

= 13a – 13b + 13c = 0 b–c–a=0

100. (b) Since, the system of linear equations has, non-trivial solution then determinant of coefficient matrix = 0 sin 3q cos 2q 2

i.e.,

= 9a - 3b - 5 (a – 3) + 1(b - 9)

= 9a - 3b - 5a + 15 + b - 9 = 4a - 2b + 6 1 1 5

D3 =

1 2 9 1 3 b

= 2b – 27 – b + 9 + 5 = b – 13

Since, the system of equations has infinite many solutions. Hence, D1 = D 2 = D 3 = D = 0 Þ a = 5, b = 13 Þ b – a = 8 102. (c) Consider the system of linear equations

i.e, a – b + c = 0 or

1 b a

1

3

7

-1

4

7

=0

sin3q(21– 28) – cos2q(7 + 7) + 2 (4 + 3) = 0 sin3q + 2cos2q – 2 = 0

x – 4y + 7z = g ...(i) 3y – 5z = h ...(ii) –2x + 5y – 9z = k ...(iii) Multiply equation (i) by 2 and add equation (i), equation (ii) and equation (iii) Þ 0 = 2g + h + k. \ 2g + h + k = 0 then system of equation is consistent. 103. (a) For non ero solution of the system of linear equations; 1 k

3

4sin3q + 4sin 2q – 3sinq = 0

3 k

-2 = 0

sinq (4sin2q + 4sinq – 3) = 0

2 4

-3

3sinq – 4sin3q + 2 – 4sin 2q – 2 = 0

sinq (4sin2q + 6sinq – 2sinq – 3) = 0 sinq [2sinq (2sinq – 1) + 3 (2sinq – 1)] = 0 sinq (2sinq – 1) (2sinq + 3) = 0 sin q = 0, sin q =

1 2

3ö æ çèQ sin q ¹ - ÷ø 2

p 5p , 6 6 Hence, for two values of q, system of equations has nontrivial solution

q=

1 1

101. (b) D = 1 2

1 3 = 2a – 9 – a + 3 + 1 = a – 5

1 3 a 5 1

1

D1 = 9 2 3 = 5(2a – 9) – 1(9a – 3b) + (27 – 2b) b 3 a = 10a - 45 - 9a + 3b + 27 - 2b = a + b -18

Þ k = 11 Now equations become x + 11y + 3 = 0 3x + 11y – 2 = 0 2x + 4y – 3 = 0 Adding equations (1) & (3) we get 3x + 15y = 0 Þ x = –5y Now put x = –5y in equation (1), we get –5y + 11y + 3 = 0 Þ = –2y

\

x

( -5 y)( -2 y)

= 10 y2 104. (c) Here, the equations are; (k + 2) x + 10y = k & kx + (k + 3)y = k – 1. These equations can be written in the form of Ax = B as y

2

=

...(1) ...(2) ...(3)

é k + 2 10 ù é x ù é k ù = ê k k + 3úû êë y úû êë k – 1úû ë

For the system to have no solution |A| = 0

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Mathematics

é k + 2 10 ù = 0 Þ (k + 2) (k + 3) – k × 10 = 0 Þ ê k + 3úû ë k Þ k2 – 5k + 6 = (k – 2) (k – 3) = 0 \ k = 2, 3 For k = 2, equations become: 4x + 10 y = 2 & 2x + 5y = 1 & hence infinite number of solutions. For k = 3, equations becomes; 5x + 10y = 3 3x + 6y = 2 & hence no solution. \ required number of values of k is 1 105. (b) The system of linear equations is: x+y+z=2 2x + y – z = 3 3x + 2y + kz = 4 As, system has unique solution.

1 1 1 So, 2 1 –1 ¹ 0 3 2 k

Þ k + 2 – (2k + 3) + 1 ¹ 0 Þk¹0 Hence, k Î R – {0} º S 106. (d) As the system of equations has no solution then D should be ero and at least one of D1, D2 and D3 should not be ero. 1 a 1

\

D= 1 2 2 =0 1 5 3 Þ –a–1=0 Þ a=–1

2

\

-l

4

4 l

2 =0

l

2

2

Þ l + 4l – 40 = 0 l has only 1 real root. 109. (b) For non-trivial solution, 3

l

1

-1

l -1 -1 = 0 1

-l

1

Þ -l(l + 1)(l - 1) = 0 Þ l = 0, +1, –1 2x1 - 2x 2 + x 3 = lx1ü ï 2x1 - 3x 2 + 2x 3 = lx 2 ý - x1 + 2x 2 = lx3 ïþ

110. (a)

Þ (2 – l)x1 – 2x2 + x3 = 0 2x1 – (3 + l) x2 + 2x3 = 0 – x1 + 2x2 – lx3 = 0 For non-trivial solution, D=0 2-l

-2

2

-(3 + l )

-1

2

i.e.

1 2 =0 -l

Þ (2 – l) [l(3 + l) – 4] + 2[–2l + 2] + 1[4 – (3 + l)] = 0 Þ l3 + l2 – 5l + 3 = 0 Þ l = 1, 1, 3 Hence l has 2 values. 111. (b) Given system of equations can be written as (a - 1) x - y - z = 0 - x + (b - 1) y - z = 0 - x - y + (c - 1) z = 0 For non-trivial solution, we have

1 3 1 D2 = 1 6 2 ¹ 0 1 b 3

Þ b¹9 1 1 1 107. (a) D = 1 a 1 = 0 a b 1

a -1

-1

-1

-1

b -1

-1

-1

-1

c -1

=0

R2 ® R2 – R3 a - 1 -1

Þ 1 [a – b] – 1 [1 – a] + 1 [b – a ] = 0 Þ (a – 1)2 = 0 Þa=1 For a = 1, First two equations are identical i.e., x + y + z = 1 To have no solution with x + by + z = 0 b=1 So b = {1} Þ It is singleton set. 108. (b) Since the given system of linear equations has infinitely many solutions. 2

-1 -c

0

b

-1

-1 c - 1

=0

C2 ® C2 – C3 a -1

0

-1

0

b+c

-c

-1

-c

c -1

=0

Apply R3 ® R3 – R1

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Determinants

a -1 0 -a

0

-1

b + c -c -c

=0

c

Þ (a - 1)[bc + c 2 - c 2 ] - 1[a(b + c)] = 0 Þ (a - 1)[bc ] - ab - ac = 0 Þ abc - bc - ab - ac = 0 Þ ab + bc + ca = abc 112. (b) Since, system of equations have no solution k +1 8 4k \ = ¹ (Q System has no solution) k k + 3 3k - 1 Þ k2 + 4k + 3 = 8k Þ k2– 4k + 3 = 0 Þ k = 1, 3

If k = 1 then

8 4.1 ¹ which is false 1+ 3 2

8 4.3 and if k = 3 then ¹ which is true, therefore k = 3 6 9 -1 Hence for only one value of k. System has no solution. 113. (b) Given system of equations is homogeneous which is x + ay = 0 y + az = 0 z + ax = 0 It can be written in matrix form as

æ1 a 0ö A =ç0 1 a÷ ça 0 1÷ è ø

Now, | A | = [1 – a(– a2)] = 1 + a3 ¹ 0 So, system has only trivial solution. Now, | A | = 0 only when a = – 1 So, system of equations has infinitely many solutions which is not possible because it is given that system has a unique solution. Hence set of all real values of ‘a’ is R – {– 1}. 114. (c)

p æ pö in ç 0, ÷ 4 è 2ø \ D1 = 2(sin a) × 0 = 0,

a=

1 sin a cos a D1 = 1 cos a sin a 1 - sin a cos a

0 sin a - cos a cos a - sin a = 0 cos a + sin a sin a - cos a 1 - sin a cos a

= (sin a – cos a)2 – (cos2 a – sin2 a) = sin2 a + cos2 a – 2 sin a . cos a – cos2 a + sin2 a = 2 sin2 a – 2 sin a . cos a = 2 sin a (sin a – cos a) Now, sin a – cos a = 0 for only

æ pö since value of sin a is finite for a Î ç 0, ÷ è 2ø Hence non-trivivial solution for only one value of a in æ pö ç 0, ÷ è 2ø

cos a sin a cos a sin a cos a sin a = 0 cos a - sin a - cos a

Þ

0 sin a cos a 0 cos a sin a = 0 2cos a - sin a - cos a

Þ 2 cos a (sin2 a – cos2 a) = 0 \ cos a = 0 or sin2 a – cos2 a = 0 æ pö But cos a = 0 not possible for any value of a Î ç 0, ÷ è 2ø \ sin2 a – cos2 a = 0 Þ sin a = – cos a, which is also not æ pö possible for any value of a Î ç 0, ÷ è 2ø Hence, there is no solution. 115. (d) Given system of equations can be written in matrix form as AX = B where

æ1 2 3ö æ6ö A = ç 1 3 5 ÷ and B = ç 9 ÷ ç2 5 a÷ çb÷ è ø è ø Since, system is consistent and has infinitely many solutions \ (ad. A) B = 0 æ 3a - 25 15 - 2a 1 ö æ 6 ö æ 0 ö a - 6 -2 ÷ ç 9 ÷ = ç 0 ÷ Þ ç 10 - a ç -1 1 ÷ø çè b ÷ø çè 0 ÷ø -1 è Þ – 6 – 9 + b = 0 Þ b = 15 and 6(10 – a) + 9(a – 6) – 2(b) = 0 Þ 60 – 6a + 9a – 54 – 30 = 0 Þ 3a = 24 Þ a = 8 Hence, a = 8, b = 15. 116. (a) Given system of equations is x + ky + 3z = 0 3x + ky – 2z = 0 2x + 3y – 4z = 0 Since, system has non-trivial solution

\

1 k

3

3 k

-2 = 0

2 3 -4

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Mathematics

Þ 1 (– 4k + 6) – k(– 12 + 4) + 3 (9 – 2k) = 0 Þ 4k + 33 – 6k = 0 Þ k =

33 2

Hence, statement - 1 is false. Statement-2 is the property. It is a true statement. 117. (d) Given system of equations is x+y+z=6 x + 2y + 3z = 10 x + 2y + lz = 0 It has unique solution. \

1 1 1 1 2 3 ¹0 1 2 l

Þ 1(2l – 6) – 1 (l – 3) + 1(2 – 2) ¹ 0 Þ 2l – 6 – l + 3 ¹ 0 Þ l – 3 ¹ 0 Þ l ¹ 3 118. (a)

x - ky + z = 0

kx + 3 y - kz = 0 3x + y - z = 0 The given that system of equations have trivial solution, 1 -k 1 \ k 3 -k ¹ 0 3 1 -1 Þ 1(-3 + k ) + k (-k + 3k ) + 1(k - 9) ¹ 0 Þ k - 3 + 2k 2 + k - 9 ¹ 0 Þ k 2 + k - 6 ¹ 0 Þ k = -3, k ¹ 2 So, the equation will have only trivial solution, when k Î R – {2, – 3} 119. (a) Given that system of equations have non- ero solution D=0 4 k 2 Þ k 4 1 =0

2 2 1 Þ 4(4 - 2) - k (k - 2) + 2(2k - 8) = 0 Þ

8 - k 2 + 2k + 4k - 16 = 0

k 2 - 6k + 8 = 0 Þ (k - 4)(k - 2) = 0 Þ k = 4, 2

120. (c)

1 2 1 D= 2 3 1 =0 3 5 2

3 2 1 Dx = 3 3 1 ¹ 0 1 5 2

Þ Given system, does not have any solution. Þ No solution

121. (d) The given equations are –x + cy + bz = 0 cx –y + az = 0 bx + ay – z = 0 Given that x, y, z are not all ero \ The above system have non- ero solution –1 c b Þ D = 0 Þ c –1 a = 0 b a –1

Þ –1(1– a2) – c(– c – ab) + b(ac + b) = 0 Þ –1 + a2 + b2 + c2 + 2abc = 0 Þ a2 + b2 + c2 + 2abc = 1 122. (a) ax + y + = a – 1; x + a y + = a – 1; x+ y+ a = a –1 a 1

1

D= 1 a

1

1

1

a

2 = a( a - 1) - 1(a - 1) + 1(1 - a) = a (a - 1)(a + 1) - 1(a - 1) - 1(a - 1) = (a - 1)[a 2 + a - 1 - 1] = (a - 1)[a 2 + a - 2] = (a – 1) [a 2 + 2a - a - 2]

= (a - 1)[a (a + 2) - 1(a + 2)] = ( a - 1) ( a + 2 ) Q Equations has infinite solutions \ D=0 Þ (a - 1) = 0, a + 2 = 0 Þ a = – 2, 1; But a ¹ 1 . \a=–2 123. (d) For homogeneous system of equations to have non ero solution, D = 0 2

1 2a a 1 3b b = 0 C ®

-

-

1 4c c Applying C2 ® C2 – 2C3 1 Þ 1

0 b

a b = 0 R3 ® R3 - R1, R2 ® R2 - R1

1 2c c

1

0

Þ 0

b

a b-a =0

0 2c c - a Þ bc – ab = 2bc – 2ac 2 1 1 Þ = + b a c \ a, b, c are in Harmonic Progression.

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Continuity and Differentiability

20

Continuity and Differentiability 5.

TOPIC Ć Continuity 1.

2.

ì 2 cos x – 1 p ,x ¹ ïï cot x – 1 4 í f(x) = ï p k, x= ïî 4 is continuous, then k is equal to:

éxù Let f (x) = x. ê ú , for –10 < x < 10, where [t] denotes the ë 2û greatest integer function. Then the number of points of discontinuity of f is equal to ______. [NA Sep. 05, 2020 (I)] If a function f (x) defined by ìae x + be - x , - 1 £ x < 1 ïï f ( x) = í cx 2 , 1 £ x £ 3 be continuous for some ï 2 ïî ax + 2cx , 3 < x £ 4

6.

(c)

3.

4.

e 2 - 3e + 13 e e 2 + 3e + 13

(b)

(d)

lim f(x) and lim f(x) exist but are not equal. (c) Both x® 4x®4+

e

lim f(x) exists but lim f(x) does not exist. (d) x® 4x®4+

e 2 - 3e - 13 e

7.

(c)

If the function

{

a | p - x | +1, x £ 5 f(x) = b | x - p | +3, x > 5

e 2 - 3e + 13

é4ù x ú = A. Let [t] denote the greatest integer £ t and xlim ®0 ê ëxû Then the function, f(x) = [x2] sin(px) is discontinuous, when x is equal to : [Jan. 9, 2020 (II)] (a)

1 (b) 1 (c) 1 (d) 2 2 éxù If f ( x) = [ x] - ê ú , x Î R, where [x] denotes the greatest ë4û integer function, then: [April 09, 2019 (II)] (a) f is continuous at x = 4.

lim f(x) exists but lim f(x) does not exist. (b) x® 4+ x®4-

[Sep. 02, 2020 (I)] (a)

A +1

(b)

A+5

A + 21

(d)

A

[April 09, 2019 (I)]

(a) 2

a, b, c Î R and f '(0) + f '(2) = e, then the value of a is :

1

æ p pö If the function f defined on ç , ÷ by è6 3ø

is continuous at x = 5, then the value of a – b is: [April 09, 2019 (II)] 2 -2 2 (b) (c) p+5 p+5 p -5 Let f : [– 1, 3] ® R be defined as

(a) 8.

(d)

2 5-p

æ 1 1ö If the function f defined on ç - , ÷ by è 3 3ø

ì x + [ x ] , - 1£ x < 1 ï f(x) = í x + x , 1 £ x < 2 ï x + [ x ] , 2 £ x £ 3, î

ì1 æ 1 + 3x ö , when x ¹ 0 ï log f(x) = í x e çè 1 - 2 x ÷ø is continuous, then k , when x = 0 ïk î is equal to __________. [NA Jan. 7, 2020 (II)]

where [t] denotes the greatest integer less than or equal to t. Then, f is discontinuous at : [April 08, 2019 (II)] (a) only one point (b) only two points (c) only three points (d) four or more points

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9.

Mathematics

Let f : R ® R be a function defined as

14.

if x £1 ì 5, ï a + bx, if 1 < x < 3 ï f ( x) = í ï b + 5 x, if 3 £ x < 5 ïî 30, x³5 if Then, f is : [Jan 09, 2019 (I)] (a) continuous if a = 5 and b = 5 (b) continuous if a = – 5 and b = 10 (c) continous if a = 0 and b = 5 (d) not continuous for any values of a and b 10. If the function f defined as f (x) =

15.

k -1 1 x e2 x - 1

11.

(b) (3, 2)

æ1 ö (c) ç , 2 ÷ è3 ø

is continuous at x = p, then k equals: [Online April 19, 2014]

(d) (2, 1)

1 ì ï 2- x Let f (x) = í ( x - 1) , x > 1, x ¹ 2 ïî k, x=2

(a) 0 16.

12.

e –1

(a) (b) e (c) The value of k for which the function

13.

(b)

2 5

(c)

3 5

(d) -

17.

2 5

, 0 £ x p î k2 cos x, ferentiable, then the ordered pair (k1, k2) is equal to:

g (a) = b, then f ¢ (b) is equal to: (a) 35.

that:

(b)

(a) f (5) + f '(5) £ 26

33.

f '(5) + f ''(5) £ 20

36.

in the interval:

ì sin(a + 2) x + sin x ; x0 ïî x 4/3

37.

[Jan. 9, 2020 (I)] (d) –2

[Jan. 7, 2020 (I)]

(a) (– ฀ , 20]

(b) [–3, 11]

(c) (– ฀, 11]

(d) [–6, 20]

Let S be the set of points where the function, f(x) = |2 – |x – 3||, xÎR, is not differentiable. Then å f(f(x)) is equal to ¾¾¾. xÎS

38.

[NA Jan. 7, 2020 (I)]

ì sin (p + 1) x + sin x ,x < 0 ï x ï q ,x=0 If f (x) = í ï 2 x+x - x ï , x >0 î x3/2

is continuous at x = 0, then the ordered pair (p, q) is equal to: [April 10, 2019 (I)]

39.

is continuous at x = 0, then a + 2b is equal to:

(c) 0

Let the function, f: [–7, 0] ® R be continuous on [ –7, 0]

all xÎ(–7, 0), then for all such functions f, f ¢(–1) + f(0) lies

(d) f (5) £ 10

ìp -1 ïï 4 + tan x, | x |£ 1 The function f ( x) = í ï 1 (| x | -1) , | x |> 1 ïî 2 is : [Sep. 04, 2020 (II)] (a) continuous on R – {1} and differentiable on R – {–1, 1}. (b) both continuous and differentiable on R – {1}. (c) continuous on R – {–1} and differentiable on R – {–1, 1}. (d) both continuous and differentiable on R – {–1}.

(b) –1

f (1) - f (c ) = f ¢(c) 1- c

and differentiable on (–7, 0). If f(–7) = –3 and f ¢ (x) d” 2, for

(b) f (5) + f '(5) ³ 28

f ( x) = 1, then f '(3) is equal to ___________. x [NA Sep. 04, 2020 (I)]

(a) 1

[Jan. 8, 2020 (II)]

(d) | f (c) – f (1)| < | f ¢(c)|

f ( x + y ) = f ( x) + f ( y ) + xy 2 + x 2 y , for all real x and y. If

32.

2 5

(c) | f (c) + f (1)| < (1 + c) |f ¢(c)|

Suppose a differentiable function f (x) satisfies the identity

x ®0

(d)

every f in S, there exists a c Î (0,1), depending on f, such

æ1 ö (c) ç , -1÷ (d) (1, 1) 2 è ø Let f be a twice differentiable function on (1, 6). If f (2) = 8,

lim

(c) 5

Let S be the set of all functions f : [0,1] ® R, which are

(b) (1, 0)

f '(2) = 5, f '(x) ³ 1 and f ''( x ) ³ 4, for all x Î (1, 6), then : [Sep. 04, 2020 (I)]

31.

(b) 1

(a) | f (c) – f (l)| < (l – c)|f ¢(c)|

æ1 ö (a) ç ,1÷ è2 ø

(c)

1 5

[Jan. 9, 2020 (II)]

continuous on [0, 1] and differentiable on (0,1). Then for

[Sep. 05, 2020 (I)]

30.

Let f and g be differentiable functions on R such that fog

æ 3 1ö (a) ç - , - ÷ è 2 2ø

æ 1 3ö (b) ç - , ÷ è 2 2ø

æ 3 1ö (c) ç - , ÷ è 2 2ø

æ5 1ö (d) ç , ÷ è2 2ø

Let f(x) = loge (sinx), (0 < x < p) and g(x) = sin–1 (e–x), (x > 0). If a is a positive real number such that a = (fog)¢ (a) and b = (fog) ( a), then: [April 10, 2019 (II)] (a) aa2 + ba + a = 0 (b) aa2 – ba – a =1 (c) aa2 – ba – a = 0 (d) aa2 + ba – a = – 2a2

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Continuity and Differentiability

40.

Let f : R ® R be differentiable at c Î R and f (c) = 0. If g (x) = f (x) , then at x = c, g is :

(a) is an empty set (b) equals {– 2, – 1, 0, 1, 2}

[April 10, 2019 (I)]

(a) not differentiable if f '(c) = 0

(c) equals {– 2, – 1, 1, 2}

(b) differentiable if f "(c) ¹ 0

(d) equals {– 2, 2}

(c) differentiable if f ' (c) = 0

48.

(d) not differentiable 41.

Let f(x) = 15 – |x – 10|; x Î R. Then the set of all values of x, at which the function, g(x) = f (f(x)) is not differentiable, is:

42.

(b) {10, 15}

(c) {5, 10, 15, 20}

(d) {10}

f (f (f (x))) + (f (x))2 at x = 1 is : 43.

44.

49.

If f (1) = 1, f¢(1) = 3, then the derivative of [April 08, 2019 (II)]

(a) 33 (b) 12 (c) 15 (d) 9 Let f be a differentiable function such that f(1) = 2 and f ¢ (x) = f (x) for all x Î R. If h (x) = f ( f (x)), then h¢ (1) is equal to : [Jan. 12, 2019 (II)] (a) 2e2 (b) 4e (c) 2e (d) 4e2 -2 £ x < 0 ìï -1, and Let f ( x ) = í 2 ïî x - 1, 0 £ x £ 2

50.

51.

(1 + 2e ) 2 4+ e

2

( 2e –1) (b)

(1 + 2e ) (c) 46.

47.

4+e

2

2 4+ e

(a)

52.

2

ìïmax {| x |, x 2 } | x|£ 2 Let f (x) = í 2 0 ) , then at x = e is dx equal to : [Jan. 11, 2019 (I)]

(a)

}

1 - x 2 . If K be the set of all points at which f

x

y2

y2

é æ 2x + 3ö ù (a) sin ê log ç è 3 - 2 x ÷ø úû ë 12 (b) ( 3 - 2x) 2

[2017]

(d)

3

(c)

89.

(b)

x + 2x it is defined) is equal to :

d2 y

at the point dx 2 [Online April 15, 2018] (c) – 2 (d) 4

x2

Let f (x) =

(a)

æ 6x x ö æ 1ö If for x Îç 0, ÷ , the derivative of tan -1 ç ÷ is è 4ø è 1 - 9x 3 ø

(a)

88.

3 loge 3

(d)

sin y = 4, then the value of

(– 2, 0) is (a) – 34

(a)

95.

If f ( x + y ) = f ( x ). f ( y )"x. y and f (5) = 2, f '(0) = 3, then f ¢ (5) is (a) 0 (b) 1

[2002] (c) 6

(d) 2

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Continuity and Differentiability

TOPIC Ė

96.

Differentiation of Infinite Series, Successive Differentiation, nth Derivative of Some Standard Functions, Leibnitz’s Theorem, Rolle’s Theorem, Lagrange’s Mean Value Theorem

For all twice differentiable functios f : R®R, with f(0) = f(1) = f’(0) = 0 [Sep. 06, 2020 (II)]

f "(0) = 0

æ p pö 2 2 If y + log e (cos x) = y, x Î ç - , ÷ , then : è 2 2ø [Sep. 03, 2020 (I)] (a) y ''(0) = 0

(b) | y '(0) | + | y ''(0) |= 1

(c) | y ''(0) |= 2

(d) | y '(0) | + | y ''(0) |= 3

If c is a point at which Rolle’s theorem holds for the æ x2 + a ö function, f ( x ) = log e ç ÷ in the interval [3, 4], where è 7x ø

a Î R, then f ²(c) is equal to: (a) -

1 12

1 12

(b)

[Jan. 8, 2020 (I)]

(c) -

1 24

(d)

3 7

1

99.

dy æ y ö 3 Let x + y = a , (a, k > 0) and + ç ÷ = 0, then k is: dx è x ø k

k

k

[Jan. 7, 2020 (I)] 3 (a) 2

4 (b) 3

2 (c) 3

1 (d) 3

100. The value of c in the Lagrange’s mean value theorem for the function f(x) = x3 – 4x2 + 8x + 11, when x Î [0,1] is: [Jan. 7, 2020 (II)] (a)

4- 5 3

(b)

4- 7 3

(c)

2 3

(d)

7-2 3

1

101. If 2x = y 5 + y

-

1 5

d2y

dy + lx + ky = 0, dx dx 2 [Online April 9, 2017] (c) 26 (d) – 26

= and (x2 – 1)

then l + k is equal to : (a) – 23 (b) – 24

d2 y dx 2

15

é x + x 2 - 1ù ëê ûú

= +x

dy dx

15

+ éê x - x 2 - 1ùú ë û

is equal to

,

then

[Online April 8, 2017]

1 , then 2b + c equals : 2 [Online April 10, 2015] (a) –3 (b) –1 (c) 2 (d) 1 105. If f and g are differentiable functions in [0, 1] satisfying

+ cx, x Î [–1, 1], at the point x =

(d) f "(x) = 0, at every point x Î (0,1)

98.

y

(a) 12 y (b) 224 y2 (c) 225 y2 (d) 225 y 104. If Rolle’s theorem holds for the function f (x) 2x3 + bx2

(b) f "(x) = 0, for some x Î (0,1)

97.

103. If

(x 2 - 1)

(a) f "( x ) ¹ 0 at every point x Î (0,1)

(c)

102. Let f be a polynomial function such that f (3x) = f ¢ (x) , f ² (x), for all x Î R. Then : [Online April 9, 2017] (a) f (b) + f ¢ (b) = 28 (b) f ² (b) – f ¢ (b) = 0 (c) f ² (b) – f ¢ (b) = 4 (d) f (b) – f ¢ (b) + f ² (b) = 10

f (0) = 2 = g(1), g(0) = 0 and f (1) = 6, then for some c Î]0,1[ [2014] (a) f ¢(c) = g¢(c) (b) f ¢(c) = 2g ¢(c) (c) 2f ¢(c) = g ¢(c) (d) 2f ¢(c) = 3g¢(c) 106. Let f(x) = x|x|, g(x) = sin x and h(x) = (gof) (x). Then [Online April 11, 2014] (a) h(x) is not differentiable at x = 0. (b) h(x) is differentiable at x = 0, but h¢(x) is not continuous at x = 0 (c) h¢(x) is continuous at x = 0 but it is not differentiable at x= 0 (d) h¢(x) is differentiable at x = 0 107. Let for i = 1, 2, 3, pi(x) be a polynomial of degree 2 in x, p¢i(x) and p¢¢i(x) be the first and second order derivatives of pi(x) respectively. Let, é p ( x ) p ¢ ( x ) p ¢¢ ( x ) ù 1 1 ê 1 ú ê ¢ A ( x ) = p2 ( x ) p2 ( x ) p2¢¢ ( x ) ú ê ú ê p3 ( x ) p3¢ ( x ) p3¢¢ ( x ) ú ë û

and B(x) = [A(x)]T A(x). Then determinant of B(x): [Online April 11, 2014] (a) is a polynomial of degree 6 in x. (b) is a polynomial of degree 3 in x. (c) is a polynomial of degree 2 in x. (d) does not depend on x. 108. If the Rolle’s theorem h olds for the function f(x) = 2x3 + ax2 + bx in the interval [– 1, 1] for the point 1 , then the value of 2a + b is: [Online April 9, 2014] 2 (a) 1 (b) – 1 (c) 2 (d) – 2 c=

EBD_8344

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Mathematics

109. If f (x) = sin (sin x) and f "(x) + tan x f¢ (x) + g(x) = 0, then g(x) is : [Online April 23, 2013] (a) cos2 x cos (sin x) (b) sin2 x cos (cos x) (c) sin2 x sin (cos x) (d) cos2 x sin (sin x) 110. Consider a quadratic equation ax2 + bx + c = 0, where x3 x2 + b + cx . 3 2 [Online May 19, 2012] Statement 1: The quadratic equation has at least one root in the interval (0, 1). Statement 2: The Rolle’s theorem is applicable to function g(x) on the interval [0, 1]. (a) Statement 1 is false, Statement 2 is true. (b) Statement 1 is true, Statement 2 is false. (c) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1. (d) Statement 1 is true, Statement 2 is true, , Statement 2 is a correct explanation for Statement 1.

2a + 3b + 6c = 0 and let g ( x ) = a

2

114. Let f be differentiable for all x. If f (1) = – 2 and f '( x) ³ 2 for x Î [1, 6], then (a) f (6) ³ 8(b) f (6) < 8

dy 2

equals :

æ d2y ö (a) - ç 2 ÷ è dx ø

-1

[2011] æ dy ö çè ÷ø dx

-3

æ d 2 y ö æ dy ö -3 ç 2 ÷ çè dx ÷ø (c) è dx ø

æ d 2 y ö æ dy ö -2 (b) ç 2 ÷ çè dx ÷ø è dx ø æd2yö (d) ç 2 ÷ è dx ø

-1

(b) loge3

(c) 2 log3e

(d) f (6) = 5

a1 ¹ 0, n ³ 2, has a positive root x = a , then the equation n -2 nan x n -1 + (n – 1) an -1 x + ......... + a1 = 0 has a positive root, which is [2005] (a) greater than a (b) smaller than a (c) greater than or equal to a (d) equal to a

116. If 2a + 3b + 6c = 0, then at least one root of the equation ax 2 + bx + c = 0 lies in the interval (a) (1, 3) (b) (1, 2) (c) (2, 3)

[2004] (d) (0, 1)

117. If f ( x) = x n , then the value of f (1) -

[2003]

f ' (1) f ' ' (1) f ' ' ' (1) ( -1) n f n (1) + + .......... is n! 3! 1! 2!

(b) 2 n (d) 0. (c) 2 n - 1 118. Let f (a) = g (a) = k and their nth derivatives (a) 1

f n (a) , g n (a) exist and are not equal for some n. Further if

112. Let f (x) = x | x | and g (x) = sin x. Statement-1 : gof is differentiable at x = 0 and its derivative is continuous at that point. Statement-2 : gof is twice differentiable at x = 0. [2009] (a) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is false. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. 113. A value of c for which conclusion of Mean Value Theorem holds for the function f (x) = loge x on the interval [1, 3] is [2007] (a) log3e

(c) f (6) < 5

115. If the equation an x n + an -1 x n -1 + ............. + a1 x = 0

d x

111.

[2005]

(d)

1 log e 2 3

lim

x ®a

f ( a ) g ( x) - f ( a ) - g ( a ) f ( x ) + f ( a ) =4 g ( x) - f ( x)

then the value of k is (a) 0 (b) 4

[2003] (d) 1

(c) 2

119. If y = (x + 1 + x 2 )n, then (1 + x2)

d2y dx2

+x

dy is [2002] dx

(a) n2y (b) – n2y (c) –y (d) 2x2y 120. If 2a + 3b + 6c = 0, (a, b, c Î R) then the quadratic equation ax2 + bx + c = 0 has [2002] (a) at least one root in [0, 1] (b) at least one root in [2, 3] (c) at least one root in [4, 5] (d) None of these

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Continuity and Differentiability

1.

(8) We know [x] discontinuous for x Î Z

éxù x f ( x ) = x ê ú may be discontinuous where is an 2 2 ë û integer. So, points of discontinuity are,

4.

æ 3ln(1 + 3x) 2ln(1 - 2 x) ö = lim ç ÷ 3x -2 x x ®0 è ø =3+ 2=5 Q f(x) will be continuous

but at x = 0 lim f ( x ) = 0 = f (0) = lim f ( x)

2.

x ®0

æ ln(1 + 3x) ln(1 - 2 x) ö = lim ç ÷ x x x ®0 è ø

x = ±2, ± 4, ± 6, ± 8 and 0

x ® 0+

æ 1 æ 1 + 3x ö ö lim f ( x) = lim ç ln ç ÷÷ x®0 è x è 1 - 2 x ø ø

(5)

x ® 0-

\

So, f (x) will be discontinuous at x = ±2, ±4, ±6 and ±8. (d) Since, function f (x) is continuous at x = 1, 3 5.

\ f (1) = f (1+ )

k = f (0) = lim f ( x ) = 5 x®0

(b) Since, f(x) is continuous, then lim f ( x)

Þ ae + be -1 = c

...(i)

p 4

f (3) = f (3+ ) lim

2 cos x - 1 =k cot x - 1

Þ 9c = 9a + 6c Þ c = 3a From (i) and (ii),

...(ii)

p x® 4

b = ae(3 - e)

...(iii)

Now by L- hospital’s rule

é ae x - be - x ê f '( x ) = ê 2cx ê 2ax + 2c ë

æ 1 ö 2ç ÷ è 2 ø =k Þ k = 1 = Þ k lim 2 p 2 ( 2 )2 x ® c osec x

-1 < x < 1

2 sin x

1< x < 3 3< x< 4

4

f '(0) = a - b, f '(2) = 4c 6.

Given, f '(0) + f '(2)= e a - b + 4c= e From eqs. (i), (ii), (iii) and (iv),

...(iv)

é4ù f ( 4 ) = [ 4] - ê ú = 4 - 1 = 3 ë4û

Þ 13a - 3ae + ae 2 = e Þa=

(a)

Q LHL = f (4) = RHL \ f (x) is continuous at x = 4

e e - 3e + 13 2

é 4 ì 4 üù é4ù lim x ê ú = A Þ lim x ê - í ý ú = A x ®0 ë x î x þ û x ®0 ë x û

7.

4 Þ lim 4 - x ìí üý = A Þ 4 – 0 = A x ®0 îxþ As, f (x) = [x2]sin(px) will be discontinuous at non-integers And, when x = A + 1 Þ x = 5 , which is not an integer. Hence, f (x) is discontinuous when x is equal to

æ éxùö (a) L.H.L. lim- ç [ x ] - ê 4 ú ÷ = 3 - 0 = 3 ë ûø x®4 è

éxù R.H.L. lim+ [ x ] - ê ú = 4 - 1 = 3 ë4û x ®4

a - 3ae + ae 2 + 12a = e

3.

æpö = fç ÷ è4ø

A +1

(d) R.H.L. lim+ b ( x - p ) + 3 = ( 5 - p ) b + 3 x ®5

f (5) = L.H.L. lim- a ( p - x ) + 1 = a ( 5 - p ) + 1 x ®5

Q function is continuous at x = 5 \ LHL = RHL (5 – p) b + 3 = (5 – p) a + 1 Þ 2 = (a – b) (5 – p) Þ a - b =

2 5- p

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8.

Mathematics

For the limit to exist, power of x in the numerator should be greater than or equal to the power of x in the denominator. Therefore, coefficient of x in numerator is equal to ero Þ 3–k=0 Þ k=3 So the limit reduces to

(c) Given function is, ì| x | +[ x], -1 £ x < 1 ï x + | x |, 1 £ x < 2 f (x) = í ï x + [ x ], 2£ x£3 î ì - x - 1, ï x, ïï = í 2 x, ï x + 2, ï ïî6,

-1 £ x < 0 0 £ x 0 Þ g '(c) = lim x®c x - c - f ( x) ; if f (x) < 0 and g '(c) = lim x ®c x - c Þ g '(c) = f '(c) = – f ‘ (c) Þ 2f '(c) = 0 Þ f '(c) = 0 Hence, g (x) is differentiable if f ¢(c) = 0

40. (c) g '(c) = lim

x ®c

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Continuity and Differentiability

41. (a) Since, f(x) = 15 – |(10 – x)| \ g(x) = f(f(x)) = 15 – |10 – [15 – |10 – x|]| = 15 – ||10 – x| – 5| \ Then, the points where function g(x) is Nondifferentiable are 10 – x = 0 and |10 – x| = 5 Þ x = 10 and x – 10 = ± 5 Þ x = 10 and x = 15, 5 42. (a) Let g (x) = f ( f ( f (x))) + (f(x))2 Differentiating both sides w.r.t. x, we get g' (x) = f '( f ( f (x))) f '( f (x)) f '(x) + 2 f (x) f '(x) g' (1) = f '( f ( f (1))) f '( f (1)) f '(1) + 2 f (1) f '(1) = f '( f (1)) f '(1) f '(1) + 2 f (1) f '(1) = 3 × 3 × 3 + 2 × 1 × 3 = 27 + 6 = 33 43. (b) Since, f ¢(x) = f(x)

ì x2 , -2 £ x < 0 ï 0 £ x 0 f (x) = sin x – x + 2(x – p) cos x

2 c =e e Then, from eqn (1)

Þ

f ¢(x) = cos x – 1 + 2(1 – 0) cos x – 2 sin (x – p) f ¢(x) = 3 cos x – 2(x – p) sin x – 1

2 x f(x) = e e

Þ

dy 1 - 2x + 2 y x × log e x dx = 0

dy 1 - 2x + 2 y dx = 0 ...(1) log e x

f ¢( x ) Then, =1 f ( x)

Þ ln |f(x)| = x + c f(x) = ±ex + c ...(1) Since, the given condition f(1) = 2 From eqn (1) f(x) = ex + c = ecex Then, f(1) = ec × e1 c Þ 2=e ×e

Then, function f(x) is differentiable for all x > 0 Case (2) x < 0

2 x f ¢(x) = e e

f(x) = – sin x + x + 2(x – p) cos x

Nowh(x) = f(f(x)) Þ h¢(x) = f ¢(f(x)) × f ¢(x) h¢(1) = f ¢(2) × f ¢(1) =

44.

2 -2 £ x < 0 ïì 1 + x - 1, Þ g(x) = í 2 2 ïî( x - 1) + | x - 1|, 0 £ x £ 2

2 2 2 e × × e = 4e e e

-2 £ x < 0 ìï -1, (d) f(x) = í 2 ïî x - 1, 0 £ x £ 2 -2 £ | x | < 0 ìï -1, Then, f(|x|) = í 2 ïî| x | -1, 0 £ | x | £ 2 2 Þ f(|x|) = x – 1, –2 £ x £ 2

f ¢(x) = – cos x + 1 – 2(x – p) sin x + 2 cos x f ¢(x) = cos x + 1 – 2(x – p) sin x Then, function f(x) is differentiable for all x < 0 Now check for x = 0 f ¢(0+) R.H.D. = 3 – 1 = 2 f ¢(0–) L.H.D. = 1 + 1 = 2 L.H.D. = R.H.D. Then, function f(x) is differentiable for x = 0. So it is differentiable everywhere Hence, k = f

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Mathematics

| x| £ 2 ïìmax{| x |, x 2 } 47. (b) Given f(x) = í x < |x | £ 4 8 2 | | 2 ïî y = x2

y = 8 - 2|x|

y = x2

y = 8 - 2|x|

y = |x| y = |x|

50. Q f(x) is not differentiable at –2, –1, 0, 1 and 2. \ S = {–2, –1, 0, 1, 2} 48. (c) Consider the function

{

2 f(x) = max - | x |, - 1 - x

}

Now, the graph of the function

x

1 2

x

1 2

From the graph, it is clear that f(x) is not differentiable at x = 0, -

1 1 , 2 2

ìï(| l | et – m) sin 2t, t > 0 =í -t ïî (| l | e – m) (– sin 2t ), t < 0 t t ïì(| l | e ) sin 2t + (| l | e – m) (2cos 2t ), t > 0 f ¢ (t) = í -t -t ïî| l | e sin 2t + (| l | e – m) (– 2cos 2t ), t < 0 As, f (t) is differentiable \ LHD = RHD at t = 0 Þ |l| . sin 2 (0) + (|l|e0 – m)2 cos (0) = |l|e–0 . sin 2 (0) – 2 cos (0) (|l|e– 0 – m) Þ 0 + (|l| – m)2 = 0 – 2 (|l| – m) Þ 4 (|l| – m) = 0 Þ |l| = m So, S º (l, m) = {l Î R & m Î [0, ¥)} Therefore set S is subset of R × [0, ¥)

ìï- x x 0 and f ²(x) < 0, then for any c Î (a, b),

f (c ) - f ( a ) is greater f (b) - f ( c )

than:

23.

[Jan. 9, 2020 (I)]

(a)

b+ a b-a

(b) 1

(c)

b-c c-a

(d)

c-a b-c

29.

é p pù Let f(x) = x cos–1 (–sin |x|), x Î ê - , ú , then which of the ë 2 2û following is true? [Jan. 8, 2020 (I)] æ p ö æ pö (a) f ’ is increasing in çè - ,0÷ø and decreasing in çè 0, ÷ø 2 2

24.

a +x

2

-

d-x b + (d - x) 2

2

, x Î R where a, b

and d are non- ero real constants. Then : [Jan. 11, 2019 (II)] (a) f is an increasing function of x (b) f is a decreasing function of x (c) f ¢ is not a continuous function of x (d) f is neither increasing nor decreasing function of x The function f defined by f(x) = x3 – 3x2 + 5x + 7, is : [Online April 9, 2017] (a) increasing in R. (b) decreasing in R. (c) decreasing in (0, ¥) and increasing in (– ¥ , 0). (d) increasing in (0, ¥) and decreasing in (– ¥, 0). Let f(x) = sin4x + cos4x. Then f is an increasing function in the interval :

ù 5p 3p ù (a) ú , ú û 8 4û

ù p 5p ù (b) ú , ú û2 8 û

æ p ö æ pö (d) f ’ is decreasing in çè - ,0÷ø and increasing in çè 0, ÷ø 2 2

ùp pù (c) ú , ú û 4 2û

ù pù (d) ú 0, ú û 4û

Let f(x) = ex – x and g(x) = x2 – x, x Î R. Then the set of all x Î R, where the function h(x) = (fog) (x) is increasing, is : [April 10, 2019 (I)]

é -1 ù (d) ê ,0ú È [1, ¥ ) ë2 û If the function f : R – {1, –1} ® A defin ed by x2 1 - x2

( x - 1) 2 (a) –7 (c) 7

32.

, is surective, then A is equal to:

[April 09, 2019 (I)] (a) R – {–1} (b) [0, 벜) (c) R – [–1, 0) (d) R – (–1, 0) Let f: [0 : 2] ® R be a twice differentiable function such that f''(x) > 0, for all xÎ(0, 2). If f(x) = f(x) + f(2 – x), then f is : [April 08, 2019 (I)] (a) increasing on (0, 1) and decreasing on (1, 2). (b) decreasing on (0, 2) (c) decreasing on (0, 1) and increasing on (1, 2). (d) increasing on (0, 2) If the function f given by f (x) = x3 – 3(a – 2)x2 + 3ax + 7, for some aÎR is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, f ( x ) - 14

31.

é 1ù (b) ê 0, ú È [1, ¥ ) ë 2û

[0,¥ )

f(x) =

27.

x 2

p 2 (c) f ’ is not differentiable at x = 0

(c)

26.

30.

Let f ( x ) =

(b) f’ (0) = –

é -1ù é 1 ö (a) ê -1, ú È ê , ¥ ÷ 2 û ë2 ø ë

25.

28.

= 0( x ¹ 1) is

[Jan. 12, 2019 (II)] (b) 5 (d) 6

33.

34.

Let f and g be two differentiable functions on R such that f ¢(x) > 0 and g¢(x) < 0 for all x Î R . Then for all x: [Online April 12, 2014] (a) f (g (x)) > f (g (x – 1)) (b) f (g (x)) > f (g (x + 1)) (c) g(f (x)) > g (f (x – 1)) (d) g(f (x)) < g (f (x + 1)) The real number k for which the equation, 2x3 + 3x + k = 0 has two distinct real roots in [0, 1] [2013] (a) lies between 1 and 2 (b) lies between 2 and 3 (c) lies between .1 and 0 (d) does not exist. Statement-1: The function x2 (ex + e–x) is increasing for all x > 0. Statement-2: The functions x2ex and x2e–x are increasing for all x > 0 and the sum of two increasing functions in any interval (a, b) is an increasing function in (a, b). [Online April 22, 2013] (a) Statement-1 is false; Statement-2 is true. (b) Statement-1is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (c) Statement-1 is true; Statement-2 is false. (d) Statement-1is true; Statement-2 is true; Statement-2 is a correct explanation for statement-1. Statement-1: The equation x log x = 2 – x is satisfied by at least one value of x lying between 1 and 2. Statement-2: The function f (x) = x log x is an increasing function in [l, 2] and g (x) = 2 – x is a decreasing function in [1, 2] and the graphs represented by these functions intersect at a point in [1, 2] [Online April 9, 2013]

EBD_8344

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35.

36.

37.

38.

39.

Mathematics

(a) Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is true; Statement-2 is true; Statement-2 is not correct explanation for Statement-1. (c) Statement-1 is false, Statement-2 is true. (d) Statement-1 is true, Statement-2 is false. If f(x) = xex(1 – x), x Î R , then f(x) is [Online May 12, 2012] (a) decreasing on [–1/2, 1] (b) decreasing on R (c) increasing on [–1/2, 1] (d) increasing on R For real x, let f (x) = x3 + 5x + 1, then [2009] (a) f is onto R but not one-one (b) f is one-one and onto R (c) f is neither one-one nor onto R (d) f is one-one but not onto R How many real solutions does the equation [2008] x7 + 14x5 + 16x3 + 30x – 560 = 0 have? (a) 7 (b) 1 (c) 3 (d) 5 The function f (x) = tan –1(sin x + cos x) is an increasing function in [2007]

p (a) æç 0, ö÷ è 2ø

æ p pö (b) ç - , ÷ è 2 2ø

(c) æç p , p ö÷ è 4 2ø

p p (d) çæ - , ÷ö è 2 4ø

41.

curve x4 e y + 2 y + 1 = 3 at the point (1, 0)?

42.

x - 3x + 3x + 3

(b) [2, ¥ )

2 x3 - 3 x 2 - 12 x + 6

1ù æ (c) ç - ¥, ú 3û è

3x 2 - 2 x + 1

(d) (– ¥ , – 4)

x3 + 6 x 2 + 6

3

2

43.

a is equal to _____. [NA Sep. 05, 2020 (II)] b If the tangent to the curve, y = ex at a point (c, ec) and the normal to the parabola, y2 = 4x at the point (1, 2) intersect at the same point on the x-axis, then the value of c is ____________. [NA Sep. 03, 2020 (II)]

44.

If y =

6

e -1 (a) e

(c)

æ 1 ö ç ÷ eè 1-e ø

(b)

æ 1 ö ç ÷ eè e-1 ø

e (d) e -1

-1 ì 3

4 dy ü at x = 0 is í cos kx - sin kx ý, then dx 5 î5 þ

___________. 45.

[NA Sep. 02, 2020 (II)]

Let the normal at a point P on the curve y2 – 3x2 + y + 10 = 0

æ 3ö intersect the y-axis at ç 0, ÷ . If m is the slope of the è 2ø tangent at P to the curve, then |m| is equal to ______. [NA Jan. 8, 2020 (I)] 46. The length of the perpendicular from the origin, on the normal to the curve, x 2 + 2xy – 3y2 = 0 at the point (2, 2) is: [Jan. 8, 2020 (II)] (a)

2

(c) 2 47.

48.

If the tangent to the curve, y = f (x) = xlogex, (x > 0) at a point (c, f(c)) is parallel to the line segement oining the points (1, 0) and (e, e), then c is equal to: [Sep. 06, 2020 (II)]

å k cos k =1

TOPIC Đ Tangents & Normals 40.

[Sep. 05, 2020 (II)] (a) (2, 2) (b) (2, 6) (c) (– 2, 6) (d) (– 2, 4) If the lines x + y = a and x – y = b touch the curve y = x2 – 3x + 2 at the points where the curve intersects the x-axis, then

A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched? [2005] Interval Function (a) (– ¥ , ¥ )

Which of the following points lies on the tangent to the

49.

(b) 4 2 (d) 2 2

(

)

x , x Î R, x ¹ ± 3 , x2 - 3 at a point (a, b) (0, 0) on it is parallel to the line 2x + 6y – 11 = 0, then : [April 10, 2019 (II)] (a) |6a + 2b| = 19 (b) |6a + 2b| = 9 (c) |2a + 6b| = 19 (d) |2a + 6b| = 11 If the tangent to the curve, y = x3 + ax – b at the point (1, –5) is perpendicular to the line, – x + y + 4 = 0, then which one of the following points lies on the curve? [April 09, 2019 (I)] (a) (–2, 1) (b) (–2, 2) (c) (2, –1) (d) (2, –2) Let S be the set of all values of x for which the tangent to the curve y = f(x) = x3 – x2 – 2x at (x, y) is parallel to the line segment oining the points (1, f(1)) and (– 1, f(– 1)), then S is equal to: [April 09, 2019 (I)]

If the tangent to the curve y =

ì1 ü (a) í ,1ý î3 þ

1 (b) ìí– , –1üý î 3 þ

ì1 ü (c) í , –1ý î3 þ

ì 1 ü (d) í– ,1ý î 3 þ

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Applications of Derivatives

50.

51.

The tangent and the normal lines at the point ( 3 , 1) to the circle x2 + y2 = 4 and the x-axis form a triangle. The area of this triangle (in square units) is : [April 08, 2019 (II)] (a)

4 3

(b)

(c)

2 3

(d)

1 3

(c)

1 3

56.

The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 – x2 such that the rectangle lies inside the parabola, is: [Jan. 12, 2019 (I)] (a) 36

53.

(d) 18 3 The tangent to the curve y = x2 – 5x + 5, parallel to the line 2y = 4x + 1, also passes through the point : [Jan. 12, 2019 (II)]

æ7 1ö (a) ç , ÷ è2 4ø

æ1 ö (b) ç , -7 ÷ è8 ø

æ 1 ö (c) ç - , 7 ÷ è 8 ø

æ1 7ö (d) ç , ÷ è4 2ø

curve y = x , ( x > 0), is: (a) (c) 54.

5 2 3 2

57.

(d)

59.

The tangent to the curve, y = xe

60. x2

(d)

1 2

7 3

8 15 8 (d) 17 (b)

If the curves y2 = 6x,9x 2 + by 2 = 16 intersect each other at right angles, then the value of b is : [2018] 7 2

(b) 4

9 (d) 6 2 Let P be a point on the parabola,x2 = 4y. If the distance of P from the centre of the circle, x2 + y2 + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is [Online April 16, 2018] (a) x + 4y – 2 = 0 (b) x + 2y = 0 (c) x + y + 1 = 0 (d) x – y + 3 = 0 If the tangents drawn to the hyperbola 4y2 = x2 + 1 intersect the co-ordinate axes at the distinct points A and B, then the locus of the mid point of AB is[Online April 15, 2018] (a) x2 – 4y2 + 16 x2 y2 = 0 (b) 4x2 – y2 + 16 x2 y2 = 0 (c) 4x2 – y2 – 16 x2 y2 = 0 (d) x2 – 4y2 – 16 x2 y2 = 0 If b is one of the angles between the normals to the ellipse,

x 2 + 3y 2 = 9 at the points (3cosq, 3 sin q) and passing through the

(– 3sin q,

point (1, e) also passes through the point: [Jan. 10, 2019 (II)]

55.

1 3

(c)

3 2 5 4

7 3

(b)

If q denotes the acute angle between the curves, y = 10 – x2 and y = 2 + x2 at a point of their intersection, then |tan q| is equal to: [Jan. 09, 2019 (I)]

(a)

[Jan. 10, 2019 (I)] (b)

1 6

4 9 7 (c) 17

58.

æ3 ö The shortest distance between the point ç , 0 ÷ and the è2 ø

5 6

(a)

(b) 20 2

(c) 32

52.

(a)

(a) (2, 3e)

æ4 ö (b) ç , 2e ÷ è3 ø

æ5 ö (c) ç , 2e ÷ 3 è ø

(d) (3, 6e)

(a)

A helicopter is flyin g along the curve given by

æ1 ö y – x3/2 = 7, (x ³ 0). A soldier positioned at the point ç , 7 ÷ è2 ø wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is: [Jan. 10, 2019 (II)]

1

(b)

2 3

3 4 3 A normal to the hyperbola, 4x2 – 9y2 = 36 meets the coordinate axes x and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the locus of P is [Online April 15, 2018] (a) 4x2 – 9y2 = 121 (b) 4x2 + 9y2 = 121 (c) 9x2 – 4y2 = 169 (d) 9x2 + 4y2 = 169 (c)

61.

2

2 cot b æ pö 3 cos q); Î ç 0, ÷ ; then is equal to sin 2q è 2ø [Online April 15, 2018]

(d)

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62.

Mathematics

The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes through the point: [2017]

æ 1 1ö (a) çè , ÷ø 2 3

63.

64.

1 . If one of its directices is x = – 4, then the 2

æ 3ö equation of the normal to it at ç1, ÷ is : [2017] è 2ø (a) x + 2y = 4 (b) 2y – x = 2 (c) 4x – 2y = 1 (d) 4x + 2y = 7 A tangent to the curve, y = f (x) at P(x, y) meets x-axis at A and y-axis at B. If AP : BP = 1 : 3 and f (a) = 1 , then the curve also passes through the point : [Online April 9, 2017]

æ 1ö æ 1 ö (c) ç 2, ÷ (d) ç 3, ÷ è 8ø è 28 ø The tangent at the point (2, –2) to the curve, x2y2 – 2x = 4 (1–y) does not pass through the point : [Online April 8, 2017] æ 1ö (a) ç 4, ÷ è 3ø (c) (– 4, – 9)

66.

The normal to the curve, x2 + 2xy – 3y2 = 0, at (1, 1) [2015] (a) meets the curve again in the third quadrant. (b) meets the curve again in the fourth quadrant. (c) does not meet the curve again. (d) meets the curve again in the second quadrant. The equation of a normal to the curve,

70.

æp ö sin y = x sin ç + y ÷ at x = 0, is : è3 ø [Online April 11, 2015]

72.

(a) 2x –

3y = 0

(b) 2x +

3y = 0

(c) 2y –

3x = 0

(d) 2y +

3x = 0

If the tangent to the conic, y – 6 = x2 at (2, 10) touches the circle, x2 + y2 + 8x – 2y = k (for some fixed k) at a point (a, b) ; then (a, b) is : [Online April 10, 2015]

æ 7 6ö (a) çè - , ÷ø 17 17

æ 4 1ö (b) çè - , ÷ø 17 17

æ 6 10 ö (c) çè - , ÷ø 17 17

æ 8 2ö (d) çè - , ÷ø 17 17

The distance, from the origin, of the normal to the curve, p , is : 4 [Online April 10, 2015]

(b) (8, 5)

x = 2 cost + 2t sint, y = 2 sint – 2t cost at t =

(d) (– 2, – 7) (a) 2

Consider

(b) 4

(c)

æ 1 + sin x f ( x ) = tan -1 ç ç 1 - sin x è

ö æ pö ÷÷ , x Î ç 0, ÷ . è 2ø ø

A normal to y = f(x) at x =

67.

69.

71.

æ1 ö (b) ç , 4 ÷ è2 ø

æ1 ö (a) ç , 24 ÷ è3 ø

65.

If the tangent at a point P, with parameter t, on the curve x = 4t2 + 3, y = 8t3 – 1, t Î R, meets the curve again at a point Q, then the coordinates of Q are : [Online April 9, 2016] (a) (16t2 + 3, –64t3 – 1) (b) (4t2 + 3, –8t3 – 2) (d) (t2 + 3, – t3 – 1) (c) (t2 + 3, t3 – 1)

æ 1 1ö (b) ç - , - ÷ è 2 2ø

æ 1 1ö æ 1 1ö (c) ç , ÷ (d) ç , - ÷ è 2 2ø è 2 3ø The eccentricity of an ellipse whose centre is at the origin is

68.

[2016]

73.

p also passes through the point: 6

æp ö (a) ç , 0 ÷ è6 ø

æp ö (b) ç , 0 ÷ è4 ø

(c) (0, 0)

æ 2p ö (d) ç 0, ÷ è 3 ø

3 . If P is 4 2 a point on C, such that the tangent at P has slope , then 3 a point through which the normal at P passes, is : [Online April 10, 2016] (a) (1, 7) (b) (3, –4) (c) (4, –3) (d) (2, 3)

Let C be a curve given by y(x) = 1 +

(d) 2 2 2 For the curve y = 3 sinq cosq, x = eq sin q, 0 £ q £ p, the tangent is parallel to x-axis when q is: [Online April 11, 2014] (a)

3p 4

(b)

p 2

74.

p p (d) 6 4 If an equation of a tangent to th e curve, y – cos(x + f), – 1 -1 £ x £ 1 + p, is x + 2y = k then k is equal to : [Online April 25, 2013] (a) l (b) 2

75.

p p (d) 4 2 The equation of the normal to the parabola, x2 = 8y at x = 4 is [Online May 19, 2012] (a) x + 2y = 0 (b) x + y = 2 (c) x – 2y = 0 (d) x + y = 6

(c)

4x - 3, x >

(c)

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Applications of Derivatives

76.

77.

78.

79.

80.

4 The equation of the tangent to the curve y = x + 2 , that x is parallel to the x-axis, is [2010] (a) y = 1 (b) y = 2 (c) y = 3 (d) y = 0 Angle between the tangents to the curve y = x 2 - 5 x + 6 at the points (2, 0) and (3, 0) is [2006] p (a) p (b) 2 p p (c) (d) 6 4 The normal to the curve [2005] x = a (cos q + q sin q ), y = a (sin q – q cos q ) at any point q is such that (a) it passes through the origin

83.

[Sep. 06, 2020 (II)]

æ 1 1ö (a) ç - , ÷ - {0} è 2 2ø

84.

æ 1 1ö æ 3 3ö (c) ç - , ÷ (d) ç - , ÷ - {0} è 2 2ø è 2 2ø If x = 1 is a critical point of the function [Sep. 05, 2020 (II)]

(a) x = 1 and x = -

2 are local minima of f. 3

p (b) it makes an angle + q with the x- axis 2

(b) x = 1 and x = -

2 are local maxima of f. 3

æ p ö (c) it passes through ç a , - a÷ è 2 ø (d) It is at a constant distance from the origin

(c) x = 1 is a local maxima and x = -

The normal to the curve x = a(1 + cos q), y = a sinq at ‘q’ always passes through the fixed point [2004] (a) (a, a) (b) (0, a) (c) (0, 0) (d) (a, 0)

85.

A function y = f ( x ) has a second order derivative

(a) ( x + 1)2

(b) ( x - 1)3

(c) ( x + 1)3

(d) ( x - 1)2

1 + sin 2 x

sin 2 x

1 + cos 2 x

sin 2 x

sin 2 x

2

cos x

2

sin x

1 + sin 2 x

Then the ordered pair (m, M) is equal to : (a) (– 3, 3) (b) (– 3, – 1) (c) (– 4, – 1) (d) (1, 3) Let AD and BC be two vertical poles at A and B respectively on a hori ontal ground. If AD = 8 m, BC = 11 m and AB = 10 m; then the distance (in meters) of a point M on AB from the point A such that MD2 + MC2 is minimum is ______. [NA Sep. 06, 2020 (I)]

2 is a local maxima of f. 3 The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x-axis and vertices C and D lie on the parabola, y = x2 – 1 below the x-axis, is : [Sep. 04, 2020 (II)]

(a)

(c) 86.

Let m and M be respectively the minimum and maximum values of [Sep. 06, 2020 (I)] cos 2 x

2 is a local minima of f. 3

(d) x = 1 is a local minima and x = -

TOPIC Ė Approximations, Maxima & Minima

82.

æ 3 3ö (b) ç - , ÷ è 2 2ø

f ( x) = (3x 2 + ax - 2 - a)e x , then :

f "( x) = 6( x - 1). If its graph passes through the point (2,1) and at that point the tangent to the graph is y = 3x – 5, then the function is [2004]

81.

The set of all real values of l for which the function æ p pö f ( x) = (1 - cos2 x ).(l + sin x) , x Î ç - , ÷ , has exactly 2 2ø one maxima and exactly minima, is: è

87.

88.

2 3 3 4 3

(b)

(d)

1 3 3 4 3 3

Suppose f(x) is a polynomial of degree four, having critical points at –1, 0, 1. If T = {x Î R | f ( x ) = f (0)}, then the sum of squares of all the elements of T is : [Sep. 03, 2020 (II)] (a) 4 (b) 6 (c) 2 (d) 8 Let f(x) be a polynomial of degree 3 such that f(–1) = 10, f(1)= –6, f(x) has a critical point at x = –1 and f ¢(x) has a critical point at x = 1. Then f(x) has a local minima at x = ________. [NA Jan. 8, 2020 (II)] Let f(x) be a polynomial of degree 5 such that x = ±1 are its critical points. If = 4, then which one of the following is not true ? [Jan. 7, 2020 (II)] (a) f is an odd function. (b) f(l) – 4f(–l) = 4. (c) x = 1 is a point of maxima and x = –1 is a point of minima of f. (d) x = 1 is a point of minima and x = –1 is a point of maxima of f.

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89.

Mathematics

If m is the minimum value of k for which the function

96.

f ( x ) = x kx - x 2 is increasing in the interval [0,3] and M is the maximum value of f in [0,3] when k = m, then the ordered pair (m, M) is equal to : [April 12, 2019 (I)] (a) (c) 90.

91.

92.

93.

(b)

(4,3 2)

(d)

(3,3 3)

(4,3 3)

pö æ The maximum value of 3cos q + 5sin ç q - ÷ for any real 6ø è value of q is: [Jan. 12, 2019 (I)] (a)

(b)

19

79 2

94.

(d) 31 34 Let P(4, –4) and Q(9, 6) be two points on the parabola, y2 = 4x and let this X be any point arc POQ of this parabola, where O is vertex of the parabola, such that the area of DPXQ is maximum. Then this minimum area (in sq. units) is: [Jan. 12, 2019 (I)] (a)

75 2

(b)

125 4

625 125 (d) 4 2 The maximum value of the function f(x) = 3x3 – 18x2 + 27 x – 40

(c)

95.

{

(1 + x2m )(1 + y2n )

is :

[Jan. 11, 2019 (II)]

}

2 on the set S = x Î R : x + 30 £ 11x is :

m+n 1 (d) 6mn 4 The maximum volume (in cu.m) of the right circular cone having slant height 3 m is: [Jan. 09, 2019 (I)] (a) 6 p (b) 3 3 p

(c) 97.

(c) 98.

4 p 3

(b) – 222 (d) 222

(d) 2 3 p

1 1 Let f (x) = x 2 + and g(x) = x - , x Î R - {-1,0,1} . 2 x x f (x) If h(x) = , then the local minimum value of h(x) is : g(x) [2018] (a) – 3 (b) -2 2

(c) 2 2 (d) 3 Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f (x) = 2x3 – 9x2 + 12x + 5 in the interval [0, 3]. Then [Online April 16, 2018] M – m is equal to (a) 1 (b) 5 (c) 4 (d) 9 100. If a right circularcone having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is [Online April 15, 2018] 99.

(b) 6 2p

(c) 6 3p (d) 8 2p 101. Twenty metres of wire is available for fencing off a flowerbed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is : [2017] (a) 30 (b) 12.5 (c) 10 (d) 25 102. A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then: [2016] (a) x = 2r (b) 2x = r (c) 2x = (p+ 4)r (d) (4– p) x= pr 103. The minimum distance of a point on the curve y = x2 – 4 from the origin is : [Online April 9, 2016] (a)

15 2

(b)

19 2

(c)

15 2

(d)

19 2

[Jan. 11, 2019 (I)] (a) – 122 (c) 122

1 (b) 2

(a) 8 3p

(c)

xm y n

The maximum value of the expression

(a) 1

(5,3 6)

Let a1, a2, a3, …. be an A. P. with a6 = 2. Then the common difference of this A.P., which maximises the product a1 a 4 a 5, is : [April 10, 2019 (II)] 3 8 (a) (b) 2 5 6 2 (c) (d) 5 3 If S1 and S2 are respectively the sets of local minimum and local maximum points of the function, f(x) = 9x4 + 12x3 – 36x2 + 25, xÎR, then : [April 08, 2019 (I)] (a) S1 = {–2}; S2 = {0, 1} (b) S1 = {–2, 0}; S2 = {1} (c) S1 = {–2, 1}; S2 = {0} (d) S1 = {–1}; S2 = {0, 2} The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is : [April 08, 2019 (II)] 2 3 (b) (a) 6 3 (c) 2 3 (d) 3

Let x, y be positive real numbers and m, n positive integers.

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Applications of Derivatives

104. Let k and K be the minimum and the maximum values of the function f(x) =

(1 + x )0.6 1 + x 0.6

in [0, 1] respectively, then

the ordered pair (k, K) is equal to : [Online April 11, 2015] (b) (2–0.4, 20.6) (a) (2–0.4, 1) (c) (2–0.6, 1) (d) (1, 20.6) 105. From the top of a 64 metres high tower, a stone is thrown upwards vertically with the velocity of 48 m/s. The greatest height (in metres) attained by the stone, assuming the value of the gravitational acceleration g = 32 m s2, is: [Online April 11, 2015] (a) 128 (b) 88 (c) 112 (d) 100 106. If x = –1 and x = 2 are extreme points of f ( x ) = a log x + b x 2 + x then

(a) a = 2, b = -

1 2

[2014]

(b) a = 2, b =

1 2

1 1 (d) a = -6, b = 2 2 107. The minimum area of a triangle formed by any tangent to

(c) a = -6, b =

x 2 y2 + = 1 and the co-ordinate axes is: 16 81 [Online April 12, 2014] (a) 12 (b) 18 (c) 26 (d) 36 108. The volume of the largest possible right circular cylinder

the ellipse

that can be inscribed in a sphere of radius = 3 is: [Online April 11, 2014] (a)

4 3p 3

(c) 4p

(b)

8 3p 3

(d) 2p

bö æ 109. The cost of running a bus from A to B, is ` ç av + ÷ , vø è where v km/h is the average speed of the bus. When the bus travels at 30 km/h, the cost comes out to be ` 75 while at 40 km/h, it is ` 65. Then the most economical speed (in km/ h) of the bus is : [Online April 23, 2013] (a) 45 (b) 50 (c) 60 (d) 40 110. The maximum area of a right angled triangle with hypotenuse h is : [Online April 22, 2013]

(a)

(c)

h2 2 2 h2 2

(b)

h2 2

(d)

h2 4

111. Let a, b Î R be such that the function f given by f (x) = ln | x | + bx2 + ax, x ¹ 0 has extreme values at x = –1 and x = 2 Statement-1 : f has local maximum at x = –1 and at x = 2. 1 -1 Statement-2 : a = and b = [2012] 2 4 (a) Statement-1 is false, Statement-2 is true. (b) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1. (c) Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1. (d) Statement-1 is true, statement-2 is false. 112. A line is drawn through the point (1,2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is : [2012] 1 (a) (b) – 4 4 1 (c) – 2 (d) 2 113. Let f: ( -¥, ¥ ) ® ( -¥, ¥ ) be defined by f(x) = x3 + 1. [Online May 26, 2012] Statement 1: The function f has a local extremum at x = 0 Statement 2: The function f is continuous and differentiable on ( -¥, ¥ ) and f ¢(0) = 0 (a) Statement 1 is true, Statement 2 is false. (b) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1. (c) Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1. (d) Statement 1 is false, Statement 2 is true. 114. Let f be a function defined by [2011RS] ì tan x , x¹0 ï f (x) = í x ï1, x=0 î

Statement - 1 : x = 0 is point of minima of f Statement - 2 : f ¢ ( 0) = 0. (a) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for statement-1. (b) Statement-1 is true, statement-2 is true; statement-2 is NOT a correct explanation for statement-1. (c) Statement-1 is true, statement-2 is false. (d) Statement-1 is false, statement-2 is true. x

æ 5p ö 115. For x Î ç 0, ÷ , define f ( x) = ò t sin t dt . Then f has è 2 ø 0 [2011] (a) local minimum at p and 2p (b) local minimum at p and local maximum at 2p (c) local maximum at p and local minimum at 2p (d) local maximum at p and 2p

EBD_8344

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Mathematics

116. Let f : R ® R be a continuous function defined by

f ( x) =

(a) The cubic has minima at

1

[2010]

e x + 2e - x

1 Statement -1 : f (c) = , for some c Î R. 3

Statement -2 : 0 < f (x) £

1

, for all x Î R

2 2 (a) Statement -1 is true, Statement -2 is true ; Statement 2 is not a correct explanation for Statement -1. (b) Statement -1 is true, Statement -2 is false. (c) Statement -1 is false, Statement -2 is true . (d) Statement - 1 is true, Statement 2 is true ; Statement -2 is a correct explanation for Statement -1. 117. Let f : R ® R be defined by f ( x) =

{ k2-x+23,x, ifif xx £>--11

If f has a local minimum at x = – 1 , then a possible value of k is [2010] 1 (a) 0 (b) 2 (c) –1 (d) 1 118. Given P(x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of P' (x) = 0. If P(–1) < P(1), then in the interval [ –1, 1] : [2009] (a) P(–1) is not minimum but P(1) is the maximum of P (b) P(–1) is the minimum but P(1) is not the maximum of P (c) Neither P(–1) is the minimum nor P(1) is the maximum of P (d) P(–1) is the minimum and P(1) is the maximum of P 119. Suppose the cubic x3 – px + q has three distinct real roots where p > 0 and q > 0. Then which one of the following holds? [2008]

(b) The cubic has minima at –

p p and maxima at – 3 3 p and maxima at 3

(c) The cubic has minima at both

p p and – 3 3

(d) The cubic has maxima at both

p p and – 3 3

p 3

x 2 + has a local minimum at 2 x (a) x = 2 (b) x = -2 [2006] (c) x = 0 (d) x = 1 121. The real number x when added to its inverse gives the minimum value of the sum at x equal to (a) –2 (b) 2 [2003] (c) 1 (d) –1

120. The function f ( x ) =

122. If the function f ( x) = 2 x3 - 9ax 2 + 12a 2 x + 1 , where a > 0 , attains its maximum and minimum at p and q respectively such that p 2 = q , then a equals

[2003]

1 (b) 3 2 (c) 1 (d) 2 123. The maximum distance from origin of a point on the curve

(a)

æ at ö æ at ö x = a sin t–b sin çè ÷ø , y = a cos t – b cos çè ÷ø , both b b a, b > 0 is (a) a – b

(b) a + b

(c) a 2 + b2

(d)

a 2 - b2

[2002]

www.jeebooks.in M-349

Applications of Derivatives

1.

2.

From (i) and (ii), 50 = 4p(10 + x)

f (t2 ) - f (t1 ) t2 - t1 t1 + t2 2at + b = a(t1 + t2 ) + b Þ t = 2 (d) Let the side of cube be a.

(c) Average speed = f '(t ) =

S = 6a 2 Þ

dS da da = 12a × Þ 3.6 = 12 a × dt dt dt

5.

Þ

dx 1 cm / min = dt 18p

[Q thickness of ice x = 5]

(b) According to the question,

...(i)

When y = 1 Þ x = 3 Diff. equation (i) w. r. t. t,

dV da æ 3 ö = 3a 2 × = 3(10)2 × ç =9 è 100 ÷ø dt dt (d) Since, function f (x) is continuous at x = 1, 3 V = a3 Þ

2x

+

\ f (1) = f (1 ) Þ ae + be -1 = c

50 = 4p(10 + 5)2

dy = – 25 at y = 1 dt By Pythagoras theorem, x2 + y2 = 4

da da Þ 12(10) = 3.6 Þ = 0.03 dt dt

3.

dx dt

Þ

dx dy + 2y =0 dt dt

...(i)

f (3) = f (3+ ) Þ 9c = 9a + 6c Þ c = 3a From (i) and (ii),

...(ii)

b = ae(3 - e)

...(iii)

é ae x - be - x ê f '( x ) = ê 2cx ê 2ax + 2c ë

y

x dx dy dx =0Þ 3 Þ x +y + (– 25) = 0 dt dt dt

-1 < x < 1 1< x < 3 3< x< 4

f '(0) = a - b, f '(2) = 4c

6.

Given, f '(0) + f '(2)= e a - b + 4c= e From eqs. (i), (ii), (iii) and (iv),

dVice = 50 dt 4 4 Vice = p(10 + r )3 - p(10)3 3 3 dV 4 dr dr Þ = p3(10 + r )2 = 4p(10 + r )2 dt 3 dt dt

Þ 13a - 3ae + ae 2 = e

4.

dx 25 Þ dt = cm/s 3 (a) Given that ice melts at a rate of 50 cm3/min.

\

...(iv)

a - 3ae + ae 2 + 12a = e

Þa=

2

e

e 2 - 3e + 13 (d) Let the thickness of ice layer be = x cm

Total volume V =

10

4 p(10 + x)3 3

dV dx = 4p(10 + x )2 dt dt Since, it is given that

dV = 50 cm3 / min dt

5 r

...(i) Substitute r = 5, ...(ii)

50 = 4p(225)

dr Þ dr = 50 1 cm/min = dt 4p(225) 18p dt

EBD_8344

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7.

Mathematics

9.

(d)

r

A

(a)

tan q = 1/2

q

120

10 m

O

3

5 m /min

B

Let OA = x km, OB = y km, AB = R (AB)2 = (OA)2 + (OB)2 – 2 (OA) (OB) cos 120

æ 1ö R2 = x2 + y2 – 2 xy ç - ÷ = x 2 + y 2 + xy è 2ø

Given that water is poured into the tank at a constant rate of 5 m3/minute.

R at x = 6 km, and y = 8 km

dv = 5m3 / min dt Volume of the tank is, \

R=

62 + 82 + 6 ´ 8 = 2 37 Differentiating equation (i) with respect to t

1 V = pr 2 h ...(i) 3 where r is radius and h is height at any time. By the diagram, tan q =

2R

=

r 1 = h 2

dh 2dr Þ h = 2r Þ = dt dt Differentiate eq. (i) w.r.t. ‘t’, we get

...(ii) 10.

(c)

dV = 5 in the above equation. dt

Volume of sphere V =

4 3 pr 3

4 288 ´ 3 p(r 3 ) Þ = r3 3 4 Þ 216 = r3 Þr= 6

Hence,

dr 1 = . dt 36

(a) Volume of sphere V =

4 3 pr 3

35 dr dr or = dt dt 4pr 2

35 = 4pr 2 .

...(i)

Surface area of sphere = S = 4pr2 dS dr dr = 4p´ 2 r ´ = 8pr . dt dt dt

...(i)

dv 4 2 dr = .3pr . dt 3 dt dr 4p = 4pr2. dt 1 dr = dt r2 Since, V = 288p, therefore from (i), we have

288p =

1 [2 ´ 8 ´ 20 + 2 ´ 6 ´ 30 + (8 ´ 30 + 6 ´ 20)] 2R

dV 4 dr = . p . 3r 2 . dt 3 dt

75p dh dh 1 5= Þ = m/min. 3 dt dt 5p 8.

dx dy æ dy dx ö dR = 2x + 2 y + ç x + y ÷ dt dt è dt dt ø dt

1 260 dR [1040] = = 2 ´ 2 37 37 dt

dV 1 æ dr dh ö = p2r h + pr 2 ÷ dt 3 çè dt dt ø Putting h = 10, r = 5 and

...(i)

dS 70 = dt r

(By using (i))

Now, diameter = 14 cm, r = 7 \ 11.

dS = 10 dt

(d) V =

dV dr 4 3 pr Þ = 4pr 2 . dt dt 3

S = 4pr2 Þ Þ 8 = 8pr

dS dr = 8pr . dt dt

1 dr dr Þ = dt dt pr

...(i)

www.jeebooks.in M-351

Applications of Derivatives

Putting the value of

dr in (i), we get dt

dV 1 = 4pr 2 ´ = 4r dt pr

Þ 12.

dV is proportional to r. dt

(c) Volume of spherical balloon = V = 4 pr 3 3 Differentiate both the side, w.r.t 't' we get,

dV æ dr ö ...(i) = 4pr 2 ç ÷ è dt ø dt \ After 49 min, Volume = (4500 – 49 ´ 72)p = (4500 – 3528)p = 972 p m3 Þ V = 972 p m3 4 3 pr 3 Þ r3 = 3 ´ 243 = 3 ´ 35 = 36 = (32)3 Þ r=9

\

972 p =

Given Putting \

13.

dV = 72 p dt

dV = 72p and r = 9, we get dt

dA dl db = -5 , =2, = -3 dt dt dt We know, A = l × b

Given :

dA db dl = l. + b. = -3l + 2b dt dt dt Þ – 5 = – 3l + 2b. When b = 2, we have 9 – 5 = – 3l + 4 Þ l = = 3m 3 16. (b) Let A = pr2.

Þ

dA dr = 2pr . dt dt 6p = 2p ( 30 ) .

Also, given

1 dr = 1mm = cm dt 10

\ A = pr2 dA dr 1 = 2 pr = 2p.50. = 10 p dt dt 10 Hence, area of plate increases in 10p cm2/hour. 14. (c) Let W = nw dW dw dn =n + w. Þ ...(i) dt dt dt 2 Given : w = t – t + 2 and n = 2t2 + 3

Þ

dw dn = 2t - 1 and = 4t dt dt \ Equation (i)

Þ

dw = (2t2 + 3) (2t – 1) + (t2 – t + 2) (4t) dt

Thus,

dW dt

= (2 + 3) (2 – 1) + (2) (4) t =1

= 5 (1) + 8 = 13

dr dt

3 dr dr 1 = Þ = = 0.1 30 dt dt 10 Thus, the rate at which the radius of the circular sheet increases is 0.1 17. (d)

Þ

u=0

f t+m

s+n

u=0

f¢ t

s

A

æ dr ö 72 p = 4 p´ 9 ´ 9 ç ÷ è dt ø

dr æ 2 ö =ç ÷ Þ dt è 9 ø (b) Let A = pr2 be area of metalic circular plate of r = 50 cm.

Þ

15. (d) Let A be the area, b be the breadth and l be the length of the rectangle.

B

v

v

As per question if point B moves s distance in t time then point A moves (s + n) distance in time (t + m) after which both have same velocity v. Then using equation v = u + at we get

v = f (t + m) = f 't Þ t =

f m f '- f

....(i)

Using equation v 2 = u 2 + 2 , as we get v2 = 2 f ( s + n) = 2 f ' s

Þs=

f n f '- f

....(ii)

1 2 Also for point B using the eqn s = ut + at , we get 2 1 2 s = f 't 2 Substituting values of t and s from equations (i) and (ii) in the above relation, we get f n f 2m 2 1 = f' f '- f 2 ( f '- f ) 2 Þ ( f '- f ) n =

1 ff ' m 2 2

EBD_8344

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Mathematics

18. (c) Let the li ard catches the insect after time t then distance covered by li ard = 21cm + distance covered by insect

19.

Þ

1 2 ft = 4 ´ t + 21 2

Þ

1 ´ 2 ´ t 2 = 20 ´ t + 21 2

Þ

dr = 50 dt

dy dy 9 20. (a) Given y 2 = 18 x Þ 2 y = 18 Þ = dx dx y dy 2dx dy =2 = Þ dt dt dx

æ æp ö æp öö = x ç p - ç - sin -1 (sin | x |) ÷ ÷ = x ç + | x | ÷ è2 øø è2 ø è

x³0 x 0 then x Î (-¥, 0) È ç , ¥÷ è 15 ø 22. (d) Since, function f (x) is twice differentiable and continuous in x Î [a, b]. Then, by LMVT for x Î [a, c]

- 1] are either both positive or

0

–ve 1 2

é 1ù x Î ê0, ú È [1, ¥) ë 2û

f ( c ) - f (a ) = f ¢(a ), a Î (a, c ) c-a

+ 1

www.jeebooks.in M-353

Applications of Derivatives

25. (c)

f ( x) =

Þ f ( - x) =

f ¢(- x) =

Þ 1 – 2a + 4 + a = 0 Þ a=5 Then, f(x) = x3 – 9x2 + 15x + 7 Now,

x2 1- x2

x2 1 - x2

= f ( x)

f ( x) - 14 =0 ( x - 1)2

2x (1 - x 2 )2

f(x) increases in x Î (10, ¥) Also f(0) = 0 and lim f ( x ) = -1 and f(x) is even function x ®± ¥

Set A = R – [–1, 0) And the graph of function f(x) is

Þ

x3 - 9 x 2 + 15 x + 7 - 14 =0 ( x - 1)2

Þ

( x - 1)2 ( x - 7) =0Þx=7 ( x - 1) 2

x

=

0

–1

x

28. (a) f (x) =

2

a +x

2

a +x 2

b + ( x – d )2

f '(x) – f '(2 – x) > 0

Þ

f '(x) > f '(2 – x)

But f ''(x) > 0 Þf '(x) is an increasing function Then, f '(x) > f '(2 – x) > 0 Þ x>2–x Þ x>1 Hence, f (x) is increasing on (1, 2) and decreasing on (0, 1). 27. (c) f(x) = x3 – 3 (a – 2) x2 + 3ax + 7, f(0) = 7

2 a 2 + x2

b2 + ( x – d )2 –

+

=

(

a2

=

(b

2

(a

2

+x

)

2 3/2

+

)

b2

(b

2

+ ( x – d )2

)

3/2

Þ

f ¢(x) > 0, ฀ x Î R

Þ

f (x) is increasing function.

>0

Hence, f(x) is increasing function. f ( x ) = x3 - 3 x 2 + 5 x + 7

For increasing f ¢ ( x ) = 3x2 - 6 x + 5 > 0

Þ xÎ R For decreasing f ¢ ( x ) = 3x2 - 6 x + 5 < 0 x= 1

Þ

f ¢(x) = 3x2 – 6(a – 2) x + 3a f ¢(1) = 0

30. (c) f (x) = sin4 x + cos4 x f '(x) = 4sin3 x cos x + 4cos3 x (– sin x) = 4sin x cos x (sin2 x – cos2 x) = – 2sin 2x cos 2x = – sin 4x

( x – d ) 2( x – d ) 2 b 2 + ( x – d )2

+ ( x – d )2

a 2 + x 2 – x 2 b2 + ( x – d )2 – ( x – d )2 + 3/2 (a 2 + x 2 )3/2 b 2 + ( x – d )2

29. (a) f (x) = 0 at x = 1

x(2 x)

(a 2 + x 2 )

–1

Þ

b + (d – x)2 2

2

f ¢(x) =

26. (c) f (x) = f (x) + f (2 – x) Now, differentiate w.r.t. x, f '(x) = f '(x) – f '(2 – x) For f (x) to be increasing f '(x) > 0

(d – x)

(x – d )

+

a 2 + x2 –

1

2

)

EBD_8344

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Mathematics

Thus statement -1 and 2 both are true and statement-2 is a correct explanation of statement 1.

f (x) is increasing when f '(x) > 0 Þ – sin 4x > 0 Þ sin 4x < 0 Þ

f ( x ) = xe x(1- x) , x Î R

35. (c)

x Î çæ p , p ù è 4 2 úû

f ' ( x ) = e x(1- x) . é1 + x – 2 x 2 ù ë û x(1- x ) é 2 . 2 x - x –1ù = -e ë û

y = sin 4x

p 4

= -2e

p 2

x(1- x )

éæ ù 1ö . êç x + ÷ ( x - 1)ú è ø 2 ë û

f ' ( x ) = -2e x(1- x ) . A

1ö æ where A = çè x + ÷ø ( x - 1) 2 31. (b) Since f ¢(x) > 0 and g¢(x) < 0, therefore f(x) is increasing function and g(x) is decreasing function. Þ f (x + 1) > f (x) and g (x + 1) < g (x) Þ g [f (x + 1)] < g [ f (x)] and f [g (x + 1)] < f [g (x)] Hence option (b) is correct. 32. (d) f (x) = 2x3 + 3x + k f'(x) = 6x2 + 3 > 0 " x Î R (Q x2 > 0) Þ f (x) is strictly increasing function Þ f (x) = 0 has only one real root, so two roots are not possible. 33. (c) Let y = x2 . e–x For increasing function, dy > 0 Þ x [(2 – x) e–x] > 0 dx Q x > 0, \ (2 – x) e–x > 0 Þ (2 – x)

1

1

36. (b) Given that f (x) = x3 + 5x + 1 \ f ' (x) = 3x2 + 5 > 0, " x Î R Þ f (x) is strictly increasing on R Þ f (x) is one one \ Being a polynomial f (x) is continuous and increasing.

f ( x ) = -¥ on R with xlim ®¥

Þ f '(x) = 7x6 + 70x4 + 48x2 + 30 > 0, " x Î R Also lim f ( x) = ¥ and lim f ( x ) = – ¥ x ®¥

f(x) = x log x, xÎ [1, 2]

1

g(x) = 2–x, x Î [1, 2] 1

2

x® – ¥

From (i) and (ii) clear that the curve y = f (x) crosses x-axis only once. \ f (x) = 0 has exactly one real root. 38. (d) Given that f (x) = tan–1 (sin x + cos x) Differentiate w.r. to x

Y

log 4

...(i)

Þ f is an increasing function on R

34. (a) f (x) = x log x, f (1) = 0, f (2) = 4 g(x) = 2 – x, g(1) = 1, g(2) = 0 log 10 > log 4 Þ 1 > log 4

O

é 1 ù \ f(x) is increasing on ê - ,1ú ë 2 û

R \ Range of f = ( - ¥, ¥) = Hence f is onto also. So, f is one one and onto R. 37. (b) Let f (x) = x7 + 14x5 + 16x3 + 30x –560

< 0 , but it is not possible ex Hence the statement-2 is false.

é 1 ù Hence, f ¢(x) is +ve in ê - ,1ú ë 2 û

x ®¥

e For 0 < x < 2, (2 – x) < 0

\

é 1 ù be opposite to the sign of A which is –ve in ê - ,1ú ë 2 û

and lim f ( x ) = ¥

>0

x

Now, exponential function is always +ve and f ¢(x) will

X

f '(x) =

1 1 + (sin x + cos x) 2

.(cos x - sin x )

...(ii)

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Applications of Derivatives

æ 1 ö 1 2. ç cos x sin x ÷ è 2 ø 2

=

(4 x3 + x 4 × y ')e y +

1 + (sin x + cos x) 2

p p æ ö 2 ç cos .cos x - sin .sin x÷ è ø 4 4

=

1 + (sin x + cos x) 2

\

f '(x) =

y - 0 = -2( x - 1) Þ 2 x + y = 2

p p p < x+ < 2 4 2

\

3p p 0 Þ cos ç x + ÷ > 0 è 4ø

Þ

=0

\ Equation of tangent; 2

Given that f (x) is increasing

Þ-

1+ y

æ dy ö Þç ÷ = -2 è dx ø(1,0)

pö æ 2 cos ç x + ÷ è 4ø 1 + (sin x + cos x )

y'

x Î[1/ 3, ¥)

a 1 = = 0.5. b 2

43. (4)

1ù æ \ f (x) is incorrectly matched with ç -¥, ú 3û è (b) The given tangent to the curve is,

For (1, 2) of y 2 = 4 x Þ t = 1, a = 1 Equation of normal to the parabola

y = x loge x

Þ x + y = 3 intersect x-axis at (3, 0)

(x > 0)

Þ

dy = 1 + log e x dx

Þ

dy ù = 1 + loge c dx úû x = c

dy = ex dx Equation of tangent to the curve y = ex Þ

(slope)

Q The tangent is parallel to line oining (1, 0), (e, e) \1 + log e c =

e-0 e -1

1 e Þ log e c = - 1 Þ log e c = e -1 e -1 1 e e -1

Þ tx + y = 2at + at 3

Þc= 4 y 41. (c) The given curve is, x × e + 2 y + 1 = 3 Differentiating w.r.t. x, we get

Þ y - e c = ec ( x - c) Q Tangent to the curve and normal to the parabola intersect at same point. \ 0 - ec = ec (3 - c) Þ c = 4. 44. (91) y=

6

ì3

4

ü

å k cos -1 íî 5 cos kx - 5 sin kxýþ

k =1

Let cos a =

3 4 and sin a = 5 5

EBD_8344

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Mathematics 6

\ y = å k cos -1{cos a cos kx - sin a sin kx} k =1

3 (a2 + 3) = (a2 – 3)2 Þ a2 = 9

6

= å k cos -1 (cos(kx + a))

And, b =

k =1 6

6

k =1

k =1

= å k (kx + a) = å (k 2 x + ak ) \

6 6(7)(13) dy = å k2 = = 91. dx k =1 6

45. (4.0)P º (x1, y1) 2yy¢ – 6x + y¢ = 0 Þ

æ 6 x1 ö y¢ = ç ÷ è 1 + 2 y1 ø

[By point slope form, y – y1 = m(x – x1)] 9 – 6y1 = 1 + 2y1 y1 = 1 x1 = ± 2

æ ±12 ö Slope of tangent (m) = ç ÷ =±4 è 3 ø \ |m| = 4 46. (d) Given equation of curve is x2 + 2xy – 3y2 = 0 Þ 2x + 2y + 2xy¢ – 6yy¢ = 0 Þ x + y + xy¢ –3yy¢ = 0 Þ y¢(x – 3y) = – (x + y) dy x + y = dx 3 y - x

- dx x - 3 y = Slope of normal = dy x - 3y

2-6 =–1 2+2 Equation of normal to curve = y – 2 = – 1 (x – 2) Þ x+y=4 \ Perpendicular distance from origin

Normal at point (2, 2) =

=

47

0+ 0- 4 2

a a Þ =6 b b

1 2 These values of a and b satisfies |6a + 2b| = 19 48. (d) y = x3 + ax – b Since, the point (1, –5) lies on the curve. Þ1+a– b= –5 Þ a – b = –6 Þ a = ±3, b = ±

...(i)

Since, required line is perpendicular to y = x – 4, then slope of tangent at the point P (1, –5) = –1 3+a=–1 a=–4 b=2 the equation of the curve is y = x3 – 4x – 2 (2, –2) lies on the curve 49. (d) y = f(x) = x3 – x2 – 2x dy - 3x 2 - 2 x - 2 dx f(1) = 1 – 1 – 2 = – 2, f(–1) = –1 –1 + 2 = 0 Since the tangent to the curve is parallel to the line segment oining the points (1, –2) (–1, 0) Since their slopes are equal

Þ 3x 2 - 2 x - 2 =

-2 - 0 2

Þ x = 1,

-1 3

ì -1 ü Hence, the required set S = í ,1ý î3 þ 50. (c) Equation of tangent to circle at point ( 3,1) is

3x + y = 4 P( 3,1)

=2 2

2

(a) Given curve is, y = Þ

Þ a2 - 3 =

æ dy ö ç ÷ è dx øat x =1 = 3 + a

\

Þ

x a2 - 3

dy = 3x 2 + a dx

æ3 ö ç 2 - y1 ÷ æ 1 + 2 y1 ö ç ÷ = -ç ÷ x è 6 x1 ø 1 ÷ ç è ø

Þ Þ \

a2 - 3 2 1 dy = 2 =- =dx (a,b) (a - 3) 2 6 3

x 2

x -3

dy ( x 2 - 3) - x (2 x ) - x 2 - 3 = = dx ( x 2 - 3)2 ( x 2 - 3) 2

O

æ 4 ö Aç ,0÷ è 5 ø

M 3x + y = 4

2

2

x +y =4

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Applications of Derivatives

æ 4 ö ,0 ÷ coordinates of the point A = ç è 3 ø

Area =

y2 = x

53. (a)

1 4 2 1 ´1 = ´ OA ´ PM = ´ sq. units 2 3 3 2

tö æ1 P ç t2, ÷ 2ø è4

51. (c) Given, the equation of parabola is, x2 = 12 – y

æ3 ö Q ç ,0 ÷ è2 ø

y (0, 12) (–t, 12–t2)

(t, 12–t2) x

1 Here the curve is parabola with a = . 4 æ t2 t ö Let P(at2, 2at) or P ç , ÷ be a point on the curve. è 4 2ø

Now, y2 = x 1 dy dy Þ2 y dx = 1 = dx = 2 x

Area of the rectangle = (2t) (12 – t2) A = 24t – 2t3

1 æ dy ö = Þçè dx ÷ø at p t

dA = 24 – 6t2 dt

\

tö 1 2ö æ æ çè y - ÷ø = -t çè x - t ÷ø 4 2

dA Put = 0 Þ 24 – 6t2 = 0 dt Þ t = ±2

+ –2

52.

1 1 3 Þy = -tx + t + t 2 4

c =-

...(i)

æ3 ö For minimum PQ, (i) passes through Q çè , 0ø÷ 2

+2

At t = 2, area is maximum = 24(2) – 2(2)3 = 48 – 16 = 32 sq. units (b) Q Tangent to the given curve is parallel to line 2y = 4x + 1 \ Slope of tangent (m) = 2 Then, the equation of tangent will be of the form y = 2x + c ...(i) 2 Q Line (i) and curve y = x – 5x + 5 has only one point of intersection. \ 2x + c = x2 – 5x + 5 x2 – 7x + (5 – c) = 0 \ D = 49 – 4(5 – c) = 0 Þ

equation of normal at P to y2 = x is,

29 4

-3 t t 3 t + + = 0 Þ –4t + t3 = 0 2 2 4 Þ t(t2 – 4) = 0 Þ t = –2, 0, 2 Q t ³ 0 Þ t = 0, 2 If t = 0, P(0, 0) Þ AP = If t = 2, P(1, 1) Þ AP =

3 2

5 2

æ3 ö 5 Shortest distance çè , 0÷ø and y = x is 2 2 54. (b) The equation of curve y = xe x

Hence, the equation of tangent: y = 2x –

29 4

Þ

dy 2 2 = e x .1 + x.e x .2 x dx

2

EBD_8344

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Mathematics 2

Since (1, e) lies on the curve y = xe x , then equation of tangent at (1, e) is

(

2

x 2 y – e = e (1 + 2 x )

)

x =1

( x - 1)

y – e = 3e(x – 1) 3ex – y = 2e So, equation of tangent to the curve passes through the

Adding eqn (i) and (ii), we get 2y = 12 Þ y = 6 Then, from eqn (i)

Differentiate equation (i) with respect to x

55. (c) f(x) = y = x3/2 + 7

Þ

y = 2 + x2 ...(ii)

x = ±2

æ4 ö point çè , 2e÷ø 3

Þ

56. (b) Since, the equation of curves are y = 10 – x2 ...(i)

dy 3 x >0 Þ dx 2 f(x) is increasing function " x > 0

dy = –2x Þ dx

æ dy ö èç dx ø÷ (2, 6) = 4 and

At (–2, 6), tan q = \ |tan q| =

(

)

(4) - ( -4) 8 8 = Þ tan q = 1 + (4)( -4) -15 15

8 15

57. (c) Let curve intersect each other at point P(x1, y1)

Y

mTP = mat P = –1

y2 = 6x

Þ

æ ö 1 3/ 2 ç x1 ÷ 3 2 ´ x ç 1 ÷ 2 1 = –1 x èç 1 2 ø÷

Þ

x12 2 - =x -1 1 3 2

Þ

-3x12 = 2x1 – 1 Þ 3 x12 + 2 x1 - 1 = 0

Þ

3x12 + 3x1 - x1 - 1 = 0

Þ

3x1(x1 + 1) – 1(x1 + 1) = 0

Þ

x1 =

1 3

P(x1, y1) X

9x2 + by 2 = 16

(Q x1 > 0)

1 ö æ1 Þ P çè 3 ,7 + ÷ 3 3ø

TP =

æ dy ö çè ÷ø dx ( -2, 6) = –4

æ ( -4) - (4) ö 8 At (2, 6) tan q = çè 1 + ( -4) ´ (4) ÷ø = 15

1 ,7 T 2

3/ 2 Let P x1 , x1 + 7

æ dy ö çè ÷ø dx ( -2, 6) = 4

Differentiate equation (ii) with respect to x dy = 2x Þ dx

(0, 7)

æ dy ö çè ÷ø dx (2, 6) = –4 and

1 1 1 7 + = 27 36 6 3

Since, point of intersection is on both the curves, then y12 = 6x1

...(i)

and 9x12 + by12 = 16

...(ii)

Now, find the slope of tangent to both the curves at the point of intersection P(x1, y1) For slope of curves: Curve (i):

æ dy ö çè ÷ø dx (x

1,y1 )

= m1 =

3 y1

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Applications of Derivatives

Equation of tangent at (x1, y1) is y = mx + c

Curve (ii):

9x æ dy ö = m2 = - 1 and çè dx ÷ø by1 (x1,y1 )

Þ y=

Since, both the curves intersect each other at right angle then, m1m 2 = -1 Þ

27x1 by12

= 1 Þ b = 27

x1

1 9 = 6 2 58. (c) Let P(2t, t2) be any point on the parabola. Centre of the given circle C = (– g, – f ) = (–3, 0) For PC to be minimum, it must be the normal to the parabola at P.

y2 - y1 t 2 - 0 = x2 - x1 2t + 3

Also, slope of tangent to parabola at P = \ Slope of normal =

dy x = =t dx 2

-1 t

t 2 - 0 -1 = 2t + 3 t Þ t 3 + 2t + 3 = 0 Þ (t + 1) (t 2 – t + 3) = 0 \ Real roots of above equation is t=–1 Coordinate of P = (2t, t 2) = (– 2, 1) Slope of tangent to parabola at P = t = – 1 Therefore, equation of tangent is: (y – 1) = (– 1) (x + 2) Þ x+y+1=0 59. (d) Equation of hyperbola is : 4y2 = x2 + 1 Þ – x2 + 4y2 = 1 x2 12

+

y2 æ1ö ç ÷ è2ø

2

=1

\ a = 1, b = 1 2

4 y12 – x12 4 y1

Therefore, y =

dy = 2 x1 dx

x dy 2 x1 Þ = = 1 dx 8 y1 4 y1

=

1 4 y1

1 x1 x+ 4 y1 4 y1

Þ 4y1y = x1 x + 1

æ 1 ö B ç 0, ÷ è 4 y1 ø

Let midpoint of AB is (h, k) \ h=

–1 2 x1

Þ x1 =

–1 1 & y1 = 8k 2h 2

2

æ 1 ö æ – 1ö Thus, 4 ç ÷ = ç ÷ + 1 è 8k ø è 2h ø Þ

1 16 k 2

Þ1=

=

1 4h2

+1

16k 2 + 16k 2 4h 2

Þ h2 = 4k2 + 16 h2 k. So, required equation is x2 – 4y2 – 16 x2 y2 = 0 60. (b) Since, x2 + 3y2 = 9 Þ 2x + 6y

Now, tangent to the curve at point (x1, y1) is given by 4 ´ 2 y1

ÞC =

x1 x1 +c 4 y1

æ –1 ö , 0 ÷ and y axis at which intersects x axis at A ç è x1 ø

\

Þ –

As tangent passes through (x1, y1) \ y1 =

y12

\ from equation (i), b = 27 ´

Slope of line PC =

x1 ×x+c 4 y1

Þ

dy =0 dx

dy – x = dx 3 y

Slope of normal is –

dx 3y = dy x

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Mathematics

æ

dx ö ÷ è dy ø(3cos q,

Þ ç–

æ dx ö & ç– ÷ è dy ø( -3sin q, =

3sin q)

=

3 3 sin q = 3 tan q = m1 3 cos q

4(x0)2 – 9(y0)2 = 36

- 9x 4y and y0 = 13 13 From equation (ii), we get 9x2 – 4y2 = 169 Hence, locus of point P is : 9x2 – 4y2 = 169

As, b is the anagle between the normals to the given ellipse then

=

So, tan b

=

x+6 62. (c) We have y = (x - 2)(x - 3)

At y-axis, x = 0 Þ y = 1 On differentiating, we get

m1 – m2 1 + m1m2

3 tan q + 3 cot q 1 – 3 tan q cot q

...(ii)

From equation (i): x0 =

3 cos q)

3 3 cos q = – 3 cot q = m2 – 3 sin q

tan b =

æ - 13x0 13 y0 ö , ...(i) So, P (x, y) º ç 4 ÷ø è 9 Q (x0, y0) lies on hyperbola, therefore

=

3 tan q + 3 cot q 1–3

3 tan q + cot q 2

dy (x 2 - 5x + 6) (1) - (x + 6) (2x - 5) = dx (x 2 - 5x + 6)2

dy = 1 at point (0, 1) dx

Þ

1 3 sin q cos q = + cot b 2 cos q sin q

\ Slope of normal = – 1 Now equation of normal is y – 1 = –1 (x – 0) Þ y–1=–x x + y= 1

Þ

1 3 1 = cot b 2 sin q cos q

æ1 1ö \ ç , ÷ satisfy it. è2 2ø

Þ

1 3 = cot b sin 2 q

Þ

2 cot b 2 = sin 2 q 3

61. (c) Given, 4x2 – 9y2 = 36 After differentiating w.r.t. x, we get 4.2.x – 9.2.y.

dy =0 dx

dy 4 x = dx 9 y -9 y So, slope of normal = 4x Now, equation of normal at point (x0, y0) is given by

Þ Slope of tangent =

y – y0 =

-9 y0 (x – x0) 4 x0

63. (c) Eccentricity of ellipse = Now, –

1 a =–4Þa=4× =2Þa=2 2 e

æ 1ö We have b2 = a2 (1 – e2) = a2 ç 1 - ÷ è 4ø 3 =3 4 \ Equation of ellipse is

=4×

x 2 y2 + =1 4 3

Now differentiating, we get

x 2y + ´ y¢ = 0 Þ y¢ = – 3x 2 3 4y

As normal intersects X axis at A, Then

Þ

æ 13x0 ö æ 13 y0 ö Aº ç , 0 ÷ and B º ç 0, 4 ÷ø 9 è ø è As OABP is a parallelogram

3 2 1 y¢ (1,3/ 2) = - ´ = 4 3 2

æ 13 y0 ö º Midpoint of AP \ midpoint of OB º ç 0, 8 ÷ø è

1 2

Slope of normal = 2

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Applications of Derivatives

65. (d) x2y2 – 2x = 4 – 4y Differentiate w.r.t. 'x'

æ 3ö \ Equation of normal at ç1, ÷ is è 2ø

2xy2 + 2y . x2 .

3 y – = 2 (x – 1) Þ 2y – 3 = 4x – 4 2

\ 4x – 2y = 1 64. (c)

Y

B

, (x y)

X

Let y = f (x) be a curve slope of tangent = f ¢ (x) Equation of tangent (Y – y) = f ¢ (x) (X – x) Put Y = 0 æ y ö X = çç x ÷ f ¢ ( x ) ÷ø è

Put X = 0

Y = y - x f ¢ (x)

Þ

æ ö y , 0 ÷÷ A = çç x ¢ f x ( ) è ø

and B = (0, y – x f ¢ (x)) \ AP : PB = 1 : 3 3æ y ö çx ÷ 4 çè f ¢ ( x ) ÷ø

Þ

x =

Þ

dy -3y -3y = Þ x= dx x ¢ f (x)

dy -3 dx C = Þ y= 3 y x x Q

\

f (a) = 1 Þ C = 1

y=

1 x3

dy (2y . x2 + 4) = 2 – 2x . y2 dx

Þ

2 - 2 ´ 2 ´ 4 -14 7 dy = = = dx 2, -2 2 ( -2 ) ´ 4 + 4 -12 6

\

Equation of tangent is

7 or 7 x - 6 y = 26 6 \ (–2, –7) does not passes through the required tangent.

A

Þ

Þ

( y + 2) = ( x - 2)

P

Þ

æ 1ö is required curve and ç 2, ÷ passing è 8ø

66. (d)

æ 1 + sin x ö f ( x) = tan –1 ç è 1– sin x ÷ø

æ ç = tan –1 ç ç ç è

2ö x xö æ xö æ çè sin + cos ÷ø ÷ 1 + tan ÷ 2 2 ÷ –1 ç 2 ÷ 2 ÷ = tan ç x x x æ ö ç 1 – tan ÷ çè sin – cos ÷ø ÷ è 2ø x 2 ø

æ æ p xö ö = tan –1 ç tan ç + ÷ ÷ è 4 2ø ø è

Þ y=

p x dy 1 + Þ = dx 2 4 2

Slope of normal =

–1 = –2 dy æ ö çè ÷ø dx

æp p pö Equation of normal at çè , + ÷ø 6 4 12

pö æp pö æ y – ç + ÷ = –2 ç x – ÷ è 4 12 ø è 6ø

y–

4p 2p = –2 x + 12 6

y–

p p = –2 x + 3 3

y = –2 x +

2p 3

æ 2p ö This equation is satisfied only by the point ç 0, ÷ è 3ø

67. (a)

1 through y = 3 x

dy dy – 2 = –4 . dx dx

Þ Þ

dy 1 2 = ´4= 3 dx 2 4 x - 3 4x – 3 = 9 x=3

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Mathematics

71. (d) x2 – y + 6 = 0

So, y = 4 Equation of normal at P (3, 4) is

2x –

3 y – 4 = – (x – 3) 2 i.e. 2y – 8 = – 3x + 9 Þ 3x + 2y – 17 = 0 This line is satisfied by the point (1, 7) 68. (d) P (4t2 + 3,8t3 – 1)

dy =4 dx ( x, y )= (2,10) equation of tangent y – 10 = 4(x – z) 4x – y + z = 0 tangent passes through (a, b) 4a – b + z = 0 Þ b = 4a + z and 2x + 2yy' + 8 – 2y' = 0

dy / dt dy = = 3t (slope of tangent at P) dt / dt dx Let Q = (4l2 + 3,8l3 – 1) slope of PQ = 3t

8t 3 - 8l3 2

2

2 x + 8 2a + 8 = =4 2 - 2 y 2 - 2b from (i) and (ii)

= 3t

a=

-t 2 \ Q [t2 + 3, – t3 – 1]. 69. (b) Given curve is x2 + 2xy – 3y2 = 0 Differentiatew.r.t. x

...(i) so,

dy dy + 2y - 6y =0 dx dx

Equation of normal at (1, 1) is y=2–x Solving eqs. (i) and (ii), we get x=1,3 Point of intersection (1, 1), (3, –1) Normal cuts the curve again in 4th quadrant.

æp ö 70. (b) Given curve is sin y = x sin ç + y÷ è3 ø Diff with respect to x, we get æp ö æp ö dy dy cos y = sin çè + y÷ø + x cos çè + y÷ø 3 3 dx dx æp ö sin ç + y÷ è3 ø æp ö cos y - x cos ç + y ÷ è3 ø dy 3 at (0, 0) = dx 2

dx = -2 sin t + 2 [ t cos t + sin t ] dt

dy = 2t sin t dx

...(ii)

dy 2t sin t = dx 2t cos t dy = tan t dx

æ dy ö =1 çè ÷ø dx t =p / 4

so the slope of the normal is – 1 At t = p /4x =

2+

p 2 2

and

y = 2 - p/ 2 2 the equation of normal is

dy Þ = dx

3 y=0

–8 2 ,b = 17 17

dy = 2 cos t - 2 [ - t sin t + cos t ] dt

æ dy ö =1 ç ÷ è dx ø(1, 1)

Þ Equation of normal is y – 0 = -

...(ii)

æ -8 2 ö ç , ÷ è 17 17 ø 72. (a) Given that x = 2 cos t + 2t sin t

t = l (or) l =

Þ 2x +

...(i)

y¢ =

4t - 4l Þ t3 – 3l2t + 2l3 = 0 (t – l) . (t2 + tl – 2l2) = 0 (t – l)2 . (t + 2l) = 0

2x + 2x

dy dy =0Þ = 2x dx dx

éy ë

2 3

(

y- 2+ (x – 0)

)

( (

2 - p / 2 2 ù = -1 éê x û ë

p 2 2

))

2 + p / 2 2 ùú û

= -x + 2 + p / 2 2

x + y = 2 2 , so the distance from the origin is 2

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Applications of Derivatives

73. (c) Given, y = 3 sin q.cos q

Þ x=

dy = 3[sin q(- sin q) + cos q(cos q)] dq

Thus x + 2y = k

dy = 3[cos 2 q - sin 2 q] = 3 cos 2q dq and x = eq sin q

Now ...(ii)

3(cos 2 q - sin 2 q) dy 3cos 2q = q = q dx e (sin q + cos q) e (sin q + cos q)

3( cos q + sin q )(cos q - sin q) dy = eq ( sin q + cos q ) dx

3(cos q - sin q) dy = dx eq

dy 2 x x dy ù = = , ú =1 dx 8 4 dx û x = 4

1 = -1 dy dx Euqation of normal at x = 4 is y – 2 = – 1 (x – 4) Þy = – x + 4 + 2 = – x + 6 Þx+ y = 6 76. (c) Since the tangent is parallel to x-axis,

Slope of normal = -

8 dy = 0 Þ 1- 3 = 0 Þ x = 2 Þ y = 3 dx x Equation of the tangent is y – 3 = 0 (x – 2) Þ y=3

dy = 0 dx

3(cos q - sin q)

dy = 2 x - 5 \ m1 = (2 x - 5)(2, 0) = -1 , dx m2 = (2 x - 5)(3, 0) = 1 Þ m1m2 = -1

77. (b)

eq or cos q – sin q = 0 Þ cos q = sin q tan p p Þ q= 4 4

i.e. the tangents are perpendicular to each other. 78. (d) Given x = a ( cos q + q sin q)

74. (d) Let y = cos (x + y) dy æ dy ö = - sin ( x + y ) ç1 + ÷ Þ dx è dx ø

Now, given equation of tangent is x + 2y = k

So,

...(i)

\

Given tangent is parallel to x-axis then

Þ Slope =

p =k 2

75. (d) x2 = 8y When, x = 4, then y = 2

dx = eq (sin q + cos q) dq Dividing (i) by (i)

Þ tan q = 1 Þ tan q =

Þ

...(i)

dx = eq cos q + sin q eq dq

0=

p and y = 0 2

-1 2

Þ

dx = a ( - sin q + sin q + q cos q ) dq

Þ

dx = aq cos q dq

...(i)

.....(i)

y = a ( sin q - q cos q) dy = a [ cos q - cos q + q sin q ] dq

dy -1 put this value in (i), we get = dx 2

dy = aq sin q dq From equations (i) and (ii) we get

Þ

.....(ii)

-1 æ 1ö = - sin ( x + y ) ç1 - ÷ 2 è 2ø

dy = tan q Þ Slope of normal = – cot q dx

Þ sin (x + y) = 1

Equation of normal at 'q ' is y – a (sin q – q cos q) = – cot q (x – a (cos q + q sin q)) Þ y sin q – a sin 2 q + a q cos q sin q

Þ x+ y =

p p Þ y= -x 2 2

p Now, - x = cos (x + y) 2

= – x cos q + a cos 2 q + a q sin q cos q Þ x cos q + y sin q = a Clearly this is an equation of straight line which is at a constant distance ‘a’ from origin.

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Mathematics

79. (d) Since, x = a (1 + cos q)

æ 3p ö M = f ç ÷ = -2 + 1 = -1 è 4ø

dx = -a sin q and y = a sin q dq dy Þ = a cos q dq Þ

So, (m, M ) = ( -3, - 1) 82. (5)

dy = - cot q. dx \ The slope of the normal at q = tan q \ The equation of the normal at q is

C

\

D

8

y - a sin q = tan q( x - a - a cos q) Þ y cos q – a sin q cos q = x sinq – a sinq – a sinq cosq

Þ x sin q - y cos q = a sin q Þ y = ( x - a ) tan q f ¢¢ ( x) = 6( x - 1). Inegrating, we get

f ¢ ( x) = 3 x 2 - 6 x + c

Slope at (2, 1) = f ¢(2) = c = 3 [Q slope of tangent at (2,1) is 3] \ f ¢ ( x ) = 3 x 2 - 6 x + 3 = 3( x - 1) 2

Inegrating again, we get f ( x ) = ( x - 1)3 + D The curve passes through (2, 1)

Let f ( x) = 2

2

sin x

cos 2 q

= 0 - sin 2 x 1

M

10–x

B

\ (MD)2 + (MC ) 2 = 64 + x 2 + 121 + (10 - x) 2 = f ( x) (say) f '( x) = 2 x - 2(10 - x) = 0 Þ 4 x = 20 Þ x = 5

f ''( x ) = 2 - 2( -1) > 0 \ f (x) is minimum at x = 5 m. 83. (d)

f ( x) = (1 - cos 2 x)(l + sin x) = sin 2 x(l + sin x)

Þ f ( x) = l sin 2 x + sin 3 x ...(i)

2l Þ x = a (let) 3 So, f (x) will change its sign at x = 0, a because there is Þ sin x = 0 and sin x = -

sin 2 x sin 2 x

1 sin 2 x 1 + sin 2 x R1 ® R1 - 2 R3 ; R2 ® R2 - 2 R3 0

x

Þ f '( x ) = sin x cos x [2l + 3sin x] = 0

Þ 1 = (2 - 1)3 + D Þ D = 0 \ f (x) = ( x – 1)3 81. (b) C1 ® C1 + C2 2 1 + sin 2 x

A

Let AM = x m

which always passes through (a, 0) 80. (b)

11

æ -p p ö exactly one maxima and one minima in ç , è 2 2 ÷ø – -p 2

- (2 + sin 2 x)

+ a

0

- (2 + sin 2 x) = -2 - 2 sin 2 x

sin x

f '( x) = -2cos 2 x = 0 Þ cos 2 x = 0 Þ x =

-

p 3p , 4 4

æ pö So, f '' ç ÷ = 4 > 0 è 4ø

Þ -1 £ -

(minima)

æ pö m = f ç ÷ = -2 - 1 = -3 è 4ø æ 3p ö f '' ç ÷ = -4 < 0 è 4ø

+ a

p 2

Now, sin x = -

f ''( x) = 4sin 2 x

+ p 2

OR

1 + sin 2 x

2

– 0

+ p 2

2l 3

2l 3 3 £ 1 Þ - £ l £ - {0} 3 2 2

Q If l = 0 Þ f ( x) = sin 3 x (from (i))

Which is monotonic, then no maxima/minima

(maxima)

æ 3 So, l Î ç - , è 2

3ö ÷ - {0} 2ø

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Applications of Derivatives

84. (d) The given function

Q f ( x ) = f (0)

f ( x ) = (3x 2 + ax - 2 - a)e x f '( x) = (6 x + a)e + (3x + ax - 2 - a)e 2

x

Þk

x

( x4 - 2 x2 ) +C = C 4

Þ x2 ( x2 - 2) = 0

f '( x) = [3x 2 + (a + 6) x - 2]e x Q x = 1 is critical point :

Þ x = 0, 2, - 2

\ f '(1) = 0

Þ T = {0, 2, - 2}

Þ (3 + a + 6 - 2) × e = 0

87. (3) Let f(x) = ax3 + bx2 + cx + d (Q e > 0)

Þ a = -7

\ f '( x) = (3x2 - x - 2)e x = (3x + 2)( x - 1)e x +

–2/3 \x = -

1

and x = 1 is point of local minima. 85. (d) Area of rectangle ABCD

A = 2 x × ( x - 1) = 2 x - 2 x

b=

-3 9 ,c=4 4

Þ

f(x) = a(x3 – 3x2 – 9x) + d

f ¢(x) =

æ d 2 Aö -12 = (12 x) Þ ç 2 ÷ = 0

At x = p, f ''( x ) < 0

116. (d) Given f ( x) =

Clear that f (x) is minimum at (–1, 1) \ f (–1) = 1 1=k+2Þ k=–1 118. (a) Given that P (x) = x4 + ax3 + bx2 + cx + d Þ P' (x) = 4 x3 + 3ax2 + 2bx + c But given P' (0) = 0 Þ c = 0 \ P(x) = x4 + ax3 + bx2 + d Again given that P ( – 1) < P (1) Þ 1–a+b+d0 Now P ' (x) = 4x 3 + 3ax 2 +2bx = x (4x 2 + 3ax + 2b) As P' (x) = 0, there is only one solution x = 0, therefore 4x2 + 3ax + 2b = 0 should not have any real roots i.e. D < 0

d2y dx 2 x = p

= + ve and

3

\ y has minimum at x = x= –

d2y dx 2 x = – p

= – ve

3

p and maximum at 3

p 3

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Applications of Derivatives

1 2 x 2 + Þ f '( x) = - 2 = 0 2 x 2 x Þ x2 = 4 Þ x = 2, – 2;

120. (a) Given f (x) =

Now, f ''( x) =

4 x3

f ''( x )] x =2 = + ve Þ f ( x ) has local min at x = 2.

1 1 dy Þ = 1x dx x2 For maxima. or minima.,

121. (c) ATQ, y = x +

1-

1 x2

= 0 Þ x = ±1

Þ x = a or x = 2a. f ''( x ) = 12 x - 18a f "(a) = – 6a < 0 \ f(x) is max. at x = a, f "(2a) = 6a > 0 \ f(x) is min. at x = 2a \ p = a and q = 2a ATQ, p 2 = q \ a 2 = 2a Þ a = 2or a = 0 but a > 0, therefore, a = 2.

123. (b) We know that distance of origin from (x, y) =

x2 + y 2

æ at ö a 2 + b2 - 2ab cos ç t - ÷ ; è bø

æ d2 yö d2 y 2 = 3 Þç 2 ÷ = 2> 0 2 dx x è dx ø x =1

=

\ y is minimum at x = 1

£ a 2 + b 2 + 2ab

122. (d)

f ( x) = 2 x3 - 9ax 2 + 12a 2 x + 1 2

2

f '( x ) = 6 x - 18ax + 12a ;

For maxima or minima.

éì ù æ at ö ü = -1ú = a + b ê í cos ç t - ÷ ý è b ø þ min êë î úû \ Maximum distance from origin = a + b

6 x 2 - 18ax + 12a2 = 0 Þ x 2 - 3ax + 2a2 = 0

EBD_8344

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Mathematics

22

Integrals Standard Integrals, Integration by

5.

TOPIC Ć Substitution, Integration by Parts

Let f ( x) = ò

x (1 + x)2

dx ( x ³ 0). Then f (3) – f (1) is equal

to : 1.

2.

3.

æ ( x -1) 2 ö t cos(t 2 )dt ÷ ç ò0 lim ç [Sep. 06, 2020 (I)] ÷ x ®1 ç ( x - 1)sin( x - 1) ÷ ç ÷ è ø 1 (a) is equal to (b) is equal to 1 2 1 (c) is equal to – (d) does not exist 2 -x

(e x + e - x )

x

6.

-x

dx = g ( x)e(e + e ) + c , If ò (e + 2e - e - 1) e where c is a constant of integeration, then g (0) is equal to: [Sep. 05, 2020 (I)] (a) e (b) e 2 (c) 1 (d) 2 2x

If

x

cos q

ò 5 + 7 sin q - 2 cos 2 qd q = A loge | B(q) | +C ,

C is a constant of integration, then 2sin q + 1 sin q + 3

(b)

2sin q + 1 5(sin q + 3)

(c)

5(sin q + 3) 2 sin q + 1

(d)

5(2 sin q + 1) sin q + 3

7.

3 p 1 + + 12 2 4

(b)

3 p 1 + 6 2 4

(c) -

3 p 1 + + 6 2 4

(d)

3 p 1 + 12 2 4

æ x ö -1 If sin -1 çç ÷÷ dx = A( x) tan ( x ) + B ( x ) + C , wher e 1 x + è ø C is a constant of integration, then the ordered pair (A(x), B(x)) can be : [Sep. 03, 2020 (II)]

ò

(a) ( x + 1, - x )

(b) ( x + 1,

x)

(c) ( x - 1, - x )

(d) ( x - 1,

x)

x æ ö The integral ò ç ÷ dx is equal to è x sin x + cos x ø (where C is a constant of integration) : [Sep. 04, 2020 (I)] x sec x +C (a) tan x x sin x + cos x x tan x +C (b) sec x + x sin x + cos x x tan x +C (c) sec x x sin x + cos x x sec x (d) tan x + +C x sin x + cos x

The integral

ò ( x + 4)8/7 ( x - 3)6/7

is equal to:

(where C is a constant of integration) 1/7

æ x-3 ö (a) ç ÷ è x+4ø

+C

1æ x-3 ö 2 çè x + 4 ÷ø

+C

dq

8.

If

ò cos2 q(tan 2q + sec 2q)

[Jan. 9, 2020 (I)]

æ x -3ö (b) - ç ÷ è x+4ø

3/7

(c)

(d) -

-1/7

+C

1 æ x-3 ö 13 çè x + 4 ÷ø

-13/7

+C

= ltanq + 2loge|f(q)| + C where

C is a constant of integration, then the ordered pair (l, f(q)) is equal to: [Jan. 9, 2020 (II)] (a) (1, 1 – tanq) (b) (–1, 1 – tanq) (c) (–1, 1 + tanq) (d) (1, 1 + tanq)

2

4.

(a) -

dx

where

B (q ) can be : A [Sep. 05, 2020 (II)]

(a)

[Sep. 04, 2020 (I)]

9.

If

cos x dx

ò sin3 x(1 + sin 6 x)2/3 = f ( x)(1 + sin

6

1/l

x)

+ c where c is

æ pö a constant of integration, then lf çè ÷ø is equal to: 3 [Jan. 8, 2020 (II)] (a) -

9 8

(b) 2

(c)

9 8

(d) –2

www.jeebooks.in M-375

Integrals

10.

11.

2 x3 - 1 ò x4 + x dx is equal to : (Here C is a constant of integration) [April 12, 2019 (I)]

The integral

(a)

x3 + 1 1 log e +C 2 x2

(c)

log e

x3 + 1 +C x

(b)

1 ( x 3 + 1)2 +C log e 2 x3

(d)

log e

x3 + 1 x2

+C

Let a Î (0, p/2) be fixed. If the integral tan x + tan a ò tan x - tan a dx = A(x) cos2a+B(x) sin2a+C, where C is a constant of integration, then the functions A(x) and B(x) are respectively : [April 12, 2019 (II)] (a)

x + a and loge sin( x + a)

(b)

x - a and loge sin( x - a)

(c)

x - a and loge cos( x - a)

(d)

x + a and loge sin( x - a)

15.

13.

14.

If

16.

1

17.

ò ( x2 - 2 x + 10) 2

x -1 ö f ( x) æ ö = A ç tan -1 çæ ÷+ ÷ + C where C is a è 3 ø x 2 - 2 x + 10 ø è constant of integration, then : [April 10, 2019 (I)] 1 (a) A = and f(x) = 3 (x – 1) 54 1 (b) A = and f(x) = 3 (x – 1) 81 1 and f(x) = 9 (x – 1) (c) A = 27 1 (d) A = and f(x) = 9 (x – 1)2 54 If f(x) is a non- ero polynomial of degree four, having local extreme points at x = –1, 0, 1; then the set S = {x Î R : f(x) = f(0)}contains exactly: [April 09, 2019 (I)] (a) four irrational numbers. (b) four rational numbers. (c) two irrational and two rational numbers. (d) two irrational and one rational number.

The integral ò sec 2/3 x cosec4/3 xdx is equal to: [April 09, 2019 (I)] 3 (a) –3 tan–1/3 x + C (b) – tan–4/3 x + C 4 (c) –3 cot–1/3 x + C (d) 3 tan–1/3 x + C (Here C is a constant of integration)

5x ò 2x dx is equal to : sin 2 (where c is a constant of integration.) [April 08, 2019 (I)] (a) 2x + sinx + 2 sin2x + c (b) x + 2 sinx + 2 sin2x + c (c) x + 2 sinx + sin2x + c (d) 2x + sinx + sin2x + c sin

dx

12.

x If ò esec (sec x tan x f ( x ) + (sec x tan x + sec 2 x)) dx = esec x f(x) + C, then a possible choice of f(x) is: [April 09, 2019 (II)] 1 1 (a) sec x + tan x + (b) sec x - tan x 2 2 1 1 (c) sec x + x tan x (d) x sec x + tan x + 2 2

18.

dx 6 3 = xf ( x ) (1 + x ) + C , where C is a If ò 3 x (1 + x 6 )2/3 constant of integration, then the function f(x) is equal to : [April 08, 2019 (II)] 3 1 1 1 (a) 2 (b) – 3 (c) – 2 (d) – 3 x 6x 2x 2x

Theintegral ò cos ( log e x )dx is equal to : (where C is a constant of integration) [Jan. 12, 2019 (I)] x ésin ( log e x ) - cos ( log e x ) ùû + C 2ë

(a)

(b) x éëcos ( log e x ) + sin ( log e x ) ùû + C

x écos ( loge x ) + sin ( log e x )ûù + C 2ë

(c)

(d) x éëcos ( log e x ) - sin ( log e x ) ùû + C 19.

The integral

3 x13 + 2 x11

ò (2 x 4 + 3x 2 + 1) 4 dx

is equal to:

(where C is a constant of integration) [Jan. 12, 2019 (II)] x4

(a)

6(2 x 4 + 3 x 2 + 1)3

If

x12 4

6(2 x + 3 x 2 + 1)3

+C

x12 +C + C (d) (2 x 4 + 3 x 2 + 1) 3 (2 x 4 + 3 x 2 + 1) 3 x4

(c)

20.

+ C (b)

ò

1 - x2 x4

m

dx = A(x) æç 1 - x 2 ö÷ + C , for a suitable è ø

chosen integer m and a function A (x), where C is a constant of integration, then (A(x))m equals : [Jan. 11, 2019 (I)] (a)

-1 27x

9

(b)

-1 3x

3

(c)

1 27x

6

(d)

1 9x 4

EBD_8344

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21.

If

Mathematics

x +1

ò

2x -1 of integration, then f(x) is equal to:

(a)

1 ( x + 1) 3

(b)

2 ( x - 4) 3

(c) 22.

dx = f ( x ) 2 x - 1 + C , where C is a constant

25.

sin 2 x cos 2 x

ò (sin5 x + cos3 x sin 2 x + sin 3x cos2 x + cos5 x)2 dx

[Jan. 11, 2019 (II)]

to:

2 ( x + 2) 3

(a)

1 (d) ( x + 4 ) 3

1

Then

ò

sin n + 1 q

26. dq is equal to:

(a)

n æ 1 ç1 ÷ n 2 - 1 è sin n - 1 q ø

n æ 1 ö (b) 2 ç1 n -1 ÷ n +1è sin qø

(c)

1 æ ö ç1 + ÷ n2 - 1 è sin n - 1 q ø n

n +1 n

+C

n +1

23.

n æ 1 ö n +C (d) 2 ç1 ÷ n - 1 è sin n + 1 q ø (where C is a constant of integration) For x2 ¹ np + 1, nÎN (the set of natural numbers), the integral [Jan. 09, 2019(I)]

òx

2 sin ( x

(a) log e (b)

( - 1) + sin 2 ( x

) dx is equal to: - 1)

28.

2 sin ( x 2 - 1) - sin 2 x 2 - 1 2

(

2

29.

)

1 sec 2 x 2 - 1 + c 2

æ x2 - 1 ö (d) log e sec çç 2 ÷÷ + c è ø (where c is a constant of integration)

5 x8 + 7 x 6

(x

2

+ 1 + 2x

7

)

2

(c)

1 2

1 2

30.

dx, ( x ³ 0 ) ,

and f(0) = 0, then the value of f(1) is: (a) –

(b) – (d)

1 1 + cot 3 x

+C

-1 1 +C +C (d) 1 + cot 3 x 3(1 + tan3 x) (where C is a constant of integration) If K

æ x – 4ö If f ç ÷ = 2x + 1, (x Î R = {1, – 2}), then ò f (x) dx is è x + 2ø equal to (where C is a constant of integration) [Online April 15, 2018] (a) 12 loge |1 – x| – 3x + c (b) – 12 loge |1 – x| – 3x + c (c) – 12 loge |1 – x| + 3x + c (d) 12 loge |1 – x| + 3x + c æ x +3ö dx = A 7 - 6 x - x 2 + B sin -1 ç ÷+C è 4 ø 7 - 6x - x (where C is a constant of integration), then the ordered pair (A, B) is equal to [Online April 15, 2018] (a) (– 2, – 1) (b) (2, – 1) (c) (– 2, 1) (d) (2, 1)

ò

2x + 5

2

4 æ 3x - 4 ö If f ç ÷ = x + 2, x ¹ - , and 3 è 3x + 4 ø

(A, B) is equal to : [Online April 9, 2017] (where C is a constant of integration)

1 log e | sec ( x 2 - 1) | + c 2

If f ( x ) = ò

(b)

ò f (x) dx = A log |1 – x | + Bx + C, then the ordered pair

2 1 2 æ x -1ö log sec ç e (c) 2 ç 2 ÷÷ + c è ø

24.

+C

æ K tan x + 1 ö tan -1 ç ÷ + C, A A è ø (C is a constant of integration), then the ordered pair (K, A) is euqal to [Online April 16, 2018] (a) (2, 3) (b) (2, 1) (c) (– 2, 1) (d) (– 2, 3)

27.

+C

-1 3(1 + tan 3 x)

tan x

+C

n +1 n

[2018]

ò 1 + tan x + tan 2 x dx = x -

[Jan 10, 2019(I)] n +1 ö n

is equal

(c)

p Let n ³ 2 be a natural number and 0 < q < 2 (sin n q + sin q) n cos q

The integral

1 4

1 4

[Jan. 09, 2019 (II)]

æ8 2ö (a) ç , ÷ è3 3ø

æ 8 2ö (b) ç - , ÷ è 3 3ø

æ 8 2ö (c) ç - , - ÷ è 3 3ø

æ8 2ö (d) ç , - ÷ è3 3ø

The integral

ò

1 + 2 cot x(cosec x + cot x)dx

pö æ [Online April 8, 2017] ç 0 < x < ÷ is equal to : 2ø è (where C is a constant of integration) (a) 2 log sin

x +C 2

(b) 4 log sin

x +C 2

(c) 2 log cos

x +C 2

(d) 4 log cos

x +C 2

www.jeebooks.in M-377

Integrals

31.

If

ò cos3 x

= (tan x) A + C(tan x) B + k , where k

37.

2sin 2x is a constant of integration, then A + B + C equals : [Online April 9, 2016] 16 5

(a)

32.

dx

ò

If

27 10

(b)

7 10

(c)

log(t + 1 + t 2 )

dt =

21 5

(d)

1 ( g (t )) 2 + C , where C is a 2

1+ t2 constant, then g(b) is equal to : [Online April 11, 2015]

(a)

1 5

(c) 2 log(2 + 5)

38.

(c) 34.

( x + 1) e

1 x

+c

(b) - xe

( x - 1) e

x+

1 x

+c

(d) xe

The integral ò

2

3

x+

1 x

1 x

x + cos3 x

39.

+c

+c

40.

)

2

dx is equal to:

[Online April 12, 2014] (a)

1

(1 + cot x) 3

3

(c)

35.

36.

sin x

(1 + cos x) 3

(b) -

+c

(d) -

+c

(

1

3 1 + tan 3 x

)

(

cos x

+c

41.

3

3 1 + sin 3 x

)

+c

æ 1 - x2 ö x cos ç ÷ The integral ò ç 1 + x 2 ÷ dx (x > 0) is equal to: è ø [Online April 11, 2014] (a) – x + (1 + x2) tan–1 x + c (b) x – (1 + x2) cot–1 x + c (c) – x + (1 + x2) cot–1x + c (d) x – (1 + x2) tan– 1 x + c

sin8 x - cos8 x

(1 - 2sin2 x cos2 x) 1 sin 2x + c 2

1 (c) - sin x + c 2

(c)

1 3 x y ( x3 ) - ò x 2 y ( x3 )dx + C 3

1é 3 x y ( x 3 ) - ò x3 y ( x3 ) dx ùû + C 3ë If the integral

1 1 1 1 (b) (c) (d) 16 16 8 8 5tan x If the ò tan x - 2 dx = x + a ln sin x - 2cos x + k, then a is equal to : [2012] (a) – 1 (b) – 2 (c) 1 (d) 2 æ x 2 + sin 2 x ö 2 If f ( x ) = ò ç ÷ sec x dx and f (0) = 0, then f (1) è 1 + x2 ø equals [Online May 19, 2012] p (b) tan1 + 1 (a) tan1 4 p p (c) (d) 1 4 4

x2 - x The integral of 3 w.r.t. x is x - x2 + x - 1 [Online May 12, 2012]

(

)

1 log x 2 + 1 + C 2

(

)

(b)

1 log x 2 - 1 + C 2

2 (d) log x - 1 + C

42.

Let f(x) be an indefinite integral of cos3x. Statement 1:f(x) is a periodic function of period p. Statement 2: cos3x is a periodic function. [Online May 7, 2012] (a) Statement 1 is true, Statement 2 is false. (b) Both the Statements are true, but Statement 2 is not the correct explanation of Statement 1. (c) Both the Statements are true, and Statement 2 is correct explanation of Statement 1. (d) Statement 1 is false, Statement 2 is true.

43.

The value of

dx is equal to:

1 (b) - sin 2x + c 2

(d) - sin 2 x + c

1 3 x y ( x3 ) - 3ò x 3 y ( x 3 )dx + C 3

2 (c) log x + 1 + C

[Online April 9, 2014] (a)

(b)

(a)

-1

ò

1é 3 x y ( x 3 ) - ò x 2 y ( x3 )dx ùû + C 3ë

(a) -

[2014]

2

sin x cos x

(sin

x+

(a)

where k is an arbitrary constant, then A is equal to : [Online April 25, 2013]

(d) log(2 + 5)

x+

f ( x 3 )dx is equal to [2013]

cos8 x + 1

x+ The integral ò æç1 + x - 1 ö÷ e x dx is equal to xø è

(a)

5

ò cot 2 x - tan 2 x dx = A cos8 x + k ,

1

33.

ò f ( x)dx = y(x), then ò x

(d)

1 log(2 + 5) (b) 2

log(2 + 5)

If

sin xdx is pö æ sin ç x – ÷ è 4ø

[2008]

EBD_8344

www.jeebooks.in M-378

Mathematics

pö æ (a) x + log | cos ç x – ÷ | + c è 4ø pö æ (b) x – log | sin ç x – ÷ | + c è 4ø

TOPIC n

pö æ (c) x + log | sin ç x – ÷ | +c è 4ø

Expressions of ex

pö æ (d) x – log | cos ç x - ÷ | + c è 4ø 44.

dx

ò cos x +

3 sin x

equals

[2007]

49.

(a) e (4e +1) (c) e (4e–1) A value of a such that

ò

a

(b) log tan æç x - p ö÷ + C è 2 12 ø 50.

dx

ò cos x - sin x

If

[2004]

æ xö log cot ç ÷ + C è 2ø 2

2

(d)

46.

is equal to

52.

æ x 3p ö log tan ç - ÷ + C è2 8 ø

æ x pö log tan ç - ÷ + C è 2 8ø 2

1

sin x

ò sin( x - a) dx = Ax + B log sin( x - a), +C, then value

of (A, B) is

47.

51.

1

1

(c)

(b) (cos a, sin a)

(c) (- sin a, cos a)

(d) (sin a, cos a)

f(x) and g(x) are two differentiable functions on [0, 2] such that f ¢¢( x ) - g ¢¢ ( x) = 0, f ¢(1) = 2 g ¢(1) = 4 f(2) = 3g(2) = 9 then f (x)–g(x) at x = 3/2 is [2002] (b) 2

(c) 10

1 2

(c) -

1 2

(d) 2

dx = g ( x)e - x + c , where c is a constant of 2

(d) 5

(c) -

(b) 1

5 2

(d) -

1 2

1 - 4 x3 f ( x) + C, where C is a e 48 constant of integration, then f (x) is equal to: [Jan. 10, 2019 (II)] (a) – 2x3 – 1 (b) – 4x3 – 1 (c) – 2x3 + 1 (d) 4x3 + 1 If

òx

5 - 4 x3

e

The integral

dx =

dx

ò

is equal to : (1 + x ) x - x 2 (where C is a constant of integration) [Online April 10, 2016] (a) -2

1+ x +C 1- x

(b) -

1- x +C 1+ x

(c) -2

1- x +C 1+ x

(d) 2

1+ x +C 1- x

[2004]

(a) (- cos a, sin a)

(a) 0

5 - x2

òx e

(a) –1

æ x 3p ö log tan ç + ÷ + C è2 8 ø 2

(b)

(b)

integration, then g(–1) is equal to : [April 10, 2019 (II)]

1

(a)

If

[Sep. 06, 2020 (II)] (b) 4e2 – 1 (d) e (2e – 1)

dx æ9ö = log e ç ÷ is : [April 12, 2019 (II)] ( x + a)( x + a + 1) è8ø

(a) –2

1 æx pö (d) log tan ç - ÷ + C 2 è 2 12 ø

45.

x (2 + log e x) dx equals:

The integral

a+1

1 x p log tan çæ + ÷ö + C 2 è 2 12 ø

2 x x .

ò1 e

48.

æx pö (a) log tan ç + ÷ + C è 2 12 ø

(c)

Integration of the Forms: òex(f(x) + f'(x))dx, òekx(df(x) + f'(x))dx, Integration by Partial Fractions, Integration of Some Special Irrational Algebraic Functions, Integration of Different

dx

53.

The integral

ò x 2 (x 4 + 1)3/4

equals :

[2015] 1

(a)

1 - (x 4 + 1) 4

+c

æ x 4 + 1ö 4 (b) - ç 4 ÷ + c è x ø

1

æ x 4 + 1ö 4 (c) ç 4 ÷ + c è x ø

1

(d) (x 4 + 1) 4 + c

www.jeebooks.in M-379

Integrals

54.

ò

The integral

dx 3 ( x + 1) 4

(x

5 - 2) 4

Evaluation of Definite Integral

is equal to :

TOPIC Đ by Substitution, Properties of Definite Integrals

[Online April 10, 2015] 1 x +1ö 4

4 (a) - çæ ÷ 3 è x - 2ø

1 x +1ö 4

(b) 4 æç è x - 2 ÷ø

+C

1

60. +C 1

ò

x

5m -1

+ 2x

4m-1

( x 2m + xm + 1)

3

56.

5m

( ) 2m ( x 5m + x 4m ) 2 ( x 2m + xm + 1)

The integral

ò

2

(b)

x

( ) ( x5m - x4m ) 2 2m ( x 2m + x m + 1)

(d)

(a) 1 63.

(b) 0

ò

equals :

(c)

xe

x

1+ x

2

+C

(d)

[Sep. 04, 2020 (II)]

7 1 1 9 (b) (c) (d) 18 9 18 2 Let {x} and [x] denote the fractional part of x and the greatest integer £ x respectively of a real number x. If n

and 10(n2 – n), (n Î N, n > 1) are

three consecutive terms of a G.P., then n is equal to _______________. [NA Sep. 04, 2020 (II)] p

65.

ò | p- | x || dx is equal to :

(a)

66.

2p2

(b) 2p2

If the value of the integral

[Sep. 03, 2020 (I)]

(c) p2

ò

(1 - x 2 )3/ 2

k is equal to :

x

dx is

k , then 6

(b) 2 3 + p (d) 3 2 - p

(c) 3 2 + p +C

p2 2

[Sep. 03, 2020 (II)]

(a) 2 3 - p

+C

(d) x2

1/ 2

0

[2005]

(log x)2 + 1

3 2

-p

2

ïì (log x - 1) ïü ò íï1 + (log x)2 ýï dx is equal to î þ

x2 + 1

(d)

tan 3 x × sin 2 3 x (2 sec 2 x × sin 2 3 x + 3 tan x × sin 6 x ) dx

ò0 {x} dx, ò0 [ x] dx

[Online April 9, 2013] (b) ln | x | + p (x) + c (d) x + p (x) + c

(b)

1 2

(a)

ò x + x7 = p( x) then, ò x + x7 dx is equal to:

+C

(c)

is equal to :

x6

x

[Sep. 04, 2020 (I)]

p /6

-1 x - x + 1 cot -1 x e dx = A( x )ecot x + C , then A(x) is If ò 2 x +1 equal to : [Online April 22, 2013] (a) – x (b) x (c) 1- x (d) 1+ x

log x

[Sep. 05, 2020 (I)]

The integral

n

(log x)2 + 1

5051 5050

p /3

2

(a)

sin x -p / 2 1 + e

dx is:

(d)

0

2 (c) - x log 1 - 2 - x2 + c (d) x log 1 - 2 + x + c

59.

5050 5051

3

2 - x 2 + 2 - x2

(a) ln | x | – p (x) + c (c) x – p (x) + c

[Sep. 06, 2020 (I)]

Then ò ( g ( x) - f ( x)) dx is equal to :

2

64.

If

dx such

62.

(a) log 1 + 2 + x 2 + c (b) - log 1 + 2 - x 2 + c

58.

)

p 3p p (b) p (c) (d) 2 2 4 Let f ( x) =| x - 2 | and g ( x) = f ( f ( x)), x Î [0, 4].

4m

2m x 2m + x m + 1

xdx

dx

1

ò

(c)

50 101

The value of

[Online April 23, 2013]

57.

5050 5049

1

ò0 (1 - x

(a)

2m x 2m + x m + 1

(c)

(b) p/2

[Online April 19, 2014] x

5049 5050

)

61.

dx = f ( x ) + c ,

then f(x) is: (a)

50 100

dx and I2 = that I2 = aI1 then a equals to :

(a)

x - 2 ö4 4 x - 2 ö4 (c) 4 çæ (d) - æç ÷ +C ÷ +C 3 è x +1 ø è x +1 ø 55. If m is a non- ero number and

1

ò0 (1 - x

If I1 =

2

67.

The integral

ò | x - 1| - x dx

is equal to ________.

0

[NA Sep. 02, 2020 (I)]

EBD_8344

www.jeebooks.in M-380

68.

Mathematics

Let [t] denote the greatest integer less than or equal to t. Then the value of

ò

2

75.

| 2 x - [3x] | dx is ___________.

1

69.

[NA Sep. 02, 2020 (II)] If for all real triplets (a, b, c), f(x) = a + bx + cx2; then

If

1

ò

f ( x ) dx is equal to:

Let f : R ® R be a continuously differentiable function 1 such that f(2) = 6 and f’(2) = . 48

ì æ 1 öü (a) 2 í3 f (1) + 2 f ç ÷ý è 2 øþ î

1ì æ 1 öü (b) 2 í f (1) + 3 f ç 2 ÷ ý è øþ î

1ì æ 1 öü (c) 3 í f (0) + f ç 2 ÷ ý è øþ î

1ì æ 1 öü (d) 6 í f (0) + f (1) + 4 f ç 2 ÷ý è øþ î

The value of

ò 0

x sin8 x dx is equal to: 8 sin x + cos8 x

(b) 2p2

(a) 2p 2

If I =

ò

1

(a)

72.

dx 2 x - 9 x + 12 x + 4 3

2

1 1 < I2 < 8 4

[Jan. 9, 2020 (I)] (d) 4p

(c) p2 , then:

76.

1 1 1 1 < I2 < < I2 < (c) (d) 16 9 6 2 If f (a + b + 1 – x) = f(x), for all x, where a and b are fixed

a

is equal to:

cot x

1 2

(b) 1

(a)

òa+1 f ( x)dx

(c)

òa-1 f ( x + 1)dx

b -1

òa-1 f ( x)dx

(d)

òa+1 f ( x + 1)dx

òe

- a| x|

dx = 5 , is :

The value of

ò éësin 2x (1 + cos3x )ùû dx , where [t] denotes 0

78.

The integral

p/3

òp/6 sec

2/3

(a) 3 - 3 (c) 37/6 - 35/6

(b) 3 - 31/3 (d) 35/3 - 31/3

2/3

4/3

p /2

79.

The value of

ò

0

(a)

p-2 8

sin 3 x dx is: sin x + cos x

(b) p - 1 4

80.

q2

ò cos

2

3q dq is equal to:

p (d) 9

p -1 2

[April 09, 2019 (II)] p 1 p - loge 2 - loge 2 (b) 2 2 4 p p 1 - loge 2 - loge 2 (c) (d) 2 4 2 If f : R ® R is a differentiable function and f (2) = 6, then

(a)

81.

f (x)

lim

x ®2

82.

ò 6

2t dt ( x - 2) is:

[April 09, 2019 (II)]

(a) 24 f ¢ (2) (b) 2 f ¢ (2) (c) 0 (d) 12 f ¢ (2) 2 - x cos x and g ( x) = loge x, (x > 0) then the If f ( x) = 2 + x cos x p 4

value of the integral

ò

g ( f ( x))dx is :

p 4

q1

p 1 + (c) 3 6

(d)

-1 2 4 The value of the integral ò x cot (1 - x + x )dx is:

[Jan. 7, 2020 (II)] 2p (b) 3

p-2 4

0

(a) loge2

5 2cot q + 4 = 0 , then sin q

(c)

[April 9, 2019 (I)]

1

[Jan. 7, 2020 (II)]

2

x cos ec 4/3 x dx is equal to : [April 10, 2019 (II)]

5/6

-1

3 4 (b) loge æç ö÷ (c) loge (d) loge æç ö÷ 2 è 2ø è 3ø If q1 and q2 be respectively the smallest and the largest values of q in (0,2p) – {p} which satisfy the equation,

(d) –1

the greatest integer function, is: [April 10, 2019 (I)] (a) p (b) –p (c) –2p (d) 2p

b +1

2

1 2

2p

77.

b -1

(b)

The value of a for which 4a

p (a) 3

(c)

[Jan. 7, 2020 (I)]

b +1

[April 12, 2019 (I)] (d) 36

(c) 12

ò02 cot x + cosec x dx = m (p + n) , then m.n is equal to :

(a) -

1 1 < I2 < 9 8

(b)

If

b

74.

(b) 24

[April 12, 2019 (I)]

[Jan. 8, 2020 (II)]

1 positive real numbers, then a + b ò x(f (x) + f (x + l))dx

73.

x®2

x

2p

71.

4t3dt = (x – 2) g (x), then lim g(x) is equal to :

(a) 18

[Jan. 9, 2020 (I)]

0

70.

f ( x)

ò6

(a) loge3

(b) logee

(c) loge2

[April 8, 2019 (I)] (d) loge1

www.jeebooks.in M-381

Integrals x

83.

( 0, 2 ) ( 2, - 2 )

(a)

Let f(x) = ò g ( t ) dt , where g is a non- ero even function. If 0

(c) x

f(x + 5) = g(x), then

ò f ( t ) dt equals : [April 08, 2019 (II)] 0

5

5

g ( t ) dt

(d) 5

ò

ò f ( x ) g ( x ) dx 0

[Jan. 12, 2019 (I)]

a

a

(a) 4 ò f ( x ) dx

(b)

0 a

0

eì x 2x ïæ x ö æ e ö üï The integral ò íç ÷ - ç ÷ ý loge x dx is equal to : è x ø þï ïè e ø 1î [Jan. 12, 2019 (II)]

(a)

1 1 -e- 2 2 e

1 1 1 (b) - + 2 e 2e2

(c)

3 1 1 - 2 e 2e 2

(d)

3 1 -e- 2 2 2e

sin 2 x ò é x ù 1 dx -2 êë p úû + 2 (where [x] denotes the greatest integer less than or equal to x) is : [Jan. 11, 2019 (I)] (a) 0 (b) sin 4 (c) 4 (d) 4 –sin 4

The value of

p /4

òp/6 sin 2 x

dx

( tan5 x + cot5 x )

(c) 88.

1 æp -1 æ 1 ö ö ÷÷ (b) 10 ç 4 - tan ç è 9 3 øø è

ò cos x

3

dx is:

[Jan 9, 2019 (I)]

(b)

2 ò f ( x ) dx is equal to :

[Jan. 09, 2019 (II)]

0

(a) 1

(b) 2

(c)

1 2

tan q 1 dq = 1 , ( k > 0 ) then the value of k 2k sec q 2 0 is: [Jan. 09, 2019 (II)] 1 (a) 4 (b) (c) 1 (d) 2 2 If

ò

p 2

94.

The value of

(a) 95.

p 2

If f (x) = (a) (b) (c) (d)

ò

sin 2 x

p 1+ 2 2

(b) 4p

x

dx is :

(c)

[2018] p 4

(d)

p 8

x

ò0 t (sin x - sin t ) dt then

[Online April 16, 2018] f ²¢(x) + f ¢(x) = cos x – 2x sin x f ²¢(x) + f ²(x) – f ¢(x) = cos x f ²¢(x) – f ²(x) = cos x – 2x sin x f ²¢ (x) + f ²(x) = sin x

b

Let I = ò ( x 4 - 2 x 2 )dx. If I is minimum then the ordered a

pair (a, b) is:

(d) 0

p /3

93.

equals :

1æp -1 æ 1 ö ö ÷÷ (d) 5 ç 4 - tan ç è 3 3 øø è

p 40

dx , where [t] denotes the [ ] + [sin x x] + 4 - p /2

ò

1

[Jan. 11, 2019 (II)] 1 æ 1 ö tan -1 ç (a) ÷ 20 è9 3ø

[Jan. 10, 2019 (II)] 6 (d) 25

4 2 4 (c) (d) 3 3 3 92. Let f be a differentiable function from R to R such that |f (x) – f (y)| £ 2|x – y|3/2, for all x, y, Î R. If f (0) = 1 then (a) 0

The value of the integral

The integral

4 (c) 5

0

2

87.

The value of

p

91.

(d) –3ò f ( x ) dx

0

86.

18 (b) 25

a

(c) 2 ò f ( x ) dx

85.

ò f ( x ) dx 0

)

greatest integer less than or equal to t, is: [Jan. 10, 2019 (II)] 1 1 (a) (7p + 5) (7p - 5) (b) 12 12 3 3 (4p - 3) (4p - 3) (c) (d) 20 10

a

is equal to:

2, 2

p /2

90.

Let f and g be continuous functions on [0, a] such that f(x) = f(a – x) and g(x) + g(a – x) = 4, then

)

x

24 (a) 25

g ( t ) dt

2, 0

f (t) dt = x 2 + ò t 2 f (t) dt, then f ¢ (1/2) is:

x +5

5

84.

g ( t ) dt

5

x +5

ò

ò

(b)

x +5

(c) 2

ò 0

x +5

g ( t ) dt

ò

(a)

If

(d)

((-

1

x

89.

(b)

[Jan 10, 2019 (I)]

EBD_8344

www.jeebooks.in M-382

96.

Mathematics

The value of integral

ò

3p 4 p 4

104. For x Î R, x ¹ 0, if y(x) is a differentiable function such

x dx is 1 + sin x

[Online April 15, 2018]

p (a) ( 2 + 1) 2 (c) 2p ( 2 - 1) 97.

1 -x e 0

If I1 = ò

(d) p 2 1 - x3 e 0

cos2 x dx and I3 = ò

[Online April 15, 2018] (b) I3 > I1 > I2 (d) I3 > I2 > I1

ò–2p sin

4

2

æ æ 2 + sin x ö ö x çç1 + log ç ÷ ÷÷ dx è 2 – sin x ø ø è

is

[Online April 15, 2018] 3 p 16

(a)

(b) 0

(c)

99.

dx

is equal to :

[2017]

p 4

(a) –1 100. Let In =

ò tan

(c) 2

(d) 4

x dx,(n > 1) . I4 + I6 = a tan5x + bx5 + C,

where C is constant of integration, then the ordered pair (a, b) is equal to : [2017]

æ 1 ö æ 1 ö æ1 ö (a) çè - , 0÷ø (b) çè - ,1÷ø (c) çè , 0÷ø 5 5 5 2

101. If

dx

ò

(b) 2

102. The integral

p 4 p 12

(tan x + cot x)3

dx equals :

15 64

(b)

2x12 + 5x 9

103. The integral ò

(x

x

(a)

(c)

+ x3 + 1

5

(

(

)

5

)

2 x5 + x3 +1 -x

(c)

5

)

x5 + x3 +1

2

2

+C

+C

3

13 32

15 256

(d)

(b)

(a)

1 3

(b) 6

(c) 7

1

1

0

0

-1

(

)

2 x5 + x3 +1 10

(d)

x

(

)

2 x5 + x3 +1

2

-

1 x

(d) 3

[Online April 9, 2016] p + log 2 2 p - log 4 (c) 2 107. The integral 4

(b) log2 (d) log4 [2015]

log x

2

ò log x 2 + log(36 - 12 x + x 2 ) dx is equal to : 2

(a) 1 (b) 6 (c) 2 (d) 4 108. Let f : R ® R be a function such that f(2 – x) = f(2 + x) and f(4 – x) = f(4 + x), for all xÎR and

ò f (x)

dx = 5. Then the value of

50

ò10 f (x)

dx is :

0

[Online April 11, 2015] (a) 125 (b) 80 (c) 100 (d) 200 109. Let f : (–1, 1) ® R be a continuous function. If

ò

f (t ) dt =

0

1 2

æ 3ö 3 x, then f ç ç 2 ÷÷ is equal to : 2 è ø [Online April 11, 2015]

(b)

110. For x > 0, let f (x) = 2

x3

e

(1 - x + x 2 )dx is equal to :

3 2 x

-x

C

4 ë û denotes the greatest integer less than or equal to x, is : [Online April 10, 2016]

[2016]

10

(d)

ò é x 2 - 28x + 196ù + [x 2 ] , where [x]

(a)

dx is equal to :

C -x e x

[x 2 ]dx

sin x

[Online April 8, 2017] 15 128

1

(c)

2

[Online April 9, 2017] (c) 3 (d) 4 8cos 2x

ò

æ1 ö (d) çè , -1÷ø 5

k then k is equal to : k +5

=

3 1 2 ( x - 2 x + 4) 2

(a) 1

(a)

10

(a)

(b) –2 n

C

(b)

[Online April 10, 2016]

2 -1 -1 106. If 2 ò tan xdx = ò cot (1 - x + x )dx , then

3 4

(d)

1 e x

x2 105. The value of the integral

1

ò 1 + cos x

The integral

3 p 8

1 Cx 3e x

ò0 tan

3p 4

1

(a)

cos2 x dx; I2 = ò

The value of the integral

1

(b) p ( 2 - 1)

p

98.

x

(where C is a constant)

1 - x2 e 0

dx; then (a) I2 > I3 > I1 (c) I2 > I1 > I3

x

that x ò y(t)dt = (x + 1) ò ty(t)dt , then y(x) equals :

1

+C

log t

ò 1+ t

(c)

3 2

3

1 dt. Then f (x) + f æç ö÷ is equal to: è xø [Online April 10, 2015]

(a)

1 ( log x ) 2 4

(b) log x

(c)

1 ( log x ) 2 2

(d)

+C

(d)

1 log x 2 4

www.jeebooks.in M-383

Integrals p

111. The integral

ò

1 + 4sin 2

0

x x - 4sin dx equals: 2 2

[2014]

(c) p - 4

1 x

value of the integral

e

b

Statement-2 :

e dt, x > 0 then the t

t

ò t + a dt, where a > 0, is: [Online April 19, 2014]

(b) e -a éë F ( x + a ) - F ( a ) ùû

sin 2 x

ò

(c) ea éë F ( x + a ) - F (1 + a ) ùû

cos 2 x

-1

ò ( f ( x ) + x ) dx

-p

(b) 0 p/2

120. The value of

-p / 2 1 + 2

x

dx is :

(a) p

(b)

p 2

(c) 4p

(b) 0

(d) -

(c) –1

p 4

ò

tan 2 x dx is equal to :

p 2

[Online April 22, 2013] (b) log 2

(c) 2 log 2

(d) log 2

e

122. If x = ò

1

0

1 2 ln

ò

(a) log 2 2

y

[Online April 11, 2014] (c) – 9 e (d) 10

(b) 10e

0

(1 + 2x )

1 + 4x

2

dt 1+ t2

(a) y

x

ò t dt , x Î R, which are parallel to the line y = 2x, are

d2y dx 2

is equal to :

(b) x 1+ y

p p p p ln2 ln2 (d) ln2 ln2 (b) (a) (c) 8 16 32 4 117. The intercepts on x-axis made by tangents to the curve,

, then

[Online April 9, 2013]

(c)

dx , equals: [Online April 9, 2014]

2

1+ y 2

(d) y 2

x

ò

123. If g (x) = cos 4t dt , then g (x + p) equals 0

g ( x) (a) g ( p) (c) g (x) – g (p)

(b) g (x) + g (p) (d) g (x) . g (p)

0

equal to :

(d)

7p / 3

121. The integral

[Online April 12, 2014]

115. If for n ³ 1, Pn = ò ( log x )n dx , then P10 – 90P8 is equal to:

y=

sin 2 x

p 4

7p / 4

is equal to:

0

116. The integral

ò

(d) -

(c) 1

[Online April 23, 2013]

p

(a) – 9

cos -1 ( t ) dt equals :

0

= p2 –

p p p (a) p (b) (c) (d) 3 6 2 114. If [ ] denotes the greatest integer function, then the integral

p 2

ò

sin ( t )dt +

[Online April 25, 2013]

æ pö t2, for all t ³ – p, then f ç - ÷ is equal to: è 3ø [Online April 12, 2014]

(a)

[2013]

a

p , the value of 2

p (a) 4

t

ò [ cos x ] dx

ò

0

éëF ( x + a ) - F (1 + a ) ùû

113. If for a continuous function f(x),

b

f ( x )dx = ò f (a + b - x )dx.

(a) Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (c) Statement-1 is true; Statement-2 is false. (d) Statement-1 is false; Statement-2 is true. 119. For 0 £ x £

(a) e éë F ( x ) - F (1 + a ) ùû a

(d) e

is equal to p/6

a

1

-a

dx

p / 6 1 + tan x

x t

112. Let function F be defined as F(x) = ò

p /3

ò

p (b) 4 3 - 4 3 2p - 4-4 3 (d) 3

(a) 4 3 - 4

(a) ± 1 (b) ± 2 (c) ± 3 (d) ± 4 118. Statement-1 : The value of the integral

[2013]

[2012]

EBD_8344

www.jeebooks.in M-384

Mathematics

124. If [x] is the greatest integer £ x, then the value of the integral æ 2 æ 2 - xöö dx is [Online May 26, 2012] ò ç éë x ùû + log ç è 2 + x ÷ø ÷ø è -0.9 (a) 0.486 (b) 0.243 (c) 1.8 (d) 0 0.9

125. The value of the integral

ò

éë x - 2 [ x ]ûù dx ,

2 2 (d) I > and J < 2 and J > 2 3 3 133. The solution for x of the equation x

[Online May 12, 2012]

(b) 2[G (p 4) - G(p 16)]

(c) p[G (1 2) - G (1 4)]

(d) G (1

x

127. If ò t f ( t ) dt = sin x - x cos x e

2) - G(1 2)

(b) 1

F(e) equals (a) 1 (b) 2

3 2

3 4

(d)

5 4

1

(c) [a ] f ([a ]) - { f (1) + f (2) + .............. f ( a)} (d) af ([ a]) - { f (1) + f (2) + .............. f ( a )} -

p 2

ò

[( x + p )3 + cos 2 ( x + 3p )]dx is equal to

[2010]

0

(a) 21

(b) 41

(c) 42

(d)

(a)

p4 32

ò [cot x] dx , where [ . ] denotes the greatest integer 0

function, is equal to : (a) 1

(b) –1 1

132. Let I = ò

0

sin x x

p (c) 2 1

dx and J = ò

0

the following is true?

cos x x

p (d) 2

(b)

p4 p p + (c) 32 2 2

ò xf (sin x)dx

p -1 4

is equal to

[2006]

0

p

p

(a) p ò f (cos x )dx

(b) p ò f (sin x )dx 0

0

(c) p 2

p /2

ò

(d) p

f (sin x)dx 6 3

1 2

ò

f (cos x )dx

0

138. The value of integral, I =ò (a)

p /2

0

[2009]

dx. Then which one of

(d)

p

137.

41

p

131.

[2006]

3p 2

1

equals

[2006]

(b) [a ] f (a) - { f (1) + f (2) + .............. f ([a ])}

136.

dx is [2011] 1+ x2 p p log 2 (b) log 2 (c) log 2 (a) (d) p log 2 8 2 130. Let p(x) be a function defin ed on R such that p¢(x) = p¢(1 – x), for all x Î [0, 1], p (0) = 1 and p (1) = 41.

ò p( x) dx

(d) 0,

greatest integer not exceeding x is (a) af (a) - { f (1) + f (2) + .............. f ([ a])}

[2011 RS] (c)

[2007] (c) 1/2

135. The value of ò [ x ] f '( x )dx , a > 1 where [x] denotes the

0

Then

(d) None of these

log t 1 dt , Then 134. Let F(x) = f (x) + f æç ö÷ ,where f ( x) = ò 1+ t è xø l

8log(1 + x)

ò

129. The value of

(b) 2 2

a

x2 , for all x Î R - { 0} , 2

x éë x 2 ùû dx is :.

(a) 0

3 2

(a)

æ pö [Online May 7, 2012] then the value of f ç ÷ is è 6ø (a) 1/2 (b) 1 (c) 0 (d) – 1/2 128. Let [.] denote the greatest integer function then the value

0

[2007]

(c) 2

(a) G (p 4) - G (p 16)

ò

p is 2

=

x

2 tan(p x 2 ) dx is equal to ò .e 14 x

of

2

2 t t -1

, x Î (0, p 2), then

12

1.5

dt

ò

where [.] denotes the greatest integer function is [Online May 19, 2012] (a) 0.9 (b) 1.8 (c) – 0.9 (d) 0 d e 126. If G( x ) = dx x

2 and J < 2 3

(b) I

0.9

(b)

3 2

x 9- x + x

(c) 2

[2008]

dx is

(d) 1

[2006]

www.jeebooks.in M-385

Integrals p

(a) a p

cos 2 x

ò

139. The value of

- p 1+ a

147. Let f(x) be a function satisfying f '(x) = f(x) with f(0)=1 and g(x) be a function that satisfies

2

1

1

p a

(c)

x 140. If I1 = ò 2 dx , I 2 =

ò f ( x ) g ( x ) dx, is

(d) 2p

3

0

2

2

x ò 2 dx and

(a) e +

e2 5 + 2 2

(b) e -

e2 5 2 2

(c) e +

e2 3 2 2

(d) e -

e2 3 - . 2 2

1

3

x I 4 = ò 2 dx then

[2005]

b

148. If f (a + b - x) = f ( x) then ò xf ( x )dx is equal to [2003]

1

a

(a) I 2 > I1 (b) I1 > I 2 (c) I3 = I 4 (d) I3 > I 4 141. Let f : R ® R be a differentiable function having f (2) = 6, æ 1 ö f '(2) = ç ÷ . Then lim x®2 è 48 ø

(a) 24 142. If f ( x) =

(b) 36 e

f ( x)

ò

6

b

3

4t dt equals x-2

(c) 12

(a)

a+b a+b b f (a + b + x )dx (b) ò f (b - x)dx ò 2 2 a a

(c)

a+b b ò f ( x)dx 2 a

[2005]

(d) 18

f ( a)

x

1+ e

ò

, I1 =

x

ò

2 ò sec tdt

f (-a )

x ®0

(a) 0

g{x (1 - x )}dx,

I2 is I1 (b) –3

p

p/2

0

0

ò xf (sin x)dx = A ò

(a) 2p

(b) p

ò

(c) –1

p 4

1 + sin 2 x

(b) 1

(c) 2

[2004]

(d) 0

151.

dx is

(b) F (t ) = 1 - te -t (1 + t )

(c) F (t ) = e t - (1 + t )

(d) F (t ) = tet .

ò

p 2 x (1 + sin x )

1 + cos 2 x

[2004]

p2 4

(a) (d) 0

dx is

(b) p 2

[2002]

(c) ero

(d)

p 2

2

ò |1- x

(b)

[2003]

(a) F (t ) = te -t

-p

2

|dx is

[2004]

152.

14 3

ò [x

2

]dx is

[2002]

0

-2

1 3

(d) 1

0

3

(a)

(c) 2

(b) 3

t

(d) 2

f (sin x )dx, then A is

(sin x + cos x )2

0

145. The value of

[2003]

F (t ) = ò f (t - y ) g ( y ) dy , then

[2004]

(c)

p/2

144. The value of I = (a) 3

is

x sin x

150. If f ( y ) = e y , g ( y ) = y; y > 0 and

then the value of

143. If

0

149. The value of lim

f (-a )

(a) 1

b-a b ò f ( x )dx . 2 a

(d) x2

xg{x (1 - x )}dx

f ( a)

and I 2 =

[2003]

0

x ò 2 dx , I3 =

0

2

[2005]

f(x) + g(x) = x 2 . Then the value of the integral

p 2

(b) 1

x

dx , a > 0, is

(c)

7 3

(d)

(a) 2 –

28 3

p/4

[2003]

153. I n =

0

ò

tan n x dx then lim n[ I n + I n + 2 ] equals n®¥

0

(a)

1 1 + n +1 n + 2

(b)

1 n +1

(c)

1 n+2

(d)

1 1 . n +1 n + 2

2

(d) - 2 - 3 + 5

2 –1

(c)

1

146. The value of the integral I = ò x(1 - x ) n dx is

(b) 2 +

2

(a) 1 2 154.

10 p

ò0

(b) 1

(c) ¥

(d) ero

| sin x | dx is

(a) 20

(b) 8

[2002]

[2002] (c) 10

(d) 18

EBD_8344

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Mathematics

TOPIC Ė

1

Reduction Formulae for Definite Integration, Gamma & Beta Function, Walli's Formula, Summation of Series by Integration

æ ( n + 1) (n + 2)...3n ö n 159. lim ç ÷ is equal to: n ®¥ è n 2n ø

(a)

155. Let a function f : [0, 5] ® R be continuous, f (1) = 3 and F be defined as: x

x

F(x) =

òt

2

g (t )dt , where g(t) =

ò f (u )du

Then for the function F, the point x = 1 is : [Jan. 9, 2020 (II)] (a) a point of local minima. (b) not a critical point. (c) a point of local maxima. (d) a point of inflection. æ (n + 1)1/3 (n + 2)1/3 (2n)1/3 ö + + + lim ..... ç ÷ is equal to : 156. n ®¥ ç 4/3 n 4/3 n 4/3 ÷ø è n [April 10, 2019 (I)] 3 4/3 3 4 4/3 (2) (2) (a) (b) 4 4 3 3 4/3 4 4 3/4 (2) (2) (c) (d) 3 2 3 n n 1 ö æ n + + ..... + ÷ is equal 157. lim ç 2 2 + 2 2 2 2 n ®¥ è n + 1 5n ø n +2 n +3 to : [Jan. 12, 2019 (II)]

p 4

158. If lim

n ®¥ ( n

(b) tan–1(3) (c)

p 2

+ 1)a -1 [ (na + 2) + .....(na + n) ]

=

1 60

for some positive real number a, then a is equal to : [Online April 9, 2017] (a) 7

(b) 8

(c)

15 2

18 e

(d)

17 2

(d)

4

dx

é1

161. nLim ê ®¥

ë n2

27 e2

is a polynomial of degree

sin 6 x

[Online May 26, 2012] (b) 5 in tan x (d) 3 in cot x

(a) 5 in cot x (c) 3 in tan x

ù 1 2 4 1 sec 2 2 + 2 sec 2 2 ............ + sec 2 1ú n n n n û

equals (a)

[2005]

1 sec 1 2

1 cosec 1 2 1 (d) tan 1 2

(b)

(c) tan 1

162.

Lim n®¥

n

163. lim

164.

[2004]

(b) e – 1

1 + 24 + 34 + ...n 4

n®¥

(a)

r

1 å n e n is r =1

(a) e + 1

(d) tan–1(2)

1a + 2a + ......... + na

(b) 3 log 3 – 2

e2

160. f ( x ) = ò

1

1

(a)

(c)

9

[2016]

n

1 5

lim n ®¥

5

(b)

(c) 1 – e - lim

1 + 23 + 33 + ...n3

(c) Zero

1 p + 2 p + 3 p + ..... + n p n p +1

[2003]

n5

n®¥

1 30

(d) e

is

(d)

1 4

[2002]

(a)

1 p +1

(b)

1 1- p

(c)

1 1 p p -1

(d)

1 p+2

www.jeebooks.in M-387

Integrals

1.

(Bonus) 1 sin( x - 1) 4 lim 2 x ®1 ( x - 1) sin( x - 1)

4.

Q

Let x – 1 = h when x ® 1 then h ® 0

lim

sin h4 h4

h® 0

´

h ´ h2 = 1 ´ 1 ´ 0 = 0 sin h

x2

ò ( x sin x + cos x) 2 dx

(a)

d ( x sin x + cos x ) = x cos x dx

x cos x æ x ö ç ÷ dx ( x sin x + cos x ) 2 è cos x ø

2.

(d)

ò (e

+ 2e x - e - x - 1) × e( e

2x

I = ò ( e 2 x + e x - 1) × e( e

= ò e x (e x + 1 - e - x ) × e (e = ò (e - e x

-x

+ 1)e

+ e- x )

x

x

x

+ e- x )

dx + e( e

dx + e

=

dx

dx + ò ( e x - e - x ) e( e

+ e- x )

(e x + e- x + x )

x

x

+ e- x )

= e( e

x

+e

-x

x

+ x)

= (e x + 1). e(e

+ e- x )

= et + e (e

x

+ e- x )

(ex + e- x )

+ e- x )

+ e(e x

+e

+ e- x )

-x

)

+C

ò 5 + 7 sin q - 2 cos

2

q

dq =

5 + 7t - 2 + 2t 2

(d)

x

ò (1 + x)2 dx ( x > 0) 2 tan 2 q × sec 2 q sec 4 q

sin 2q +C 2

f ( x) = q -

1 2t + 1 1 2sin q + 1 = ln + C = ln +C 5 5 sin q + 3 t +3

tan q 1 + tan 2 q

+ C = tan -1 x -

Now f (3) - f (1) = tan -1 ( 3) -

1 2sin q + 1 and A = 5 2(sin q + 3)

B(q) 5(2sin q + 1) = A (sin q + 3)

d q = ò 2sin 2 qd q

1 2 tan q Þ f ( x) = q - ´ +C 2 1 + tan 2 q

ò

Þ

- x sec x + tan x + C x sin x + cos x

= q-

dt

1 t+ dt 1 1 2 +C Þ = ln 2 æ 7 ö2 æ 5 ö2 5 t + 3 çt + ÷ -ç ÷ è 4ø è4ø

\ B (q ) =

=

I =ò

(d) Let sin q = t Þ cos q d q = dt cos q

x é -1 ù + sec2 x dx cos x êë x sin x + cos x úû ò

Put x = tan 2 q Þ 2 x dx = 2 tan q sec 2 q d q

+C

So, g (x) = 1 + ex and g (0) = 2 3.

x sin x + cos x é -1 ù dx ê 2 cos x ë x sin x + cos x úû

=

+C

5. x

x é -1 ù cos x êë x sin x + cos x úû

dx

Let e x + e - x + x = t Þ (e x + e - x + 1)dx = dt = ò et dt + e (e

I

II

(No any option is correct)

=

6.

x +C 1+ x

3 1 - tan -1 (1) + 1+ 3 2

3 p 1 + 12 2 4

æ x ö -1 (a) I = sin -1 ç x × 1 dx ç 1 + x ÷÷ dx = tan è ø

ò

ò

I

II

EBD_8344

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Mathematics

= x tan -1 x - ò = x tan

1 1 × .x dx + C 1+ x 2 x

9.

(Put x = t 2 Þ dx = 2t dt ) = x tan -1 x - ò

t

2

I =ò

dt + C

1+ t2

cos x dx

ò sin 3 x(1 + sin6 x)2 3

= f (x) (1 + sin6x)1/l + c If sin x = t then, cos x dx = dt

1 t × 2t dt x- ò +C 2 1+ t 2

-1

(d) Let I =

dt 2 t3 1 + t 6 3

= x tan -1 x - t + tan -1 t + C Put 1 +

= x tan -1 x - x + tan -1 x + C

(

)

1

= r3 Þ

t6

...(i)

dt 2

1 ö3 t ç1 + 6 ÷ è t ø dt t7

=

-1 2 r dr 2

= ( x + 1) tan -1 x - x + C -

Þ A( x) = x + 1 Þ B( x) = - x 7.

(a) I = ò

dx

1

-6 7

1

dx

( x + 4)2

f(x) = –

x-3 7 =t , x+4 Differentiate on both sides, we get ( x + 4)

dx = 7t 6dt 2

10.

Hence, I = ò t -6t 6 dt = t + C = æç ÷ +C è x+4ø (c) I = ò =ò

(c) Given integral, I =

Þ I=

dq

= log

cos 2 q(tan 2q + sec 2q)

11.

sec2 q 1 + tan 2 q 2

+

2 tan q

dq

ò

(2 x 3 - 1)dx 4

x +x

(2 x - x -2 )dx x 2 + x -1

sec 2 q(1 - tan 2 q) (1 + tan q)2

du

òu

= log | u | + c = log| x2 + x–1 | + c

x3 + 1 +c x

(b) Given integral sin( x + a )

tan x + tan a

ò tan x - tan adx = ò sin( x - a) dx

2

1 - tan q 1 - tan q

[from eqn. (i)]

Put x2 + x–1 = u Þ (2x – x–2)dx = du 1 x - 3 ö7

8.

1 cosec2x and l = 3 2

æpö \ lf ç ÷ = -2 è3ø

Let

7

1

1 æ sin 6 x + 1 ö 3 1 =- ç (1 + sin 6 x ) 3 + c ÷ +c =2 çè sin 6 x ÷ø 2sin 2 x

( x + 4)8/7 ( x - 3)6/7

æ x-3ö = òç ÷ è x+4ø

1 r 2 dr 1 =- r +c 2 ò r2 2

Let x – a = t Þ dx = dt dq

=

sec 2 q(1 - tan q) dq 1 + tan q Let tanq = t Þ sec2q dq = dt, then =ò

2 ö æ 1- t ö æ I = òç dt = ò ç -1 + dt ÷ 1 + t ÷ø è 1+ t ø è = – t + 2 log (1 + t) + C = – tanq + 2 log (1 + tanq) + C Hence, by comparison l = – 1 and f (x) = 1 + tanq

ò

sin(t + 2a) dt = ò [cos 2a + sin 2a. cot t ]dt sin t

= cos 2a . t + sin 2a . log | sin t | + c = (x – a) . cos 2a + sin 2a . log | sin (x – a) | + c dx

12.

(a) Let I =

Let (x – 1)2 = 9 tan2 q Þ tan q =

dx

ò ( x2 - 2 x + 10)2 = ò (( x - 1)2 + 9) 2 x -1 3

...(i)

www.jeebooks.in M-389

Integrals

After differentiating equation ...(i), we get 2 (x – 1) dx = 18 tan q sec2q dq \ I=

15.

òe

sec x

18tan q sec2 qd q

ò 2 ´ 3tan q ´ 81sec4 q

I=

1 1 1 cos 2 qd q = ´ (1 + cos 2q)d q ò 27 27 2 ò

I=

1 ì sin 2q ü íq + ý+c 54 î 2 þ

(

\ f (x) = sec x + tan x + C

16.

= ò (3 - 2(1 - cos 2 x ) + 2cos x) dx = ò (1 + 2 cos x + 2cos 2 x ) dx

4

2 x6 ) 3

dx 2

x 7 (1 + x -6 ) 3

1

1

6 3 1 = - (1 + x -6 ) 3 + C = - 1 (1 + x ) + C 2 2 2 x

sec 2 xdx 4

tan 3 x Put tan x = Þ sec2 x dx = d

=-1

dx

ò

2 1 æ 1 ö t dt Now, I = ò ç - ÷ 2 = - t + C 2 2 è ø t

(a) I = sec 3 x.cosec 3 dx ò

Þ I =òz

(d) Let,

dx æ 1 ö 2 Put 1 + x–6 = t3 Þ – 6–7 dx = 3t2 dt Þ 7 = ç - ÷ t dt è 2ø x

\ S = {0, - 2, 2}

4 3

= x + 2 sin x + sin 2x + c

x3 (1 +

x2 æ x2 ö Þ 2 çç 2 - 1÷÷ = 0 Þ x = 0,0, 2, - 2 è ø

-

sin 3 x + sin 2 x dx sin x

[Qsin 2x = 2 sin x cos x and sin 3x = 3 sin x – 4 sin3x]

17.

æx x ö Q f(x) = f(0) Þ K çç 4 - 2 ÷÷ = 0 è ø

ò

x 5x 2cos .sin 2 2 dx ò x x 2cos .sin 2 2

= ò (3 - 4sin 2 x + 2cos x)dx

æ x4 x 2 ö Þ f(x) = K çç 4 - 2 ÷÷ + C (using integration) è ø Þ f(0) = C 2

æ 5x ö sin ç ÷ è 2 ø dx (c) ò = æxö sin ç ÷ è 2ø

=

1 and f (x) = 3 (x – 1) 54 (d) Since, function f(x) have local extreem points at x = –1, 0, 1. Then f(x) = K (x + 1) x (x – 1) = K (x3 – x)

I=ò

)

f ( x ) = ò ( sec x tan x ) + sec 2 x dx

we get: A =

2

( ( g ' ( x ) f ( x ) ) + f '( x ) ) dx = e g ( x ) ´ f ( x ) + C

Our comparing above equation by equation (i),

é -1 æ x - 1 ö ù f ( x) Compare it with A ê tan ç b ÷ + 2 ú+c, è ø x - 2 x + 10 û ë

14.

2

g x Q òe ( )

1 é -1 æ x - 1 ö 3( x - 1) ù I = 54 ê tan ç 3 ÷ + 2 ú+c è ø x - 2 x + 10 û ë

4

(sec x tan x f ( x ) + (sec x + tan x + sec x )) dx = esec x f ( x ) + C ...(i)

é æ x -1 ö ù 2ç ê ÷ ú 1 ê -1 æ x - 1 ö 1 3 ø ú + ´ è +c tan ç ÷ 2ú I = 54 ê è 3 ø 2 æ x -1 ö ú ê 1+ ç ÷ êë è 3 ø úû

13.

(a) Given,

-1

z3 × dz = + C Þ I = -3(tan x ) 3 + C æ -1 ö ç ÷ è 3 ø

1 2 x3

x(1 +

1 6 3 x )

Hence, f (x) = -

+C

1 2x3

EBD_8344

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18.

Mathematics

(c) Let the integral, I = ò cos(ln x) dx Þ I = cos(ln x ).x - ò

3

1æ 1 ö2 = - ç 2 - 1÷ + C ø 3è x

- sin(ln x ) .x dx x

3

1 1 = - × 3 × (1 - x 2 ) 2 + C 3 x

= x cos(ln x) + ò sin(ln x)dx

(

)

3 -1 1 - x2 + C 3 3x Compare both sides,

cos(ln x ) .x dx = x cos (ln x) + sin(ln x).x – ò x

=

= x cos(ln x) + sin(ln x) . x – I Þ A(x) = -

x [cos (ln x) + sin (ln x )] + C 2

Þ (A(x))3 =

Þ I=

19.

3x13 + 2 x11 dx = (b) I = ò (2 x4 + 3x2 + 1) 4

ò

3x13 + 2 x11 3 1ö æ x16 ç 2 + 2 + 4 ÷ è x x ø

4

dx

21.

Let 2 +

I=

I=

ò

Let

I=

+C =ò

x+4 3

Hence, f (x) =

1

22.

n n (a) Let, I = ò (sin q - sin q) cos qd q n+1 sin q

Let sin q = u Þ cos q dq = du

1- x dx x4 2

1 n n \ I = ò (u - u) du n +1 u

1 -1 x2 dx x3

1

1 ön æ 1 ç1 ÷ = u - n (1 - u1-n ) n du = è u n -1 ø du ò ò un

1 - 1 = u2 x2

Þ -

(t 2 + 3) t 3 3t dt = + + C 2 6 2

= f (x). 2 x –1 + C

1 x12 +C 6 (2 x4 + 3x 2 + 1)3 2

ò

1

dt -4+1 2 = -1 t +C 2 -4 + 1 t4

( 1- x )

2 x -1 = t

3

-

m

x +1 dx 2x – 1

ò

2 = (2 x –1) + 3 (2 x –1) 2 + C 6 2 æ x + 4ö = 2 x –1 çè ÷ +C 3 ø

-1 1 1 ´ +C 3 2 (-3) æ 3 1ö çè 2 + 2 + 4 ÷ø x x

(a) A( x)

=

ò

-1 27x 9

\ 2x – 1 = t2 Þ dx = tdt

3 1 æ 3 2ö + 4 = t, -2 ç 3 + 5 ÷ dx = dt 2 èx x ø x x

Then, I =

(c) Let I = Put

3 2 + x3 x5 dx ò 4 I= æ 3 1ö çè 2 + 2 + 4 ÷ø x x

20.

1 and m = 3 3x3

Þ 2I = x(cos (ln x) + sin (ln x)) + C

2 2u du = x3 dx

Let 1 – u1 – n = v Þ –(1 – n)u–n du = dv

dx = –u du x3

Þ u–n du = m

3

u A( x) æç 1 - x 2 ö÷ + C = ò (-u 2 )du = - + C è ø 3

dv n -1

www.jeebooks.in M-391

Integrals

\ I=ò

1 vn

n = 2 v n -1

= 23.

25.

1 +1 vn

1 × dv × = n -1 1 +1 n -1 n

n +1 n

=ò n +1 ö n

æ +C = ç1 ÷ n2 - 1 è u n -1 ø n

n+1 ö n

n æ 1 ç1 - n-1 ø÷ è n -1 sin q 2

1

+C

sin2 xcos2 x (sin5x+ cos3xsin2x+ sin3xcos 2 x + cos5x)2 sin 2 x cos2 x [(sin 2 x + cos 2 x)(sin 3 x + cos3 x)]2

sin 2 x cos 2 x 3

+C

2sin( x 2 - 1) - 2sin( x 2 - 1)cos( x 2 - 1) dx 2sin( x 2 - 1) + 2sin(x 2 - 1)cos( x 2 - 1)

\ I= 26.

ÞI=

1 - cos( x - 1) dx 1 + cos( x 2 - 1) 2

æ x - 1ö Þ I = ò x tan ç 2 ÷ dx , è ø

= x-ò

2 æ x 2 - 1ö 1 2 æ x - 1ö or I = ln sec ç 2 ÷ + c = 2 ln sec ç 2 ÷ + c è ø è ø

Þ f(x) = ò

- dt = ò - t -2 dt = t -1 + c t2

Þ f(x) =

1 + c, f (0) = 0 Þ c = 0 2 + x -7 + x -5

\ f(1) =

1 4

dt 1 1 t2 + t + +1 4 4

æ 1ö æ 3ö ÷ ç t + ÷ + çç è 2 ø è 2 ÷ø 2

2

æ 2 tan x + 1 ö tan -1 ç ÷+C 3 3 è ø \ A = 3 and K = 2

5x + 7 x dx (2 + x -7 + x -5 )2

Let 2 + x–7 + x–5 = t Þ (–7x–8 – 5x–6)dx = dt

sec2 xdx 1 + tan x + tan 2 x

dt

ÞI= x-

-8

27.

dx

tan x

æ 1ö çt + ÷ 2 -1 2 ÷+C tan ç ÞI= x3 ç 3 ÷ ç ÷ è 2 ø

5 x8 + 7 x 6 dx, x ³ 0 ( x 2 + 1 + 2 x7 )2

5 x8 + 7 x 6 dx = ò 14 -5 x ( x + x -7 + 2)2 -6

(1 + tan 3x) 2

tan x + 1 + tan 2 x (1 + tan 2 x) dx ò 1 + tan x + tan 2 x tan x + 1 + tan 2 x

\I= x-ò

2x x2 - 1 dx = dt =t Þ 2 2

(a) f(x) = ò

ò

tan 2 x × sec 2 x

Put tan x = t Þ sec2 x . dx = dt

\ I = ò | tan(t ) | dt = ln | sec t | + C

24.

dx = ò

dx

ò 1 + tan x + tan 2 x dx

ÞI= x-ò

2

Now let

2

dx

-1 1 dt 1 +C =- + C = ò 2 3 t 3t 3(1 + tan3 x)

(a) Let I =

(\ sin 2 q = 2sin q cos q) Þ I=ò x

3

(sin x + cos x) Now, put (1 + tan3x) = t Þ 3 tan2x sec2x dx = dt

(c, d ) Consider the given integral I=ò x

(a) Let I

2

(b) Suppose,

x–4 = y Þ x – 4 = y (x + 2) x+2

Þ x (1 – y) = 2y + 4 Þ x =

2y + 4 1– y

æ 2y + 4 ö So, f (y) = 2 ç ÷ +1 è 1– y ø æ 2x + 4 ö 3x + 9 Now, f (x) = 2 ç ÷ +1 = 1 – x 1– x è ø

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3 ( x + 3) 3 (x –1 + 4) 12 = = – 3+ 1– x 1– x 1– x

ò

\ ò f (x) dx = – 12 loge |1 – x| – 3x + c (a) Q 7 – 6x – x2 = 16 – (x + 3)2

Þ I =

ò

and

d (7 – 6x – x2) = – 2x – 6 dx

Þ I =

ò cosec x + cot x . dx

So,

ò

2x + 5 7 - 6x - x

dx = ò

2

1 16 - ( x + 3)2

2x + 6 7 - 6x - x2

Þ I = log 1 - cos x + C

dx

Þ I = log 2sin 2

4 3x - 4 ö (b) f çæ ÷ = x + 2, x ¹ 3 è 3x + 4 ø Consider

3x - 4 =t 3x + 4

31.

Þ 3x - 4 = 3tx + 4t Þ x=

4t + 4 +2 3 - 3t

Þ I = 2log sin

x + C1 2

dx

ò cos3 x

(a)

dx

=

ò 2 cos4 x

4sin x cos x Let tan x = t2 Þ sec2 x = 1 + t4 sec2 x dx = 2t dt sec 4 x dx

ò2

tan x

=

ò

2 tan x

t5 (1 + t 4 )2t dt = ò (1 + t 4 ) dt = t + + k 2t 5

ò

2x - 10 ò f ( x ) dx = ò 3x - 3 dx

=

1 tan x + tan 5 / 2 x + k ét = tan x ù ë û 5

dx 2x dx - 10ò = ò 3x - 3 3x - 3

A=

1 1 5 ,B= ,C= 5 2 2

2 x -1 2 dx 10 dx dx + ò = ò 3 x -1 3 x - 1 3 ò x -1

A+B+C=

16 5

(d) Let I =

ò

2x 8 - ln ( x - 1) + C 3 3

32.

log(t + 1 + t 2 ) 1 + t2

8 2 Here, A = - , B = 3 3

put u = log(t + 1 + t 2 )

æ 8 2ö \ (A, B) = ç - , ÷ è 3 3ø

du =

(a) Let, I = Þ I=

ò

ò

1 + 2 cot x cosec x + 2 cot 2 x . dx

sin 2 x + 2cos x + 2cos2 x sin 2 x

tan x

sec 2 x (sec2 x dx )

2x - 10 3x - 3

=

30.

x + log 2 + C 2

=

Þ f (x) = \

x +C 2

Þ I = log sin 2

=

10 - 2t 3 - 3t

Þ f (t) =

1 + cos x . dx sin x

Þ I = log cosec x - cot x + log sin x + C

dx

æ x+3ö = -2 7 - 6 x - x 2 - sin -1 ç ÷+C è 4 ø Therefore, A = – 2, & B = – 1. 29.

1 + 2cos x + cos2 x . dx sin x

Þ I=

=

28.

Mathematics

dt

é t + 1+ t2 ù ú = .ê t + 1 + t 2 êë 1 + t 2 úû 1

\ I = ò u du =

. dx Since, I =

u2 +c 2

1 [g(t)]2 + c 2

1 1+ t2

dt

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Integrals

\ g(t) = log (t + 1 + t 2 ) 35.

Put t = 2 g (b) = log (2 + 5) 33.

= òe

= x.e

x+ 1 x

= x.e = xe 34.

\I=2

1ö æ (d) Let I = ò ç1 + x - ÷ e è xø x+ 1 x

x+

x+

1 x

1 x

x+ 1

1ö æ -ò x ç 1 - ÷ e è x2 ø

x+ 1

x dx

1 ö x+ 1 æ + ò ç x - ÷ e x dx è xø

é x2 ù 1 x2 + 1 1 1 dx + ò dx ú + c I = 2 ê tan -1 x - ò 2 2 x2 + 1 2 1+ x êë 2 úû é x2 ù 1 1 I = 2 ê tan -1 x - ò 1.dx + tan -1 x ú + c 2 2 ëê 2 ûú

+C

sin 2 x cos 2 x

ò (sin 3 x + cos3 x)2 dx

é x2 ù x 1 I = 2 ê tan -1 x - + tan -1 x ú + c 2 2 2 ëê ûú 2 -1 -1 tan tan x x + x x + c I=

ö ÷ dx è cos3 x (1 + tan 3 x) ø÷ sin x. cos x

or I = - x + ( x 2 + 1) tan -1 x + c

2

æ sin x.sec 2 x ö = òç ÷ dx è (1 + tan3 x ) ø

36.

(b) Let I =

Put 1 + tan 3 x = t dt = 3tan 2 x sec2 x dx or dx = \I=

I=

ò

sin 2 x.sec4 x t

2

dt

=

3tan 2 x sec 2 x

2

3 tan x sec x

1 dt 1 -2 = ò t dt 3 ò t2 3

1 é t -2+1 ù -1 é1ù +c I = 3 ê -2 + 1ú + c = 3 êë t úû êë úû 1 3(1 + tan 3 x)

ò

(sin 4 x + cos 4 x)(sin 4 x - cos 4 x)

=

dt 1 sin 2 x.sec4 x ´ = 3ò 2 2 t sin x sec 4 x

or I = -

ò

=

2

+c

sin8 x - cos8 x

ò 1 - 2sin 2 x cos 2 x dx

(sin 4 x )2 - (cos 4 x)2

dt

´

1 sin 2 x.sec 4 x dt ´ 3ò sin 2 x t2 ´ sec2 x cos 2 x

\I=

x dx

é x2 1 x2 + 1 - 1 ù I = 2 ê tan -1 x - ò 2 dx ú + c 2 x +1 ëê 2 ûú

1 ö x+ 1 æ 1 ö x+ 1 æ - ò ç x - ÷ e x dx + ò ç x - ÷ e x dx è xø è xø

2

ò çç

-1

æ 1 - x2 ö ç ÷ dx è 1 + x2 ø

é x2 x2 ù 1 -1 ´ dx ú + c I = 2 ê tan x - ò 1 + x 2 2 úû êë 2

æ sin x.cos x ö ò çè sin3 x + cos3 x ÷ø dx

æ

I=

ò IIx .tanI

-1

é ù æ d ö I = 2 ê tan -1 x ò xdx - ò ç (tan -1 x÷ ò xdx)dx ú è dx ø ë û

2

I=

ò x cos

Applying Integration by parts

x dx

1 ö x+ 1 æ dx + ò ç x - ÷ e x dx è xø

(b) Let I =

(a) Let I =

ò

1 - 2sin 2 x cos 2 x

1 - 2sin 2 x cos2 x

dx

[(sin 2 x + cos 2 x )2 - 2sin 2 x cos 2 x ] [(sin 2 x + cos 2 x ][sin 2 x - cos 2 x] 1 - 2sin 2 x cos 2 x

= - ò cos 2x dx = 37.

dx

(c) Let Let I =

- sin 2 x 1 + c = - sin 2 x + c 2 2

ò f ( x)dx = y ( x)

òx

5

dx

f ( x3 )dx

put x3 = t Þ 3x2dx = dt 1 3 · x 2 · x 3 · f ( x3 ) · dx 3ò 1 1 = ò tf (t )dt = éët ò f (t )dt - ò f (t )dt ùû 3 3

I=

EBD_8344

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Mathematics

1é t y (t ) - ò y (t )dt ùû 3ë 1 3 3 2 3 = éë x y ( x ) - 3ò x y ( x )dx ùû + c 3 1 3 3 2 3 = x y ( x ) - ò x y ( x )dx + c 3

=

38.

(a) Let I = ò

=

cos 2 x sin 2 x sin 2 x cos 2 x

cos 2 2 x - sin 2 2 x 2 cos 4 x = sin 2 x cos 2 x sin 4 x

2 cos 2 4 x 2cos 2 4 x . sin 4 x dx = ò dx 2cos 4 x 2 cos 4 x sin 4 x 1 1 cos8 x 1 + k = - .cos 8 x + k = ò sin 8 x dx = 2 2 8 16 1 Now, - .cos 8 x + k = A cos 8 x + k 16 1 Þ A=16 sin x 5 5 tan x cos x dx dx = ò (d) ò sin x tan x - 2 -2 cos x cos x æ 5sin x ö dx = òç ´ è cos x sin x - 2cos x ÷ø 5sin x dx =ò sin x - 2 cos x æ 4sin x + sin x + 2cos x - 2cos x ö = òç ÷ø dx è sin x - 2cos x

\ I=ò

39.

ò

ò

1+ x

sin 2 x cos 2 x dx

2

2

x sec 2 x + tan 2 x 1 + x2

(

dx

)

x 2 1 + tan 2 x + tan 2 x

=

ò

=

ò

=

ò 1 + x2

=

ò

1+ x

dx

2

x 2 + tan 2 x(1 + x2 ) 1 + x2 x2

dx

dx + ò tan 2 x dx

x2 +1 -1 1 + x2

(

)

dx + ò sec 2 x - 1 dx

dx 2

+ ò sec 2 xdx - ò dx

1+ x = – tan–1 x + tan x + c Given : f (0) = 0 Þ f (0) = – tan –10 + tan0 + c Þ c = 0 \ f (x) = – tan–1x + tan x Now, f (1) = – tan–1(1) + tan 1 = tan 1 – 41.

p 4

x2 - x (a) Let I = ò dx x3 - x 2 + x - 1 = ò

x ( x - 1) x 2 ( x - 1) + ( x - 1)

dx = ò

x dx x2 + 1

=

1 2 x dx ò 2 ( x 2 + 1)

Let x2 + 1 = t Þ 2x dx = dt

dx sin x - 2 cos x ( sin x - 2 cos x ) + 2 ( cos x + 2sin x ) dx =ò ( sin x - 2 cos x )

cos x + 2sin x = ò dx + 2ò dx = I1 + I2 sin x - 2 cos x cos x + 2sin x where, I1 = ò dx and I 2 = 2ò dx sin x - 2 cos x Let sin x – 2cos x = t Þ (cos x + 2sin x) dx = dt dt \ I 2 = 2ò = 2 ln t + C = 2 ln (sin x–2cos x) + C t Hence, I1 + I 2 = dx + 2ln ( sin x - 2cos x ) + c

ò

x 2 sec 2 x +

= ò 1 dx - ò

( sin x - 2 cos x) + ( 4 sin x + 2 cos x )

sin x - 2cos x æ cos x + 2sin x ö dx + 2ò ç dx =ò è sin x - 2cos x ÷ø sin x - 2cos x

æ x 2 + sin 2 x ö 2 (a) Let f (x) = ò ç ÷ sec x dx è 1 + x2 ø

=

cos 8x + 1 dx cot 2 x - tan 2 x

Now, Dr = cot 2x – tan 2x = =

40.

\I=

(

)

1 log x 2 + 1 + c 2 where ‘c’ is the constant of integration. (d) Statement - 2: cos3x is a periodic function. It is a true statement. Statement - 1

=

42.

1 dt 1 ò = log t + c 2 t 2

æ cos 3x 3cos x ö + Given f ( x) = ò cos3 x dx = ò ç ÷ dx è 4 4 ø =

1 sin 3 x 3 1 3 + sin x = sin 3x + sin x 12 4 4 3 4

Now, period of

1 2p sin 3x = 12 3

= x + 2ln |(sin x – 2 cos x)| + k Þ a = 2

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Integrals

Period of

3 sin x = 2p 4

æ p x pö log tan ç + + ÷ + C è 4 2 8ø

1

=

2

L.C.M. ( 2p, 2p ) 2p = 2p Hence period of f(x) = HCF of 1,3 = ( ) 1

43.

Thus, f(x) is a periodic function of period 2p. Hence, Statement - 1 is false. sin xdx (c) Let I = 2 ò pö æ sin ç x – ÷ è 4ø Let x –

é æ p xö ù êQ ò sec x dx = log tan çè + ÷ø ú 4 2 û ë

46.

p = t Þ dx = dt 4

= x–

47.

p pö æ + log sin ç x – ÷ + c1 è 4 4ø

(c) I = ò Þ I =ò

sin( x - a + a) dx sin( x - a)

ò sin( x - a) dx =ò

sin( x - a ) cos a + cos( x - a)sin a dx sin( x - a )

= (cos a) x + (sin a) logsin( x - a) + C Comparing with Ax + B logsin(x – a) + c \ A = cosa, B = sin a (d) Q f ''(x) – g¢¢ (x) = 0 Integrating, f ¢ (x) – g¢ (x) = c;

Þ f ¢ (1)– g¢ (1) = c Þ 4 – 2 = c Þ c = 2.

pö æ pö æ = x + log sin èç x – ÷ø + c çè where c = c1 – ÷ø 4 4

44.

sin x

(b)

= ò {cos a + sin a cot( x - a )}dx

æ pö sin ç t + ÷ è 4 ø = 2 æ sin t + cos t ö dt Þ I = 2ò dt ò ç sin t ÷ø 2 è sin t Þ I = ò (1 + cot t )dt = t + log |sin t| + c1

æ x 3p ö log tan ç + ÷ + C è2 8 ø 2

1

=

\ f ¢ (x) – g¢ (x) = 2;

Integrating, f (x) – g (x) = 2x + c1

dx

Þ f (2) – g(2) = 4 + c1 Þ 9 – 3 = 4 + c1; Þ c1 = 2 \ f (x) – g(x) = 2x + 2 At x = 3/2, f (x) – g(x) = 3 + 2 = 5.

cos x + 3 sin x

dx é1 ù 3 2 ê cos x + sin x ú 2 2 ë û

48.

2

(c) I = ò e x x x (2 + loge x ) dx 1

2

=

dx 1 1 dx = .ò pö p 2 2ò é p ù æ sin ç x + ÷ êësin 6 cos x + cos 6 sin x úû è 6ø

Þ I=

1

2

= ò e x [ x x + x x (1 + loge x)] dx 1

Q ò e x ( f ( x) + f '( x)) dx = e x f ( x) + c

1 pö æ . cosec ç x + ÷ dx è 2ò 6ø

2

We know that

\ I = ëée x x x ûù 1

ò cosec x dx = log | (tan x / 2) | + C

= e2 ´ 4 - e ´ 1 = 4e2 - e = e(4e - 1)

\ I=

45.

I = ò e x x x [1 + (1 + loge x)] dx

(a)

1 . log tan 2 dx

æx pö çè + ÷ø + C 2 12

ò cos x - sin x =ò

dx 1 æ 1 ö 2ç cos x sin x÷ è 2 ø 2

1 pö dx æ = ò sec çè x + 4 ÷ø dx pö æ 2 2 cos ç x + ÷ è 4ø

a+1

49.

ò

(a)

a

dx ( x + a )( x + a + 1)

a+1

=

1 é 1 ù ê x + a - x + a + 1 ú dx [Using partial fraction] ë û a

ò

a+1

æ ( x + a) ö ÷ = log ç è ( x + a + 1) ø a

= log

æ 2a + 1 2a + 1 ö . = log ç ÷ è 2a + 2 2a ø

9 (Given) 8

EBD_8344

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Mathematics

(2a + 1)2 9 = Þ So, a(a + 1) 2

2

\

2

8a + 8a + 2 = 9a + 9a

Þ a2 + a – 2 = 0 Þ a = 1, – 2 50.

=

2

(c) Let, 1 = ò x2 .e - x dx -1 t 2 t 2 .et dt e (t - 2t + 2)c = ( -2) 2

\ g ( x) =

51.

(b) I =

54.

-1 4 -5 ( x + 2 x 2 + 2) Þ g (-1) = 2 2

(b)

ò

5 -4x ò x e dx 3

Put 1 +

x =tÞ

Þ 1=

2dt

2

( x - 2)

1 2 x

55.

x 2 (x 4 + 1)3/ 4

(b) ò

dx = dt

2

,

x +1 =t x-2

put

( x - 2 )2 =

dt dx

=-

-1 dt 1 -3 dt = - ò t lt = 3 ò t 3/ 4 3 4 3

x 5m -1 + 2 x 4 m -1 (x

2m

m

+ x + 1)

3

x - m -1 + 2 x -2 m -1 (1 + x - m + x -2 m )3

dx = ò

x5 m -1 + 2 x 4 m -1 x

6m

(1 + x - m + x -2 m )3

dx

Put t = 1 + x–m + x–2m dt = - mx - m -1 - 2mx -2 m -1 dx dt = ( x - m -1 + 2 x -2 m -1 ) dx Þ -m

\

\ ò =

=

2-t 1- x +c =– 2 +c =– 2 t 1+ x dx

( x - 2 )5/ 4

é -3 +1 ù 1/ 4 1 ê t 4 ú -4 é x + 1 ù +c ú = = ê ê ú 3 ê -3 ú 3 ë x - 2û +1 ëê 4 ûú

t 2t - t 2 1 -1 Again put t = Þ dt = 2 dz z 2 -1 dz - dz z2 Þ I=2 ò = 2ò 1 2 1 2z - 1 z z z2 2 = – 2 2 z -1 + c = – 2 - 1 + c t

(b) I = ò

3/ 4

dx

I=

53.

( x + 1)

( x - 2)

1 q 1 q q ò 48 qe d q = 48 [q e - e ] + C

ò

dx

-3

1 -4 x3 e (-4 x 3 - 1) + C 48 Then, by comparison f(x) = –4x3 – 1 dx (c) I = ò (1 + x ). x 1 - x

+C

3/ 4

æ x +1 ö ç ÷ è x-2ø

dq Þ x2 dx = 12

52.

ò

1/ 4

dx

Put –4x3 = q Þ –12x2 dx = dq

I=

-1 ´ 4(1 + y)1/ 4 = -(1 + x -4 )1/ 4 + C 4

æ x 4 + 1ö = –ç 4 ÷ è x ø

Put – x2 = t Þ – 2x dx = dt 1= ò

-1 x 3dy -1 dy = ò I= 4 ò 3 3/ 4 4 (1 + y)3/ 4 x (1 + y)

\

dx

=

ò x 3 (1 + x -4 )3/ 4

–4

Let x = y Þ –4x–3 dx = dy Þ dx =

-1 3 x dy 4

56.

x5 m -1 + 2 x 4 m -1 (x

2m

m

+ x + 1)

3

dx =

1 -3 1 +C ò t dt = -m 2mt 2

1 2m(1 + x - m + x -2 m ) 2

x 4m 2m( x 2 m + x m + 1) 2

f ( x) =

(b) I =

ò

+C

+C

x 4m 2 m( x 2 m + x m + 1) 2 x dx

2 - x 2 + 2 - x2

Put t = 2 - x 2 ,

dt 1 = . (-2 x) dx 2 2 - x 2

dx

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Integrals

Þ – t dt = x dx

(-t ) dt

\ I=ò

= - log 57.

=-ò

t2 + t

Þ

1 dt = - log | t + 1| t +1

Þa=

2 - x2 + 1 + c

x 2 - x + 1 cot -1 x dx .e x2 + 1 Put x = cot t Þ – cosec2 t dt = dx Now, 1 + cot2 t = cosec2 t

e (cot t - cot t + 1)

I=ò

\

2

t

61.

(1 + cot t )

0

ò e (cot t - cosec t

2

-1

(a)

=

x6

ò

x+ x

7

(d)

0

6

x(1 + x )

-1

x

p/2

+C

=

ò (1 + (log x)2 )2

dx = ò

dx =

ò

(1 + x ) - 1 x (1 + x 6 )

dx

62.

1 + ( log x ) - 2 log x 2

2 ù2

é1 + ( log x ) ë û

é 1 2t ù t = òe ê ú dt 2 (1 + t 2 ) 2 ûú ëê1 + t

60.

+c =

x

1

= f (x) × e x + cù û

63.

I 2 = ò (1 - x ) 0

1

=

dx - ò x x (1 - x50 )100 dx 4244 3 0 14 II 1

50 101 1 (1 - x ) é x ù (1 - x50 )101 ú - ò I 2 = I1 + ê dx 0 5050 ë 5050 û0

I 2 = I1 + 0 -

I2 5050

1 + esin x

dx =

p 2

f ( x) < 2 f ( x) ³ 2

x I2 > I1 (c) Let p 2 p – 2

dx

ò 1 + cos x

Using ò f (x)dx = ò f (a + b - x) dx

For x Î (0, 1) Þ x > x2 or – x < – x2 and x2 > x3 or – x2 < – x3

-x

(c) I =

p 4

1 - x3 e dx 0

I3 =

3p 4

2

1 - x2 e 0

4

[By Gamma function]

p = ( 2 + 1) 2 (d) Given: 1 -x e 0

sin 4 x . dx

2I = 4

ép p p p ùü - ê sec - ln sec + tan ú ý 4 4 4 4 ûþ ë

97.

x öö ÷ ÷ . dx .....(2) x ø ÷ø

æ æ 2 + sin x ö ö sin x çç1 + log ç ÷ ÷÷ dx ..... (1) è 2 – sin x ø ø è 4

pù é 3p I = – (cot x)3pp/ 4/ 4 = – ê cot - cot ú = 2 4 4û ë 100. (c) In =

ò tan

n

x dx, n > 1

Let I = I4 + I6 = ò (tan 4 x + tan 6 x)dx = ò tan 4 x sec2 x dx

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Integrals

Let tan x = t Þ sec2 x dx = dt \ I = ò t 4 dt =

t5 +C 5

1 5 tan x + C Þ On comparing, we have 5 1 a= ,b=0 5

=

101. (a) Let I = ò

dx

2

((x - 1a) 2 + 3)3 / 2

1

Let; x - 1 = 3 tan q 2 Þ dx = 3 sec . dq

Þ I=ò

0

x

3 sec 2 q dq

p6

æ çè

(

3 tan q

2 ö 3/2

1 p6 1 p 6 sec2 q dq = ò cos q dq = ò0 3 3 0 sec3 q 1 1 1 1 p6 = [sin q]0 = ´ = 3 2 6 3

=

102. (a)

cos 2 x

ò

p æ 1 ö ç ÷ 12 è sin 2 x ø

=

1 4

3

=

p 4

2 ò cos 2 x ´ sin 2 x . sin ( 2 x) dx

p 12

p 4

ò

sin 4 x . (1 - cos 4 x ) dx

p 12

p ép ù 4 ê ú 1 1 4 = ê ò sin 4 x - ò sin 8 x ú 4êp 2p ú êë12 úû 12 p 4

1 é cos4 x cos8x ù + 4 16 úûp

= ê4ë 103. (d)

12

1 é15 ù 15 = ê ú= 4 ë 32 û 128

ò ( x5 + x3 + 1)3 dx

Dividing by x15 in numerator and denominator

ò

x

3

+

5 x6

ò

ty (t) dt +

1

ò

x

ò y(t )dt + x[ y( x) - y(1)] 1 x

= ò ty (t )dt + x[ xy ( x) - y (1)] + xy( x) - y(1) x

x

1

1

Diff. again w.r. to x y (x) – y (a) = xy (x) – y (a) + 2x y (x) + x2y1 (x) (1 – 3x) y (x) = x2y1 (x)

y1 ( x) 1 - 3x = y( x) x2 1dy 1 - 3x 1 = Þ ln y = – – 3 ln x y dx x2 x 1 ln (y x3) = – x yx3 = – e–1/x -

y=

e

1 x

x3

or y =

ce

- 1x

x3

é x2 ù ë û ò é x 2 - 28 x + 196 ù + é x2 ù dx .....(a) 4 ë û ë û

10

105. (d) I =

dx

æ 1 1ö çè1 + 2 + 5 ÷ø x x

b

3

ty (t) dt

1

Differentiate w.r. to x.

2 x12 + 5 x9

2

y (t) dt = x

1

x

2 ò y(t )dt = ò ty(t )dt + x y( x) -y (1)

k =1 p 4

x

1

1 k = Þ k + 5 = 6k 6 k +5

Þ

ò

104. (d) x

) + ( 3 ) ÷ø 2

1 1 Let 1 + 2 + 5 = t x x 5 ö æ –2 5 ö æ 2 Þ ç dx = dt Þ ç 3 + 6 ÷ dx = - dt – è x 3 x 6 ÷ø èx x ø This gives, 2 5 dx + – dt 1 x3 x6 = ò 3 = +C ò 3 t 2 t2 1 1ö æ + + 1 çè ÷ x 2 x5 ø 1 x10 = + C = +C 2 2( x5 + x3 + 1) 2 æ 1 1ö 2ç1+ 2 + 5 ÷ è x x ø

Use

òf

a

b

(a + b – x) dx =

ò f (x) dx

a

EBD_8344

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Mathematics

Þ f (x) = f (x + 4) Hence period is 4

é( x - 14) 2 ù ë û I= ò dx 2ù é 2ù é ( 14) x + x 4 ë û ë û 10

.....(b)

50

(a) + (b)

14

ò

Consider

f ( x)dx = 10

10

ò

f ( x )dx = 10 [5 + 5] = 100

10

109. (d) Let f : (–1, 1) ® R be a continuous function

10 é ( x - 14) 2 ù + é x 2 ù

ë û ë û dx 2I = ò é 2 ù é 2ù 4 ë x û + ë ( x - 14) û

sin x

Let

ò

f (t)dt =

0

3 x 2

10

ò dx Þ 2I = 6

2I =

Þ I=3

f(sin x).

4

1 1 p 106. (b) 2ò tan -1 x dx = ò æç - tan -1 (1 - x + x 2 ) ö÷ dx 0 0è 2 ø 1

1p

0

0

2 ò tan -1 x dx = ò 1

ò0 tan

-1

2

Let, I1 =

1

ò0 tan

-1

Þ f (sin x). cosx =

dx - ò tan -1 (1 - x + x 2 ) dx

(1 - x + x 2 ) dx =

1

put x =

0

1 p - 2 ò tan -1 xdx 0 2

æ 3ö fç ÷ = è 2 ø

1/x

4

2

+ log(36 - 12x + x )

2 log x

4

2

+ log(6 - x) 2

2 log(6 - x)

dx

1 2

d

2

ln f(x) = ò + 1 d ( ) 1

+ log x 2

dx .

...(i)

x

dx

...(ii)

x é ( ln ) 2 ù ( ln x ) 2 ln x æ 1ö ú = f ( x) + f ç ÷ = ò d = ê è xø 2 êë 2 úû 1 1 p

1

x

log(6 - x) 2

I=ò

dt = -

2

log x

I=ò

ln t

ò 1+t dt

1 Let t =

log x 2

2 log x

3

1 110. (c) f æç ö÷ = èxø

p ép 1 ù - 2 ê - log 2ú = log 2 2 4 2 ë û

2

p 3

æ 3ö 1 f ç ÷. = 3 è 2 ø 2 2

xdx

p 1 x p 1 dx = - log 2 = - ò0 4 4 2 1 + x2 By equation (a)

4

3 2

pö p æ 3 f çè sin ÷ø .cos = 3 3 2

.....(a)

1 1 1 -1 x dx = ëé (tan x ) x ûù - ò0 0 1 + x2

107. (a) I = ò

d 3 (sin x) = dx 2

2I = ò dx = [ x ]2 = 2 4

2

I=1 108. (c) Let f : R ® R be a function such that f (2 – x) = f (e + x) Put x = 2 + x we get f (–x) = f (4 + x) = f (4 – x)

111. (b) Let I =

ò

1 + 4 sin 2

0 p /3

x x - 4sin dx = 2 2

p

p

x

ò 2 sin 2 - 1 dx 0

xö x ö æ æ = ò ç 1 - 2 sin 2 ÷ dx + ò ç 2 sin 2 - 1÷ dx è ø è ø 0 p/3 x 1 x p é êQ sin 2 = 2 Þ 2 = 6 ë p x 5p 5p ù Þx= , = Þx= > pú 3 2 6 3 û

www.jeebooks.in M-407

Integrals p /3

xù é = ê x + 4cos ú 2 û0 ë

æ p 3 3 pö +4 - 4 +ç0 - p + 4 + ÷ ç 3 2 2 3 ÷ø è p = 4 3-43 =

112. (d) F(x) =

x

ò1

et dt , x > 0 t

et ò1 t + a dt Put t + a = z Þ t = z – a; dt = dz for t = 1, z = 1 + a for t = x, z = x + a x+a

ò1+ a

\I=

-a = e ò

x+a

1+ a

e

z -a

dz

z

ez x + a et dz º e - a ò dt 1+ a z t

= e

é 1+ a et ù x + a et dt + ò dt ú ê-ò 1 1 t t úû ëê

= e [ - F (1 + a) + F ( x + a)] = e [ F ( x + a) - F (1 + a)]

113. (a) Let t

Þ

ò-p

Þ

t

ò-p

t

ò-p

( f ( x ) + x)dx = p2 – t2

f ( x )dx + ò

t -p

xdx = p2 – t2

æ t 2 p2 ö f ( x )dx + ç - ÷ = p2 – t2 2ø è2

3 2 2 (p - t ) ò-p 2 differentiating with respect to t

Þ

t

f ( x )dx =

d é t 3 d 2 2 f ( x) dx ùú = (p - t ) dt êë ò-p û 2 dt

dt d - f ( -p) (-p ) = – 3t dt dt f (t) = – 3t f (t ).

æ pö æ pö f ç - ÷ = -3 ç - ÷ = p è 3ø è 3ø

...(1) p

ò

[cos( p - x )]dx =

ò

[ - cos x ]dx ...(2)

0

0

On adding (1) and (2), we get p

p

0

0

ò [cos x]dx + ò [ - cos x]dx p

2I =

ò

0

2I =

[cos x ] + [ - cos x]dx p

ò0

(Q [ x ] + [ - x] = -1 if x Ï Z )

- 1dx p

2I = - x 0 = -p -p 2 e

ò (log x )

115. (c) Pn =

n

dx

1

put log x = t then x = et and dx = et dt Also, when x = 1, then t = log 1 = 0 and when x = e, then t = loge e = 1 1

\ Pn =

òt

n t

. e dt

0 1

(By the definition of F(x)) -a

[cos x]dx

p

I=

\ P10 =

-a

ò

0

ÞI=

é x + a et ù et -a 1 dt + ò dt ú I = e êò1 a 1 + t t úû ëê -a

114. (d) Let I =

2I =

x

Let I =

p

p

x é ù + ê -4 cos - x ú 2 ë ûp / 3

Now, P10

1

10 t ò t e dt and P8 =

òt

0

0

8 t

e dt

1

1

0

0

10 t 8 t – 90P8 = ò tI IIe dt - 90ò t e dt

1

1 9 t 1 8 t 10 t P10 – 90 P8 = éët e ùû - 10ò t e dt - 90ò t e dt 0 0 0 P10 – 90 P8

1 1 é 1 d 9 t ù 9 8 t t ê ú 10 ( ) 90 e t e dt t e dt = ò ò dt ò ú ò t e dt êë 0 0 0 û 1 8 t ù 1 8 t é P10 – 90 P8 = e - 10 ê e - 9ò0 t e dt ú - 90ò0 t e dt ë û 1

8 t 8 t P10 – 90 P8 = e - 10e + 90ò t e dt - 90ò t e dt 0

\ P10 - 90 P8 = -9e 1/ 2

ln(1 + 2 x)

1/ 2

ln(1 + 2 x) dx or 2 ò 1 + (2 x)2 dx 1 + 4 x 0 0 Put 2x = tanq

116. (c) Let I =

\

ò

2dx sec 2 qd q = sec 2 q or dx = dq 2

EBD_8344

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Mathematics

also when x = 0 Þ q = 0

x

p 1 and when x = Þ q = 45°or 2 4 p 4

ò

\I=

0

ln(1 + tan q) 2

1 + tan q

´

sec 2 qd q 2

1 ln(1 + tan q) ´ sec 2 qd q 2 ò 1 + tan 2 q 0

(Q 1 + tan2 q = sec2 q) p

14 I = ò ln(1 + tan q)d q 2

0

±2

\ Equation of tangent is y – 2 = 2(x – 2) or y + 2 = 2(x + 2) Þ x-intercept = ± 1. p /3

p /3

p é ù tan - tan q ú ê 1 4 I = ò ln ê1 + ú dq p 2 ê 1 + tan ´ tan q ú 0 4 ë û p 4

1 é 1 - tan q ù dq I = ò ln ê1 + 2 ë 1 + tan q úû 0 p 4

é1 + tan q + 1 - tan q ù 1 ln ê ú dq 2ò 1 + tan q ë û 0 p

14 é 2 ù I = ò ln ê ú dq 2 ë1 + tan q û 0 p

14 I = ò [ln 2 - ln(1 + tan q)]d q 2 p 4

1 1 ln 2.d q - ò ln(1 + tan q)d q 2ò 2 0

0

p 4 1 -I I = ln 2q (from eq. (i)) 2 0 1 æp ö I + I = ln 2 çè - 0÷ø 2 4 1 p 2I = ´ ´ ln 2 2 4 p p 2I = ln 2 or I = ln 2 8 16

dx 1 + tan x p /3

dx

ò

=

ò

=

æp ö tan ç - x÷ è2 ø

p /6 1 +

(Using the property of definite integral)

I=

ò

p /6

é 1 æp öù ln ê1 + tan ç - q÷ ú d q è øû 2ò 4 ë 0

0 p 4

t dt = ± 2

0

118. (d) Let, I =

p 4

I=

ò

Points y =

p 4

I=

dy =2 dx Þ x=±2

But from y = 2x, \

...(i)

0

ò t dt , x ÎR

dy therefore = x dx

Þ |x|=2

p 4

I=

117. (a) Since, y =

tan x dx

1 + tan x p /6

…(i)

Also, given p /3

I=

tan x dx

ò

…(ii)

1 + tan x p /6

By adding (i) and (ii), we get p /3

2I=

ò

dx

p /6

1 ép pù p = , 2 êë 3 6 úû 12 Statement-1 is false

Þ I=

b

Q

ò

b

f ( x )dx = ò f (a + b - x ) dx

a

a

It is fundamental property. Statement -2 is true. 119. (a) Consider sin 2 x

ò

-1

sin ( t ) dt +

0

cos2 x

ò

cos -1 ( t ) dt

0

Let I = f (x) after integrating and putting the limits. f ¢(x) = sin -1 sin 2 x (2 sin x cos x) - 0 + cos-1 cos2 x (-2 cos x sin x) - 0 \ f ¢(x) = 0 Þ f (x) = C (constant) p Now, we find f (x) at x = 4 1/2

\ I=

ò 0

sin -1 t dt +

1/2

ò cos

-1

t dt

0

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Integrals 1/2

ò (sin

=

-1

t + cos -1 t ) dt =

1/2

0

0

\ f ( x) =

p /2

ò

ò

Þ I=

sin 2 x

– p/2 1 + 2 p/2

x

2

sin x

-x -p/2 1 + 2 p /2

ò

Þ I=

-p /2

y

p p dt = = C 2 4

1+ t2 dy 1 . Þ 1= 2 dx 1+ y y( x) é dI(x) êQ If I( x) = ò f (t ) dt , then = f {y( x)} . dx ê f ( ) x ë 0

p 4

dx

1+ 2

x

{

...(i)

p p dx , by replacing x by æç - - x ö÷ è2 2 ø

2 x .sin 2 x

dt

ò

122. (a) x =

p 4

\ Required integration = 120. (d) I =

ò

dx

...(ii)

ò

2I =

-p /2

1 Þ I= 4 =

1 4

Þ I=

1 sin x dx = 2

p /2

ò

2

(1 - cos 2 x ) dx

d2y

Þ

dx

2

=

1 2 1+ y

123. (b, c) g ( x + p ) =

-p /2

p/2

x +p

ò

cos 4t dt

x

p+ x

0

x

éæ p sin p ö æ p sin (-p) ö ù ÷ú êçè 2 + 2 ÷ø - çè - 2 + øû 2 ë

1 ép pù p + = 4 êë 2 2 úû 4

p+ x

ò

ò

0

p

cos 4t dt = ò cos 4t dt = g (x) + g (p) = g (x) – g (p)

x

0

(QFrom graph of cos 4t, g (p) = 0) 0.9

tan 2 x dx

124. (d)

7 p /3

tan x dx = - log cos x 7 p /4

ì 2 æ 2 - xö ü í[ x ] + log ç ý dx è 2 + x ÷ø þ î -0.9

ò

0.9

=

ò

0.9

[ x 2 ] dx +

-0.9

7p 7p ù é = - ê log cos - log cos 3 4 úû ë 7p 7p - log cos 4 3 é p öù æ 7p ù é ê cos çè 2p - 4 ÷ø ú ê cos 4 ú = log ê ú ú = log ê ê cos æ 2p + p ö ú ê cos 7p ú ç ÷ 3 û ë 3 ø ûú è ëê

=0+

æ pö ç ÷ 4 = log ç ÷ p÷ ç ç ÷ 3ø è

æ 2 ö = log ç ÷ = log 2. è 2ø

1 2 1 2

ö ÷ ÷ ÷ ÷ ø

ò

-0.9 0.9

ò

-0.9

= log cos

æ ç cos = log ç çç cos è

ò

p

cos 4t dt = g ( x ) + ò cos 4t dt

(it is clear from graph of cos 4t)

7 p /3 7 p /4

dy y = . 1+ y2 = y 2 dx 1+ y

= ò cos 4t dt +

7 p /4

ò

. 2y .

0

7 p /3

=

2

sin 2 x ù é êë x + 2 úû -p / 2

121. (d) Let I =

{

dy = 1 - y2 dx

Adding equations (i) and (ii), we get p /2

}

æ 2 - xö dx log ç è 2 + x ÷ø

æ 2-x ö log ç ÷ dx è 2+ x ø

Put x = – x Þ f (x) = log

2-x 2+ x

(2 - x) 2+ x = – log = – f (x) 2-x 2+ x So, it is an odd function, hence Required integral = 0. and f (–x) = log

a

125. (d) Since

ò [ x] = 0 where 0 £ a £ 1 0

0.9

\

}

ù d d y ( x ) - f {f( x )} . f( x ) ú dx dx û

ò [ x - 2[ x]]dx = 0 0

EBD_8344

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Mathematics a

d e tan x æ pö , x Î ç 0, ÷ G ( x) = è 2ø dx x

126. (a) Let

1

1

=8

Let px2 = t Þ 2px dx = dt

=8

p

4

\I= ò p

0

4

1 p 1 p , t = and x = , t = 2 4 4 16

p /4

p /4

é 1 - tan q ù é 2 ù ò log êë1 + 1 + tan q úû d q = 8 ò log êë1 + tan q úû d q 0 0

p /4

ò

=8

p æ pö æ pö e tan t dt = G (t ) 4p = G çè 4 ÷ø - G çè 16 ÷ø t

[log 2 - log(1 + tan q)]d q

0 p14

p14

0

0

= 8log 2

16

16

0

é æp öù log ê1 + tan ç - q÷ ú d q è4 øû ë

ò

2 2p x 2 2 2 e tan px .dx Now, I = ò e tan px × dx = ò 2 1 px 1 x

When x =

a

f ( x) dx = ò f (a - x ) dx

0

p /4

2 4

ò

Applying

ò 1d q - 8 ò

I = 8 × (log 2)[ x]0p /4 - 8

x2 127. (d) Let ò t f ( t ) dt = sin x - x cos x 2 e By using Leibnit rule, we get x

p Now, put x = , we get 6

2

2

1.5

1

129. (d) I = 8 log(1 + x ) dx ò 1+ x2 0 Put x = tan q, \ dx = sec 2 qd q p /4 \ I = 8 log(1 + tan q ) .sec 2 q dq ò 1 + tan 2 q 0 p /4

ò 0

log(1 + tan q )dq

130. (a) p '( x ) = p '(1 - x )

...(i)

...(i)

0 1

1.5

é 2ù 1.5 = ò x.0 dx + ò xdx + ò 2 xdx = 0 + ê x ú + é x 2 ù ë û 2 0 1 2 ë 2 û1 1 1 = ( 2 - 1) + ( 2.25 - 2) = + 0.25 2 2 1 1 3 = + = 2 4 4

I =8

[From equation (i)]

1

2

2

p I = 8 × × log 2 - I 4 Þ 2I = 2p log 2 \ I = p log 2

Let I = ò p ( x)dx

2 2 2 2 ò x éë x ùû dx = ò x[ x ]dx + ò x éë x ùû dx + ò x éë x ùû dx 0 0 1

1

log(1 + tan q) d q

Þ p ( x ) + p (1 - x) = 42

1 1 p æ pö Þ f ç ÷ = sin - 1 = - 1 = – è 6ø 6 2 2 1

ò

Þ p( x) = - p(1 - x ) + c at x = 0 p(0) = – p(1) + c Þ 42 = c Now, p( x ) = - p(1 - x) + 42

p æ pö p p p . f ç ÷ = .sin 6 è 6ø 6 6 6

1.5

p /4 0

ù d é d éx x2 ù êsin x - x cos x - ú ê ò t f ( t ) dt ú = dx ë e 2 úû û dx ëê Þ x f(x) – e f(e). 0 = x sin x – x

128. (c)

log(1 + tan q) d q

Þ I = ò p(1 - x)dx

...(ii)

0

Adding eqn. (i) and (ii), 1

2 I = ò (42) dx Þ I = 21 0

131. (c) Let I = =

p

ò0 [cot x] dx

p

1

....(i) p

ò0 [ cot (p - x) ] dx = ò0 [- cot x] dx

....(ii)

Adding eqn s (i) & (ii), We get 2I =

p

ò0 ([ cot x] + [ - cot x]) dx p

= ò ( -1)dx 0

[ Q[ x] + [- x] = -1, if x Ïz and [x] + [–x] = 0, if x Î z] = [ - x ]0p = -p

Þ I=–

p 2

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Integrals

132. (b) We know that Þ

sin x 1

Þ

ò

x

1

x

0

Þ

1

ò

Þ

ò

dx
0. If y (p) = p, then æpö æpö y '' ç ÷ + y ç ÷ is equal to : 2 è ø è2ø

xe(x – 1)

52.

53.

(c) y = (d) y = x ln x + x The normal to a curve at P(x, y) meets the x-axis at G. If the distance of G from the origin is twice the abscissa of P, then the curve is a [2007] (a) circle (b) hyperbola (c) ellipse (d) parabola. dy = y (log y – log x + 1), then the solution of the dx equation is [2005]

57.

2+

p 2

(b) 1 +

p p2 + 2 4

(c)

2+

p p2 + 2 4

(d) 1 +

p 2

If for x ³ 0, y = y(x) is the solution of the differential equation, (x + 1)dy = ((x + 1)2 + y – 3)dx, y(2) = 0, then y(3) is equal to _____.

58.

[NA Jan. 09, 2020 (I)]

Let y = y(x) be the solution curve of the differential

(

2 equation, y - x

) dydx = 1, satisfying y(0) = 1. This curve

æ xö (a) y log ç ÷ = cx è yø

æ yö (b) x log ç ÷ = cy è xø

intersects the x-axis at a point whose abscissa is:

æ yö (c) log ç ÷ = cx è xø

æ xö (d) log ç ÷ = cy è yø

(a) 2 – e

(b) – e

(c) 2

(d) 2 + e

The solution of the equation

d y dx

2

= e -2 x

[Jan. 7, 2020 (II)]

[2002]

59.

æ 1ö 2 Consider the differential equation, y dx + ç x - y ÷ dy = 0 . è ø

(a)

e -2 x 4

(b)

e -2 x + cx + d 4

If value of y is 1 when x = 1, then the value of x for which y = 2, is : [April 12, 2019 (I)]

(c)

1 -2 x e + cx 2 + d 4

(d)

1 -4 x e + cx + d 4

(a)

TOPIC Đ d 2 y

60.

5 1 + 2 e

(b)

1 1 + 2 e

(d)

dy = (tan x - y )sec2 x , x Î dx

dx 2 Let y = y(x) be the solution of the differential equation

dy æ pö cos x + 2 y sin x = sin 2 x , x Î ç 0, ÷ . dx è 2ø If y (p / 3) = 0, then y (p / 4) is equal to : [Sep. 05, 2020 (II)] (a)

2- 2

(b)

(c)

2 -2

(d)

2+ 2

1 2

3 1 2 e

3 - e 2 If y = y(x) is the solution of the differential equation (c)

55.

[Sep. 04, 2020 (I)]

(a)

If x

2

54.

Let y = y (x) be the solution of the differential equation,

æ p pö ç - 2 , 2 ÷ , such that y (0) = 0, è ø

æ pö then y ç - ÷ is equal to : è 4ø

[April 10, 2019 (I)]

(a) e – 2

(b)

1 e

(d)

(c)

2+

1 -e 2 1 -2 e

-1

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Differential Equations

61.

Let y = y(x) be the solution of the differential equation,

66.

dy æ 2 x + 1 ö 1 -2 -2 x +ç ÷ y = e , x > 0, where y (1) = e , then dx è x ø 2 : [Jan. 11, 2019 (I)]

æ p pö dy + y tan x = 2 x + x 2 tan x , x Î ç - , ÷ , such that y (0) dx è 2 2ø = 1. Then : [April 10, 2019 (II)] (a)

æpö æ pö y 'ç ÷ + y ' ç - ÷ = - 2 è4ø è 4ø

(c)

æpö æ pö yç ÷ - yç - ÷ = 2 4 è ø è 4ø

(d) 62.

2 æpö æ pö p +2 yç ÷ + yç- ÷ = è4ø è 4ø 2

(b)

(b)

y ( log e 2 ) =

log e 2 4

67.

Let f be a differentiable function such that f ¢(x) = 7 -

3 f ( x) , 4 x

The solution of the differential equation x

æ1ö (x > 0) and f (1) ¹ 4. Then lim x f ç ÷ : + x ®0 è xø [Jan. 10, 2019 (II)]

(x ¹ 0) with y(1) = 1, is:

(a) exists and equals

dy + 2y = x2 dx [April 09, 2019 (I)]

y(1) = (a)

(b) y = x + 1 5 5 x2

dy + 2x(x2 + 1)y = 1 such that y(0) = 0. If dx

68.

a

p , then the value of ‘a’ is : [April 08, 2019 (I)] 32

1 4

(b)

1 2

æ1ö dy + 2 y = x 2 satisfying y(1) = 1, then y ç ÷ is equal to: dx è2ø [Jan. 09, 2019 (I)] 1 7 (a) (b) 64 4 49 13 (c) (d) 16 16 69. Let y - y(x) be the solution of the differential equation x

sin x

æ pö dy + y cos x = 4x, x Î (0, p) . If y ç ÷ = 0 , then è2ø dx

æ pö y ç ÷ is equal to : è6ø

dy + y = x log x, (x > 1). If 2y(2) = log 4 –1, then y(e) is e e dx equal to : [Jan. 12, 2019 (I)]

(a)

(d)

x

(a)

-

e 2

(b)

-

e 2

e e2 (d) 4 4 If a curve passes through the point (1, – 2) and has slope

x2 - 2 y of the tangent at any point (x, y) on it as , then the x curve also passes through the point :[Jan. 12, 2019 (II)]

(c) (–1, 2)

(b) (d)

( ) ( - 2,1)

-8 2 p 9 3

[2018] (b)

8 - p2 9

4 2 4 p - p2 (d) 9 3 9 Let y = y (x) be the solution of the differential equation (c)

2

(c)

(a) (3, 0)

(b) exists and equals 4.

(c) does not exist. (d) exists and equals 0. If y = y(x) is the solution of the differential equation,

1 16 Let y = y(x) be the solution of the differential equation,

(c) 1

4 . 7

3

3 2 1 3 x2 + 2 (d) y = x + 2 4 4x 4 4x Let y = y(x) be the solution of the differential equation,

(x2 + 1)2

65.

y ( log e 2 ) = loge 4

æ1 ö y ( x ) is decreasing in ç ,1÷ è2 ø (d) y (x) is decreasing in (0, 1)

(c) y =

64.

(a)

(c)

æpö æ pö y¢ ç ÷ - y ¢ ç - ÷ = p - 2 è4ø è 4ø

4 3 1 (a) y = x + 2 5 5x

63.

If y(x) is the solution of the differential equation

70.

dy + 2 y = f (x), where f (x) = dx

ì1, x Î [0, 1] í î0, otherwise

æ 3ö If y (0) = 0, then y ç ÷ is è 2ø (a)

3, 0

(c)

e2 – 1 2e 1 2e

3

[Online April 15, 2018] (b) (d)

e2 – 1 e3 e2 + 1

2e 4

EBD_8344

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71.

The curve satisfying the differential equation, ydx–(x + 3y2) dy = 0 and passing through the point (1, 1), also passes through the point : [Online April 8, 2017] (a)

72.

Mathematics

æ1 1ö ç ,- ÷ è4 2ø

(b)

æ1 1ö (c) ç , - ÷ è 3 3ø The solution

of

(a)

æ 1 1ö ç- , ÷ è 3 3ø

æ1 1ö (d) ç , ÷ è4 2ø the differential

x sec x + tan x

(b) y = 1 +

(b) x2 – 1

x -1

x x2 - 1 (d) x2 - 1 x The general solution of the differential equation dy 2 + y = x2 is [Online May 19, 2012] dx x

(c) 78.

equaiton (a)

dy y tan x p + sec x = , where 0 £ x < , and y(0) = 1, is dx 2 2y 2 given by : [Online April 10, 2016] (a) y2 = 1 +

1 2

x sec x + tan x

x x (d) y2 = 1 – sec x + tan x sec x + tan x Let y(x) be the solution of the differential equation dy (x log x) + y = 2x log x, (x ³ 1). Then y (e) is equal to: dx [2015] (a) 2 (b) 2e (c) e (d) 0

74.

75.

dy + y tan x = sin 2x and y(0) = 1, then y(p) is equal to: dx [Online April 19, 2014] (a) 1 (b) – 1 (c) – 5 (d) 5 The general solution of the differential equation,

(a)

(b)

80.

81.

(a) (c) 77.

dy + 2 xy = 4 x 2 is dx

(1 + x 2 ) y = x3 (1 + x 2 ) y = 3x3

[Online April 25, 2013] (b) (d)

3(1 + x 2 ) y = 2 x3 3(1 + x 2 ) y = 4 x3

The integrating factor of the differential equation

(

)

dy x2 - 1 + 2 xy = x is dx

(b)

3-

[Online May 26, 2012]

1 e + y e 1

1 ey 1 ey (c) 1 + (d) 1 - + y e y e Solution of the differential equation p cos x dy = y (sin x - y ) dx, 0 < x < is 2 (a) y sec x = tan x + c (b) y tan x = sec x + c (c) tan x = (sec x + c) y (d) sec x = (tan x + c) y Solution of the differential

[2010]

equation

= 0 is ydx + ( x + x y )dy

(a)

[2004]

log y = Cx

(b)

-

y cot x = tan x + c

(d) y cot x = x + c y tan x = cot x + c The equation of the curve passing through the origin and satisfying the differential equation (1 + x 2 )

2 e y e

1 y

2

1 + log y = C xy

1 1 =C (d) + log y = C xy xy The solution of the differential equation

(c)

(c) 76.

4-

1 y

1

æ dy ö sin 2x ç - tan x ÷ - y = 0 , is : dx è ø [Online April 12, 2014]

y tan x = x + c

x2 4

æ 1ö y 2 dx + ç x - ÷ dy = 0. If y (1) = 1, then x is given by : yø è

If

(a)

y = cx 3 -

(b)

y = cx 2 +

(c) y = 1 – 73.

x2 4

x3 x3 (d) y = cx -2 + 5 5 Consider the differential equation [2011RS]

(c) 79.

y = cx -3 -

82.

(1 + y 2 ) + ( x - e tan

(a)

xe 2 tan

-1

y

-1

= e tan

(b) ( x - 2) = ke 2 tan (c) (d)

2 xe tan

xe tan

-1

-1

y

y

y

)

dy = 0 , is dx

-1

-1

= e 2 tan

y

+k

y -1

y

+k

= tan -1 y + k

[2003]

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Differential Equations

1.

\ Eqn is (x)2 + (y – a)2 = a2 where (0, a) is the centre of circle.

(a) Since, x2 = 4b(y + b) x2 = 4by + 4b2 2x = 4by’

Þ x 2 + y 2 + a 2 - 2ay = a2

x Þ b= 2 y' So, differential equation is

æ xö 2x .y + ç ÷ x = y' è y' ø

2

2x + 2 y

2

2

x2 y2 + =1 a 2 b2 Since, it passes through (0, 3)

(c)

\

x+ y

Þ

0 9 + =1 a 2 b2

dy dy -a =0 dx dx dy dx

dy dx

=a

Put value of 'a' in eqn (1), we get

Þ b2 = 9

é dy ù ê y dx + x ú 2 2 x + y - 2y ê ú =0 ê dy ú ë dx û

\ eq. of ellipse becomes: x2 y2 + =1 9 a2

dy dy - 2 y2 - 2 xy = 0 dx dx

differential w.r.t.x, we get;

Þ ( x2 + y 2 )

2 x 2 y dy + =0 9 dx a2

Þ ( x2 + y 2 - 2 y 2 )

Þ

y x

æ dy ö – 9 ç ÷= 2 è dx ø a

2

y d y + x dx 2

x

dy = 2xy dx

dy = 2xy º g ( x) y dx Hence, g(x) = 2x (b) Statement -1 : y2 = ± 4ax 2 2 Þ (x - y )

Again differentiating w.r.t.x, we get;

3.

dy dy - 2a =0 dx dx

Þ x+ y

x(y¢)2 = 2yy¢ + x 2.

2

...(1) Þ x + y - 2ay = 0 Differentiate both side w.r.t 'x', we get

4.

dy –y dy dx =0 dx x2

Þ

Þ xyy" + x(y')2 – yy' = 0 (c) Since family of all circles touching x-axis at the origin

dy = 4a dx Thus both statements are true but statement-2 is not a correct explanation for statement-1. (d) Statement - 1 dy Given differential equations are + y 2 = x and dx

Statement -2 : y2 = 4ax Þ 2 y

5. (0, a)

1 dy dy 1 = ± 2a . Þ µ dx y dx y

d2y

(0, 0)

+ y = sin x dx 2 Their degrees are 1. Both have equal degree. Also, Statement - 2 is the correct explanation for Statement - 1.

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6.

(c) We have y = c1e c2 x Differentiate it w.r. to x Þ

y ¢ = c1c2 e c2 x = c2 y

Þ

y¢ = c2 Differentiate it w.r. to x y

Þ

7.

Mathematics

10.

= 0 Þ y ¢¢ y = ( y ¢)2 y2 (c) Let the centre of the circle be ( h, 2) \ Equation of circle is

11.

dy dx

(d) Ax + By = 1 Differentiate w.r. to x dy Ax + By =0 dx Again differentiate w.r. to x

12.

2 ìï d2y dy æ dy ö üï x í - By 2 - B ç ÷ ý + By =0 è ø dx dx dx îï þï Dividing both sides by –B, we get

d2y

2

dy æ dy ö + xç ÷ - y =0 2 è dx ø dx dx Therefore order 2 and degree 1. xy

y 2 = 4a( x - h),

(c)

Differentiating 2 yy1 = 4a Þ yy1 = 2a Again differentiating, we get Þ y12 + yy2 = 0 Degree = 1, order = 2.

13.

14.

2 æ 4d3yö æ d yö 1 3 + =ç (c) ç ÷ ÷ d xø è è d x3 ø

3

2 æ d3yö æ d yö Þ ç1 + 3 = 16 ç ÷ ÷ è d xø è d x3 ø Order = 3, degree 3

3

1 + x 2 × 1 + y 2 = - xy

(a)

...(i) ....(ii)

ò

dy dx

1 + x2 y dx = - ò dy x 1 + y2

Let x = tan q Þ dx = sec 2 qd q

2

d2y

æ dy ö + Bç ÷ = 0 2 è dx ø dx From (ii) and (iii) A + By

x + yy ¢ dy dy - 2a =0 Þa= y¢ dx dx

Þ ( x 2 - y 2 ) y ' = 2 xy

dy .y = 0 Þ x2 + y2 – 2x2 – 2x dx

9.

......(1)

Þ ( x 2 + y 2 ) y '- 2 xy - 2 y 2 y ' = 0

2

2

........ (iii)

æ x + yy ¢ ö Putting in (1) we get, x 2 + y 2 - 2 ç y=0 è y ¢ ÷ø

( y – 2) ç ÷ + ( y – 2) = 25 è dx ø Þ (y – 2)2 (y')2 = 25 – (y –2)2 (a) General equation of circles passing through origin and having their centres on the x-axis is x2 + y2 + 2gx = 0 ...(i) On differentiating w.r.t x, we get dy dy ö æ 2x + 2y . + 2g = 0 Þ g = – ç x + y ÷ è dx dx ø Putting in (i) ì æ dy ö ü x2 + y2 + 2 í - ç x + y ÷ ý . x = 0 è dx ø þ î

2

x 2 + y 2 - 2ay = 0 Differentiate w.r. to x,

(c)

2x + 2 y

2

Þ y2 = x2 + 2xy

........ (ii)

( y - 2 xy ') 2 = 4 yy '3 Hence equation (iii) is of order 1 and degree 3.

…(1) ( x – h) 2 + ( y – 2) 2 = 25 Differentiating with respect to x, we get dy 2( x – h) + 2( y – 2) = 0 dx dy Þ x – h = –( y – 2) dx Substituting in equation (1) we get

8.

........ (i)

Þ y 2 = 2 yy ' ( x + yy ') On simplifying, we get

y ¢¢ y - ( y ¢ ) 2

2 æ dy ö

2 (c) y = 2c( x + c ) Differentite it w.r. to x 2 yy ' = 2c.1 or yy ' = c [On putting value of c from (ii) in (i)]

....(iii)

Þò Þò

sec3 qd q 2y = -ò dy tan q 2 1 + y2 sin 2 q + cos 2 q sin q × cos 2 q

dq = - 1+ y2

Þ ò (tan q × sec q + cosecq ) d q = - 1 + y 2

Þ sec q + log e | cosec q - cot q | = - 1 + y 2 + C \ 1 + x 2 + loge

1 + x2 - 1 = - 1 + y2 + C x

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Differential Equations

Þ 1+ y2 + 1 + x2 =

15.

1 æ 1 + x 2 + 1ö ln ç ÷ +C 2 çè 1 + x 2 - 1÷ø

2 (a) Q y = æç x - 1ö÷ cosec x èp ø

1 Þ x - (ln( y + 3x )) 2 = C 2

18.

Þ lim

é2 æ 2 ù ö = cosec x ê - ç x - 1÷ cot x ú è ø p ëp û dy 2 - cosec x = y cot x dx p It is given that,

dy 2 - cosec x = - yp( x ) dx p By comparision of (i) and (ii), we get p(x) = cot x Þ

16.

(b)

Þ f ( x) = xf '( x)

ò

...(i)

f '( x) 1 dx = ò dx f ( x) x

log e f ( x) = log e x + loge C Q f (1) = e

Þ f ( x ) = Cx , ...(ii)

Þ C = e; so f (x) = ex

When f ( x ) = 1 = ex Þ x =

5 + e x dy × = -e x 2 + y dx

19.

1 e

æ y 2 + 1ö e x dx dy = ò çè y 2 ÷ø ò ex + 1

(c)

ex

dy

ò 2 + y = - ò 5 + e x dx

Þ y-

Þ log e | 2 + y | × log e | 5 + e x |= log e C Þ| (2 + y)(5 + e ) |= C x

1 = log e | e x + 1| +c y

Q Passes through (0, 1). Q y (0) = 1

\ c = - log e 2

C = 18.

æ e x + 1ö Þ y 2 - 1 = y log e ç ÷ è 2 ø

\ (2 + y) × (5 + e ) = 18 x

When x = loge13 then (2 + y) × 18 = 18

æ e x + 1ö Þ y 2 = 1 + y log e ç ÷ è 2 ø

Þ 2 + y = ±1 \ y = -1, - 3 20.

\ y (ln13) = -1 17.

lim

2tf 2 ( x) - 2 x 2 f (t ) × f '(t ) =0 t®x 1 Using L'Hospital's rule

dy 2 æ2 ö = cosec x - ç x - 1÷ cosec x × cot x èp ø dx p

Þ

t 2 f 2 ( x ) - x 2 f 2 (t ) =0 t-x t® x

(a)

x3dy + xy dyx = 2 y dx + x 2 dy

(b)

(a) Let y + 3x = t

Þ ( x3 - x 2 )dy = (2 - x) y dx

dy dt +3= dx dx Putting these value in given differential equation

Þ

Þ

Þò

dt t = dx loge t log e t dt = ò dx Þò t

Þ

2

dy 2-x dx = 2 y x ( x - 1)

(loge t ) = x-C 2

Let

dy 2- x dx =ò 2 y x ( x - 1) 2- x x 2 ( x - 1)

=

...(i)

A B C + + x x2 x - 1

Þ 2 - x = A( x - 1) + B( x - 1) + Cx 2 Compare the coefficients of x, x2 and constant term. C = 1, B = –2 and A = –1

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Mathematics

dy 1 ü ì -1 2 = òí - 2 + ý dx y x - 1þ îx x

Þ ln y = - ln x +

22.

2 + ln | x - 1 | + C x

\ y (2) = e

Þ v+ x

[Q log e = 1]

Þ 1 = - ln 2 + 1 + 0 + C

Þ C = ln 2

Þ

Þ ln y = - ln x +

2 + ln | x - 1| + ln 2 x

Þ

1 Þ ln y (4) = - ln 4 + + ln 3 + ln 2 2

23.

3 1/ 2 e 2 (c) The given differential equation is Þ y (4) =

dy

- 2x = loge x - 1 y

= tan

( - cos x)dx 2 + sin x

1 1 Þ y= 2 1 + log e 2

(a) f ¢(x) = tan–1 (sec x + tan x)

= tan

2 + sin x dy = - cos x, y > 0 y + 1 dx dy cos x dx Þ =y +1 2 + sin x Integrate both sides,

ò y +1 = ò

-2 -2 x = loge x + c Þ = log e x + c v y

Hence, put x =

æ 3ö 1 æ3 ö Þ ln y (4) = ln ç ÷ + = ln ç e1/ 2 ÷ è 2ø 2 è2 ø [Q log m + log n = log(mn)]

dv v2 dv dx Þò2 2 =ò =v+ x dx 2 v

Put x = 1, y = 2, we get c = –1

At x = 4,

21.

dy 2 xy + y 2 = dx 2 x2 It is homogeneous differential equation. \ Put y = vx

(a)

-1 æ 1 + sin x ö

ç ÷ = tan è cos x ø

æ æp öö ç 1 - cos ç 2 + x ÷ ÷ è ø÷ ç sin æ p + x ö ÷ ç ÷ ÷ ç è2 ø ø è

-1 ç

æ ç

ö æp xö 2sin 2 ç + ÷ ÷ è4 2ø ÷ ç 2sin æ p + x ö cos æ p + x ö ÷ ç ÷ ç ÷÷ ç è4 2ø è 4 2øø è

-1 ç

ln | y + 1|= - ln | 2 + sin x | + ln C

æ æ p x öö p x = tan -1 ç tan ç + ÷ ÷ = + è 4 2 øø 4 2 è

Þ ln | y + 1| + ln | 2 + sin x |= ln C

Integrate both sides, we get

Þ ln | ( y + 1)(2 + sin x) |= ln C

ò ( f ¢( x) ) dx = ò çè 4 + 2 ÷ø dx

æp

Q y (0) = 1 Þ ln 4 = ln C Þ C = 4 \ ( y + 1)(2 + sin x ) = 4

\y=

Q

2 - sin x 2 - sin p Þ y (p ) = =1 2 + sin x 2 + sin p

C = 0 Þ f ( x) =

Þ a =1

Now, dy dx

So,

dy (2 + sin x )( - cos x) - (2 - sin x) × cos x = dx (2 + sin x )2 = 1 Þ b = 1.

x =p

p x2 x+ +C 4 4 f(0) = 0

f ( x) =

4 -1 2 + sin x

Þy=

24.

f (1) =

x2 p x+ 4 4

p +1 4

(d) The given differential equation, dy xy = dx x 2 + y 2

Ordered pair (a, b) = (1, 1).

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Differential Equations

Put y = vx Þ Then, v + x Þ

1 + v2 v3

27.

dy dv =v+ x dx dx

ey

\

1 dv = - dx x

I.F. = e ò

Þ

-1 æ 1 ö + ln v = - ln x + c 2 çè v 2 ÷ø

28. -

2 y2

é êQ v = ë

= - ln y + c

yù x úû

Þ

x = ± 3e

29.

So, x = 3e 25.

(

= e- x

)

( ydx - xdy ) y x2

æ yö = xdx Þ - yd ç ÷ = xdx èxø

2 y æ yö 1 y - .d ç ÷ = dx Þ - æç ö÷ = x + c 1 x èxø 2è x ø

Þ 2x3 + cx2 + y2 = 0 [Here, c = 2c1] (c) cos xdy – (sin x) y dx = 6x dx Þ

dy 1 d = sin -1 f ( x ) (b) dx 2 dx 2y = sin –1 f(x) + C = sin–1 (sin(2tan –1x)) + C

ò d ( y cos x ) = ò 6 x dx Þ

æ æ 2p ö ö æpö Þ 2 ç ÷ = sin -1 ç sin ç ÷ ÷ + C 6 è ø è è 3 øø

Putting x =

p p = +C \ C=0 3 3

(10 ) ´ æç

æ æ -2p ö ö for x = - 3 , 2y = sin–1 ç sin ç ÷÷ + 0 è è 6 øø

So, from (1) y cos x = 3x 2 -

1- y2

+

dx 1 - x2

Now, put x =

= 0 Þ sin –1y + sin –1x = c

1 p 3 At x = , y = Þ c= 2 2 2 Þ sin–1y = cos–1x æ 1 öö æ 1 ö -1 æ Hence, y ç ÷ = sin ç cos ç ÷÷ 2ø 2 øø è è è æ æ 1 öö 1 = sin ç p - cos -1 ç ÷÷ = 2 øø 2 è è

p and y = 0 in eq. (1), we get 3

-p2 1 ö 3p 2 +C ÞC ÷= 9 3 è2ø

-p -p Þ y= 3 6 (c) The given differential eqn. is dy

y cos x = 3 x 2 + C ...(1)

æpö Given, y ç ÷ = 0 è3ø

Þ 2y =

26.

-1.dx

Put x = 0, y = 0, then we get c = 1 ey – x = x + 1 y = x + loge(x + 1) Put x = 1 \ y = 1 + loge2 (b) Given differential equation can be written as, y2dx – xydy = x3dx Þ

1 When x = 1, y = 1, then - = c 2 Þ x2 = y2(1 + 2 ln y) At y = e, x2 = e2(3) Þ

é y dy y xù êëQ e dx - e = e úû

dt - t = ex dx

t (e - x ) = ò e x .e- x dx Þ ey – x = x + c

-1 æ 1 1ö ò çè v3 + v ÷ødv = ò x dx

Þ

dy dt = dx dx

dv vx 2 v = 2 2 2 = dx x + v x 1+ v2

Þ

x2

(a) Let ey = t

y 30.

p in the above equation, 6

3 3p2 p2 = Þ 2 36 3

(1) Given

p2 3

-p2 3 y -3p2 = Þ y= 2 12 2 3

dy 2 y = dx x 2

Integrating both sides,

ò

dy dx = 2ò 2 y x

2 Þ ln | y | = - + C ...(i) x Equation (i) passes through the centre of the circle x2 + y2 – 2x – 2y = 0, i.e., (1, 1)

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Mathematics

\ C=2

31.

\

-1 1 ln |1 - 3 y | = tan x – 1 – ln 3 3 3

2 Now, ln | y | = - + 2 x x ln | y | = – 2 (1 – x) Þ x ln | y | = 2 (x – 1) (b) The given differential equation

Put, x = –

dy = (x – y)2 dx

Þ

¼(1)

Let x – y = t Þ 1 – Þ

dy dt = dx dx

dy dt =1– dx dx

dt ö æ çè 1 – ÷ø = (t)2 dx Þ 1 – t2 =

33.

dt dt Þ ò dx = ò 1– t 2 dx

æ y2 ö = - ò dx x ÷ø

y2 = –x + C ...(1) x Since, the above curve passes through the point (1, 1)

1– x + y 1– y + x

dy = sec2 x(1 – 3y) dx dy

1 - ln |1 - 3 y | = tan x + C ...(i) 3

Q

æpö 4 yç ÷ = è4ø 3

Þ

p 1 - ln |1 - 4| = tan + C 3 4

Þ

1 1 - ln 3 = C + 1 Þ C = –1 – ln 3 3 3

(Given)

12 = –1 + C Þ C = 2 1 Now, the curve (1) becomes y2 = –x2 + 2x Þ y2 = –(x – 1)2 + 1 (x – 1)2 + y2 = 1 The above equation represents a circle with centre (1, 0) and centre lies on x-axis. (a) f(xy) = f(x).f(y) ...(1) Put x = y = 0 in (1) to get f(0) = 1 Put x = y = 1 in (1) to get f(1) = 0 or f(1) = 1 f(1) = 0 is reected else y = 1 in (1) gives f(x) = 0 imply f(0) = 0. Hence, f(0) = 1 and f(1) = 1 By first principle derivative formula, Then,

1 dy 3 y= + (a) Given, dx cos 2 x cos 2 x

Þ

(a) (x2 – y2)dx + 2xy dy = 0 y2dx – 2xydy = x2dx 2xydy – y2dx = –x2dx d(xy2) = –x2 dx

ò d çè

1 1– 1 – 1 ln + c Þ c = –1 2 1 –1+1

ò (1 - 3 y) = ò sec 2 x dx

1 1 1 - y = 6 Þ - y = e6 Þ y = ± e6 3 3 3

æ y2 ö d ç ÷ = –dx è xø

Q The given condition y(1) = 1

Þ

ln

xd ( y 2 ) - y 2 d ( x ) = –dx x2

1 x – y –1 Þ –x = ln +c 2 x – y +1

Hence, 2(x – 1) = – ln

1 æ pö 1 - ln |1 - 3 y | = tan ç - ÷ - 1 - ln 3 3 3 è 4ø

Þ

1 t –1 ln Þ –x = +c 2 ´1 t +1

–1 =

p 4

1 = -1 - 1 - ln 3 3 Þ ln |1 – 3y| = 6 + ln 3

Now, from equation (1)

32.

in eq. (i), we get

34.

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Differential Equations

h® 0

æ ç lim f ( x) ç = h®0 ç çè

ö æ hö f ç1 + ÷ - f (1) ÷ è xø ÷ h ÷ ÷ø

Þ f ¢(x) =

f ( x) f ¢ (1) x

Þ

(2 + sin x) (y + 1) = C At x = 0, y = 1 we have (2 + sin 0) (1 + 1) = C Þ C=4 4 Þ y+ 1 = 2 + sin x 4 -1 y= 2 + sin x

f ( x + h) - f ( x) h

f ¢(x) = lim

f ¢( x ) k = Þ ln f(x) = k ln x + c f ( x) x

f(1) = 1 Þ ln1 = k ln 1 + c Þ c = 0 Þ ln f(x) = k ln x Þ f(x) = xk but f(0) = 1 Þ k=0 \ f(x) = 1

37.

xdy - ydx y2

35.

1 x x2 - = + C as y(1) = –1 Þ C = 2 y 2

æ 1ö æ 3ö 1 3 y ç ÷ + y ç ÷ = +1 + + 1 = 3 è 4ø è 4ø 4 4

(b) (x2 – y2) dx + 2xydy = 0 dy y 2 - x 2 = dx 2 xy Let y = vx

Hence, y =

38.

Þ

v+x

Þ

x

æ -1 ö 4 Þfç ÷= è 2 ø 5 x +1 2

t 2 f ( x ) - x 2 f (t ) =1 t-x t® x Applying L.H. rule

(d) Let L = lim

2t f ( x) - x 2 f ¢ (t ) =1 1 t® x 2x f (x) – x2 f ¢(x) = 1 solving above differential equation, we get 2 2 1 f (x) = x + 3 3x

dv v 2 x 2 - x 2 dv v 2 - 1 = Þ v+x = 2 dx 2v dx 2vx

dv - v 2 - 1 = dx 2v

Put x =

2vdv dx =x v2 + 1 After integrating, we get ln | v2 + 1| = – ln | x | + lnc

Þ

y2 c +1= x x2 As curve passes through the point (1, 1), so 1 + 1 = c Þc=2 x2 + y2 – 2x = 0, which is a circle of radius one.

36.

-2x

L = lim

dy dv =v+ x dx dx Þ

= xdx

æxö Þ ò -d ç ÷ = ò xdx è yø

dy = f(x) = 1 Þ y = x + c, y(0) = 1 Þ c = 1 dx Þ y=x+1

\

4 1 4 æ pö - 1 = -1 = Now y ç ÷ = p 3 3 2 è ø 2 + sin 2 (b) y (1 + xy)dx = xdy

(b) We have (2 + sinx)

dy + (y + 1) cos x = 0 dx

d (2 + sin x)(y + 1) = 0 Þ dx On integrating, we get

3 2 2

39.

3 2 27 + 4 31 æ3ö 2 æ3ö 1 2 = f ç ÷ = ´ç ÷ + ´ = + = 2 9 18 18 è2ø 3 è2ø 3 3 (b) Given differential equation is ydx – (x + 2y2) dy = 0 Þ ydx – xdy – 2y2dy = 0 ydx - xdy Þ = 2dy y2 æ xö Þ d ç ÷ = 2dy è yø

Integrate both the side x = 2y + c Þ y using f(–1) = 1, we get

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Mathematics

c=1 x Þ = 2y + 1 y put y = 1, we get f(a) = 3 40.

(a)

dy ( x + 2 ) = x 2 + 4x - 9 x ¹ -2 dx

t

p(t ) - 200 = e 2 k 2 Using given condition p(t) = 400 – 300 et/2

Þ

42.

dy y æ yö = + fç ÷ è xø dx x

dy dv = x +v dx dx

x 2 + 4x - 9 dy = dx x+2

or

x 2 + 4x - 9 ò dy = ò x + 2 dx

from (1) & (2), x

13 ö æ y = òç x + 2 ÷dx x+2ø è 1 dx y = ò (x + 2) dx - 13 ò x+2

or,

x2 + 2x – 13 log | x + 2 | + c 2 Given that y = (0) = 0 0 = – 13 log 2 + c

ò

...(2)

dv æ 1ö + v = v + fç ÷ è vø dx

dx dv = x æ 1ö fç ÷ è vø

Integrating both sides, we get

y=

dx = x

ò

dv dv Þ ln x + c = ò æ 1ö æ 1ö fç ÷ fç ÷ è vø è vø

(where c being constant of integration) x But, given y = is the general solution ln | cx |

x2 y= + 2x – 13 log |x + 2| + 13 log 2 2 y(–4) = 8 – 8 – 13 log 2 + 13 log 2 = 0

so that

(c) Given differential equation is dp (t ) 1 = p (t ) - 200 2 dt By separating the variable, we get é1 ù dp(t) = ê p(t ) - 200ú dt 2 ë û dp(t ) Þ = dt 1 p (t ) - 200 2 Integrate on both the sides, d ( p (t )) = ò dt ò1 p (t ) - 200 2 1 dp(t ) = ds Let p(t ) - 200 = s Þ 2 2 d p (t ) = dt So, ò 1 æ ö ò ( ) 200 p t ç ÷ è2 ø 2ds = ò dt Þ 2 log s = t + c Þ ò s æ p (t ) ö - 200÷ = t + c Þ 2log ç è 2 ø

...(1)

æ yö Let çè ÷ø = v so that y = xv x

dy x 2 + 4x - 9 = dx x+2

41.

(d) Given

x 1 = = ln|cx| = v y

ò

dv æ 1ö fç ÷ è vø

Differentiating w.r.t v both sides, we get æ xö æ 1ö y2 -1 fç ÷ = Þ f = ç ÷ è vø è yø v2 x2 2

when 43.

2

x æ yö æ 1ö æ -1ö = 2 i.e. f(2) = - ç ÷ = - ç ÷ = ç ÷ è xø è 2ø è 4ø y

(c) Given, Rate of change is

dP = 100 - 12 x dx

Þ dP = (100 – 12 x ) dx By integrating

ò dP = ò (100 - 12

x ) dx

P = 100x – 8x3/2 + C Given when x = 0 then P = 2000 Þ C = 2000 Now when x = 25 then P = 100 × 25 – 8 × (25)3/2 + 2000 = 4500 – 1000 Þ P = 3500

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Differential Equations

44.

Þ

ò

Putting in (i)

1 dy =1- 2 dx x 1 ö æ ç1 - 2 ÷ dx è x ø

(a) Slope = dy = ò

1 + C , which is the equation of the curve x æ 7ö since curve passes through the point ç 2, ÷ è 2ø 7 1 \ = 2+ +C Þ C=1 2 2 1 \ y = x + +1 x 1 -3 +1 = when x = – 2, then y = -2 + 2 -2 (d) Given differential equation is

47.

dy cos x dx =ò 1+ y 2 + sin x Þ log (1 + y) = log (2 + sin x) + log C Þ 1 + y = C (2 + sin x) Given y(0) = 2 3 Þ 1 + 2 = C[2 + sin 0] Þ C = 2 æ pö Now, y ç ÷ can be found as è 2ø 9 pö 3æ 1 + y = ç 2 + sin ÷ Þ 1 + y = 2 2è 2ø 7 Þ y= 2 æ pö 7 Hence, y ç ÷ = è 2ø 2 (b) Equation of tangent at P dy Y–y= ( X - x) dx ò

x2

2

46.

0 = 900 - 50e 2 \ t1 = 2ln 18 (c) Given differential equation is

(2 + sin x) dy . = cos x (1 + y ) dx which can be rewritten as dy cos x dx = 1 + y 2 + sin x Integrate both the sides, we get

2

Now, y 2 e - y / x = C satisfies the given diff. equation \ It is the solution of given diff. equation. Thus, statement-2 is also true. (a) Given differential equation is dp (t ) = 0.5 p(t ) - 450 dt dp(t ) 1 = p(t ) - 450 Þ dt 2 dp (t ) p (t ) - 900 = Þ dt 2 dp(t ) = - ëé900 - p ( t ) ûù Þ 2 dt dp(t ) = -dt Þ 2 900 - p(t ) Integrate both the side, we get

Þ

p(t ) = 900 - 50e 2 t1

x é xù æxö = ê1 - ú » F ç ÷ èzø z2 z ë z û Hence, statement-1 is true. dx x = dz z

900 - p (t ) = 50e 2

let p (t1) = 0

dz 2z z = = dx 2( xz - x 2 ) xz - x 2

Now,

Þ

t

dy y3 = dx 2( xy 2 - x 2 ) By substituting z = y2, we get diff. eqn. as 2

é æ 900 - p ( t ) ö ù 2 êln ç ÷ø ú = t 50 ë è û t

Þ y = x+

45.

Þ

48.

Y B (0, y – xdy/dx )

P

dp(t ) - 2ò = dt 900 - p(t ) ò Let 900 – p (t) = u Þ – dp (t) = du du = dt Þ 2 ln u = t + c ...(i) u ò Þ 2ln [900 – p(t)] = t + c Given t = 0, p (0) = 850 2 ln (50) = c 2ò

( x, y)

O A(x – y)/(dy/dx), 0) Y¢

y dy / dx xdy Y-intercept = y – dx Since P is mid-point of A and B

X-intercept = x -

X

EBD_8344

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Mathematics

y xdy = 2 x and y = 2y dy dx - xdy -y =y = x and Þ dy dx dx dx dy Þ =x y lny = – lnx + lnc c y= x Since the above line passes through the point (2, 3). \ c=6 6 Hence y = is the required equation. x dV (t ) = -k (T - t ) (a) dt x-

49.

52.

dx ( X - x) dy Coordinate of G at X axis is (X, 0) (let) \ 0 – y = – dx ( X - x ) dy dy Þ y = X -x dx dy Þ X=x+y dx dy ö æ \ Co-ordinate of G çè x + y , 0÷ø dx Given distance of G from origin = twice of the abscissa of P. Q distance cannot be –ve, therefore abscissa x should be +ve dy dy \ x+ y = 2 x Þ y dx = x dx Þ ydy = xdx

Y–y=–

Þ ò dV (t ) = -k ò (T - t )dt

V (t ) =

k (T - t )2 +c 2

at t = 0, V (t) = I I =

kT 2 +c 2

kT 2 2 k Þ V (t ) = I + (t 2 - 2tT ) 2 k 2 k V (T ) = I + (T - 2T 2 ) = I - T 2 2 2

On Integrating, we have

Þc=I -

50.

(d)

dy = y+3 Þ dx

53.

dy

ò y + 3 = ò dx

xdy = y (log y – log x + 1) dx

Put y = vx dy xdv xdv =v+ Þ v+ = v (log v + 1) dx dx dx dv dx xdv = v log v Þ ò =ò v log v x dx Put log v = z 1 dz dx dv = dz Þ ò = v z ò x ln z = ln x + ln c æ yö x = cx or log v = cx or log ç ÷ = cx. è xø

Þ ln y + 3 = x + ln5 Put x = ln 2 , then ln |y + 3| = ln 2 + ln5 Þ ln | y + 3 | = ln 10 \ y + 3 = ±10 Þ y = 7, – 13 dy x + y y = = 1+ dx x x

It is homogeneous differential eqn. dy dv =v+x Putting y = vx and dx dx we get dv dx v+x = 1+ v Þ ò = dv x ò dx Þ v = ln x + c Þ y = xlnx + cx

(c)

dy y æ æ yö ö = log ç ÷ + 1÷ è xø ø dx x çè

Given y (0) = 2, \ l n 5 = c

(d)

y 2 x2 = + c1 2 2

Þ x2 – y2 = –2c1 \ the curve is a hyperbola

Þ ln y +3 = x+ c

51.

As y(1) = 1 \ c = 1 So solution is y = x lnx + x (b) Equation of normal at P(x, y) is

54.

(b)

-2 x = e-2 x ; on integration dy = e +c; dx -2 dx

d2y 2

Again integrate we get y = 55.

(c)

e -2 x + cx + d 4

dy + 2 y tan x = 2 sin x dx

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Differential Equations

ò I.F. = e

2 tan xdx

Q \

= sec 2 x

The solution of the differential equation is

3 é ù -3 Then, y = (1 + x) ê x + ë 1 + x úû

ò

y × I.F. = I.F. ´ 2sin x dx + C

ò

Þ y × sec 2 x = 2sin x × sec 2 x dx + C

Þ y sec2 x = 2sec x + C When x =

3 é ù -3 = 3 Now, at x = 3, y = (1 + 3) ê3 + ë 1 + 3 úû

...(i)

p , y = 0; then C = – 4 3

58.

\ From (i), y sec 2 x = 2sec x - 4

Þy= 56.

2sec x - 4 sec 2 x

I.F. =

1 - ò dx e x

=

æpö Þ yç ÷ = 2 -2 è4ø

= y2e y – 2(y.e y – e y) + C Þ x.ey = y2ey – 2yey + 2ey + C Þ x = y2 – 2y + 2 + C. e–y As y(0) = 1, satisfying the given differential eqn, \ put x = 0, y = 1 in eqn. (i)

0 = 1- 2 + 2 +

Q y (p) = p Þ C = 1

2 p æ pö p + y = x 2 sin x + x Þ y ç ÷ = è 2ø 4 2

59.

p æ pö y '' = 2 sin x - x 2 sin x Þ y '' ç ÷ = 2 è 2ø 4

2

p2 p2 p p æ pö æ pö \ y '' ç ÷ + y ç ÷ = 2 + + = 2+ è 2ø è 2ø 4 4 2 2

C=–e y = 0, x = 0 – 0 + 2 + (– e) (e–0) x=2–e (b) Consider the differential equation,

1 dx æ 1 ö Þ dy + çç 2 ÷÷ x = 3 y y è ø 1

dy æ y -3ö = (1 + x ) + ç ÷ dx è 1+ x ø

\ x.e

1 3 dy y = (1 + x) dx (1 + x ) (1 + x)

Put -

I.F. = e

\

1 dx (1+ x )

=

1 (1 + x )

d æ y ö 3 ç ÷ = 1dx è 1 + x ø (1 + x )2

é ù 3 y = (1 + x) ê x + + Cú (1 + x ) ë û

1

dy I.F. = ò y 2 e =e y

(c) (x + 1)dy = ((x + 1)2 + (y – 3))dx = 0 Þ

C e

æ 1ö y2dx + ç x - ÷ dy = 0 yø è

y ' = 2 x sin x + x 2 cos x + 1

57.

dx + Px = Q, where P = 1, Q = y2 dy

x.e y = ò ( y 2 )e y . dy = y 2 .e y - ò 2 y.e y .dy

1 x

y = x sin x + C x

dx + x = y2 dy

Now, I.F. = e ò1.dy = e y

æ yö \ ò d ç ÷ = ò ( x cos x + sin x)dx è xø Þ

(a) The given differential equation is Comparing with

dy y - = x( x cos x + sin x) dx x

(a)

At x = 2, y = 0 0 = 3(2 + 1 + C) Þ C = – 3

Þ

-

1 y

= òe

1 y

1 y3

dy + c

1 1 = u Þ 2 dy = du y y -

x.e

1 y

= - ò ueu du + c = -ueu + eu + c

1

Þ

-

1

- æ 1 ö x.e y = e y ç + 1 ÷ + c èy ø

At y = 1, x = 1

...(i)

EBD_8344

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Mathematics

æpö æ pö Þ y¢ ç ÷ – y¢ ç - ÷ = p - 2 è4ø è 4ø

1

1 = 2 + ce Þ c = -

1 æ 1ö 1 Þ x = ç1 + ÷ - e y e yø e è

3 1 On putting y = 2, we get x = 2 e

60.

62.

dy + y sec2 x = sec 2 ´ tan x dx Given equation is linear differential equation.

2

ò dx differential equation, then I .F = e x = x 2 Now, the solution of the linear differential equation

(a)

sec IF = e ò

2

xdx

dy 2 + y = x y(1) = 1 (given) dx x Since, the above differential equation is the linear

(c)

y ´ x 2 = ò x3dx

= e tan x

Þ y.etanx = ò e tan x sec 2 ´ tan xdx

Þ yx 2 =

Put tan x = u = sec2x dx = du Q y (1) = 1

yetan x = ò euu du Þ yetan x = ueu – eu + c

3 1 \1´ 1 = + C Þ C = 4 4 \ Solution becomes.

Þ yetan x = (tan x – 1) etan x + c Þ y = (tan x – 1) + c. e– tan x \ y (0) = 0 (given) Þ 0 = – 1 + c Þ c = 1 Hence, solution of differential equation,

æ pö yç- ÷ = – 1 – 1 + e = – 2 + e è 4ø 61.

(d) Given differential equation is,

dy + y tan x = 2 x + x 2 tan x dx Here, P = tan x, Q = 2x + x2 tan x tanxdx = eln|sec x| =| sec x | I.F. = e ò

\ y (sec x) = ò (2 x + x 2 tan x )sec xdx =

òx

2

tan x sec xdx + ò 2 x sec xdx = x2sec x + c

Given y (0) = 1 Þ c = 1 \ y = x2 + cos x ...(i) Now put x =

p -p and x = in equation (i), 4 4

2 2 1 1 æpö p æ pö p yç ÷ = + + and y ç - ÷ = 2 2 è 4 ø 16 è 4 ø 16

æpö æ pö Þ yç ÷ - yç- ÷ = 0 è4ø è 4ø dy = 2 x - sin x dx

p 1 æpö p 1 æ pö \ y¢ ç ÷ = and y¢ ç - ÷ = - + 2 2 2 è4ø 2 è 4ø

x4 +C 4

y=

63.

3 x2 + 4 4 x2

(d) (1 + x2)2 Þ

dy + 2x (1 + x2) y = 1 dx

dy æ 2 x + dx çè 1 + x 2

1 ö ÷y = (1 + x 2 )2 ø

Since, the above differential equation is a linear differential equation \

2x

I.F. = ò 1+ x2 dx = elog(1+ x 2 ) = 1 + x2 e

Then, the solution of the differential equation Þ y (1 + x2) =

dx

ò 1 + x2 + c

Þ y (1 + x2) = tan–1 x + c If x = 0 then y = 0 (given) Þ 0=0+c Þ c=0 Then, equation (1) becomes,

...(1)

Þ y (1 + x2) = tan–1 x Now put x = 1 in above equation, then 2y =

p 4

Þ

æ p ö p 2ç ÷= è 32 a ø 4

pù é êë a y (1) = 32 úû

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Differential Equations

Þ

a=

Then, equation of curve

1 4

y=

1 16 (c) Consider the differential equation,

Þ a=

64.

Since, above curve satisfies the point. Hence, the curve passes through

dy y + = logex dx x

66.

IF = e ò x dx = x

\

yx =

ò x ln x dx

2x -2 x y( x) × x e2 x = ò x e × e dx + c

2

Þ y(e) =

\

1 -2 2 1 e × e ×1 = + c Þ c = 0 2 2

\

y(x) =

x 2 e -2 x × x 2

x -2 x y(x) = × e 2 Differentiate both sides with respect to x,

x2 - 2 y x

dy x 2 - 2 y = dx x

y¢(x) =

e -2 x æ1 ö (1 - 2 x ) < 0 " x Î ç , 1÷ è2 ø 2

æ1 ö Hence, y(x) is decreasing in çè , 1÷ø 2

dy 2 + y =x dx x 2

x2 +c 2

1 -2 Given, y(1) = e 2

e e e - = 2 4 4

(b) Q Slope of the tangent = Q

y ( x) × e 2 x × x =

x x ln x 2 4

Þ y=

65.

= ò x dx + c

x2 x2 ln x 2 4

Þ xy =

(c) Given differential equation is,

æ 1ö ò ç 2+ ÷ dx è ø

x x × ln x - + c 2 4 Given, 2y(2) = loge 4 – 1. \ 2y = 2 ln 2 – 1 + c Þ ln4 – 1 = ln4 – 1 + c i.e. c = 0 Þ xy =

67.

dx

I.F. = e ò x = e2ln x = x 2

(b) Let y = f(x)

Solution of equation

dy æ 3 ö +ç ÷ y =7 dx è 4 x ø

y × x2 = ò x × x 2dx

I.F. =

x4 +C 4 \ curve passes through point (1, –2)

x2y =

(1)2 (–2) = Þ C=

14 +C 4

-9 4

)

3, 0 .

x IF = e = e2x + ln x = xe2x Complete solution is given by

x2 1 x2 - ò × dx 2 x 2

2

(

dy æ 1ö + ç 2 + ÷ y = e–2x, x > 0 è dx xø

1

Q

Þ xy = ln x ×

9 x2 4 4 x2

e

ò

3 dx 4x

3

= e4

ln x

æ 3ö ç ÷

= xè 4 ø

Solution of differential equation 3

3

y × x 4 = ò 7 × x 4 dx + C 7 7

x4 3 7× + C = 4x4 + C y × x4 = æ 7ö çè ÷ø 4

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Mathematics 3

y = 4x + Cx - 4

Then, ysin x = 2x 2 3

æ 1ö 4 Þ f ç ÷ = + Cx 4 è xø x 7ö æ æ 1ö lim+ x × f ç ÷ = lim+ ç 4 + Cx 4 ÷ = 4 è xø x®0 è x ®0 ø

Þ

68.

(c) Since, x Þ

p2 is the solution 2

2 p2 ö 8p2 æpö æ p \ y ç 6 ÷ = çç 2. 36 - 2 ÷÷ 2 = - 9 è ø è ø

70.

(a) When x Î [0, 1], then Q y (0) = 0 Þ y (x) =

dy + 2 y = x2 dx

dy 2 + y =x dx x

Here, y (1) =

1 1 –2 x – e 2 2

1 1 –2 e 2 – 1 – e = 2 2 2e 2

2

I.F. = e ò x dx = e2 ln x = e lnx 2 = x2. Solution of differential equation is:

When x Ï [0, 1], then

y × x2 = ò x × x 2dx

Q y (1) =

x4 + C ...(1) 4 y(1) = 1 71.

3 C= 4

\

Þ

x4 3 y×x = + 4 4

69.

\

3 x2 + y= 4 4 x2

2x 2 + c ...(2) sin x

y(x) =

Q

æp ö eq. (2) passes through ç ,0 ÷ è2 ø p2 +C 2

p2 2 Now, put the value of C in (1) ÞC= -

= e - ln y =

solution is

1 y

x 1 = 3 y . dy y ò y

x = 3y + c y which passes through (1, 1) \ 1 = 3 + c Þ c = –2 \ solution becomes Þ x = 3y2 – 2y

(b) Consider the given differential equation the sinxdy + ycosxdx = 4xdx Þ d(y.sinx) = 4xdx Integrate both sides Þ y.sinx = 2x2 + C ...(1) Þ

1 - ò dy y

Þ

æ 1ö 1 49 yç ÷ = + 3 = è 2 ø 16 16

Þ0=

dx x dx x - = 3y = + 3y Þ dy y dy y

if = e

2

\

e2 – 1 e2 – 1 2 –2 e2 – 1 Þ = c e Þ C2 = 2 2 2

(b) y dx – x dy – 3y2 dy = 0

Then, from equation (1)

\

dy + 2 y = 0 Þ y = c e –2x 2 dx

2 æ e 2 – 1 ö –2 x æ3ö e –1 \ y (x) çç ÷÷ e Þ y ç ÷ = è2ø 2e3 è 2 ø

y × x2 = Q

dy 1 + 2 y = 1Þ y = + C1e –2 x dx 2

72.

æ 1 1ö which also passes through ç - , ÷ è 3 3ø dy y tan x + sec x = (d) dx 2 2y dy 2y + y2 sec x = tan x dx dt dy Put y2 = t Þ 2y = dx dx dt + t sec x = tan x dx sec xdx In(sec x + tan x ) I.f = e ò = sec x + tan x =e

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Differential Equations - cosec2x Now, integrating factor (I.F) = e ò

dt (sec x + tan x) + t sec x (sec x + tan x) dx = tan x(sec x + tan x)

or, I.F = e

ò d (t (sec x + tan x)) = ò tan x (sec x + tan x) dx

73.

x x Þ y2 = 1 – sec x + tan x sec x + tan x

(a) Given,

log(log x)

=e y. logx = ò 2 log xdx + c

I.F. = e

(c) Let

\ y cot x =

76.

2sin x cos x dx + c cos x

77.

y (sec x) = –2cos x + c ...(1) Given y(0) = 1 \ put x = 0 and y = 1, we get 1 (sec 0) = – 2 cos 0 + c Þc=1+2Þc=3 \ from eqn (1), we have y sec x = –2 cos x + 3 ...(2) To find y (p), put x = p in eqn (2), we get y (secp) = – 2 cos p + 3 y = – 2 (–1) (–1) + 3 (–1) = –2 – 3 = – 5

æ dy ö (d) Given, sin 2 x ç - tan x ÷ - y = 0 è dx ø dy y + tan x = dx sin 2 x dy or, - y cosec2 x = tan x dx

(d) Given differential equation is

\ Solution is 4x2 ´1 + x 2 + C y(1 + x2) = ò 1 + x2 4 x3 +C Þ y(1 + x2) = 3 Þ Required curve is 3y (1 + x2) = 4x3 (Q C = 0)

y (sec x) = 2ò sin x dx + c

75.

cot x dx + c

This is linear diff. equation 2x dx ò 2 + 1 x2 I.F = e = elog (1+ x ) = 1 + x 2

tan x dx

ò

ò tan x . ò 1.dx + c

dy + 2 xy = 4 x 2 dx dy æ 2 x ö 4 x2 +ç y = Þ dx è 1 + x2 ÷ø 1 + x2

= e - log cos x = sec x I.F = eò Required solution is

y (sec x) =

) -1

(1 + x 2 )

dy + y tan x = sin 2x dx

ò sin 2 x sec x dx + c

tan x

\ y cot x = x + c

= log x

y (sec x) =

(

ò Q(I.F.) dx + c

\ y cot x =

y logx = 2[x log x – x] + c Put x = 1, y.0 = –2 + c c=2 Put x = e y loge = 2e(log e – 1) + c y(e) = c = 2 74.

log

1

y(I. F.) =

dy æ 1 ö + y=2 dx çè x log x ÷ø

1 ò x log x dx

= e

= cot x tan x Now, general solution of eq. (1) is written as =

t (sec x + tan x) = sec x + tan x – x t=1–

1 - log|tan x| 2

or,

(b) Given differential equation is ( x 2 - 1)

dy x 2x + .y = 2 dx x 2 - 1 x -1 This is in linear form. Þ

Integrating factor = ò

78.

2x

dt where t = x2 – 1 e t

dx = ò

e x -1 = elog t = x2 – 1 Hence, required integrating factor = x2 – 1. (d) Given differential equation is 2

dy 2 + .y = x2 dx x This is of the linear form. 2 \ P = , Q = x2 x 2

...(1)

dy + 2 xy = x dx

I.F = e

ò x dx

2

= e log x = x2

EBD_8344

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Mathematics

Solution is

79.

ò I.F. = e

x5 +c y.x 2 = ò x 2 .x 2 dx + c = 5 3 x y= + cx -2 5 1 dx x (c) dy + 2 = 3 y y It is linear differential eqn. 1 1 ò 2 dy – y =e y I.F. = e -

1 y

So

x.e

Let

-1 =t y

Þ Þ Þ

1 y2

1 y

e 3

1 y

xe

81.

=e

dx x dx x = - - x2 Þ + = -x2 , dy y dy y It is Bernoulli form. Divide by x2 æ 1ö dx + x -1 ç ÷ = -1. x -2 dy è yø dx dt we get, = dy dy dt æ 1 ö æ 1ö dt t =1 - + t ç ÷ = -1 Þ dy çè y ÷ø dy è y ø

put x -1 = t , - x -2

-

1 y

+

1 e y

-

1 y

-

1 y

1

+

80.

I.F = e

1 1 dy dt =tÞ= y y 2 dx dx Putting in (i) dt - - t tan x = - sec x dx dt Þ + (tan x )t = sec x dx

1 ò - dy y

= e - log y = y -1

\ Solution is t ( y -1 ) = ò ( y -1 )dy + C 1 1 1 Þ . = log y + C Þ log y =C x y xy

x = 1+

(d) cos x dy = y (sin x - y ) dx dy = y tan x - y 2 sec x dx 1 dy 1 - tan x = - sec x ...(i) y 2 dx y Let

It is linear differential eqn. in t.

1 -y e +c y

+c

1 + c.e1/ y y Given y(l) = 1 1 \ c=e 1 1 Þ x = 1 + - .e1/ y y e Þ

(b) ydx + ( x + x 2 y )dy = 0

Þ

dy

I = -ò tet dt = et - tet = e 1 y

= e log|sec x| = sec x

Solution : t secx = ò sec x sec x dx 1 Þ sec x = tan x + c y

dy = dt

-

tan x dx

82.

(c)

(1 + y 2 ) + ( x - e tan

-1

y

)

dy =0 dx

-1

dx x e tan y + = 2 dy (1 + y ) (1 + y 2 ) It is form of linear differential equation. Þ

I .F = e x(e tan

x(etan

ò

-1

-1

1 (1+ y 2 ) y

y

\ 2 xe tan

)=ò

)= -1

y

dy

=e e tan

-1

1+ y

e2 tan 2

tan -1 y

y

2

-1

y

= e2 tan

e tan

-1

y

dy

+C

-1

y

+k

é e2 x ù 2x êQ e dx = ú 2 úû êë

ò

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Vector Algebra

25

Vector Algebra 5.

TOPIC Ć

Algebra of Vectors, Section Formula, Linear Dependence & Independence of Vectors, Position Vector of a Point, Modulus of a Vector, Collinearity of Three points, Coplanarity of Three Vectors & Four Points, Vector Inequality

\$i + l \$j + k\$ , \$j + l k\$ and l\$i + k\$ is minimum, then l is equal

to : (a) 6.

1.

Let a, b, c Î R be such that a 2 + b 2 + c 2 = 1. If

2p ö 4p ö æ æ p a cos q - b cos ç q + ÷ = c cos ç q + ÷ , where q = , 3 3 9 è ø è ø    then the angle between the vectors ai + bj + ck and

bi + cj + ak is : (a) 2.

p 2

(b)

(c)

p 9

(d) 0

Let the position vectors of points 'A' and 'B' be i + j + k and 2i + j + 3k, respectively. A point 'P' divides the line

7.

3.

4.

[NA Jan. 9, 2020 (I)] r r Let a = 3\$i + 2 \$j + 2k\$ and b = \$i + 2 \$j - 2k\$ be two vectors. If r r r r a vector perpendicular to both the vectors a + b and a - b has the magnitude 12 then one such vector is : [April 12, 2019 (I)]

( ) 4 ( 2\$i + 2 \$j - k\$ )

( ) 4 ( -2\$i - 2 \$j + k\$ )

1 3

(b)

1 3

(c)

(d) - 3

3

r Let a Î R and the three vectors a = ai\$ + \$j + 3k\$ , r r b = 2\$i + \$j - a k\$ an d c = a\$i - 2 \$j + 3k\$ . Then the set r r r [April 12, 2019 (II)] S = (a : a, b and c are coplanar)

is singleton is empty contains exactly two positive numbers contains exactly two numbers only one of which is positive r If a unit vector a makes angles p 3 with i\$, p 4 with \$j and q Î (0, p) with k\$ , then a value of  is:

segment AB internally in the ratio l :1 (l > 0) . If O is the uuur uuur uuur uuur region and OB × OP - 3 | OA ´ OP |2 = 6, then l is equal to ___________. [NA Sep. 02, 2020 (II)] r ur If the vectors, p = (a + 1)i + aj + ak, q = ai + (a +1) j + ak r and r = ai + aj + (a + 1)k (a Î R) are coplanar and ur r r r 3( p . q) 2 - l | r ´ q |2 = 0, then the value of l is _____.

[April 12, 2019 (I)]

(a) (b) (c) (d)

[Sep. 03, 2020 (II)] 2p 3

If the volume of parallelopiped formed by the vectors

[April 09, 2019 (II)] 5p 5p 2p p (b) (c) (d) 6 12 3 4 The sum of the distinct real values of m, for which the

(a) 8.

vectors, m i + j + k, i + m j + k, i + j + m k, are co-planar,

9.

is : [Jan. 12, 2019 (I)] (a) –1 (b) 0 (c) 1 (d) 2 r r Let a = i + 2 j + 4k, b = i + l j + 4k and

r c = 2i + 4 j + l 2 - 1 k be coplanar vectors. Then the

(

)

r r non- ero vector a ´ c is :

[Jan. 11, 2019 (I)]

(a) 4 2\$i + 2 \$j + 2k\$

(b) 4 2\$i - 2 \$j - k\$

(a) –10i - 5 j

(b) –14i - 5 j

(c)

(d)

(c) –14i + 5 j

(d) –10i + 5 j

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10.

Mathematics

Let

3i + j, i + 3 j and bi + (1 - b ) j respectively be the

17.

position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is

11.

12.

13.

14.

3 2

, then the sum of

all possible values of b is : [Jan. 11, 2019 (II)] (a) 4 (b) 3 (c) 2 (d) 1 ur r r r r r Let a = (l - 2) a + b and b = (4l - 2) a + 3b be two r r given vectors where vectors a and b are non-collinear.. r ur The value of l for which vectors a and b are collinear, is: [Jan. 10, 2019 (II)] (a) – 4 (b) – 3 (c) 4 (d) 3 r Let u be a vector coplanar with the vectors r r r r a = 2i + 3 - k and b =  + k . If u is perpendicular to a r r r and u × b - 24 , then | u |2 is equal to : [2018] (a) 315 (b) 256 (c) 84 (d) 336 Let ABC be a triangle whose circumcentre is at P. If the r r r a +b+c r r r position vectors A, B, C and P are a, b, c and 4 respectively, then the position vector of the orthocentre of this triangle, is : [Online April 10, 2016] r r r æ a + b + cö r (a) - ç (b) ar + b + cr 2 ÷ø è r r r æ a + b + cö r (c) ç (d) 0 2 ÷ø è uuur uuur If the vectors AB = 3i + 4k and AC = 5i – 2 j + 4k are the sides of a triangle ABC, then the length of the median through A is [2013]

18.

®

16.

3 2

(b)

2 3

(c) –

3 2

(d) –

2 3

19.

20.

forces acting along AB, BC with magnitudes

(a)

®

c = ri + j + ( 2r - 1) k are three vectors such that c is ®

AB 2 + AC 2 2

( AB) ( AC )

(c) 21.

1 1 + AB AC

2

(b) ( AB )( AC ) AB + AC (d)

If C is the mid point of AB and P is any point outside AB, then [2005] uuur uuur uuur (a) PA + PB = 2 PC uuur uuur uuur (b) PA + PB = PC uuur uuur uuur (c) PA + PB + 2 PC = 0 uuur uuur uuur (d) PA + PB + PC = 0 Let a, b and c be distinct non- negative numbers. If the vectors ai + aj + ck , i + k and ci + cj + bk lie in a plane,

If a = i - 2 j + 3k , b = 2i + 3 j - k and ®

1 and AB

1 respectively is the force along AD , where D is the AC foot of the perpendicular from A onto BC. The magnitude of the resultant is [2006]

22.

®

[2011RS] (b) 0 (c) – 1 (d) – 2 r The vector a = a i + 2 j + bk lies in the plane of the r r vectors b = i + j and c = j + k and bisects the angle r between b and cr . Then which one of the following gives possible values of a and b? [2008] (a) a = 2, b = 2 (b) a = 1, b = 2 (c) a = 2, b = 1 (d) a = 1, b = 1 ABC is a triangle, right angled at A. The resultant of the (a) 2

(b) 72 (c) 33 (d) 45 18 r r If a and b are non-collinear vectors, then the value of a r r r for which the vectors u = (a - 2)a + b and r r r v = (2 + 3a)a - 3b are collinear is : [Online April 23, 2013] (a)

If the pi\$ + \$j + k\$ , \$i + q \$j + k\$ and i + j + rk ( p ¹ q ¹ r ¹ 1) vector are coplanar, then the value of pqr - ( p + q + r ) is

(a)

15.

r r r Let a, b, c be three non- ero vectors which are pairwise r r r r r non-collinear. If a + 3b is collinear with c and b + 2c is r r r r collinear with a , then a + 3b + 6c is : [2011RS] r r r r r (a) a (b) c (c) 0 (d) a + c

®

parallel to the plane of a and b , then r is equal to [Online May 19, 2012] (a) 1 (b) – 1 (c) 0 (d) 2

then c is (a) the Geometric Mean of a and b (b) the Arithmetic Mean of a and b (c) equal to ero (d) the Harmonic Mean of a and b

[2005]

www.jeebooks.in M-461

Vector Algebra

23.

r r r If a, b , c are non-coplanar vectors and l is a real number,,

24.

then the vectors a + 2b + 3c , lb + 4c and (2l - 1)c are non coplanar for [2004] (a) no value of l (b) all except one value of l (c) all except two values of l (d) all values of l r r r Let a, b and c be three non- ero vectors such that no two r r of these are collinear. If the vector a + 2b is collinear r with cr and b + 3cr is collinear with ar (l being some nonr ero scalar) then ar + 2b + 6cr equals [2004] r r r (c) lc (d) la (a) 0 (b) lb Consider points A, B, C and D with position vectors

25.

7i - 4 j + 7 k, i - 6 j + 10k , - i - 3 j + 4 k an d

26.

If

a2

1 + a3

b

b2

1 + b3 = 0

c

c2

1 + c3

and

31.

32.

33.

27.

34. (1, a, a 2 ),

35.

(d) 1

The vectors AB = 3i + 4 k & AC = 5i - 2 j + 4k are the sides of a triangle ABC. The length of the median through A is [2003] (a)

288

(b)

18

(c)

72

(d)

33

other Vector, Component of a Vector 28.

36.

If a and b are unit vectors, then the greatest value of ® ®

® ®

3 | a + b | + | a - b | is ______. [NA Sep. 06, 2020 (I)] ®

29.

®

If® x ®and ®y be two non- ero vectors such that ® ® ® | x + y | = | x | and 2 x + l y is perpendicular to y , then the value of l is ______. [NA Sep. 06, 2020 (II)]

(b)

7 6 3

(c)

5 7

(d)

5 3 3

r A vector a = a i + 2 j + b k (a, b ÎR) lies in the plane of r r r the vectors, b = i + j and c = i – j + 4 k . If a bisects r r [Jan. 7, 2020 (I)] the angle between b and c , then: r r (a) a · i + 3 = 0 (b) a · i + 1 = 0 r  r (c) a · k + 2 = 0 (d) a · k + 4 = 0 r r  b = 4 i + (3 - l ) j + 6 k and Let a = 2 i + l1 j + 3 k, 2 r c = 3i + 6 j + (l3 - 1) k be three vectors such that r r r r b = 2a and a is perpendicular to c Then a possible

[Jan. 10, 2019 (I)]

æ 1 ö (b) ç - , 4, 0 ÷ è 2 ø

(a) (1, 3, 1)

Scalar or Dot Product of two Vectors,

®

7 6 6

value of (l1, l2, l3) is:

TOPIC n Projection of a Vector Along any ®

Let the volume of a parallelopiped whose coterminous r r edges are given by u = i + j + lk, v = i + j + 3k and r w = 2i + j + k be 1 cu. unit. If q be the angle between the r r edges u and w , then cos q can be: [Jan. 8, 2020 (I)] (a)

[2003] (c) –1

The proection of the line segment oining the points (–1, 2, 3) and (3, –2, 10) is ______ . [NA Jan. 9, 2020 (I)]

[2003]

vectors

[NA Sep. 02, 2020 (I)]

(1, –1, 3) and (2, –4, 11) on the line oining the points

(1, b, b2 ) and (1, c, c 2 ) are non- coplanar, then the product

abc equals (a) 0 (b) 2

r r r r r Let the vectors a, b , c be such that | a |= 2, | b |= 4 and r r r | c |= 4. If the proection of b on a is equal to the r r r r proection of c on a and b is perpendicular to c , then r r r the value of | a + b - c | is _______. [NA Sep. 05, 2020 (II)] r r r Let a, b and c be three unit vectors such that r r r r r r r r | a - b |2 + | a - c |2 = 8. Then | a + 2b |2 + | a + 2c |2 is equal to _________.

5i - j + 5k

respectively. Then ABCD is a (a) parallelogram but not a rhombus (b) square (c) rhombus (d) rectangle. a

30.

æ1 ö (c) ç , 4, - 2 ÷ (d) (1, 5, 1) è2 ø r r Let a = i + j + 2 k, b = b1 i + b 2 j + 2k r and c = 5i + j + 2k be three vectors such that the r r r proection vector of b on a is a . r r r If ar + b is perpendicular to c , then | b | is equal to: [Jan. 09, 2019 (II)] (a)

32

(b) 6

(c)

22

(d) 4

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37.

Mathematics

In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively

 -i + 3 + pk and 5i + q - 4k , then the point (p, 3i +  - k,

38.

q) lies on a line : [Online April 9, 2016] (a) making an obtuse angle with the positive direction of x–axis (b) parallel to x–axis (c) parallel to y–axis (d) making an acute angle with the positive direction of x–axis uuur uuur In a parallelogram ABD, AB = a, AD = b and uuur uuur uuur AC = c, then DA. AB has the value :

rr r 3 ( p.q ) r ( p.p )

r

44.

r r Let a and b be two unit vectors. If the vector s r r c = aˆ + 2 bˆ and d = 5 aˆ - 4bˆ are perpendicular to each other,,

then the angle between aˆ and bˆ is : (a)

p 6

(b)

1 2 1 2 (a + b2 – c2) (d) (b + c2 – a2) 3 2  y and  are three unit vectors in three-dimensional If x, space, then the minimum value of 2

2

(a)

3 2 ®

40.

41.

(b) 3 ®

2

® ®

46.

(a)

(a) 2i + j + 5k

(d) 2i + 3 j + 3k uuur r uuuur r Let ABCD be a parallelogram such that AB = q , AD = p r and ÐBAD be an acute angle. If r is the vector that coincide with the altitude directed from the vertex B to the r side AD, then r is given by : [2012] (c) 2i - j + 5k

43.

®

®

®

[Online May 12, 2012]

2 j + k

(b)

3i + 2 j - 2 k

5

(c)

17

3i + 2 j + 2k

(d)

17

47.

2i + 2 j - 2k 3

ABCD is parallelogram. The position vectors of A and C are respectively, 3i + 3 j + 5k and i - 5 j - 5k . If M is the midpoint of the diagonal DB, then the magnitude of the ®

®

proection of OM on OC , where O is the origin, is [Online May 7, 2012]

p p p p (a) (b) (c) (d) 3 6 4 2 r r r Let a = 2\$i - \$j + k\$ , b = \$i + 2 \$j - k\$ and c = \$i + \$j - 2k\$ be three r r vectors. A vector of the type b + lc for some scalar l,

2 is : 3 [Online April 9, 2013] (b) 2i + 3 j - 3k

p 4

A unit vector which is perpendicular to the vector

® ®

r whose proection on a is of magnitude

(d)

If a + b + c = 0, a = 3 , b = 5 and c = 7 , then the

2i + 2 j - k is

then the angle between the vectors a\$ and c\$ is : [Online April 22, 2013]

42.

p 3

2i - j + 2k and is coplanar with the vectors i + j - k and

(d) 6

If a = 2, b = 3 and 2 a - b = 5 , then 2 a + b equals: [Online April 9, 2014] (a) 17 (b) 7 (c) 5 (d) 1 r If a\$ , b\$ and c\$ are unit vectors satisfying a\$ - 3 b\$ + c\$ = 0 ,

[2012]

angle between a and b is [Online May 19, 2012] p p p p (a) (b) (c) (d) 3 6 4 2

[Online April 12, 2014] (c) 3 3

(c)

®

1 (b) (a2 – b2 + c2) 2

x + y + y +  +  + x

p 2 ®

45.

(c)

39.

( p . p) r r r r 3 (p.q) r (d) r = -3q - r r p (p .p)

r r r r (p.q) r (c) r = q - r r p (p.p)

[Online April 11, 2015] 1 2 (a) (a + b2 + c2) 2

r r

( p .q ) (b) rr = -qr + r r pr

(a) r = 3q - r r p

(a) 7 51 48.

49.

7

(b)

50

(c) 7 50

(d)

7 51

r r If the vectors a = i - j + 2k , b = 2i + 4 j + k% and r c = li + j + m k are mutually orthogonal, then (l, m) = (a) (2, –3) (b) (–2, 3) [2010] (c) (3, –2) (d) (–3, 2) r r r r r The non- ero vectors are a , b and c are related by a = 8b r r and c = –7b . Then the angle between ar and cr is [2008] (a) 0

(b)

p 4

(c)

p 2

(d) p

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Vector Algebra

50.

The values of a, for which points A, B, C with position vectors

2i - j + k , i - 3 j - 5k

and

ai - 3 j + k

respectively are

the vertices of a right angled triangle with C = (a) 2 and 1 (c) – 2 and 1 51.

57.

p are 2 [2006]

58.

(b) – 2 and – 1 (d) 2 and – 1

Let u , v , w be such that | u |= 1,| v |= 2, | w |= 3. If the proection v along u is equal to that of w along u and v , w are perpendicular to each other then | u - v + w | equals

52.

[2004]

59.

(a) 14 (b) 7 (c) 14 (d) 2 r r r r r r a , b , c are 3 vectors, such that a + b + c = 0 , r rr rr rr r r a = 1, b = 2, c = 3, then a.b + b .c + c .a is equal to

53.

® ® ® rr rr rr | a.b + b .c + c.a | , given that a + b + c = 0

(a) 25

(b) 50

(c) –25

54.

60.

[2002]

(d) –50 ® ® ®

® ® ®

If sdaa a , b , c are vectors such that a + b + c = 0 and ®

®

®

TOPIC Đ 55.

56.

[2002] (c) 45

61.

If the volume of a parallelopiped, whose coterminus edges r r are given by the vectors a = i\$ + \$j + nk\$ , b = 2\$i + 4 \$j - nk\$ r and c = i\$ + n \$j + 3k\$ ( n ³ 0 ) , is 158 cu.units, then: r r (b) b . c = 10

(c) n = 7

(d) n = 9

r r r Let x0 be the point of local maxima of f ( x) = a × (b ´ c), where r r r a = xi - 2 j + 3k, b = -2i + xj - k and c = 7i - 2 j + xk. r r r r r r Then the value of a × b + b × c + c × a at x = x0 is : [Sep. 04, 2020 (I)] (a) – 4 (b) – 30 (c) 14 (d) – 22

(c) -

1 2

(d) –1

r r r Let a , b and c be three unit vectors such r r that ar + b + cr = 0. if r r r r r r l = a × b + b × c + c × a and r r r r r r r d = a ´ b + b ´ c + c ´ a , then r the ordered pair, l, d is equal to: [Jan. 7, 2020 (II)]

æ 3 r rö (b) ç - ,3c ´ b ÷ è 2 ø

æ 3 r rö (d) ç - ,3a ´ b ÷ è 2 ø r r r r r Let a = 3i\$ + \$ and b = 2i\$ – \$ + 3k\$ . If b = b1 – b2 , where r r ur ur b1 is parallel to a and b2 is perpendicular to a , then r r b1 ´b2 is equal to: [April 09, 2019 (I)] (a) –3i\$ + 9\$ + 5k\$ (c)

[Sep. 05, 2020 (I)]

r r (a) a . c = 17

1 2

æ 3 r rö (c) ç ,3b ´ c ÷ è2 ø

(d) 90

Vector or Cross Product of two vectors, Area of a Parallelogram & Triangle, Scalar & Vector Tripple Product

(b)

æ 3 r rö (a) ç ,3a ´ c ÷ è2 ø

®

(b) 30

3 2

( )

®

| a |= 7,| b |= 5,| c |= 3 then angle between vector b and c is (a) 60

p r r r r and c is . If b is perpendicular to the vector b × c , 3 r r r then | a ´ (b ´ c) | is equal to _____.[NA Jan. 9, 2020 (II)] r r Let a = i - 2 j + k and b = i - j + k be two vectors. If cr r r r r is a vector such that b ´ cr = b ´ ar and c . a = 0, then r r [Jan. 8, 2020 (II)] c . b is equal to:

(a) -

[2003] (a) 1 (b) 0 (c) –7 (d) 7 r r r If | a |= 5, | b |= 4, | c |= 3 thus what will be the value of

r If a = 2i + j + 2k, then the value of r r r | i ´ (a ´ i) |2 + | j ´ (a ´ j ) |2 + | k ´ (a ´ k) |2 is equal to _______________. [NA Sep. 04, 2020 (II)] r r r Let b , b an d c be three vectors such that r r r r r | a | = 3, | b | = 5, b . c = 10 and the angle between b

62.

(b) 3i\$ – 9\$ – 5k\$

1 ( –3i\$ + 9\$ + 5k\$ ) 2

(d)

1 \$ \$ \$ ( 3i – 9  + 5k ) 2

The magnitude of the proection of the vector 2\$i + 3 \$j + k\$ on the vector perpendicular to the plane containing the vectors \$i + \$j + k\$ and \$i + 2 \$j + 3k\$ , is : [April 08, 2019 (I)] (a)

3 2

(b)

6

(c) 3 6

(d)

3 2

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63.

Mathematics

r r Let a = 3i + 2 j + xk and b = i – j + k , for some real x. r r Then a ´ b = r is possible if : [April 08, 2019 (II)] 70. 3 3 0

Then P(EC2 Ç E3C / E1 ) is equal to :

31 (b) 61

(a)

8 17

and P(E1 Ç E 2 Ç E 3 ) = 0.

(c) 2.

(b)

(c) 5.

1.

2 3

8.

(a)

3 16

(b)

1 8

(c) -

3 16

(d) -

1 8

www.jeebooks.in Probability

9.

M

In a workshop, there are five machines and the probability 1 of any one of them to be out of service on a day is . 4 If the probability that at most two machines will be out of

14.

3

æ 3ö service on the same day is ç ÷ k, then k is equal to: è 4ø [Jan. 7, 2020 (II)] (a)

17 8

(b)

17 4

17 (d) 4 2 For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the

(c)

10.

15.

4 , then the probability 5 that he is unable to solve less than two problems is : [April 12, 2019 (II)]

16.

201 æ 1 ö (a) 5 çè 5 ÷ø (c) 11.

12.

13.

54 æ 4 ö 5 çè 5 ÷ø

49

316 æ 4 ö (b) 25 çè 5 ÷ø

49

(d)

164 æ 1 ö 25 çè 5 ÷ø

(a)

1 11

(b)

1 10

(c)

1 12

(d)

1 17

1 1 1 1 , , and respectively. If all hit at the target 2 3 4 8 independently, then the probability that the target would be hit, is: [April 09, 2019 (I)] 25 192

(b)

7 32

(c)

1 192

(d)

25 32

17.

48

Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is : [April 10, 2019 (II)] (a) 5 (b) 6 (c) 8 (d) 7 Four persons can hit a target correctly with probabilities

(a)

(c)

48

Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is: [April 10, 2019 (I)]

Let A and B be two non-null events such that A Ì B . Then, which of the following statements is always correct? [April 08, 2019 (I)] (a) P(A|B) = P(B) – P(A) (b) P(A|B) ³ P(A) (c) P(A|B) £ P(A) (d) P(A|B) = 1 The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is : [April. 08, 2019 (II)] (a) 5 (b) 3 (c) 4 (d) 2 In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to : [Jan. 12, 2019 (I)] (a)

candidate can solve any problem is

- 523

200 5

6

225

(d)

150 65 175

6 65 In a game, a man wins ` 100 if he gets 5 or 6 on a throw of a fair die and loses ` 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is : [Jan. 12, 2019 (II)] (a)

5

(b)

400 loss 9

(b) 0

400 400 gain (d) loss 3 3 Two integers are selected at random from the set {1, 2, ...., 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is : [Jan. 11, 2019 (I)]

(c)

18.

7 2 3 1 (b) (c) (d) 10 5 5 2 An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, ..., 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is: [Jan 10, 2019 (I)]

(a) 19.

13 15 19 19 (b) (c) (d) 36 72 72 36 If the probability of hitting a target by a shooter, in any 1 shot, is , then the minimum number of independent shots 3 at the target required by him so that the probability of 5 hitting the target at least once is greater than , is: 6 [Jan. 10, 2019 (II)] (a) 3 (b) 6 (c) 5 (d) 4

(a)

20.

EBD_8344

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- 524

21.

22.

Mathematics

Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then P(X = 1) + P(X = 2) equals: [Jan 09, 2019 (I)] (a) 49/169 (b) 52/169 (c) 24/169 (d) 25/169 An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red is: [Jan. 09, 2019 (II)] (a)

21 49

(b)

27.

that both E and F happen is neither E nor F happens is

(a)

28.

29.

3 3 (d) 4 10 Let A, B and C be three events, which are pair-wise

independence and E denotes the complement of an event

(a) P (A) + P ( B ) 25.

26.

30.

(b) P ( A ) – P ( B )

(c) P ( A ) – P (B) (d) P ( A ) + P ( B ) A player X has a biased coin whose probability of showing heads is p and a player Y has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If X starts the game, and the probability of winning the game by both the players is equal, then the value of ‘p’ is[Online April 15, 2018] (a)

1 3

(b)

1 5

(c)

1 4

(d)

2 5

(b)

3 2

1 5 (d) 3 12 Three persons P, Q and R independently try to hit a target.

31.

by P or Q but not by R is : [Online April 8, 2017] 21 9 (a) (b) 64 64 15 39 (c) (d) 64 64 An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is : [Online April 8, 2017] 255 127 (b) (a) 256 128 63 1 (c) (d) 64 2 Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ? [2016] (a) E1 and E3 are independent. (b) E1, E2 and E3 are independent. (c) E1 and E2 are independent. (d) E2 and E3 are independent. If A and B are any two events such that P(A) = P(A Ç B) =

equal to :

7 (a) 55

(a) 11 20

(b) 5 17

(c) 8 17

(d) 1 4

(c)

12 55

(d)

14 55

2 and 5

3 , then the conditional probability,, 20

If two different numbers are taken from the set (0, 1, 2, 3, ......., 10), then the probability that their sum as well as absolute difference are both multiple of 4, is : [2017] 6 (b) 55

3 1 , and 4 2

5 respectively, then the probability that the target is hit 8

(c)

E. If P (A Ç B Ç C) = 0 and P (C) > 0, then P[( A Ç B ) |C] is equal to. [Online April 16, 2018]

P(E) 1 , then a value of is : P(F) 2 [Online April 9, 2017]

If the probabilities of their hitting the target are

(c)

24.

4 3

1 and the probability that 12

(c)

27 49

26 32 (d) 49 49 23. A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is : [2018] 2 1 (a) (b) 5 5

Let E and F be two independent events. The probability

P(A | A¢ È B¢)) , where A' denotes the complement of A, is [Online April 9, 2016]

www.jeebooks.in Probability

32.

M

Let X be a set containing 10 elements and P(X) be its power set. If A and B are picked up at random from P(X), with replacement, then the probability that A and B have equal number elements, is : [Online April 10, 2015]

(2

10

(a)

20

(b)

10

(2

2

10

(c)

)

-1

2

)

-1

20

C10

(

)

1 , 6

Let A and B be two events such that P A È B =

34.

1 1 and P A = , where A stands for the 4 4 complement of the event A. Then the events A and B are [2014] (a) independent but not equally likely. (b) independent and equally likely. (c) mutually exclusive and independent. (d) equally likely but not independent. Let A and E be any two events with positive probabilities:

(

)

is more than

220

33.

P AÇ B =

k, so that the probability of hitting the target at least once

38.

210

(d)

20

C10

( )

39.

(a) 3 (b) 5 (c) 2 (d) 4 Three numbers are chosen at random without replacement from {1,2,3,..8}. The probability that their minimum is 3, given that their maximum is 6, is : [2012] 3 1 (a) (b) 8 5 2 1 (c) (d) 5 4 Let A, B, C, be pairwise independent events with P (C) > 0

( )

c (a) P B – P ( B )

( ) ( )

c c (c) P A - P B

40.

41.

(c) Statement-1 is true, Statement-2 is false (d) Statement-1 is false, Statement-2 is true A, B, C try to hit a target simultaneously but independently. Their respective probabilities of hitting the targets are

36.

37.

the target k times (k, a given number). Then the minimum

( ) ( )

c c (b) P A + P B

( )

c (d) P A - P ( B )

(a) P (C | D ) ³ P (C )

(b) P (C | D ) < P (C )

P( D ) P (C )

(d) P (C | D ) = P (C )

(c) P (C | D) =

(b) Both the statements are false

not by C is : [Online April 23, 2013] (a) 21/64 (b) 7/8 (c) 7/32 (d) 9/64 Given two independent events, if the probability that 26 exactly one of them occurs is and the probability that 49 15 none of them occurs is , then the probability of more 49 probable of the two events is : [Online April 22, 2013] (a) 4/7 (b) 6/7 (c) 3/7 (d) 5/7 2 The probability of a man hitting a target is . He fires at 5

)

If C and D are two events such that C Ì D and P(D) ¹ 0, then the correct statement among the following is [2011]

[Online April 19, 2014]

3 1 5 , , . The probability that the target is hit by A or B but 4 2 8

[Online April 9, 2013]

(

Statement - 2: P(A/E) ³ P ( A Ç E )

35.

7 , is : 10

c c and P ( A Ç B Ç C ) =0. Then P A Ç B / C . [2011RS]

Statement - 1: P(E/A) ³ P(A/E) P(E)

(a) Both the statements are true

- 525

One ticket is selected at random from 50 tickets numbered 00,01,02,...,49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is ero, equals: [2009] (a)

1 7

(b)

5 14

1 1 (d) 50 14 It is given that the events A and B are such that

(c) 42.

P ( A) =

(a)

1 6

1 1 2 , P( A | B ) = and P ( B | A) = . Then P(B) is 4 2 3 [2008]

(b)

1 3

2 1 (d) 3 2 Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is [2007] (a) 0.2 (b) 0.7 (c) 0.06 (d) 0.14.

(c)

43.

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- 526

44.

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is [2005] (a)

(c)

45.

Mathematics

2 9

(b)

8 9

(d)

49.

1 9

50.

7 9

Let A and B be two events such that P ( A È B ) =

1 , 6

1 1 and P ( A) = , where A stands for 4 4 complement of event A. Then events A and B are [2005] P( A Ç B) =

51.

In a bombing attack, there is 50% chance that a bomb will hit the target. At least two independent hits are required to destroy the target completely. Then the minimum number of bombs, that must be dropped to ensure that there is at least 99% chance of completely destroying the target, is _________. [NA Sep. 05, 2020 (II)] 1 . The least 10 number of shots required, so that the probability of his 1 hitting the target at least once is greater than , 4 is ___________. [NA Sep. 04, 2020 (I)]

The probability of a man hitting a target is

A random variable X has the following probability distribution: X

The probability that A speaks truth is

4 , while the 5

3 . The probability that they contradict 4 each other when asked to speak on a fact is [2004] 4 1 (a) (b) 5 5 7 3 (c) (d) 20 20 A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem

is

4

5

2K

K

2K

5K2

3 4 2 (c) 3

TOPIC n

1 2 1 (d) 3

52.

53.

(b)

Random Variables, Probability Distribution, Bernoulli Trails, Binomial Distribution, Poisson Distribution

Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, is ______. [NA Sep. 05, 2020 (I)]

7 12

[Jan. 9, 2020 (II)] (b)

1 36

1 23 (d) 6 36 Let a random variable X have a binomial distribution with k mean 8 and variance 4. If P(X d” 2) = 16 , then k is equal 2 to: [April 12, 2019 (I)] (a) 17 (b) 121 (c) 1 (d) 137

(c)

1 1 1 , and . Probability that the problem is solved is 2 3 4 [2002]

(a)

48.

3

Then, P(X > 2) is equal to: (a)

K

2

(b) equally likely but not independent

probability for B is

47.

2

P(X)

(d) mutually exclusive and independent

:

1

(a) equally likely and mutually exclusive (c) independent but not equally likely

46.

:

54.

A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/ loss (in Rs.) of the person is : [April 12, 2019 (II)] 1 1 (a) gain (b) loss 2 4 1 (c) loss (d) 2 gain 2 A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then æ ö mean of X ç ÷ is equal to:[Jan. 11, 2019 (II)] standard deviation of X è ø

(a) 4

(b) 4 3

(c) 3 2

(d)

4 3 3

www.jeebooks.in Probability

55.

M

A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is: [2017] 6 12 (b) (c) 6 (d) 4 25 5 An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of this experiment is : [Online April 10, 2016] 496 192 (a) (b) 729 729 240 256 (c) (d) 729 729 If the mean and the variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than or equal to one is : [Online April 11, 2015]

(a)

56.

57.

1 15 3 (c) (d) 16 16 4 If X has a binomial distribution, B(n, p) with parameters n and p such that P(X = 2) = P (X = 3), then E(X), the mean of variable X, is [Online April 11, 2014]

(a) 58.

9 16

59.

(a)

60.

63.

(b) 3 – p

64.

17

13

11

65.

é 1ù (b) ê0, ú ë 2û æ 1 3ù (d) ç , ú è 2 4û 1ö æ In a binomial distribution B ç n, p = ÷ , if the probability è 4ø 9 of at least one success is greater than or equal to , then 10

66.

n is greater than:

[2009]

A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is [2007] (a) 8/729 (b) 8/243 (c) 1/729 (d) 8/9. At a telephone enquiry system the number of phone calls regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10 minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is [2006]

6 e

(b)

5 6

(c)

6 55

(d)

6

5 e5 A random variable X has Poisson distribution with mean 2. [2005] Then P (X > 1.5) equals 3 3 (c) 1 (d) 2 2 e e e The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is [2004] 2

2

28 256

(b) 0

(b)

219 256

(c)

128 256

(d)

37 256

A random variable X has the probability distribution: X: 1 2 3 4 5 6 7 8 p(X): 0.2 0.2 0.1 0.1 0.2 0.1 0.1 0.1

For the events E = {X is a prime number } and

67.

æ 3 11 ù (a) ç , ú è 4 12 û æ 11 ù (c) ç ,1ú è 12 û 61.

1 (d) log 4 – log 3 10 10

(a)

10

(b) 5 (c) 5 (d) 5 3 3 35 3 Consider 5 independent Bernoulli’s trials each with probability of success p. If the probability of at least one 31 , then p lies in the failure is greater than or equal to 32 interval [2011]

4 (c) log 4 – log 3 10 10

(a)

p (d) 3

(c)

9 (b) log 4 – log 3 10 10

(a)

(b)

p 2 A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers ust by guessing is: [2013]

(a) 2 – p

62.

1 (a) log 4 + log 3 10 10

- 527

68.

F = { X < 4}, the P( E È F ) is [2004] (a) 0.50 (b) 0.77 (c) 0.35 (d) 0.87 The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively, then P (X = 1) is [2003] (a)

1 4

(b)

1 32

(c)

1 16

(d)

1 8

A dice is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is [2002] (a) 8/3 (b) 3/8 (c) 4/5 (d) 5/4

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1.

- 528

Mathematics

(d) Probability of sum getting 6, P ( A) = Probability of sum getting 7, P ( B ) =

So,

5 36

æB ö Pç 1÷ = è Nø

6 1 = 36 6

P(A wins) = P ( A) + P( A) P ( B ) P ( A) + P ( A) × P( B ) P( A) P( B ) P( A) + .....

Þ

5 æ 31ö æ 30 ö æ 5 ö + ç ÷ ç ÷ ç ÷ + .....¥ 36 è 36 ø è 36 ø è 36 ø

Þ

2 ö 5 æ 155 æ 155 ö +ç + ........¥÷ ç1 + ÷ 36 è 216 è 216 ø ø

5 30 Þ 36 = 61 61 216

2.

3.

æ Nö æ Nö P ( B1 ) ´ P ç ÷ + P ( B2 ) ´ P ç ÷ è B1 ø è B2 ø

1 20 1 ´ 8 2 30 = = 3 = . 1 20 1 15 1 15 17 ´ + ´ + 2 30 2 20 3 40

5.

(

)

C C æ E C Ç E3C ö P éë E1 Ç E2 Ç E3 ùû (d) P ç 2 = ÷ E1 P ( E1 ) è ø

=

a ö æ çèQ S¥ = ÷ 1- r ø

P ( E1 ) - P[ E1 Ç ( E2 È E3 )] P( E1 )

[Q P( A Ç BC ) = P( A) - P( A Ç B)] =

P ( E1 ) - P[( E1 Ç E2 ) È ( E1 Ç E3 )] P ( E1 )

=

P( E1 ) - [ P( E1 Ç E2 ) + P( E1 Ç E3 ) - P( E1 Ç E2 Ç E3 )] P( E1 )

E1 Ç E2 = (4, 4)

=

P ( E1 ) - P ( E1 Ç E2 ) - P( E1 Ç E3 ) + 0 P ( E1 )

æE ö 1 Pç 2 ÷ = è E1 ø 9

= 1 - P ( E2 ) - P( E3 )

(d) E1 [the event for getting score a multiple of 4] = (1, 3), (3, 1), (2, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4) & (6, 6) E2 [4 has appeared atleast once] = (1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5) & (4, 6)

=

(a) Total outcomes = 9(104) Favourable outcomes

6.

Probability =

36 ´ 30 + 9 ´ 15

4 ´ 30 + 15

(b) P(second A – card appears before the third B– card) + P (BABA)

1 1 1 1 1 1 11 + + + + + = 4 8 8 16 16 16 16 (b) A and B are independent events. =

7.

1 1 1 ¢) 3 - 3 . 6 1 Ç ( A P A B æ ö So, P ç ÷ = = = 1 3 P ( B¢) è B¢ ø 6

1 2 and P (non-prime number) Q P ( B1 ) = P ( B2 ) =

1 20 1 15 = ´ + ´ 2 30 2 20

or

[Q P( A Ç B ) = P( A) × P( B)]

P( E3C ) - P( E2 )

135

= = 4 9 ´ 104 104 10 (b) Let B1 and B2 be the boxes and N be the number of non-prime number.

æ Nö æNö = P( B1 ) ´ P ç ÷ + P ( B2 ) ´ P ç ÷ è B1 ø è B2 ø

P( E2C ) - P( E3 )

= P (AA) + P (ABA) + P (BAA) + P (ABBA) + P (BBAA)

= 9C2 (25 - 2) + 9C1 (24 - 1) = 36(30) + 9(15)

4.

æ Nö P ( B1 ) ´ P ç ÷ è B1 ø

8.

(b)

0

1

2

3

4

5

1 P( k ) 32

12 32

11 32

5 32

2 32

1 32

k

k = No. of times head occur consecutively

www.jeebooks.in Probability

M

Now expectation

14.

= å xP(k ) = (-1) ´

5 2 1 1 + 4 ´ + 5´ = 32 32 32 8 (a) Required probability = when no machine has fault + when only one machine has fault + when only two machines have fault. 5

4

2

æ3ö æ 1 öæ 3 ö æ1ö æ 3ö = 5C0 ç ÷ + 5C1 ç ÷ç ÷ + 5C2 ç ÷ ç ÷ 4 4 4 è ø è øè ø è4ø è4ø =

15.

æ3ö æ 3 ö 17 =ç ÷ ´k =ç ÷ ´ 8 è4ø è4ø 17 \ k= 8 10. (c) Let p is the probability that candidate can solve a problem and q is the probability that candidate can not not solve a problem. 4 1 and q = (Q p + q = 1) 5 5 Probability of solving either 50 or 49 problem by the candidate = 50C50 . p50 . q0 + 50C49 . p49 . q1 = p49 [p + 50q] 49

æ 4 ö æ 4 50 ö 54 æ 4 ö . = ç ÷ .ç + ÷ = 5 çè 5 ÷ø è5ø è5 5 ø 11.

4

12.

n

13.

P (B) £ 1

(c) Let, p is probability for getting head and is probability for getting tail.

9 9 Þ 1 - P ( x = 0) ³ 10 10

2n ³ 10 Þ n ³ 4 Þ nmin = 4

16.

(d) Since, the experiment will end in the fifth throw. Hence, the possibilities are 4 * * 4 4, * 4 * 4 4, * * * 4 4 (where * is any number except 4)

æ 1 ö æ 5ö æ 5ö æ 1ö æ 1 ö Required Probability = çè ÷ø çè ÷ø çè ÷ø çè ÷ø çè ø÷ 6 6 6 6 6 3

17.

æ 5ö æ 1ö æ 5ö 1 1 5 1 + ç ÷ ç ÷ ç ÷ æç ö÷ æç ö÷ + æç ö÷ æç ö÷ è 6 ø è 6ø è 6 ø è 6ø è 6ø è 6ø è 6ø 25 + 25 + 125 175 = 5 = 65 6 2 1 (b) Probability of getting 5 or 6 = P(E) = = 6 3 1 2 Probability of not getting 5 or 6 = P(E) = 1 - = 3 3 E will consider as success.

Event

n

99 1 æ1ö æ1ö Þç ÷ < 1- ç ÷ > Þn³7 2 100 2 100 è ø è ø Hence, minimum value is 7. (d) P (at least one hits the target) = 1 – P (none of them hits the target)

1 1 , q =1- p = 2 2

n 1 9 1 1 9 æ1ö 1 - n C0 ç ÷ ³ Þ n £ 1- Þ n £ 10 10 2 10 2 2 è ø

(a) Let, A = At least two girls B = All girls

æ1ö ç ÷ è4ø 1 1 4 4 = æ 1 ö 4 æ 1 ö = 16 - 1 - 4 = 11 4 1 - C0 ç ÷ - C1 ç ÷ è2ø è2ø (d) According to the question,

Q

P ( x ³ 1) ³

49

æ B ö P ( B Ç A) P( B) = Pç ÷ = P ( A) P ( A) è Aø

æ A ö P ( A) Pç ÷ = è B ø P( B)

p = P (H) =

3

p=

Þ

æ Aö Þ P ç ÷ ³ P(A) è Bø

3

243 405 270 918 459 27 ´ 17 + + = = = 1024 1024 1024 1024 512 64 ´ 8 3

(b) Q A Ì B; so A Ç B = A

æ A ö P ( A Ç B) Now, P ç ÷ = P( B ) èBø

1 12 11 + (-1) ´ + ( -1) ´ 32 32 32 +3 ´

9.

- 529

Success Success Success in No success in Ist in IInd IIIrd in IIIrd attempt attempt attempt attempt

Probability

1 3

Gain/loss

100

2 1 ´ 3 3

2 2 1 ´ ´ 3 3 3

50

His expected gain/loss

æ 1 öæ 1 öæ 1 öæ 1 ö = 1 - ç1 - ÷ç1 - ÷ç1 - ÷ç 1 - ÷ è 2 øè 3 øè 4 øè 8 ø

1 2 8 ´ (-150) = ´ 100 + ´ 50 + 3 9 27

1 2 3 7 7 25 =1- ´ ´ ´ =1 - = 2 3 4 8 32 32

=

900 + 300 - 1200 =0 27

0

2 2 2 ´ ´ 3 3 3

–150

2

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18.

Mathematics

(c) Probability of getting sum of selected two numbers is even

23.

5

C2 +5C2 11 C2 Probability of getting sum is even and selected numbers = P( E1 ) =

5

are also even P(E2) =

P(Outcome is tail) =

æR ö æR ö = P(R1) ´ P ç 2 ÷ + P(B1 ) ´ P ç 2 ÷ R è 1ø è B1 ø æ 4 6ö æ 6 4ö 2 = ç ´ ÷+ç ´ ÷ = è 10 12 ø è 10 12 ø 5

1 2

1 2

P(7 or 8 is the sum of two dice) = P(7 or 8 is the number of card) =

4 6 and P(B1 ) = 10 10

æR ö 6 æR ö 4 and P ç 2 ÷ = And, P ç 2 ÷ = R 12 è B1 ø 12 è 1ø Now, required probability

C2 C2

11

P(R1 ) =

\

5 æE ö 10 2 C2 = = . Hence, P ç 2 ÷ = 6 5 15 10 5 + è E1 ø C2 + C2

19.

(a) Let Rt be the even of drawing red ball in tth draw and Bt be the event of drawing black ball in tth draw. Now, in the given bag there are 4 red and 6 black balls.

24.

(c) Here, P ( A Ç B |C) =

6 5 11 + = 36 36 36

=

1 1 2 + = 9 9 9

P( A Ç B Ç C) P (C )

.

P (C ) – P ( A Ç C – P ( B Ç C ) + P ( A Ç B Ç C )) P (C )

A

1 11 1 2 Required probability = ´ + ´ 2 36 2 9

B

1 æ 11 + 8 ö 19 = = ç 2 è 36 ÷ø 72 20.

(c) Let the number of independent shots required to hit the target at least once be n, then n

P (A Ç B Ç C ) C

n

5 æ 2ö 1 æ 2ö 1- ç ÷ > ç ÷ < è 3ø 6 è 3ø 6

21.

Hence, the above inequality holds when least value of n is 5. (d) X = number of aces drawn \ P(X = 1) + P(X = 2)

é P ( A ) . P (C ) + P ( B ). P ( C ) ù =1– ê ú P (C ) ë û

25.

4ü ì 4 48 48 4 ü ì 4 =í ´ + ´ ý+í ´ ý 52 52 52 52 52 52 î þ î þ

p p 4p + + ... = . 4 16 3 Similarly Y wins if the outcome is one of the following: TH, TTTH, TTTTTH, ... Therefore, the probability that Y wins is p+

24 1 25 + = 169 169 169 (d) Let G represents drawing a green ball and R represents drawing a red ball So, the probability that second drawn ball is red

=

22.

æ Rö æ Rö = P (G ) × P çè ÷ø + P ( R ) P çè ÷ø G R 2 6 5 4 = ´ + ´ 7 7 7 7 12 + 20 = 49 32 = 49

= 1 – P (A) – P (B) = P ( A ) – P (B)( Q P (A Ç B Ç C) = 0) (a) If the outcome is one of the following: H, TTH, TTTTH, ..., then X wins. As subsequent tosses are independent, so the probability that X wins is

1 - p 1 - p 1 - p 2(1 - p) + + = 2 8 32 3 Since, the probability of winning the game by both the players is equal then, we have 4 p 2(1 - p) 1 = Þ p= 3 3 3 26.

(b) Let A º {0, 1, 2, 3, 4, ......., 10} n (S) = 11C2 = 55 where 'S' denotes sample space Let E be the given event \ E º {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}

www.jeebooks.in Probability

M

Þ n (E) = 6 n(E) 6 \ P(E) = n(S) = 55 27.

(a)

And P(E1 Ç E2 Ç E 3) = 0 ¹ P (E1) . P(E2) . P(E3) Þ E1, E2, E3 are not independent.

1 P ( E Ç F) = P (E ) . P (F) = 12 1 P EÇF =P E .P F = 2

(

)

31.

Þ Þ Þ Þ Þ

x+y=

(

1 1 5 = 2 12 12

7 12

\

P (A / (A' Ç B')) =

33.

12 + 9 21 = 64 64 (b) Required probability = 1 – {P (All Head) + P (All Tail)}

ì 1 ü 127 = 1- í ý = î128 þ 128 (b) P(E1) =

1 1 1 ; P(E 2 ) = ; P(E 3 ) = 6 6 2

P(E1Ç E 2 ) =

1 1 1 , P(E 2 Ç E3 ) = , P(E1Ç E3 ) = 36 12 12

P ( A - ( A Ç B) )

(

P AÈ B

)

=

5 17

C0

) + ( 10C1 ) + ( 10 C2 ) 2

2

2

+ ....... +

( 10 C10 )

2

C10

220

(a) Given, P( A È B ) =

1 1 5 Þ P( A È B ) = 1 - = 6 6 6

1 1 3 Þ P ( A) = 1 - = 4 4 4 We know, P ( A È B ) = P( A) + P ( B ) - P ( A Ç B) 5 3 1 = + P( B ) Þ 6 4 4 1ö æ çèQ P( A Ç B) = ÷ø 4 1 Þ P(B) = 3 Q P ( A) ¹ P ( B ) so they are not equally likely.

(a) Required probability =

ì1 ü = 1- í 7 ý î2 þ

5 20

P ( A) =

14 3 x 13 4 = = = or 13 4 y 14 3

1ü ì1 = 1- í 8 + 8 ý î2 2 þ

5 3

Required probability is

10

20

=

=

30.

B

210

æ 3 ö æ 1 ö æ 3ö æ 1 ö æ 1 ö æ 3ö æ 3 ö æ 1 ö æ 3ö ç 4 ÷ ç 2÷ ç8÷ + ç 4÷ ç 2÷ ç8÷ + ç 4÷ ç 2 ÷ ç8÷ è øè øè ø è øè øè ø è øè øè ø

29.

A

)

P (A – (A Ç B)) =

(

1ù é êQ x . y = 12 ú ë û

1 1 ,x= 3 4 1 1 and y = , y = 4 3

28.

3 20

\

32. (d)

Þ x=

\

)

17 Þ P AÈ B = 20 A Ç (A' È B') = A – (A Ç B)

1 2

7 1 x+ = 12x 12 12x2 – 7x + 1 = 0 2 12x – 4x – 3x + 1 = 0 (4x – 1) (3x – 1) = 0

2 8 3 = ; P(A Ç B) = 5 20 20

P AÇ B = 1 –

1 Þ (1 – P(E)) (1 – P(F)) = 2 Let P(E) = x P(F) = y

Þ 1 – x – y=

(b) P(A) =

(

( ) ( )

Þ 1 – x – y + xy =

- 531

3 1 1 ´ = = P( A Ç B) 4 3 4 So A & B are independent. (a) Let A and E be any two events with positive probabilities. Consider statement-1 :

Also P ( A) ´ P( B ) =

34.

P(E / A) ³ P(A / E)P(E)

P(E Ç A) ...(1) P(A) P(E Ç A) × P(E) RHS : P(A/E). P(E) = P(E) LHS : P (E/A) =

= P(A Ç E) ...(2) Clearly, from (1) and (2), we have

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- 532

Mathematics

P(E/A) ³ P(A Ç E) Thus, statement-1 is true. Similarly, statement-2 is also true. 35.

36.

(a) P(A or B but not by C) = P((A È B) Ç C ) = P(A È B) × P(C) = [P(A) + P(B) – P(A Ç B)] × P(C)

42 49 8 and ab = 49

38.

...(i)

39.

2 k 2 é 3 æ3ö æ 3ö ù 7 ê1 + + ç ÷ + ...... + ç ÷ ú > è 5 ø û 10 5 ë 5 è5ø

Þ

æ3ö 1- ç ÷ 2 7 è5ø ´ > 3 5 10 15

Þ

3 æ3ö ç ÷ < 10 è5ø

Þ

k

7 æ3ö 1- ç ÷ > 10 è5ø

Þ k³3

(b) Given three numbers are chosen without replacement from = {1,2,3,.....,8} Let Event F : Maximum of three numbers is 6. E : Minimum of three numbers is 3. This is the case of conditional probability We have to find P (minimum) is 3 when it is given that P (maximum) is 6. 2 C æ E ö P (E Ç F ) 2 1 = 5 1 = = s\ P çè ÷ø = F P (F) 10 5 C2 c c (d) P(A Ç B /C) =

((

)

P Ac Ç B c Ç C

=

=

P (C )

) = P (( A È B) Ç C ) c

P (C )

P (( S - A È B) Ç C ) P (C )

P (( S - A - B + A Ç B ) Ç C ) P ( C)

P(C) – P(A Ç C) – P(B Ç C) + P(A Ç B Ç C) = P(C)

P(C) – P(A).P(C) – P(B)P(C) + 0 P(C) = 1 - P ( A) - P ( B ) [Q P(AC) = 1–P(A)]

...(ii)

=

40.

14 49 From (iii) and (iv), 4 2 a= ,b= 7 7

\ a–b=

Þ

Hence minimum value of k = 3

...(iii)

(a – b)2 = (a + b)2 – 4ab =

2 3 2 æ3ö 2 æ3ö 2 7 + ´ + ç ÷ ´ + ...... + ç ÷ . > 5 5 5 è5ø 5 è 5 ø 5 10

k

Now, P(A and not B) + P(not A and B) =

a+b=

(a)

k

é 3 1 3 1 ù 3 æ 6 + 4 - 3 ö 3 21 ÷´ = = ê + - ´ ú ´ = çè ë 4 2 4 2û 8 8 ø 8 64 (a) Let the probability of occurrence of first event A, be ‘a’ i.e., P(A) = a \ P(not A) = 1 – a And also suppose that probability of occurrence of second event B, P(B) = b, \ P(not B) = 1 – b

26 49 26 Þ P(A) × P(not B) + P(not A) × P(B) = 49 26 Þ a × (1 – b) + (1 – a) b = 49 26 Þ a + b – 2ab = 49 15 And P(not A and not B) = 49 15 Þ P (not A) × P(not B) = 49 15 Þ (1 – a) × (1 – b) = 49 15 Þ 1 – b – a + ab = 49 34 Þ a + b – ab = 49 From (i) and (ii),

k

2

37.

= P ( Ac ) - P ( B ) (a) We know,

æ C ö P (C Ç D) P(C ) Pç ÷ = = è Dø P ( D) P ( D)

42 42 4 ´ 8 196 ´ = 49 49 49 2401

...(iv)

Hence probability of more probable of the two events =

41. 4 7

Where, 0 £ P ( D ) £ 1 , hence æ Cö P ç ÷ ³ P (C ) è Dø (d) Given that 1 1 3 P = Þ q =1- = 4 4 4 9 and P (x ³ 1) ³ 10

[Q C Ì D]

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M

Þ 1 – P (x = 0) ³ 0

9 10

47. n

Þ

9 æ1ö æ3ö 1 - nC0 ç ÷ ç ÷ ³ 10 è4ø è4ø

Þ

1-

9 æ 3ö ³ç ÷ 10 è 4 ø n

Þ 42.

43.

1 n³ log10 4 - log10 3

1 (b) Given that P(A) = 1/4, P(A/B) = , P(B/A) = 2/3 2 By conditional probability, P(A Ç B) = P(A) P(B/A) = P(B)P(A/B) 1 2 1 1 Þ ´ = P ( B ) ´ Þ P( B ) = 4 3 2 3 (d) Given that P(I) = 0.3 and P(II) = 0.2 \ P( I ) = 1 – 0.3 = 0.7 \ The required probability

= P ( I Ç II ) = P( I ).P( II ) = 0.7 × 0.2 = 0.14 44.

1 (b) Probability of particular house being selected = 3 P (all the persons apply for the same house) 1 æ1 1 1ö = ç ´ ´ ÷3 = . 9 è 3 3 3ø

45.

1 1 1 , P ( A Ç B ) = and P ( A) = 4 6 4 5 3 Þ P ( A È B ) = , P( A) = 6 4 Also Þ P ( A È B ) = P ( A) + P ( B ) - P ( A Ç B )

(c)

P ( A È B) =

5 3 1 1 - + = 6 4 4 3 3 1 1 Þ P ( A) P ( B ) = - = = P( A Ç B) 4 3 4 Hence A and B are independent but not equally likely. Þ P( B) =

46.

(c) A and B will contradict each other if one speaks truth and other false . So , the required probability

4 æ 3 ö æ 4ö 3 P(AÇ B ) + P ( A Ç B) = 5 çè 1 - 4 ÷ø + çè1 - 5 ÷ø 4 4 1 1 3 7 = ´ + ´ = 5 4 5 4 20

1 1 , P (B) = and 3 2

_ _ _ 1 ; P ( AUBUC ) = 1 - P ( A) P ( B ) P (C ) 4 1 2 3 3 æ 1ö æ 1ö æ 1ö = 1 - ç1 - ÷ ç 1 - ÷ ç1 - ÷ = 1 - ´ ´ = 2 3 4 4 è 2 ø è 3ø è 4ø (11) Probability of getting at least two 3's or 5's in one trial

P (C) =

n

æ 3ö æ 1ö Þ ç ÷ £ç ÷ è 4ø è 10 ø Taking log at the base 3/4, on both sides, we get æ3ö æ1ö n log3/4 ç ÷ ³ log3/4 ç ÷ è4ø è 10 ø - log10 10 -1 = Þ n ³ - log3/4 10 = æ 3ö log10 3 - log10 4 log10 ç ÷ è 4ø

(a) Given that P(A) =

48.

2

2

3

æ 2ö æ 4ö æ 2ö æ 4ö æ 2ö = 4C2 ç ÷ ç ÷ + 4C3 ç ÷ ç ÷ + 4C4 ç ÷ è 6ø è 6ø è 6ø è 6ø è 6ø 33

=

34

=

11 27

æ 11 ö E (x) = np = 27 ç ÷ = 11. è 27 ø 49.

(11.00) Let 'n' bombs are required, then 1

æ1ö æ1ö 1 - nC1 × ç ÷ ç ÷ è2ø è2ø

n -1

0

n

99 æ1ö æ1ö - n C0 ç ÷ ç ÷ ³ 2 2 100 è ø è ø

n +1 1 ³ n Þ 2 n ³ 100( n + 1) Þ n ³ 11 100 2 (3.00) Þ

50.

- 533

p=

1 9 ,q= 10 10

æ 9ö P (not hitting target in n trials) = ç ÷ è 10 ø æ 9ö P (at least one hit) = 1 - ç ÷ è 10 ø

n

n

1 æ 9ö n Q1 - ç ÷ > Þ (0.9) < 0.75 è 10 ø 4

\ nminimum = 3. 51. (d)

å P( K ) = 1

Þ 6K2 + 5K = 1

6K2 + 5K – 1 = 0 6K2 + 6K – K – 1 = 0 Þ (6K – 1) (K + 1) = 0 Þ

K=

1 (K = – 1 reected) 6

P(X > 2) = K + 2K + 5K2 =

1 2 5 6 + 12 + 5 23 + + = = 6 6 36 36 36

n

4

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52.

Mathematics

(d) Given mean µ = 8 and variance s2 = 4 Þ µ = np = 8 and s2 = npq = 4.

56.

1 3 p(x > 5) = p(x = 5) + p(x = 6)

Q

1 1 p + q = 1 Þ q = , p = and n = 16 2 2

QP (X £ 2) =

5

216

16 16 16 k æ1ö æ1ö æ1ö C0 ç ÷ +16 C1 ç ÷ +16 C2 ç ÷ = 16 \ 2 è2ø è2ø è2ø Þ k = (1 + 16 + 120) = 137 (c) Let X be the random variable which denotes the Rs gained by the person. Total cases = 6 × 6 = 36. Favorable cases for the person on winning ` 15 are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) i.e., 6 cases.

6 1 = 36 6 Favorable cases for the person on winning ` 12 are (6, 3), (5, 4), (4, 5), (3, 6) i.e., 4.

\

57.

P (X = 15) =

P (X = – 6) =

54.

4

58.

55.

–6 13 18

X . P (X )

5 2

4 3

-13 3

5

1 4

4 13

=-

Þ

59.

nP nP = nPQ Q

3 4 = 48 = 4 3

(b) We can apply binomial probability distribution We have n = 10 15 3 = p = Probability of drawing a green ball = 25 5 3 2 Also q = 1 - = 5 5 3 2 12 Variance = npq = 10 × ´ = 5 5 5

nC 2

p2 (1 – p)n–2 = nC3 (p)3 (1 – p)n–3

n! p 2 (1 - p ) n n! p 3 (1 - p ) n . . = 2!( n - 2)! (1 - p ) 2 3!(n - 3)! (1 - p )3

1 p 1 = . Þ 3 (1 – p) = p (n – 2) n - 2 3 1- p Þ 3 – 3p = np – 2p Þ np = 3 – p Þ E(X) = mean = 3 – p (Q mean of B (n, p) = np) (c) p = p (correct answer), q = p (wrong answer) 1 2 Þ P= , q = , n =5 3 3 By using Binomial distribution Required probability Þ

1 2

30 3 10 1 = , Q(red ball) = = , n = 16 40 4 40 4

Mean of X = standard deviation of X

=

12

å X .P( X ) = 2 + 3 - 3

P (X = 2) = nC2 (p)2 (1 – p)n–2

Given P (X = 2) = P (X = 3) Þ

1 9

(b) P(white ball) =

(b) Since X has a binomial distribution, B (n, p) and P (X = 3) = nC3 (p)3 (1 – p)n–3

1 6

16 ´

0

1 2ö 256 æ 2ö æ = ç ÷ ç6 ´ + ÷ = è 3ø è ø 729 3 3 (d) Let mean = np = 2 ...(1) and variance = npq = 1 ...(2) On solving eqn (1) and (2), we get 1 1 q = and p = 2 2 From eqn (1), we have n=4 P (x ³ 1) = 4C1p1q3 + 4C2p2q2 + 4C3p3q + 4C4p4

\

P (X )

Hence, E (X) =

6

1 15 æ 1ö = 1 – 4C0p0q4 = 1 - ç ÷ = 1 - = è 2ø 16 16

26 13 = 36 18 15

X

1

5

4 1 = \ P (X = 12) = 36 9 Remaining cases = 36 – 6 – 4 = 26

\

p + 2p = 1 Þ p =

æ 2ö æ1ö 6 æ2ö æ1ö 6 = C5 ç ÷ ç ÷ + C5 ç ÷ ç ÷ 3 3 è ø è ø è 3ø è 3ø

k

16

53.

(d) Let p(F) = p Þ p(S) = 2p

4

æ 1ö 2 5 æ 1ö P(x ³ 4) = C4 ç ÷ · + C5 ç ÷ è 3ø 3 è 3ø 2 1 11 = 5· 5 + 5 = 5 3 3 3

5

5

60.

(b) Given that p (at least one failure) ³ Þ 1 – p (no failure) ³ 5 Þ1 - p ³

31 32

31 32

31 32

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M

Given mean (m) = 5 P (at most 1 phone call)

1 1 Þ p£ 32 2 But p ³ 0

5 Þp £

61.

é 1ù Hence p lies in the interval ê0, ú . ë 2û (d) Given that 1 1 3 P = Þ q =1- = 4 4 4 9 and P (x ³ 1) ³ 10 9 Þ 1 – P (x = 0) ³ 10 0

64.

9 æ1ö æ3ö 1 - nC0 ç ÷ ç ÷ ³ 4 4 10 è ø è ø

Þ

1-

æ lö = 1 - e-l - e-l ç ÷ = 1 - 3 . è 1!ø e2

n

65.

Þ 62.

6

66.

67.

P( X = r ) =

1 1 , p = ,n = 8 2 2 7

\ Probability of getting score 9 exactly twice

63.

2

P ( EUF ) = P ( E ) + P ( F ) - P ( E Ç F ) = 0.62 + 0.50 - 0.35 = 0.77 (b) Given that np = 4 and npq = 2 Þ q =

Number of trial = 3 2

æ 1ö æ 1ö P(2 success) = 8C2 ç ÷ ç ÷ è 2ø è 2ø 28 28 = 8 = 256 2 (b) P(E) = P ( 2 or 3 or 5 or 7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62

P ( F ) = P (1 or 2 or 3) = 0.15 + 0.23 + 0.12 = 0.50 P ( E Ç F ) = P(2 or 3) = 0.23 + 0.12 = 0.35 We know that

4 1 = P(score 9) = 36 9

æ 1ö æ 1ö = 3C2 × ç ÷ . ç1 - ÷ = 3! ´ 1 ´ 1 ´ 8 è 9ø è 9ø 2! 9 9 9 3.2! 1 1 8 8 = ´ ´ ´ = 2! 9 9 9 243 (d) From poission distribution

1 and n = 8 2

\

1 n³ log10 4 - log10 3

(b) The sample space of pair of fair dice is thrown, S = (1, 1), (1, 2) (1, 3) .... = 36 Sum 9 are (5, 4), (4, 5), (6, 3), (3, 6)

(a) Given that mean = np = 4 and variance = npq = 2 Þ p=q=

n

æ 3ö æ 1ö Þ ç ÷ £ç ÷ è 4ø è 10 ø Taking log at the base 3/4, on both sides, we get æ3ö æ1ö n log3/4 ç ÷ ³ log3/4 ç ÷ 4 è ø è 10 ø - log10 10 -1 = Þ n ³ - log 3/4 10 = æ 3ö log 3 10 - log10 4 log10 ç ÷ è 4ø

= P ( X £ 1) = P ( X = 0) + P( X = 1) 6 = e -5 + 5 ´ e -5 = e5 (c) From poission distribution, probability of getting k successes is lk P ( x = k ) = e -l k! Given mean (l) = 2 P ( x ³ 2) = 1 - P ( x = 0) - P ( x = 1)

n

Þ

9 æ 3ö ³ç ÷ 10 è 4 ø

- 535

68.

1 1 1 æ 1 ö æ 1ö P( X = 1) = 8C1 ç ÷ ç ÷ = 8. 8 = 5 = è 2 ø è 2ø 32 2 2 (d) The experiment follows binomial distribution with n = 5, p = 3/6 = 1/2. q = 1 – p = 1/2.; \ Variance = npq = 5/4.

e-m mr r!

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28

Properties of Triangles TOPIC Ć

Properties of Triangle, Solutions of Triangles, Inscribed & Enscribed Circles, Regular Polygons

b+c c+a a+b = = for a DABC with usual 11 12 13

5.

Given

6.

cos A cos B cos C = = , then the ordered triad a b γ [Jan. 11, 2019 (II)] (a, b, g) has a value : (a) (7, 19, 25) (b) (3, 4, 5) (c) (5, 12, 13) (d) (19, 7, 25) With the usual notation, in DABC, if ÐA + ÐB = 120 ,

notation. If

1.

Let A(3, 0, –1), B(2, 10, 6) and C (1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then cos (ÐGOA) (O being the origin) is equal to : 1 (a) 2 15

[April 10, 2019 (I)] (b)

1 1 (d) 6 10 30 The angles A, B and C of a triangle ABC are in A.P. and

(c) 2.

a : b = 1 : 3 . If c = 4 cm, then the area (in sq.cm) of this triangle is : [April 10, 2019 (II)] 2 (a) (b) 4 3 3 4 3 If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is : [April 08, 2019 (II)] (a) 5 : 9 : 13 (b) 4 : 5 : 6 (c) 3 : 4 : 5 (d) 5 : 6 : 7 In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. if

(c) 2 3 3.

4.

triangle, then the circumradius of the triangle is : [Jan. 11, 2019 (I)] 3 y 2 c (c) 3

(b) (d)

c 3 y

3 + 1 and b =

(a) 7 : 1 (c) 9 : 7 7.

8.

(d)

x 2 - c 2 = y , where c is the length of the third side of the

(a)

a=

1 15

9.

10.

3 - 1, then the ratio ÐA : ÐB, is: [Jan. 10, 2019 (II)] (b) 5 : 3 (d) 3 : 1

a = 2 + 3 and ÐC = 60 . Then the ordered b pair (ÐA, ÐB) is equal to : [Online April 10, 2015] (a) (45 , 75 ) (b) (105 , 15 ) (c) (15 , 105 ) (d) (75 , 45 ) ABCD is a trape ium such that AB and CD are parallel and BC ^ CD. If ÐADB = q, BC = p and CD = q, then AB is equal to : [2013]

In a DABC,

(a)

( p 2 + q 2 ) sin q p cos q + q sin q

(c)

p cos q + q sin q

p2 + q2 2

2

(b) (d)

p 2 + q 2 cos q p cos q + q sin q

( p 2 + q 2 ) sin q ( p cos q + q sin q) 2

b+c c+a a+b = = , then cosA is 11 12 13 equal to [2012] (a) 5/7 (b) 1/5 (c) 35/19 (d) 19/35 In a DPQR, If 3 sin P + 4 cos Q = 6 and 4 sin Q + 3 cos P = 1, then the angle R is equal to : [2012]

If in a triangle ABC,

(a)

5p 6

(b)

p 6

(c)

p 4

(d)

3p 4

3

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Properties of Triangles

11.

For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement among the following is [2010]

17.

r 2 = R 3

(c) There is a regular polygon with

3 r = 2 R

18.

19.

[2005]

13.

(b) A. P.

(c) A.P -G..P.

(d) H. P

In a triangle ABC, let ÐC =

p . If r is the inradius and R is 2

the circumradius of the triangle ABC, then 2 (r + R) equals [2005]

14.

(a) b + c

(b) a + b

(c) a + b + c

(d) c + a

The

sides of a

triangle are

1 + sin a cos a for some 0 < a
r2 > r3 (b) a < b < c

(c) a > b and b < c

(d) a < b and b > c

The sides of a triangle are 3x+4y, 4x+3y and 5x+5y where x, y > 0 then the triangle is [2002] (a) right angled

(b) obtuse angled

(c) equilateral

(d) none of these

A ray of light coming from the point (2, 2 3) is incident at an angle 30 on the line x = 1 at the point A. The ray gets reflected on the line x = 1 and meets x-axis at the point B. Then, the line AB passes through the point: [Sep. 06, 2020 (I)] 1 ö æ (a) ç 3, ÷ è 3ø

æ 3ö (b) ç 4, - ÷ 2 ø è

(c) (3, - 3)

(d) (4, - 3)

The angle of elevation of the summit of a mountain from a point on the ground is 45. After climbing up on km towards the summit at an inclination of 30 from the ground, the angle of elevation of the summit is found to be 60. Then the height (in km) of the summit from the ground is: [Sep. 06, 2020 (II)]

(b) are in A.P

p p and ÐABE = , then the area of the D ABC 6 3 [2003] 8 (b) 3 32 (d) 3 3

[2002]

(a) a >b > c

22.

æ A ö 3b ÷= , then the sides è2ø 2

(c) are in G..P (d) are in H.P In a triangle ABC, medians AD and BE are drawn. If AD = 4, ÐDAB =

æp ö

(d) a cot ç 2 n ÷ .

The angle of elevation of the top of a hill from a point on the hori ontal plane passing through the foot of the hill is found to be 45. After walking a distance of 80 meters towards the top, up a slope inclined at an angle of 30 to the hori ontal plane, the angle of elevation of the top of the hill becomes 75. Then the height of the hill (in meters) is ______. [Sep. 06, 2020 (I)]

[2003]

(a) satisfy a + b = c

ø

21.

If in a DABC a cos 2 ç ÷ + c cos 2 ç a, b and c

16.

20.

sin a, cos a and

(d) 60 æCö è2ø

è

TOPIC n Heights & Distances

(b) 90

(c) 120

æp ö

(b) a cot ç n ÷

If in a DABC , the altitudes from the vertices A, B, C on opposite sides are in H.P, then sin A, sin B, sin C are in (a) G. P.

æp ö

a

(c) 2 cot ç 2 n ÷ è ø

r 1 = (d) There is a regular polygon with R 2

12.

æp ö

a

(a) 4 cot ç 2 n ÷ è ø

1 r (a) There is a regular polygon with = R 2 (b) There is a regular polygon with

The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a , is [2003]

(a)

(c)

3 -1 3 +1

1 3 -1

(b)

(d)

3 +1 3 -1

1 3 +1

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23.

24.

Mathematics

Two vertical poles AB = 15 m and CD = 10 m are standing apart on a hori ontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m) above the line AC is : [Sep. 04, 2020 (I)] (a) 20/3 (b) 5 (c) 10/3 (d) 6 The angle of elevation of a cloud C from a point P, 200 m above a still lake is 30 . If the angle of depression of the image of C in the lake from the point P is 60 , then PC (in m) is equal to : [Sep. 04, 2020 (II)] (a) 100

(b) 200 3

(a) 9 (1 + 3 )

30.

31.

(c) 400

25.

(d) 400 3 ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot – 1( 3 2 ) and cosec–1 ( 2 2 ) respectively, then the height of the tower (in metres) is : [April 10, 2019 (I)] 100 (a) 3 3

26.

27.

(c) 20 (d) 25 If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30 and the angle of depression of reflection of the cloud in the lake from P be 60 , then the height of the cloud (in meters) from the surface of the lake is: [Jan. 12, 2019 (II)] (a) 60 (b) 50 (c) 45 (d) 42 Consider a triangular plot ABC with sides AB = 7 m, BC = 5 m and CA = 6 m. A vertical lamp-post at the mid point D of AC subtends an angle 30 at B. The height (in m) of the lamp-post is: [Jan. 10, 2019 (I)] (a)

3 21 2

(c) 2 21 28.

(b)

33.

34.

(c) 10 3 (d) 20 3 Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If ÐBPC = b, then tan b is equal to : [2017] 4 9

(b)

6 7

1 2 (d) 4 9 A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30 . After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60 . Then the time taken (in minutes) by him, from B to reach the pillar, is: [2016] (a) 20 (b) 5 (c) 6 (d) 10 The angle of elevation of the top of a vertical tower from a point A, due east of it is 45 . The angle of elevation of the top of the same tower from a point B, due south of A is 30 . If the distance between A and B is 54 2 m, then the height of the tower (in metres), is : [Online April 10, 2016]

(a) 108

(b) 100 3

(c) 50 2 (d) 100 A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a hori ontal road. If it takes 18 min, for the angle of depression of the car to change from 30 to 45 , then after this, the time taken (in min) by the car to reach the foot of the tower, is. [Online April 16, 2018]

(b) 10 2

(c)

2 21 3

(d) 7 3

hori ontal ground, 3km above it, is observed at an elevation of 60 from a point on the ground. If, after five seconds, its elevation from the same point, is 30 , then the speed (in km/hr) of the aeroplane is [Online April 15, 2018] (a) 1500 (b) 750 (c) 720 (d) 1440 A tower T1 of height 60 m is located exactly opposite to a tower T2 of height 80 m on a straight road. From the top of T1, if the angle of depression of the foot of T2 is twice the angle of elevation of the top of T2, then the width (in m) of the road between the feet of the towers T1 and T2 is [Online April 15, 2018]

(a)

PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45 , 30 and 30 , then the height of the tower (in m) is : [2018] (a) 50

29.

(b) 10 5

9 ( 3 – 1) 2

(c) 18 (1 + 3 ) (d) 18 ( 3 – 1) An aeroplane flying at a constant speed, parallel to the

(a) 20 2 32.

(b)

35.

(b) 36 3

(c) 54 3 (d) 54 (which are the ex-radii) then [2002] If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot of the tower, are 30 , 45 and 60 respectively, then the ratio, AB : BC, is : [2015] (a) 1: 3

(b) 2 : 3

(c)

(d)

3 :1

3: 2

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Properties of Triangles

36.

37.

Let 10 vertical poles standing at equal distances on a straight line, subtend the same angle of elevation at a point O on this line and all the poles are on the same side of O. If the height of the longest pole is ‘h’ and the distance of the foot of the smallest pole from O is ‘a’; then the distance between two consecutive poles, is : [Online April 11, 2015]

AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60 . He moves away from the pole along the line BC to a point D such that CD =7 m. From D the angle of elevation of the point A is 45 . Then the height of the pole is [2008]

(a)

h cos a – a sin a 9 sin a

(b)

h sin a + a cos a 9 sin a

(a) 7 3 1 m 2 3 –1

(c)

h cos a – a sin a 9 cos a

(d)

h sin a – a cos a 9 cos a

(c)

A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45° . It flies off hori ontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30° . Then the speed (in m/s) of the bird is [2014] (a) 20 2 (c) 40

38.

39.

(

(b) 20

)

2 -1

(d) 40

(

3 -1

(

3- 2

40.

) )

41.

The angle of elevation of the top of a vertical tower from a point P on the hori ontal ground was observed to be a. After moving a distance 2 metres from P towards the foot of the tower, the angle of elevation changes to b. Then the height (in metres) of the tower is:[Online April 11, 2014] (a)

(c)

2 sin a sin b sin ( b - a )

2sin ( b - a ) sin a sin b

sin a sin b (b) cos ( b - a )

(d)

cos ( b - a ) sin a sin b

42.

7 3 ( 3 –1)m 2

(b) 7 3 ( 3 + 1)m 2

(d)

7 3 2

1 3 +1

m

A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB (= a) subtends an angle of 60 at the foot of the tower, and the angle of elevation of the top of the tower from A and B is 30 . The height of the tower is [2007] (a) a / 3

(b) a 3

(c) 2a / 3

(d) 2a 3

A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is 60 and when he retires 40 meters away from the tree the angle of elevation becomes 30 . The breadth of the river is [2004] (a) 60 m

(b) 30 m

(c) 40 m

(d) 20 m

The upper

3 th portion of a vertical pole subtends an 4

angle tan -1

3 at a point in the hori ontal plane through its 5

foot and at a distance 40 m from the foot. A possible height of the vertical pole is [2003] (a) 80 m (b) 20 m (c) 40 m (d) 60 m.

EBD_8344

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1.

Mathematics

(b) G is the centroid of DABC.

3.

M

æ 3 + 2 + 1 0 + 10 + 2 -1 + 6 + 1 ö , , ÞGº ç ÷ º (2, 4, 2) 3 3 è 3 ø G

(b) Let the sides of triangle are a > b > c where Given A = 2C Q A + B + C = p and A = 2C Þ B = p – 3C ...(i) Q a, b, c are in A.P. Þ a + c = 2b Þ sin A + sin C = 2 sin B ...(ii) Þ sin A = sin (2C) and sin B = sin 3C From (ii), sin 2C + sin C = 2 sin 3C Þ (2cos C + 1) sin C = 2 sin C ( 3 – 4 sin 2C) Þ 2cos C + 1 = 6 – 8 (1 – cos2C) Þ 8cos2C – 2cos C – 3 = 0 3 1 or cos C = 4 2 Q C is acute angle

Þ cos C =

O

OG =

q

A

4 + 16 + 4 , OA = 9 + 1 , AG = 1 + 16 + 9

24 + 10 - 26 (OG ) 2 + (OA) 2 - ( AG ) 2 = \ cos q = 2 24 10 2(OG )(OA)

and sin A = 2 sin C cos C = 2 ×

8 4 1 = = = 2 8 ´ 3 ´ 2 ´ 5 4 15 15

2.

3 7 Þ sin C = 4 4

Þ cos C =

(c) Given that A, B, C, are in A.P. Þ 2B = A + C

p Now, A + B + C = p Þ B = 3 1 Area = (4 x )sin 60° = 2

7 3 3 7 ´ = 4 4 8

3 7 4 7 7 5 7 ´ = 4 4 16 16 Þ sin A : sin B : sin C :: a : b : c is 6 : 5 : 4 (b) Let two sides of triangle are a and b. sin B =

4.

a+b=x ab = y x2 – c2 = y Þ (a + b)2 – c2 = ab

3x

Þ (a + b – c) (a + b + c) = ab

B

Þ 2(s – c) (2s) = ab Þ 4s(s – c) = ab



A

60°

3x

x

C

16 + x 2 - 3x 2 8x Þ 4x = 16 – 2x2 Þ x2 + 2x – 8 = 0 Þx=2 [Q x can’t be negative] Now cos 60 =

Hence, area = 2 3 sq. cm

Þ

s ( s - c) 1 = ab 4

Þ

cos 2

c 1 = 2 4

Þ cos c = -

1 Þ c = 120 2

\ Area of triangle is, D =

1 3 ab (sin120°) = ab 2 4

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Properties of Triangles

5.

Q

R=

Q

R=

(a) Let

abc 4D

p2 + q2 AB = sinq sin( p - (q + a ))

abc 3 ab

=

c 3

b+c c + a a+b = = = k (Say). 11 12 13

\

b + c = 11k, c + a = 12k, a + b = 13k

\

a + b + c = 18k

\

a = 7k, b = 6k and c = 5k

\

cosA =

36k 2 + 25k 2 – 49k 2 2.30k

2

=

=

æ ççQ cos a = è

1 5

9.

q p +q 2

2

and sina =

\

cos A cosB cosC = = 7 19 25

cos A =

2

49k + 25k – 36k

2

2.35k 2 49k 2 + 36k 2 – 25k 2 2.42k

2

=

19 35

=

5 7

Hence, required ordered triplet is (7, 19, 25). (a) ÐA + ÐB = 120 ...(1) Þ ÐC = 180 – 120 = 60

æ A - Bö a - b æ C ö tan ç cot è 2 ø÷ = a + b èç 2 ø÷ =

2 1 (cot 30°) = ´ 3 =1 2 3 3

ÐA - Ð B p = (Ð is angle) 4 2 Þ ÐA – ÐB = 90 ...(2) From eqn (1) and (2) ÐA = 105 , ÐB = 15 Then, ÐA : ÐB = 7 : 1

Þ

(b)

sin A = 2+ 3 sin B

sin (105°)

= 2 + 3 cos15° = 2 + 3 sin (15° ) sin15°

(a) From Sine Rule

ö ÷ p + q ÷ø p

2

2

(b) In a triangle ABC.

cosA : cosB : cosC = 7 : 19 : 25

and cosC =

8.

( p 2 + q 2 ) sin q q sin q + p cos q

\

and cosB =

7.

p 2 + q 2 sin q sin q cos a + cos q sin a

b+c c+a a+b = = =K 11 12 13 Þ b + c = 11 K, c + a = 12 K, a + b = 13 K On solving these equations, we get a = 7K, b = 6K, c = 5 K Now we know,

2

6.

AB =

Let

10.

b2 + c2 - a 2 36 K 2 + 25K 2 - 49 K 2 1 = = 2bc 2 ( 6 K )( 5K ) 5

(b) Given that 3 sin P + 4cos Q = 6 ...(i) 4 sin Q + 3cos P = 1 ...(ii) Squaring and adding (i) & (ii) we get 9 sin2 P + 16cos2Q + 24 sin P cos Q + 16 sin2Q + 9cos2 P + 24 sin Q cos P = 36 + 1 = 37 Þ 9 (sin 2P + cos2P) + 16 (sin 2 Q + cos2 Q) + 24 (sinP cosQ + cosP sinQ) = 37 Þ 9 + 16 + 24 sin ( P + Q) = 37 [Q sin2q + cos2q = 1 and sin A cos B + cos A sin B = sin (A + B)] 1 Þ sin ( P + Q ) = 2 p 5p Þ P + Q = or 6 6 Þ

R=

p 5p or 6 6

(Q P + Q + R = p)

5p p then 0 < Q, P < 6 6 1 Þ cos Q < 1 and sin P < 2 11 Þ 3 sin P + 4 cos Q < 2 But given that 3 sin P + 4 sin Q = 6 p So, R = 6

If

R=

EBD_8344

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(b) Let O is centre of polygon of n sides and AB is one of the side, then by figure

1 sin 2 a + cos 2 a - 1 - sin a cos a =2 2 sin a cos a \ C = 120°

=

O

p/n R

A

cos

15. r

13.

B

p r = n R

3 r 1 1 , = , R 2 2 2 for n = 3, 4, 6 respectively. (b) Let altitudes from A, B and C be p1, p2 and p3 resp. 1 1 1 \ D = p1a = p2 b = p3b 2 2 2 Given that, p1, p2, p3, are in H.P. 2D 2 D 2 D Þ are in H.P.. , , a b c 1 1 1 Þ , , are in H.P.. a b c Þ a, b, c are in A.P. By sine formula Þ K sin A, K sin B, K sin C are in AP Þ sinA, sinB, sinC are in A.P. (b) We know that for the circle circumscribing a right triangle, hypotenutse is the diameter Q ÐC = 90 c \ 2R = c Þ R = 2 1 ´ a ´b D also r = = 2 s a+b+c 2 ab Þ r= a+b+c 2ab 2ab + ac + bc + c 2 \ 2r + 2 R = +c = a+b+c a+b+c

=

14.

2ab + ac + bc + a 2 + b 2 a+b+c

A ö 3b ÷= 2ø 2

a[cos C + 1] + c[cos A + 1] = 3b (a + c) + (a cos C + c cos B) = 3b We know that, b = a cos C + c cos B a + c + b = 3b or a + c = 2b or a, b, c are in A.P.

Þ

12.

2 æ Cö 2æ (b) Given that, a cos ç ÷ + c cos ç è 2ø è

16.

A

(d)

30 E

90 P

60 B

C

D

We know that median divides each other in ratio 2 :1 2 8 4 AP = AD = ; PD = ; Let PB = x 3 3 3 8/3 8 o tan 60 = or x = x 3 3

1 8 16 = Area of DABD = ´ 4 ´ 2 3 3 3 3 16 32 \ Area of DABC = 2 ´ = 3 3 3 3 [Q Median of a D divides it into two D¢s of equal area.] 17.

a æ pö a æ pö (c) We know that, tan ç ÷ = ; sin ç ÷ = è n ø 2r è n ø 2R Þ

r=

a p a p cot ; R = cosec 2 n 2 n

r+R =

pù aé p cot + cosec ú n nû 2 êë

O p n

(Q c2 = a2 + b2)

( a + b ) 2 + ( a + b )c = = (a + b) a+b+c (c) Let a = sin a, b = cos a and

c = 1 + sin a cos a Clearly a and b < 1 but c > 1 as sina > 0 and cosa > 0 \ c is the greatest side and greatest angle is C. a 2 + b2 - c2 We know that, cos C = 2ab

R

11.

Mathematics

A

r N a

B

p ù p é é ù cos + 1ú 2cos 2 ê ú p aê a a n n 2 = ê = = cot p úú 2 êê p p úú 2ê 2 2p sin 2sin cos n û 2n 2n û ë ë

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Properties of Triangles

18.

In rt. DCDE,

D D D ,r = and r3 = s -a 2 s -b s-c

(a) We know that, r1 = Given that,

D D D > > r1 > r2 > r3 Þ ; s-a s-b s-c Þ s–a c

19.

cos C =

a 2 + b2 - c 2 2ab

h- y h-x

h - 40 h - 40 3

Þ (2 + 3)(h - 40 3) = h - 40 Þ 2h - 80 3 + 3h - 120 = h - 40 Þ h + 3h = 80 + 80 3

Þ ( 3 + 1)h = 80( 3 + 1)

y P(2, 2 3)

P '(0, 2 3)

\ h = 80 m

Q 30

22.

(c) Q ÐDCA = ÐDAC = 30° \ AD = DC = 1 km

A 30 60

x'

x Þ x = 40 3 80

Þ (2 + 3) =

(3 x + 4 y )2 + (4 x + 3 y )3 - (5 x + 5 y ) 2 2(3x + 4 y)(4 x + 3 y)

(c)

cos 30° =

tan 75° =

-2 xy = 0, c = 5x + 5y is the largest side \ C is the largest angle . Now cos C =

sin 30° =

B

O

A

120 x

15

45 y'

C

x=1

15 30

D

60

F

In DDEA, Slope of AB = tan120 = - 3 \ Equation of line AB (i.e. BP') :

y - 2 3 = - 3( x - 0)

In DCDF, sin 30° =

Þ 3x + y = 2 3 \ Point (3, - 3) lies on line AB. 21.

AE 3 = sin 60° Þ AE = km AD 2

(80) Let height (AB) = h m, CD = x m and ED = y m

\ EB = DF =

DF 1 Þ DF = km CD 2

1 km 2

\ Height of mountain = AE + EB

A

æ 3 1 ö æ 3 + 1ö =ç + ÷ =ç ÷ km è 2 2ø è 2 ø (h – y) h E 45 C

x

80 m

30

75

1 3 -1

km

F

h–x y

=

y D h–x B

E B

EBD_8344

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23.

(d)

Mathematics

tan 30° =

B D 15

Þ x = 3h Now, in right DPC'D

P 10

A

Let PE ^ AC and

AE m = EC n

Q DAEP ~ DACD,

m m+n = PE 10

Þ PE =

tan 60° =

...(ii)

h + 400 x

Þ 3x = h + 400 Þ 3h = h + 400

C

E m:n

h h 1 Þ = x 3 x

25.

[From (ii)]

Þ h = 200 So, PC = 400 m [From (i)] (3) Let the height of the vertical tower situated at the mid point of BC be h.

L

10m m+ n

...(i)

Q DCEP ~ DCAB,

n m+n = PE 15

15n m+ n From (i) and (ii), Þ PE =

...(ii) In DALM, cot A =

3 10m = 15n Þ m = n 2 So, PE = 6

AM LM

Þ3 2 =

AM Þ AM = 3 2h h

In DBLM, C

24.

(c)

h P

x

30 60

200 m

D 200 m

x A

B

BM BM Þ 7 = Þ BM = 7h LM h In DABM by Pythagoras theorem AM2 + MB2 = AB2 \ AM2 + MB2 = (100)2 Þ 18h2 + 7h2 = 100 × 100 Þ h2 = 4 × 100 Þ h = 20 (2) Let height of the cloud from the surface of the lake be h meters.

cot B =

26.

(h + 200) m

C'

Here in DPCD, sin 30° =

h Þ PC = 2h PC

...(i)

\

In DPRQ:

www.jeebooks.in M-545

Properties of Triangles

tan 30 = \

MN = PM In DPMQ we have :

h - 25 PR

PR = (h – 25) 3

\

h + 25 PR

29.

h + 25 3

PR =

...(ii)

(200)2 - ( 3 h)2 = h Þ h = 100m (a) Here; Ð DOA = 45 ; Ð DOB = 60 Now, let height of tower = h.

27.

C

O

Then, from eq. (i) and (ii), (h – 25) 3 =

(200)2 - ( 3 h)2 From (2), we get :

MP =

...(i)

and in DPRS : tan 60 =

...(2)

30°

h + 25 3

45°

\ h = 50 m (b) Let the height of the lamp-post is h. D

A

In DDOA, tan (Ð DOA) = DA h Now, in D DOB

Þ tan 45 = By Appolonius Theorem,

tan (Ð DOB) =

2 æ æ AC ö ö 2 ç BD 2 + ç ÷ = BC2 + AB2 ÷ ç è 2 ø ÷ø è

Þ tan 60 =

Þ 2(m2 + 32) = 25 + 49 Þ m = 2 7 tan 30 =

DA OD

Þ h = DA

BD OD

BD h

Þ BD =

\ speed for the distance BA =

h BD

\ required time taken

1 2 21 = Þ h=2 7´ 3 3

28.

B

=

30.

45

200

BD – AD ( 3 – 1) h = 18 18

h ´ 18 18 AD = = = 9 ( 3 + 1) speed ( 3 – 1) h 3 –1

(d) P

3 h.

(d) For D OA, A, OA1 =

3 = 1 km. tan 60°

200

N Q

30

90

h M

Let height of tower MN = h In DQMN we have

tan 30° =

R

For D OB1, B, OB1 =

MN QM

\ QM = 3h = MR Now in DMNP

30

...(1)

3 = 3 km. tan 30°

As, a distance of 3 – 1 = 2 km is covered in 5 seconds. Therefore the speed of the plane is 2 ´ 3600 = 1440 km / hr 5

EBD_8344

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31.

Mathematics

(d) Let the distance between T1 and T2 be x

As tan (a + b) =

D

Þ

q

E

C

tan a + tan b 1 - tan a tan b

tan a + tan b 1 = 1 - tan a tan b 2

2q

60

T1

80

T2

A

AB é ù êQtan(a + b) = AP ú ê ú ê tan( a + b) = 1 [ From(1) ] ú ëê ûú 2

B

From the figure EA = 60 m (T1) and ÐDEC = q and Now in DDEC,

33.

DB = 80 m (T2) ÐBEC = 2q

DC 20 = AB x and in DBEC,

tan q =

...(1) tan 60° =

BC 60 = CE x We know that tan 2q =

tan 2q =

...(2)

æ 20 ö 2ç ÷ è x ø 2 æ 20 ö 1- ç ÷ è x ø

h A

(d)

Since AP = 2AB Þ Let ÐAPC = a \

AB 1 = AP 2

...(i)

60 a

\ Time taken to cover a =

AC 1 AB 1 = = tan a = AP 2 AP 4

(Q C is the mid point) (\ AC = Þ tan a =

30 x

B From (1) and (2) 3a = x + a Þ x = 2a Here, the speed is uniform So, time taken to cover x = 2 (time taken to cover a)

Þ x2 = 1200 Þ x = 20 3

32.

h h Þ 3= a a

Þ h = 3a

2 tan q 1 - (tan q)2

60 Þ = x

(b)

1 + tan b 2 1 = Þ 4 \ tan b = 1 9 2 1 - tan b 4 h tan 30° = x+a 1 h Þ = Þ 3h = x + a x +a 3

1 4

B

34. 1 AB ) 2

10 minutes = 5 minutes 2

(d) Let AP = x BP = y H ÞH=x x H tan 30 = Þ y = 3H y

tan 45 =

x2 + (54 2)2 = y2 H 2 + (54 2)2 = 3H 2

C

A

b a

2 (54 2)2 =2H

P

54 2 =

2H

54 = H

www.jeebooks.in M-547

Properties of Triangles

35.

(c)

Let OC = x and CD = y

P

45

30

h

60

B

A

20 x 20 In right DBOD, tan 30° = x+ y

In right DAOC, tan 45° =

15 15

C

Þ

B1

38.

a

h P

A10

A1 A2 A3

h h h h1 = 2 = 3 = ......... = 10 = tan a . a2 a3 a10 a1

...(1)

(Q a10 = a + 9d)

Þ h = a tan a + 9d tan a

a sin a cos a =d sin a 9 cos a

h-

h cos a - a sin a 9 sin a (b) Given that height of pole AB = 20 m

Þ d=

37.

A

O

45 30 x

B 20

20 C

y

B

x

AB PB sin a h or = cos a x+2 Þ (x + 2) sin a = h cos a

tan a =

h=

...(2)

Þ h = (a + 9d) tan a where d is distance between poles

C

Given : In DABP

Þ

and a1 = a Þ h1 = a tan a

h - a tan a =dÞ 9 tan a

b

a 2m

Since, h 10 = h = a10 tan a

Þ

Hence, speed = 20( 3 - 1) m/s (a) Let AB be the tower of height 'h'.

h

DOA1B1, DOA2B2, DOA3B3, ........, DOA10B10 all are similar triangles. Þ

20 = Þ 20 + y = 20 3 3 20 + y

A

h1 h2 h3

O

20 x+ y

1

B10

B3

=

So, y = 20( 3 - 1) m and time = 1s (Given)

h Þ PA = 2h PA 2h h Þ PC = and in DCPQ, sin60 = 3 PC 2h \ AB : BC = 2h : = 3 :1 3

B2

1 3

Also in DAPQ, sin30 =

(a)

...(ii)

From (i) and (ii), we have x = 20 and

Q

Q PB bisects ÐAPC, therefore AB : BC = PA : PC

36.

...(i)

x sin a + 2sin a cos a

Now, In DABC, tanb =

...(1) AB BC

sinb h h cos b = Þx= ...(2) cosb x sin b Putting the value of x in eq. (2) to eq. (1), we get Þ

h cos b sin a 2 sin a + sin b 1 h= cos a

Þ h=

h cos b.sin a + 2sin a sin b sin b.cos a

Þ h (sinb.cosa – cosb.sina) = 2 sina.sinb Þ h [sin (b – a)] = 2 sin a.sin b Þ h=

2sin a.sin b sin (b - a )

D

Let O be the point on the ground such that ÐAOC = 45

EBD_8344

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39.

Mathematics

h = x

(b) In right, D ABC tan 60 = A

In DOAC, tan 30° =

3

Þ h=

41.

h 1 h = Þ 3 a a

a 3

(d)

T

h

y

60 B

Þ

x=

x

45 C

h

h

h=x+7Þh–

3

(i)] Þ

40.

h=

P

7 3 3 –1

h =1 x+7

=7

tan 30° =

[From

3 +1

´

C

h

30 a 60 30

R

x

....(1)

y x + 40 Þy= x + 40 3

From (1) and (2), 42.

3x =

x + 40

a

O

3

....(2)

Þ x = 20m

­

(c)

3 h 4

¯ ­

b a

h 4

q

¯

40 m

æ 3ö q = a + b , b = tan -1 ç ÷ è 5ø or b = q – a Þ

a

Q

3 +1

7 3 ( 3 + 1)m Þ h= 2 (a) In the DAOB given that ÐAOB = 60 and OA = OB = radius \ ÐOBA = ÐOAB = 60 \ D AOB is a equilateral triangle. Þ OA = OB = AB = a Let the height of tower is h m.

A

60

30 40m

In right DQTR y tan 60 = Þ y = 3x x In right DPTR

....(i)

3

In right, D ABD = tan 45 = Þ

D

7m

tan b =

tan q - tan a 1 + tan q.tan a

h h 3 40 160 = or h h 5 1+ . 40 160 h2 – 200h + 6400 = 0 Þ h = 40 or 160 metre \ possible height = 40 metre

B

www.jeebooks.in Mock Test-1

MT-1

1 x - 4 y + 4 +1 = = is 1 2 2

PART-I (Multiple Choice Questions) 1.

In a box containing 100 bulbs, 10 are faulty. The probability that from a sample of 5 bulbs none are defective. æ 1ö (a) ç ÷ è 10 ø

2.

5

æ 9ö (b) ç ÷ è 10 ø

(c)

9 10

5

(d)

1 5

If 2 sec 2a = tan b + cot b then one of the values of (a + b) = p 2

(a) p

(b)

The value of

å r n Cr -r 1 =

5

3.

5

n

(c)

p 4

6.

(d) None 7.

C

r =1

(a) 5 (n – 3) (b) 5 (n – 2) (c) 5n

(d) 5 (2n – 9)

3

4.

14

C7 +

å 17-i C6 =

8.

i =1

(a) 5.

16

C7

(b)

17

C7

(c)

17

The angle between the two lines

C8

(d)

16

9. C8

x +1 y + 3 - 4 = = & 2 2 -1

æ4ö (a) cos -1 ç ÷ è9ø

æ3ö (b) cos -1 ç ÷ è9ø

æ2ö (c) cos -1 ç ÷ è9ø

æ1ö (d) cos -1 ç ÷ è9ø

The contrapositive of (p Ú q) Þ r is (a) r Þ (p Ú q)

(b) ~ r Þ (p Ú q)

(c) ~ r Þ ~ p Ù ~ q

(d) p Þ (q Ú r)

If (2, 3, 5) are ends of the diameter of a sphere x2 + y2 + 2 – 6x – 12y – 2 + 20 = 0. Then coordinates of the other end are (a) (4, 9, –3) (b) (4, 3, 5) (c) (4, 3, –3) (d) (4, –3, 9) If f(x) = | x – 2| and g(x) = f (f(x)) then for x > 10, g'(x) equal (a) –1 (b) 0 (c) 1 (d) 2x – 4 If a, b, c are in A.P., b, c, d are in G.P. and c, d, e are in H.P. then a, c, e are in (a) A.P. (b) G.P. (c) H.P. (d) None

EBD_8344

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Mock Test-1 15

æ

1 ö ÷÷ x2 ø

10.

The coefficient of x10 in the expansion of çç 3x 2 -

11.

(a) 10! 5! 3 (b) – (d) 7! 8! 3 310 (c) 10! 5! 10! 5! The mean of discrete observations y1, y2, y3,...,yn is given by

è

15!

n

n

(a)

i=1

(b)

n

i =1 n

(c)

S y if i

i=1

Si

n

(d)

ò ( x - b) (a)

dx

i =1 n

17.

2 ( x - a) (b - x ) + c (b) a -b

a -b (x - a ) b - x (d) None of these. 2 The spheres x 2 + y 2 + 2 + x + y + - 1 = 0 and

(c)

13.

18.

x 2 + y2 + 2 + x + y + - 5 = 0 (a) intersect in a plane (b) intersect in five points (c) do not intersect (d) None of these

14. 15.

The domain of the function f (x) = exp( 5x - 3 - 2x 2 ) is (a) [3/2, ¥) (b) [1, 3/2] (c) (–¥, 1] (d) (1, 3/2) The value of the determinant 1 a a2 cos (n –1) x cosnx cos (n + 1) x sin (n –1) x

sinnx

sin (n + 1) x

(b) cos x = 0

(c) a = 0

(d) cos x =

If AB = 0, then for the matrices 2 é cos q sin qù and A = ê cos q ú sin 2 q ûú ëêcos q sin q

p 2 (b) an odd multiple of p p (c) an even multiple of 2 (d) 0 If f'(x) = (x – 1) (x – 2) (x – 3) then f(x) in monotonically increasing in (a) x < 1 (b) x > 3, x < 1 (c) x > 3, 1 < x < 2 (d) x < 1, 2 < x < 3 An inverted cone is 10 cm in diameter and 10 cm deep. Water is poured into it at the rate of 4cm3/min. When the depth of water level is 6 cm, then it is rising at the rate

19.

(a)

9 cm 3 / min . 4p

(b)

1 cm 3 / min . 4p

(c)

1 cm 3 / min . 9p

(d)

4 cm 3 / min . 9p

The equation of tangent to 4x2 – 9y2 = 36 which are perpendicular to straight line 5x + 2y – 10 = 0 are æ

is ero, if

1 + a2 2a

(a) an odd number of

is

x-a +c b-x

(a) sin x = 0

é cos 2 f cos f sin fù B= ê ú , q – f is sin 2 f ûú ëêcos f sin f

S y if i i =1

(x - a)(b - x )

2 a-b

16.

S fi

i=1

12.

8

n

n

S yi

S yi

15!

15! 35

15!

10

is

(a) 5 (y–3) = 2çç x è

117 ö÷ 2 ÷ø

(b) 2y – 5x + 10– 2 18 = 0

www.jeebooks.in Mock Test-1

MT-3 22.

(c) 2y – 5x – 10– 2 18 = 0 (d) None of these

ò

20.

23.

ö æ1 e 2 x sec 2 ç e 2 x ÷ dx is equal to : log p 2 ø è3 log p

24.

1

(a)

(b)

3

Three persons A, B, C throw a die in succession. The one getting 'six' wins. If A starts then the probability of B winning is The eccentricity of the ellipse represented by 25x2 + 16y2 – 150x – 175 = 0 is If

p4 1 1 , then the value of + 4 + 4 + ..... + ¥ = 90 1 2 3 1

4

3 y 1 1 + 4 + 4 + ......¥ is π , then k is: 1 3 5 k

1

(c)

1

3 3 2

(d)

4

2 3

25.

The area enclosed by the curve y = x5, the x-axis and the ordinates x = –1, x = 1 in sq. units is

The probability of getting the sum more than 7 when a pair of dice is tossed is

RESPONSE SHEET 1.

a

b c d

2.

a

b c d

3.

a

b c d

4.

a

b c d

5.

a

b c d

6.

a

b c d

7.

a

b c d

8.

a

b c d

9.

a

b c d

10.

a

b c d

11.

a

b c d

12.

a

b c d

13.

a

b c d

14.

a

b c d

15.

a

b c d

16.

a

b c d

17.

a

b c d

18.

a

b c d

19.

a

b c d

20.

a

b c d

21.

22.

23.

24.

25.

EBD_8344

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Solutions Mock Test -1

HINTS & SOLUTIONS MOCK TEST-1

1.

(b)

1 10

p=

q=

9 \ Probability that none are defective = æç ö÷ è 10 ø

2.

(c)

8.

9 10

(c) For x > 10, f (x) = x – 2. Therefore, g(x) = x – 2 – 2 = x – 4 \ g¢(x) = 1.

5

9.

(b) a, b, c in A.P. Þ a + c = 2b;

\

p Þ sin2b = cos2a = sin (90-2a) Þ a + b = 4 n

3.

(b)

n

=

Cr

Cr -1

r. n r - 1. n - r + 1 n - r + 1 = . = r. n - r n n-r

2ce c+e

a + c 2ce ´ = c2 Þ (a + c)e = (c + e)c Þ c2 = ae. 2 c+e

Therefore, in G.P.

10.

(b)

n - r +1 = n - r +1 n -r

æ 2 1 çç 3x x2 è

15

ö ÷÷ ø

æ 1 Tr +1 =15 C r (3x 2 )15-r çç è x2 15 15-r r 30- 2 r - 2 r (-1) x = Cr 3

5

\

c, d, e in H.P. Þ d =

Þ bd = c2;

1 tan 2 b + 1 2 = = sin b cos b b tan cos 2a

b, c, d in G.P..

å = n + (n - 1) + (n - 2) + (n - 3) + (n - 4)

ö ÷÷ ø

r

r =1

= 5n - 10 = 5(n - 2)

Therefore, 30 – 4r = 10 Þ r = 5.

3

4.

(b)

14

=

5.

C7 +

15

å 17-i C 6 =14 C 7 +14 C 6 +15 C 6 +16 C 6 i =1 15

C7 +

C 6 + 16 C 6 =16 C 7 +16 C 6 =17 C 7

(a) a1 = 2, b1 = 2,c1 = –1 and a2 = 1, b2 = 2, c2 = 2 cos q =

a12 + b12 + c12 a 22 + b2 2 + c22

4 = =± . 9 4 + 4 +1 1+ 4 + 4 (c) Contrapositive of p Þ q is ~ q Þ ~ p \ contrapositive of (p Ú q) Þ r is ~ r Þ ~ (p Ú q) i.e. ~ r Þ (~ p Ù ~ q)

7.

11.

(a)

12.

(a) I =

a1a 2 + b1b 2 + c1c 2

2+ 4- 2

6.

Therefore, T6 = -15 C 5 310 =

(a) Let the co-ordiante of other ends are (x, y, ). x+2 = 3Þ x = 4 2

dx ( x - a )(b - x )

Put x = a sin 2 q + b cos 2 q [see the standard substitutions] dx = 2(a - b) sin q cos q dq Also, ( x - a) = (b - a ) cos 2 q ( x - b) = ( a - b) sin 2 q

\I =

The centre of sphere is C(3, 6, 1) Therefore,

ò ( x - b)

-15! 10 3 . 10! 5!

=

2( a - b) sin q cos q dq

ò (a - b) sin 2 q (b - a) sin q cos q

2 dq 2 = cos ec 2 q dq ò 2 b - a sin q b - a ò

y+3 +5 = 6 Þ y = 9 and = 1 Þ = -3 2 2

www.jeebooks.in Solutions Mock Test -1

=

MT-5 By expanding (1 + a2 – 2a cos x) [cos nx sin (n + 1) x – sin nx cos (n + 1) x]= 0 Now, (1 + a2 – 2a cos x) sin (n + 1 – n) x = 0

2 2 (- cot q) + C = cot q + C b-a a -b

Now , x = a sin 2 q + b cos 2 q 2

Þ (1 + a 2 - 2a cos x)sin x = 0

2

Þ x cos ec q = a + b cot q

sin x = 0 or cos x =

Þ x (1 + cot 2 q) = a + b cot 2 q

2 x-a ; \I= a-b b-x

\cot q = 13.

14.

(b) We have, f (x) = exp i.e. , f (x) = e

5x - 3 - 2 x2

Þ cos x > 1 It is not possible.

16.

)

écos 2 q cos 2 f + cos q cos f sin q sin f =ê 2 2 ëê cos q sin q cos f + sin q cos f sin f

5 x - 3 - 2 x 2 should be +ve.

cos 2 q cos f sin f + cos q sin q sin 2 fù ú cos q cos f sin q sin f + sin 2 q sin 2 f ûú

5x - 3 - 2 x2 ³ 0

Þ 2 x2 - 5x + 3 £ 0

(By taking –ve sign common)

écos q cos f cos q sin fù = cos( q - f) ê ú ë sin q cos f sin q sin f û

Þ 2 x( x - 1) - 3( x - 1) £ 0 Þ (2 x - 3)( x - 1) £ 0

Since, AB = 0, \ cos( q - f) = 0

Þ 2x - 3 £ 0

or

3 Þ x£ 2

or

\ 1£ x £

15.

x -1³ 0

é 3ù i.e., x Î ê1, ú ë 2û

3 2

1

cos(n - 1) x cos nx cos ( n + 1) x =0 sin (n - 1) x sin nx sin ( n + 1) x

2

1 + a - 2a cos x 0 0

a

17. 18.

(d) Let y be the level of water at time t and x the radius of the surface and V, the volume of water. We know that the volume of cone =

a2

a

p 2 (c) When x > 3, f '(x) > 0; when 2 < x < 3, f '(x) < 0; when 1 < x < 2, f '(x) > 0; when x < 1, f '(x) < 0.

\ q - f is an odd multiple of

x ³1

3 Hence, domain of the given function is [1, ]. 2 (a) Given determinant

Þ

\ sin x = 0 (a) We have, é cos 2 q cos q sin qù é cos 2 f cos f sin fù AB = ê úê ú 2 sin q úû êëcos f sin f sin 2 f úû êëcos q sin q

5 x - 3- 2 x 2

For Domain of f (x), i.e.,

(

æ 1 + a2 ö

As a ¹ 1 \ç ÷ >1 è 2a ø

x -a +C b- x

(c) As the given spheres both have same cenre and different radii therefore they are concentric and they do not have any point in common. Hence they do not intersect.

1 + a2 2a

a

2

cos nx cos ( n + 1) x = 0 sin nx sin ( n + 1) x

1 p (radius)2 × height 3

C

D 5cm

1 \ V = px 2 y. Let ÐBAD = a 3 R

M

BD 5 1 Þ tan a = = = . AD 10 2

y

Again, from right angled DAMR, we have

By applying C1 ® C1 + C3 – 2 cos x C2

A

x

B

10 cm

MT-6

Solutions Mock Test -1

tan a =

21.

MR x 1 x y = ; Þ = ; \x = . AR y 2 y 2

(0.42) Sum of 7 can be obtained when (2,6), (3,5) (3, 6), (4, 4), (4,5), (4, 6)(5, 3)(5, 4) (5, 5)(5, 6)(6, 2)(6, 3)(6, 4)(6,5)(6, 6) 15 5 = 36 12 = 0.4167 = 0.42

\ Probability of sum > 7 =

2

1 p 1 æyö \V = p.x 2 y = p.ç ÷ .y = y 2 3 3 è2ø 12

......(1).

22.

By question, the rate of change of volume

(0.33) P(E E) + P(E E E E E)

dV . = 4 cub.cm./min. dt We have to find out the rate of increase of water-level

4

=

dy . dt Differentiating (1) with respect to t, we get

i.e.

3 ù 30 5 é æ 5ö ê1 + ç ÷ ......ú = = 0.329 = 0.33 è ø 36 ê 6 úû 91 ë (0.6)The ellipse

=

23.

dV p dy p dy dy 16 = .3y 2 . ; \ 4 = y 2 . ; \ = . dt 12 dt 4 dt dt py 2

Þ

dy 16 4 = = cub.cm. / min . dt p6 2 9p (d) Slope of the equations 4x2 – 9y2 = 36

When y = 6 cm,

19.

8x - 18y

24.

4x dy dy 4x or m1 = =0Þ = 9y dx dx 9 y

14

5 2

4 x -5 10x . ´ = -1 Þ y = 9y 2 9

(a)

I=

ò

gives

\

+

1 54

+ ..... + ¥ +

1 æ1 1 1 1 ö ç + + + + ....¥÷ø 24 è 14 24 34 44

p4 90

p 4 1 æ p4 ö 1 1 1 + + + + + ¥ = .... – ç ÷ 90 16 è 90 ø 14 34 54 7 4 1

=

25.

When x = log p , t = e 2 log \I =

3 2

1 34

ö æ1 e 2 x sec 2 ç e 2 x ÷ dx log p 2 ø è3

Put e 2 x = t Þ 2 e 2 x dx = dt

=

p4 1 1 1 + 4 + 4 + 4 + ....¥ = 90 1 2 3 4 4

log p

When x = log p 2 , t = e

ò

p1 æ1 ö sec 2 ç t ÷ p2 è3 ø 2

p

16 3 =1- e2 Þ e = . 25 5

1 1 1 1 p4 p4 + + + + + ¥ + ´ = .... 16 90 90 14 34 54 74

y=

2 log p 2

\

1

10 x in 4 x 2 - 9 y 2 = 36 9 imaginary roots resulting in no tangents.

20.

+

( x - 3) 2 y 2 + =1 16 25

1

=

Therefore, for the perpendicularity, m1m2 = –1

Putting

(96)

1

Slope of the straight line, 5x + 2y – 10 = 0 is m 2 = -

Now,

8

5 1 æ 5ö 1 æ 5ö 1 ´ + ç ÷ ´ + ç ÷ ...¥ 6 6 è 6ø 6 è 6ø 6

=

=e

log p 2

=p p

t ù 1 1é dt = 2 . 1 êë tan 3 .úû p 2 3

1 ù p pù 3 é é ú= ê tan 3 - tan 6 ú = 2 ê 3 3û ë û ë

p = 2

15 æ p4 ö p 4 . ç ÷= 16 è 90 ø 96

(0.33) Required area 1

ò

1

1

ò

ò

= | y | dx = | x 5 | dx = 2 | x 5 | dx -1 1

-1

0 1

éx6 ù 2 1 = 2 x 5 dx = 2ê ú = = 6 6 3 ê ú ë û0 0

ò

3

EBD_8344

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www.jeebooks.in Mock Test-2

MT-7

2 PART-I (Multiple Choice Questions) 1.

2.

3.

Equation of straight line ax + by + c = 0 where 3a+4b+ c = 0, which is at maximum distance from (1, –2), is (a) 3x + y – 17 = 0 (b) 4x + 3y – 24 = 0 (c) 3x + 4y – 25 = 0 (d) x + 3y – 15 = 0 ì 10 - x 2 if - 3 < x < 3 ï Given f (x) = í ï2 - e x -3 if x ³ 3 î The graph of f (x) is – (a) continuous and differentiable at x = 3 (b) continuous but not differentiable at x = 3 (c) differentiable but not continuous at x = 3 (d) neither differentiable nor continuous at x = 3 The solution of the equation 2 = | | + 2i, where is a complex number, is –

(a) =

(b) =

6. 7.

8.

9.

3 +i 3

3 ±i (d) None of these 3 The equation (5x – 1)2 + (5y – 2)2 = (l 2 – 4l + 4) (3x + 4y – 1) 2 represents an ellipse if l Î (a) (0, 1] (b) (–1, 2) (c) (2, 3) (d) (–1, 0) (c) =

4.

3 -i 3

5.

10.

The straight line y = m (x – a) meets the parabola y2 = 4ax in two distinct points for – (a) all m Î R (b) all m Î [–1, 1] (c) all m Î R – {0} (d) None of these p Ú (p Ù q) is equivalent to – (a) q (b) p (c) ~p (d) ~q 3 æ ö x cos x sin x Find ò esin x ç ÷ dx cos 2 x è ø sin x sin x –e sec x + C (a) x e (b) x ecos x – esin x sec x + C (c) x2 esin x + esin x sec x + C (d) 2x esin x – esin x tan x + C The function f : [2, ¥) ® (0, ¥) defined by f (x) = x2 – 4x + a, then the set of values of ‘a’ for which f(x) becomes onto is (a) (4,¥) (b) [4, ¥) (c) {4} (d) f If a and b are the real roots of the equation x2 – (k – 2) x + (k2 + 3k + 5) = 0 (k Î R). Find the maximum and minimum values of (a2 + b2). (a) 18, 50/9 (b) 18, 25/9 (c) 27, 50/9 (d) None of these The sum of the coefficient of all the terms in the expansion of (2x – y + )20 in which y do not appear at all while x appears in even powers and appears in odd powers is – 2 20 - 1 320 - 1 (a) 0 (b) (c) 219 (d) 2 2

EBD_8344

www.jeebooks.in MT-8 11.

12.

13.

14.

15.

All the five digit numbers in which each successive digit exceeds is predecessor are arranged in the increasing order. The (105)th number does not contain the digit (a) 1 (b) 2 (c) 6 (d) All of these Three people each flip two fair coins. The probability that exactly two of the people flipped one head and one tail, is– (a) 1/2 (b) 3/8 (c) 5/8 (d) 3/4 r r r If a, b, c are non-coplanar unit vector such that 1 r r r r r a ´ (b ´ c) = (b + c) then the angle between the 2 r r vectors a, b is (a) 3p/4 (b) p/4 (c) p/8 (d) p/2 The greatest and the least value of |z1 + z2| if z1 = 24 + 7i and |z2| = 6 respectively are (a) 25, 19 (b) 19, 25 (c) –19, –25 (d) –25, –19 Let P = (–1, 0), Q = (0, 0) and R = (3, 3 3 ) be three point. The equation of the bisector of the angle PQR is (a) (c)

16.

Mock Test-2

3 x+ y = 0 2

1 p , 2 4 (c) 1, 1

(d) x +

ò0

(a) p

(b)

p 2 (d) None of these

Let f : R ® R and fn (x) = f (fn–1 (x))

(a) 5

0

cos -1 t dt is

p 4

(c)

(d) 1

2ax + sin By + cos B = 0 ,

If a is real and

21.

Box contains 2 one rupee, 2 five rupee, 2 ten rupee and 2 twenty rupee coin. Two coins are drawn at random simultaneously. The probability that their sum is Rs. 20 or more, is

22.

The value of the definite integral,

q2

where q2 = 23.

24.

25.

dq

ò 1 + tan q =

q1

(b) 15

sin B = 0 and a point P is taken on the longest side of the

p 2

cos 2 x

n ³ 2, n Î N,

(c) 10 (d) Both (a) and (b) A triangle ABC satisfies the relation 2 sec 4C + sin2 2A +

sin -1 t dt + ò

x + cos By + sin B = 0 , – x + sin By – cos B = 0, then the set of all values of a for which the system of linear equations has a non-trivial solution, is – (a) [1,2] (b) [–1, 1] (c) [1,¥] (d) [2–1/2, 21/2]

(b) 1,

the roots of equation f3(x) f2(x) f (x) – 25f2(x) f (x) + 175 f (x) = 375. Which also satisfy equation f (x) = x will be

18.

20.

sin 2 x

The value of

If ò (a, b) is : (a)

17.

19.

(b) x + 3 y = 0

3 y = 0. 2 2 1 + sin xdx = -4 cos (ax + b) + C then the value of

3x + y = 0

triangle such that it divides the side in the ratio 1 : 3. Let Q and R be the circumcentre and orthocentre of D ABC. If PQ : QR : RP = 1 : a : b, then the value of a2 + b2. (a) 9 (b) 8 (c) 6 (d) 7

501p K

p 1003p . The value of K and q1 = 2008 2008

equals The expansion of (1 + x)n has 3 consecutive terms with coefficients in the ratio 1 : 2 : 3 and can be written in the form nCk : nCk+1 : nCk+2. The sum of all possible values of (n + k) is – The mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new standard deviation of the resulting observations. lim

x ®0

tan x tan x - sin x sin x x3. x

equals

www.jeebooks.in Mock Test-2

MT-9

RESPONSE SHEET 1.

a

b c d

2.

a

b c d

3.

a

b c d

4.

a

b c d

5.

a

b c d

6.

a

b c d

7.

a

b c d

8.

a

b c d

9.

a

b c d

10.

a

b c d

11.

a

b c d

12.

a

b c d

13.

a

b c d

14.

a

b c d

15.

a

b c d

16.

a

b c d

17.

a

b c d

18.

a

b c d

19.

a

b c d

20.

a

b c d

21.

22.

23.

24.

25.

EBD_8344

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Solutions Mock Test -2

HINTS & SOLUTIONS MOCK TEST-2 1.

(d) It passes through a fixed point (3, 4) Slope of line oining (3, 4) and (1, –2) is –6/–2 = 3 \ Slope of required line = –1/3 1 Equation is y – 4 = - (x - 3) 3

(b)

x=

1

i.e.,

(Q x ³ 0) \ =

3 2

4.

i.e., x = ±

8.

9.

\ Emin = 19 -

2

2

10.

6.

m ¹ 0; 16a2 + 16a2m2 > 0 which is true \ m Î R – {0} (b) p Ú (p Ù q) is equivalent to p.

7.

(a)

òe

æ x cos3 x - sin x ö çç ÷÷ dx cos2 x è ø

m.

E = (a + b) 2 - 2ab

121 171 - 121 50 = = 9 9 9

(a)

20! 20! p (2x)p (- y)q ( )r = 2 (-1)q x p yq r p!q!r! p!q!r! p + q + r = 20, q = 0 p + r = 20 (p is even and r is odd). even + odd = even (never possible) Coefficient of such power never occur \ coefficient is ero

æ y + am ö 2 2 y2 = 4a ç è m ÷ø i.e., my – 4ay – 4a m = 0

(c)

}

Emax occurs when k = – 4 Emax = 19 – 1 = 18

is an ellipse. If 0 < | l – 2 | < 1 i.e., l Î (1, 2) È (2, 3)

sin x

}

E = (k – 2)2 – 2 (k2 + 3k + 5) = –k2 – 10k – 6 E = – (k2 + 10k + 6) = – [(k + 5)2 – 19] = 19 – (k + 5)2 \ Emin occurs when k = – 4/3

3 3 +i = +i 3 3

5.

d (sec x)

4ö æ çè k + ÷ø (k + 4) £ 0 3

1

1ö æ 2ö 3x + 4y - 1 æ çè x - ÷ø + çè y - ÷ø = | l - 2 | 5 5 5

tan x sec x dx

(k – 2)2 – 4 (k2+ 3k + 5) ³ 0 (k2 + 4 – 4k) – 4k2 – 12k – 20 ³ 0 – 3k2 – 16k –16 ³ 0 ; 3k2 + 16k + 16 £ 0

1

2

2

sin x

Now E = a 2 + b2 ;

1ö 2ö æ æ æ 3x + 4y - 1ö 2 (c) çè x - ÷ø + çè y - ÷ø = (l - 4l + 4) çè ÷ø 5 5 5

i.e.,

)–òe

sin x

= x esin x – esin x sec x + C (d) f (x) = x2 – 4x + a always attains its minimum value. So its range must be closed. So, a = {f} (a) For real roots D ³ 0

2x = x 2 + y2 and 2y = 2 i.e., y = 1 3x2 = 1

sin x

òe

{

2(x + iy) = x 2 + y2 + 2i

4x2 = x2 + 1

ò x d (e

x cos x dx –

– esin x sec x - ò esin x sec x cos x dx

10 - (3 - h)2 - 1 1 + (6h - h 2 ) - 1 = - lim h ®0 h ®0 -h -h -6 6h - h 2 = = -3 = lim 2 h ®0 - h( 1 + 6h - h 2 + 1) Hence, f ' (3+) ¹ f ' (3–) (b)

=

sin x

{

f (3 + h) - f (3) f ¢(3+ ) = lim h ®0 h æ e h - 1ö (2 - e h ) - 1 = lim = - lim ç ÷ = -1 h®0 h®0 è h ø h f (3 - h) - f (3) f ¢(3- ) = lim h®0 -h

= lim

3.

òe

= x esin x - ò esin x dx

x + 3y – 15 = 0 2.

=

11.

(a) Starting with 1

23456789

= 8C4 = 70 Starting with 2 = 7C4 = 35 Total = 105 (105)th number 26789

3456789

www.jeebooks.in Solutions Mock Test -2 12.

MT-11

(b) n = 3, P (success) = P (HT or TH) = 1/2 1 Þ p = q = and r = 2 2

Þ

1 3 æ 1ö P (r = 2) = 3C2 ç ÷ . = è 2ø 2 8 1 1    - (a.b)c  = (a) (a.c)b b + c 2 2

15.

p and angle between 4

3p a and b = 4 (a) |z1 + z2| £ |z1| + |z2| = |24 + 7i| + 6 = 25 + 6 = 31 Also, |z1 + z2| = |z1 – (–z2)| ³ ||z1| – |z2|| Þ |z1 + z2| ³ | 25 – 6| = 19 Hence the least value of | 1 + 2| is 19 and the greatest value is 25. (c) The coordinates of points P, Q, R are (–1, 0), (0, 0),

(3,3 3) respectively.. Slope of QR =

Þ q=

æx

ò 2 sin çè 2 + 4 ÷ø dx = –4 cos(ax + b) + C

Þ a= 17.

(d) f2 (x) = f (f (x)) = f (x) = x Þ x3 – 25x2 + 175x – 375 = 0 (x – 5) (x2 – 20x + 75) = 0 (x – 5)2 (x – 15 ) = 0 Þ x = 5, 15

3 Y

1 p , b= 2 4

f3 (x) = f (f2 (x)) = f (x) = x

18.

Þ tan q = 3

R (3, 3 3)

(a) 2 sec 4C + sin22A + sin B = 0 A = 45 , B = 90 and C = 45 Let AQ = a, then BP =

M

p 3

PQ =

Þ ÐRQX =

p 3

\ ÐRQP = p -

X'

æx pö Þ -4 cos ç + ÷ = -4 cos(ax + b) + C è2 4ø

1 and a .b = 2 2

Þ angle between a and c =

14.

Þ

1

= \ a.c

p

x

= –4 cos(ax + b) + C

2

13.

p

æ

ò 2 çè cos 4 sin 2 + sin 4 cos 2 ÷ø dx

2p / 3 P (–1, 0)

p 2p = 3 3

p /3 Q (0, 0)

a , 2

a and QR = a 2

X C

Y'

P Q

Let QM bisects the ÐPQR , 2p \ Slope of the line QM = tan =– 3

3

\ Equation of line QM is (y – 0) = – 3 (x – 0) Þ y= –

16.

(a) Given Þ

Þ

ò ò

ò

3x Þ

A

R,B

\ PR = a 2 +

a2 5a = 4 2

3 x + y= 0

2 1 + sin x dx = -4cos(ax + b) + C

x xö æ 2 ç sin + cos ÷ dx = -4cos(ax + b) + C 2 2ø è x 1 xö æ 1 2. 2ç sin + cos ÷ dx 2 2ø 2 è 2

= –4 cos(ax + b) + C

\ 1: a : b =

a 5a :a : = 1: 2 : 5 2 2

\ a = 2 and b = 5 19.

(c) Let I1 = ò

sin 2 x 0

\ a 2 + b2 = 9

sin -1 t dt

Put t = sin 2 u Þ dt = 2 sin u cos udu Þ dt = sin 2udu

\I1 =

x

ò0 u sin 2u du

EBD_8344

www.jeebooks.in MT-12

Solutions Mock Test -2 cos 2 x

cos -1 t dt

ò0

Let I2 =

n

23.

Put t = cos 2 v Þ dt = -2 cos v sin vdv Þ dt = - sin 2 v dv

ò

\I 2 = =-

x v ( - sin 2 v )dv p 2

x

ò p u sin 2 udu

=-

ò

(18)

n

Ck

=

C k +1

k +1 1 = n-k 2 2k + 2 = n – k n – 3k = 2

or

x v sin 2 vdv p 2

[change of variable]

2 x

ò0

\ I = I 1 + I2 = p 2

u sin 2 udu - ò

x

ò

x u sin p 2

Similarly,

2 udu

òp

òp

2

2

p 2

p 4

ò

= u sin 2udu = 0

20.

Ck +1

n

Ck + 2

2 3

3k + 6 = 2n – 2k – 2 [Integrate by parts]

2n – 5k = 8

............. (2)

From (1) and (2)

2a

sin B

cos B

1

cos B

sin B

-1

sin B - cos B

n = 14 and k = 4 \ n + k = 18

=0

24.

(12) Let the observations be x1, x2, x3, x4, x5 and x6, so

Þ a 2 [- cos 2 B - sin 2 B] - sin B [ - cos B + sin B]

6

+ cos B [sin B + cos B] = 0

their mean x =

Þ -a 2 + sin 2B + cos 2B = 0 Þ a Î[-1,1] 21.

=

k+2 2 = n - k -1 3

(b) For non-trivial solution, D=

n

.......... (1)

n! (k + 2)!(n - k - 2)! 2 = . (k + 1)! (n - k - 1)! n! 3

x

= u sin 2udu + u sin 2udu - u sin 2udu 0

1 n! (k + 1)!(n - k - 1)! 1 Þ = 2 k! (n - k)! n! 2

å xi i =1

6

6

\ P (1) = 1 – P (Total value is < 20)

= 1-

22.

Now, their mean = x =

C2 - C2 2

8

= 1-

C2

14 1 1 = 1- = 28 2 2

q2

q1

q2

ò

=

3 ´ 48 = 24 6

i =1

6

6

=

q2

tan q dq

ò 1 + tan q

q1

dq 1 + tan q

dq = q2 - q1 =

q1

1002p 501p ÞI= 2008 2008

32 å (x i - 8) 2 i =1

6

9 ´ Variance of old observations = 9 × 42 = 144 1 Thus, standard deviation of new observations = Variance = 144 = 12

=

dq = ò æp ö q1 1 + tan ç - q÷ è2 ø

\ 2I =

i =1

å (3x i - 24) 2

=

q2

and also I = ò

å 3x i

6 Variance of new observations 6

p (2008) q1 + q2 = 2

\ I=

i =1

On multiplying each observation by 3, we get the new observations as 3x1, 3x2, 3x3, 3x4, 3x5 and 3x6.

(0.5)Let A be the event such that sum is Rs. 20 or more

6

6

= 8 Þ å xi = 48

25.

(0.75) lim

(tan x)3/ 2 [1 - (cos x)3/ 2 ]

. x 3/ 2 .x 2 1 - cos3 x 1 . = 1 × xlim ®0 x2 1 + (cos x) 3/2 x ®0

=

1 1 3 . (1 + cos x + cos 2 x) = 2 2 4

Hence, K = 2008.