SAT II Mathmatics level 2: Designed to get a perfect score on the exam.

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SAT II Mathmatics level 2: Designed to get a perfect score on the exam.

Citation preview

       

Dr. John Chung’s

 

SAT II

                       

Mathematics Level 2

           

Good Luck!

         

Dr. John Chung's SAT II Math Level 2

1

Dear Beloved Students,  With this SAT II Subject Test Math Level 2 Third Edition, I like to thank all students who sent me  email to encourage me to revise my books. As I said, while creating this series of math tests has  brought great pleasure to my career, my only wish is that these books will help the many students  who are preparing for college entrance. I have had the honor and the pleasure of working with  numerous students and realized the need for prep books that can simply explain the fundamentals  of mathematics.  Most importantly, the questions in these books focus on building a solid  understanding of basic mathematical concepts. Without understanding these solid foundations, it  will be difficult to score well on the exams. These book emphasize that any difficult math question  can be completely solved with a solid understanding of basic concepts.  As the old proverb says, “Where there is a will, there is a way.” I still remember vividly on fifth‐ grader who was last in his class who eventually ended up at Harvard University seven years later. I  cannot stress enough how such perseverance of the endless quest to master mathematical  concepts and problems will yield fruitful results.  You may sometimes find that the explanations in these books might not be sufficient. In such a  case, you can email me at [email protected]  and I will do my best to provide a more detailed  explanation. Additionally, as you work on these books, please notify me if you encounter any  grammatical or typographical errors so that I can provide an update version.  It is my great wish that all students who work on these books can reach their ultimate goals and  enter the college of their dreams.  Thank you.  Sincerely,      Dr. John Chung   

2

 

Contents 61 Tips

         

TIP 01  Identical Equation 

Page 06 

Tip 02  Remainder Theorem 

Page 07 

Tip 03  Factor Theorem 

Page 08 

Tip 04  Sum and Product of the roots 

Page 09 

Tip 05  Complex Number 

Page 11 

Tip 06  Conjugate Roots 

Page 12 

Tip 07  Linear Function 

Page 13 

Tip 08  Distance from a Point to a line 

Page 14 

Tip 09  Distance from a Point to a Plane 

Page 15 

Tip 10  Quadratic Function  

Page 16 

Tip 11  Discriminant 

Page 17 

Tip 12  Circle 

Page 18 

Tip 13  Ellipse 

Page 19 

Tip 14  Parabola 

Page 21 

Tip 15  Hyperbola 

Page 23 

Tip 16  Function 

Page 25 

Tip 17  Domain and Range of a Composite Function 

Page 26 

Tip 18  Piecewise‐Defined Function 

Page 28 

Tip 19  Odd and Even Function 

Page 29 

Tip 20  Combinations of functions 

Page 30 

Tip 21  Periodic Functions 

Page 31 

Tip 22  Inverse Functions 

Page 32 

Tip 23  The Existence of an Inverse Function  

Page 33 

Tip 24  Leading Coefficient Test (Behavior of Graph) 

Page 34 

Dr. John Chung's SAT II Math Level 2

3

4

Tip 25  Arithmetic Sequence 

Page 35 

Tip 26  Geometric Sequence 

Page 36 

Tip 27  Exponential Functions 

Page 37 

Tip 28  Logarithmic Functions 

Page 39 

Tip 29  Basic Trigonometric Identities 

Page 41 

Tip 30  Circle(Trigonometry) 

Page 43 

TIP 31  Reference Angles and Cofunctions 

Page 44 

Tip 32  Trigonometric Graphs 

Page 45 

Tip 33  Inverse Trigonometric Functions 

Page 46 

Tip 34  Sum and difference of angles 

Page 48 

Tip 35  Double Angle Formula 

Page 49 

Tip 36  Half Angle Formula 

Page 50 

Tip 37  Trigonometric Equation 

Page 51 

Tip 38  The Law of Sines 

Page 58 

Tip 39  The Law of Cosines 

Page 54 

Tip 40  Permutation 

Page 55 

Tip 41  Combination 

Page 56 

Tip 42  Dividing Group 

Page 57 

Tip 43  Binomial Expansion Theorem 

Page 58 

Tip 44  Sum of Coefficients of a Binomial Expansion 

Page 59 

Tip 45  Binomial Probability 

Page 60 

Tip 46  Probability with Combinations 

Page 61 

Tip 47  Heron’s Formula 

Page 62 

Tip 48  Vectors in the Plane 

Page 63 

Tip 49  Interchange of Inputs 

Page 65 

Tip 50  Polynomial Inequalities 

Page 67 

Tip 51  Rational Inequalities 

Page 69 

Tip 52  Limits 

Page 71 

Tip 53  Rational Function and Asymptote 

Page 73 

Tip 54  Parametric Equations 

Page 76 

Tip 55  Polar Coordinates 

Page 77 

Tip 56  Matrix 

Page 79 

Tip 57  Inclination Angle 

Page 81 

Tip 58  Angle between Two Lines 

Page 82 

Tip 59  Intermediate Value Theorem 

Page 83 

Tip 60  Rational Zero Test 

Page84 

Tip 61  Descartes Rule of Sign 

Page 85 

         

Practice Tests 

     

Test 01  Test 02  Test 03  Test 04  Test 05  Test 06  Test 07  Test 08  Test 09  Test 10  Test 11  Test 12 

                       

Page 087  Page 113  Page 141  Page 167  Page 195  Page 223  Page 249  Page 277  Page 303  Page 329  Page 355  Page 381 

 

Dr. John Chung's SAT II Math Level 2

5

Tips TIP 01

Identical Equation

An identical equation is an equation which is true for all values of the variable. 10 x  5 x  15 x is an identical equation because it is always true for all real x.

10 x  5  15 is an algebraic equation because it is true for x  1 only.

 

Identical equation has infinitely many solutions. In an identical equation, the expressions of both sides are exactly same. Example 1: 5 x  5  5 x  5  ax  b  0x  0 Example 2: ax  b  0 for all real values of x

 a  0 and b  0

PRACTICE 1. If ax  b  3 x  2 is always true for all real x , what are the values of a and b ?

2. If a(x  1)  b( x  1)  x  9 is true for all real values of x , what are the values of a and b ?

3. If x2  2 x  6  a( x  1)2  b( x  1)  c is true for all real x , where a, b, and c are constants, what are the values of a, b, and c ?

EXPLANATION 1. The coefficients must be equal. 2. ax  a  bx  b  (a  b)x  a  b , Coefficients must be equal. a  b  1 and a  b  9 Therefore, a  5 , b   4 .

(a  b)x  (a  b)  x  9

3. Since x2  2 x  6  ax2  (b  2a) x  a  b  c , then a  1, b  2a  2, and a  b  c  6 . Therefore, a  1 , b  4 , and c  1.

Answer: 1. a  3, b  2

6

2. a  5, b  4

3. a  1, b  4, c  1

Tips Remainder Theorem

TIP 02

When polynomial P(x) is divided by (x  a) , the remainder R is equal to P(a) . Polynomial P(x) can be expressed as follows.

P(x)  ( x  a)Q( x)  R The identical equation is true for any value of x , especially x  a . Therefore, P(a)  R Example: If P(2)  5, then you can say that “When polynomial P(x) is divided by (x  2) , the remainder is 5.

PRACTICE 2 1. If a polynomial f ( x)  2 x  3x  5 is divided by (x 1) , what is the remainder?

3 2 2. If a polynomial g ( x)  x  2 x  2 x  3 is divided by ( x 1)( x  2) , then what is the remainder?

EXPLANATION

2 1. R  f (1)  2(1)  3(1)  5  4

2.

g( x)  (x 1)( x  2)Q( x)  ax  b

When divided by degree 2 polynomial, the remainder is represented by ax  b

At x  1 At x  2

g(1)  8  a  b g(2)  23  2a  b

Therefore, a  15 and b  7 .

The remainder is 15 x  7

Also the remainder can be obtained using long-division or synthetic division.

Dr. John Chung's SAT II Math Level 2

7

Tips Factor Theorem

TIP 03

If p(a)  0, then p( x) has a factor of ( x  a). Polynomial p( x) can be expressed with a factor of (x  a) as follows.

p(x)  (x  a)Q(x), where Q(x) is quotient. If (x  a) is a factor of p( x) , then the remainder after division should be 0. Example: If p(5)  0, then p( x) has a factor of ( x  5).

PRACTICE 1.

If a polynomial P( x)  x 2  kx  8 has a factor of (x  2) , then what is the value of constant k ?

2.

If a polynomial f ( x)  x3  ax2  bx  1 has a factor of ( x 2  1) , what are the values of a and b ?

EXPLANATION 1.

Using the factor theorem,

P(2)  0 2.



2 2  2k  8  0

 k 2

Since x3  ax 2  bx  1  ( x  1)( x  1)Q( x) ,

f (1)  a  b  2  0 and f (1)  a  b  0 Therefore, a  1 and b  1 .

Answer: 1. k  2

8

2. a  1 and b  1

Tips TIP 04

Sum and Product of the Roots

For a polynomial P( x)  an xn  an1 xn1  an2 xn2    a1 x  ao  0 Sum of the roots  

an 1 an

Product of the roots 

ao  1n , where n is the degree of the polynomial an

Example 1: P( x)  ax2  bx  c  0 , r and s are the roots of the quadratic equation. b a

Sum of the roots: r  s  

Difference of the roots: (r  s)2  (r  s)2  4r  s If r and s are real and r  s , then

r  s  (r  s )2  4rs

 rs

Product of the roots: r  s 

b2 4c b 2  4ac   a a2 a

c c (  1) 2  a a

Example 2: P( x)  Ax3  Bx2  Cx  D  0 Sum of the three roots  

B A

Product of the three roots = 

D D  13   A A

PRACTICE 1. If the roots of a quadratic equation 2 x 2  5 x  4  0 are  and  , what is the value of

1





1



?

2. What is the sum of all zeros of a polynomial function P( x)  2 x7  3x3  5x2  4 ?

3. What is the product of all zeros of g ( x)  3x7  5x3  3x2  x  2 ?

4. If one of the roots of a quadratic equation is 2  i , what is the equation?

Dr. John Chung's SAT II Math Level 2

9

Tips EXPLANATION

1. Since     

1 1   5 5 4    . and       2 , therefore    4 2 2

2. Because the coefficient of x6 is 0, the sum of the roots  

3. Because the product of all zeros is

an 1 0  0 an 2

ao 2 2 (1) n   17  an 3 3

b c x 0. a a b b The sum of the roots  (2  i )  (2  i )  4     4 a a c c The product of the roots  (2  i )(2  i )  5   5 a a

4. The quadratic equation can be defined by x 2 

Therefore, the equation is x2  4x  5  0. Proof Sum of the two roots:

b  b 2  4ac b  b 2  4ac 2b b    2a 2a 2a a Product of the two roots:



2  b  b 2  4ac  b  b 2  4ac   b   b  4ac       2a 2a 4a 2    4ac c   4a 2 a 2

Answer: 1.

10

5 4

2. 0

3.

2 3

4. x2  4 x  5  0



2

Tips TIP 05 i  1 5 i  i   

Complex Number i 2  1 6

i  1

i3  i 7

i  i

Im

i 4  1  i8  1   

z  a  bi

b

z

Complex number = Real numbers + Imaginary numbers  3  4i  a

Pure Imaginary number = only Imaginary numbers (4i) Complex plane

1. a  bi is complex number, where a and b are real numbers. (Standard form) 2. a  bi is the conjugate of a  bi .

 a  bi  a  bi 

3. If a  bi  c  di , then a  c and b  d . (Equality of Complex Numbers) 4. a  bi  a 2  b2 (Distance from the origin)

PRACTICE 1. What are the additive inverse and multiplicative inverse of the complex number 3  i ?

2. What is the value of 3  4i ?

3. If a  b  ( a  b )i  6  4 i , what are the values of a and b ?

EXPLANATION 1. Additive inverse: (3  i )  ( a  bi )  0   3  i. 2.

Multiplicative inverse:

1 3  i 

 3  i  3  i 



3i 10

3  4i  32  (4)2  25  5

3. a  b  6 and a  b  4 . Therefore, a  1and b  5. Answer: 1. 3  i ,

3 1  i 10 10

Dr. John Chung's SAT II Math Level 2

2. 5

3. a  1, b  5

11

Re

Tips TIP 06

Conjugate Roots

If a polynomial function P ( x ) has one variable with real coefficients, and a  bi is a root with a and b real numbers, then its conjugate a  bi is also a root of P ( x ) . 1. If x2  x  1  0 , the roots are

1 i 3 1 i 3   and . 2 2 2 2

2. A polynomial x 2  5  0 has the roots of i 5 and i 5 .

PRACTICE 1. If one of the roots of a quadratic equation f ( x )  0 is 3  2i , what is the quadratic equation?

2. If one of the roots of mx 2  (4m  1) x  k  0 is 1  2i , where m and k are real numbers, then what is the value of k ?

EXPLANATION 1. Using the roots

 x  (3  2i)  x  (3  2i)   0



 x  3  2i  ( x  3  2i )  0



 x  3  2   2i  2  0

Therefore, x2  6 x  13  0 . b c x 0 a a b c Sum:    3  2i   (  3  2i )   6 and Product:   3  2i  3  2i   13. a a

Using sum and product of the roots.

x2 

The equation is x2  6 x  13  0 (4 m  1) 1   1  2i  (  1  2i )   2 ,  m  2 m k 5 Product:  (  1  2i )(  1  2i )  5 , k  5m . Therefore, k  . 2 m

2. Sum: 

Answer: 1. x2  6 x  13  0

12

2. k 

5 2

Tips TIP 07

Linear Function

For two linear functions:

y  m1 x  b1 and y  m2 x  b2 1. If m1  m2 and b1  b2 , then these two lines are parallel. (Inconsistent) 2. If m1  m2 and b1  b2 , then these two lines coincide. (Dependent) 3. If m1  m2  1 , then these two lines are perpendicular. 4. If m1  m2 , these two lines are intersecting. (Consistent)

PRACTICE 1. What is the equation of the line which is equidistant from two points A (4, 0 ) and B (0, 2) ?

2. If the two lines 2 x  3 y  2  0 and 3 x  ky  1  0 are perpendicular, then k 

3. If the two lines 2 x  ay  1 and ax  ( a  4) y  2 are parallel, then a 

EXPLANATION 1. The midpoint   2, 1 and the slope of AB  

1 . The perpendicular line has slope of 2 and passes through 2

(2, 1). Therefore, the equation is y  2 x  3. 2. The product of the slopes is 1. 3. The slopes are equal. 

2 3   1 3 k

 k  2

2 a    a  4  a  2   0  a a4

a  4 or 2. But the y -intercepts are

1 2 . At a  4 , the two y-intercepts are equal (coincide). Therefore, a  2 and a a4

Answer: 1. y  2 x  3

Dr. John Chung's SAT II Math Level 2

2. k  2

3. a  2

13

Tips Distance from a Point to a Line

TIP 08

Distance D from a point  x1 , y1  to a line ax  by  c  0 :

 (x , y ) 1

D

ax1  by1  c a 2  b2

Distance between two points D  ( x2  x1 )2  ( y2  y1 )2

1

D

ax  by  c  0

Distance between two straight lines is the minimum distance which is perpendicular to the lines.

PRACTICE 1. What is the distance from a point  7, 9 to a line 12 x  5 y  0 ?

2. What is the distance from the origin to a line 3 x  4 y  8 ?

3. What is the distance between two parallel lines 3 x  y  12 and mx  2 y  4 ?

EXPLANATION

1. D 

12(7)  5(9) 122  (5)2



39 3 13

2. Origin  0, 0 , and the equation of the line is 3 x  4 y  8  0 . Therefore, D 

3(0)  4(0)  8 2

2



8 . 5

3 4 3. Since they are parallel, m  6 . Choose any point on 3 x  y  12 .  (4, 0) Distance between (4, 0) and 6 x  2 y  4  0 is D

6(4)  2(0)  4 62  22

Answer: 1. 3

14

2.

 10 8 5

3.

10

Tips Distance from a Point to a Plane

TIP 09

Distance from point A( x2 , y2, z2 ) to point B( x1 , y1 , z1 ) in space :

D  ( x2  x1 )2  ( y2  y1 )2  ( z2  z1 )2 Distance from a point  x1 , y1 , z1  to a plane ax  by  cz  d  0 :

D

ax1  by1  cz1  d a 2  b2  c 2

Distance from the origin to a point ( a , b , c ) is D  a 2  b 2  c 2 .

PRACTICE 1. What is the distance from a point (1, 2,  3) to a plane 3 x  4 y  12 z   2 ?

2. What is the length of the diagonal of a rectangular solid with dimensions 3, 4, and 12?

3. What is the distance between point A (1,  1, 2) and point B (3, 4, 1) ?

EXPLANATION 1. D 

3(1)  4(2)  12( 3)  2 2

2

3  4  12

2



39 3 13

2. D  32  42  122  13 3. D 

 3  12   4  1

Answer: 1. 3

2

 1  2  30

2. 13

Dr. John Chung's SAT II Math Level 2

2

3.

30

15

Tips Quadratic Function

TIP 10

Polynomial function of x with degree n is defined as follows.

P( x)  an xn  an1 x n1  an2 xn2     a2 x2  a1 x  ao 1. If f ( x)  ao , then f ( x) is a constant function. 2. If f ( x)  ax 2  bx  c, then f ( x) is a quadratic function. If a  0, f has a minimum at x  

b . 2a

If a  0, f has a maximum at x  

b . 2a

3. f ( x)  ax 2  bx  c : standard form. 4. f ( x)  a( x  x1 )( x  x2 ) : factored form, where x1 and x2 are the roots of f ( x)  0 5. f ( x )  a  x  h   k : vertex form 2

Axis of symmetry: x  h ,

Vertex: (h, k )

PRACTICE 1. If f ( x)   x2  6 x  8 , what are the coordinates of the vertex?

2. If a manufacturer of game computers has daily production costs of C (n)  1200  24n  0.5n 2 , where C is the total cost, in dollars, and n is the number of units produced, how many game computers should be produced each day to minimize cost?

EXPLANATION 1. Axis of symmetry x  

b 6   3 , y  f (3)  32  6(3)  8  1 2a 2(1)

2. Axis of symmetry n  

 24  2(0.5)

 24 , C has a minimum at n  24 .

Therefore, the minimum cost C (24)  1200  24(24)  0.5(242 )  912 Answer: 1.

16

 3, 1

2. 24

Tips Discriminant

TIP 11

Discriminant determines the nature of the roots of a quadratic equation ax2  bx  c  0. Roots 

 b  b 2  4 ac , Discriminant D  b 2  4 ac 2a

1. If D  0 , then the roots are real and unequal. 2. If D  0 , then the roots are real and equal. 3. If D  0 , then the roots are imaginary. (No real roots)

PRACTICE 1. If a quadratic equation 2 x2  kx  3  0 have imaginary roots, what is the value of k ?

2. If the roots of x2  (k  1) x  4  0 are real and equal, what is the value of k ?

3. If y  3x 2  2 x  k is positive for all x , then what is the smallest integral value of k ?

Explanation 1. D  k 2  24  0



 k  2 6  k  2 6   0 . Therefore, 2

2. D   k  1  16  0  ( k  5)( k  3)  0 . 2

6x2 6 .

k  5 or 3.

3. That means “no x-intersections,” or “imaginary roots.” D  4  12 k  0

 12 k  4  k 

Answer: 1. 2 6  k  2 6

Dr. John Chung's SAT II Math Level 2

1 , The smallest integer k is 1. 3

2. k  3 or  5

3. k  1

17

Tips Circle

TIP 12

A circle is the locus of points equidistant from a given point, known as the center. The standard equation of a circle whose center is at the point  h, k  is

 x  h 2   y  k 2  r 2 ,

r  radius

PRACTICE 1. What is the area of a circle whose equation is x 2  4 x  y 2  2 y  11 ?

2. The graph of the equation x 2  y 2  2ax  4 y  2a 2  0 represents a circle. What is the greatest possible integer value of a ?

3. What is the circumference of a circle whose equation is x 2  y 2  6 y  16 ?

EXPLANATION 1. The standard equation is  x  2    y  1  16 , 2

2

r 2  16

Therefore, the area is  r 2  16 . 2. The standard equation is  x  a    y  2    a 2  4 2

To form a circle, r 2  a2  4  0

2

 a 2  4  0   a  2  a  2   0

Since 2  a  2 , the greatest integer a is 1. 3. The standard equation is x 2   y  3   25 . r  5 . 2

Therefore, the circumference of the circle is 2 r  10 . Answer: 1. 16

18

2. 1

3. 10

Tips Ellipse

TIP 13

An ellipse is the set of all points  x, y  in a plane, the sum of whose distances from two distinct fixed point (foci) is constant. The standard equation of an ellipse with center  h, k  is If a  b ,

 x  h

 2



2

 y  k

2

1

2

b

: Major axis is horizontal

a b Vertex  1. The center of the ellipse is at point (h, k ) 2. Length of major axis is 2 a. Length of minor axis is 2b.

 Vertex

If a  b ,

 x  h 2



 y  k 2

focus

2

1

: Major axis is vertical

b

c

2.

a b The center of the ellipse is at point (h, k) Length of major axis is 2b. Length of minor axis is 2 a.

3.

If c is the length from the center to the focus, then c 2  b2  a 2 .

1.

  Vertex c focus



3. If c is the length from the center to the focus, then c 2  a 2  b2 .

2

 h, k 



a

a





 h, k 

a



  Vertex

PRACTICE 1. What is the center of an ellipse whose equation is x2  4 y 2  6 x  8 y  9  0 ?

2. What is the length of the major axis of the ellipse whose equation is

3.

x2 y 2  1? 5 27

The ellipse is given by 4 x2  y 2  36 . What are the coordinates of the foci?

4. If a line y  x  k is tangent to an ellipse whose equation is x 2 

Dr. John Chung's SAT II Math Level 2

y2  1 , what is the value of k ? 4

19

Tips EXPLANATION

1. The standard equation is  x  3  4  y  1  4 2

2



 x  32   y  12 4

1

1.

Center (3,  1) 2. a 2  27  a  3 3 ,

Therefore major axis is 2a  6 3.

y

x2 y 2   1 . (Major axis is vertical) 9 36

3. The standard equation is

f

a 2  36 and b2  9 2

c

2

c   b  a   27  3 3





O

Therefore the coordinates of foci is f 0,  3 3 . 4. Substitute. 4 x 2   x  k   4 2

 5 x 2  2 kx  k 2  4  0

f

Since the line is tangent to the ellipse, its discriminant must be 0.

D  4k 2  4(5)(k 2  4)  0  16k 2  80  k 2  5

y

Therefore, k   5 .

y x 5

 y x 5 x

O

 Answer: 1.

20

 3,  1

2. 6 3



3. 0,  3 3



4.  5

b a

x

Tips Parabola

TIP 14

A parabola is the set of all points in a plane that are equidistant from a fixed line, the directrix, and a fixed point, the focus. ( p is the distance from the vertex to the focus.) 1.

The vertex form of the equation with vertex at the origin (0, 0) is Vertical axis 1 2 y x 4p

y

y

1 2 x 4p

y

Directrix y p

Focus  (0, p )

 ( x, y )

O

x

x

O

 ( x, y )  Focus (0,  p )

Directrix y  p

The vertex form of the equation of a parabola with vertex at  h, k  are as follows. 1  x  h 2  k (Graph opens upward) 4p

y

2.

y

1  x  h 2  k (Graph opens downward) 4p

Horizontal axis 1 2 y 4p

x

x

1 2 y 4p

Directrix y x  p

O

y Focus ( p, 0)  

( x, y )

x

Focus (  p , 0) 

Directrix x p

O

x

( x, y ) 

The vertex form of the equation of a parabola with vertex at  h, k  are as follows.

x

1 1  y  k 2  h (Graph opens to the right) x    y  k 2  h (Graph opens to the left) 4p 4p

Dr. John Chung's SAT II Math Level 2

21

Tips PRACTICE 1. What are the focus and directrix of the parabola whose equation is given by y 

1 4

2. Find the focus of the parabola given by y   x 2  x 

1 2 x ? 16

1 . 4

3. Find the vertex form of the equation of the parabola with the vertex at (1,0) and the focus at (2, 0).

EXPLANATION 1.

2.

y

1 2 x  4  4

focus (0,4) and directrix y  4

1 2 1 x  4x  , the vertex form is  4 4 1 5 5 2 y  x  2    p  1 and vertex  2,  , graph opens downward. 4 4 1 4  y

5   1  Therefore, the focus  h, k  p    2,  1   2,  4   4 

y

3. Because the axis of symmetry is horizontal, passing through 1, 0  and  2, 0  .

directrix x0

The vertex form of the equation is 1 2 x  y  0   1 , where the vertex is at 1,0  and p  2  1  1. 4p Therefore the equation is 1 x  y2  1 4

(1, 0) O

Answer: 1. f  0, 4  , y  4 2.

22

 2, 0.25

3. x 

1 2 y 1 4



(2, 0)



Focus

x

Tips TIP 15

Hyperbola

A hyperbola is the set of points in a plane, the difference of whose distances from two distinct foci is constant.

The standard form of the equation with the center at  0, 0 is x2 y2  1 a 2 b2 Transverse axis is horizontal

y 2 x2  1 b2 a 2 Transverse axis is vertical

y

y

c

 (  c, 0)

b a

 (c, 0)

 (0, c ) a b c x

x

 (0, c ) Focus ( c,0) : c 2  a 2  b2 Asymptotes: y  

Focus (0, c) : c 2  a 2  b2

b x a

Asymptotes: y  

b x a

The standard equation of a hyperbola with center at  h, k  is

 x  h

2



a2

 y  k

2

b2

Dr. John Chung's SAT II Math Level 2



 y  k

2

b2

 x  h a2

1

Transverse axis is horizontal

1

Transverse axis is vertical

2

23

Tips PRACTICE 1. If the equation of a hyperbola is given by

x2 y2   1 , what are the equations of its asymptotes? 25 16

2. Find the asymptotes and foci of the hyperbola whose equation is

y 2 x2   1. 16 4

3. The equation of a hyperbola is defined by 9 x2  y 2  36 x  6 y  18  0 . Find the center.

EXPLANATION 1. Asymptotes: y  

b 4 x . Since a  5 and b  4 , the asymptotes is y   x . a 5

b x. a

2. Asymptotes: y  

a  2 and b  4 . Therefore, the asymptotes is y  

4 x 2



y   2 x.

For the foci (0, c) :

c   a 2  b2   16  4  2 5



 



3. Perfect squared form: 9 x 2  4 x  4  y 2  6 y  9  18  36  9

9  x  2    y  3  9  2

2

Center at  2,  3 4 Answer: 1. y   x 5

24

 x  2  2  y  3 2 1





9

1

2. y  2x , f 0,  2 5



3.

 x  2  2  y  3 2 1



9

 1 , Center  2,  3

Tips Function

TIP 16

A function f from a set X to a set Y is a relation that assigns to each element in set X exactly one element in the set Y . Domain is the set of X (input).

Range is the set of Y (output).

1. x is the independent variable and y is the dependent variable. Input x

Input x

Function f

Function f

Output f ( x)

Output f ( x)

PRACTICE 1. What are the domain and range of the function f ( x )  16  x 2 ?

2. What is the domain of the function f ( x ) 

x  10 ? x  15

3. What is the range of the function f ( x)  3 x  5  4 ?

EXPLANATION 1. 16  x 2  0

 x  4  x  4   0   4  x  4 x  15   10  x  15   x  15 or



2. x  10  0 but

interval notation: 10,15  15, 

3. Minimum of f is  4. You can a graphing calculator Answer: 1. Domain : 4  x  4 , Range : 0  y  4

2. 10  x  15   15  x

3. y  4

Dr. John Chung's SAT II Math Level 2

25

Tips TIP 17

Domain and Range of a Composite Function

Domain of a composite function is the intersection of domains of the starting and final function. Range of a composite function is the range of final function restricted by starting function. Example: f ( x) 

1 x : , g ( x)  x2 x3

g ( x)  starting function,

f ( x)  second function

f  g ( x)   final function

Domain of starting function g ( x) is x  3. (All real numbers except 3)

f  g ( x)  

1



x3  Domain of final function is x  2. 3( x  2)

x 2 x3 Therefore, the domain of f  g ( x)  is all real numbers x except 2 and 3. Now let’s find the range. First find the range of the final function. From the function, we can find two asymptotes.

 x  2 : Vertical asymptote  x3 1   : Horizontal asymptote  y  xlim  3( x  2) 3  y

But when x  3, the function is undefined.

O

3

1 y x 3

x2

 Therefore, the range of f  g ( x)  is  , 0    0, 

26

1 1 1     ,   .  All real numbers y except 0 and 3  3 3 

Tips PRACTICE 1. If f ( x)  x and g ( x)  x  1, what is the domain and range of  g  f  ( x)?

2. If f ( x ) 

1 1 and g ( x )  , what is the domain and range of  g  f  ( x)? x x 1

EXPLANATION 1. Domain of f ( x) is x  0. x  1 : Domain of  g  f  ( x) is x  0. There, the domain of the final function is x  0.

 g  f  ( x) 

The range of the final function is y  1. y

x

1 2. Domain of f ( x) is x  0 . (All real numbers except 0)

 g  f  ( x)  1

1

x 1 1 x 

x Domain of  g  f  ( x) is x  1. Therefore, actual domain is all real numbers except 0 and 1. Range of  g  f  ( x) :

y

Vertical asymptote: x  1   x  1 Horizontal asynptote: y  xlim  1  x From the first function f ( x) ,

x0 

 g  f  (0)

x y  1

is undefined.

Therefore, the range of g  f ( x)  is all real numbers except y  1 and y  0.

Dr. John Chung's SAT II Math Level 2

x 1

27

Tips TIP 18

Piecewise-Defined Function

A piecewise-defined function is a function that is defined by two or more equations over a specified domain.

f ( x)  x is a piecewise-defined function as follows.  x, when x  0 f ( x)  x    x, when x  0

PRACTICE 1. What are the domain and range of the piecewise-defined function as follows?

2  x  1, f ( x)    x  1,

x0 x0

EXPLANATION

1. The graph shows that the domain is all real x and the range is y  1. y

y  x2  1

y  x  1( x  0)

(0,1) x

O

 (0,  1) Answer: 1. Domain: All real x , Range: y  1

28

 Interval notation:Domain  ,   , Range[1, )

Tips Odd and Even Functions

TIP 19

A function f is even if f ( x)  f ( x) .

A function f is odd if f ( x)   f ( x) . y

y

  x, y 

  x, y 

  x, y 



x

x

O

O

  x,  y  Symmetric about the y-axis

 Symmetric about the origin

n 1. For f ( x)  x , if n  even , then f ( x) is even.

if n  odd , then f ( x) is odd.

PRACTICE Determine whether each function is even, odd, or neither. 1.

f ( x)  x3  2 x

2. g ( x)  x 4  2 x2  5 3. h( x)  x3  1

EXPLANATION 1. Since f ( x)   f ( x) , f ( x) is odd. 2. Since g ( x)  g ( x) , g ( x) is even. 3. h( x) is neither , because h( x)  h( x) or h( x)  h( x) . Answer: 1. Odd

2. Even

Dr. John Chung's SAT II Math Level 2

3. Neither

29

Tips Combinations of Functions

TIP 20 Sum:

f

 g  ( x)  f ( x)  g ( x)

Difference:

f

 g  ( x )  f ( x )  g ( x)

Product:

 fg  ( x)  f ( x)  g ( x)

Quotient:

f f ( x) , g ( x)  0   ( x)  g ( x) g

Compositions:

 f  g  ( x)  f  g ( x) 

PRACTICE 1. If f ( x)  2x  3 and g ( x)  2x  3 , then  fg  (4)  2. If f ( x)  log3  x  3 and g ( x)  log3  x  3 , then  f  g  (6) 





3. If f ( x)  log 2 x 2  3x  2 and g ( x)  log 2  x  2  , then  f  g  (9)  4. If f ( x)  e x and g ( x)  3ln( x  3) , then f  g (5)  

EXPLANATION 1. Since fg   2 x  3  2 x  3   4 x 2  9 ,  fg  (4)  55 . Or, f (4)  11 and g (4)  5 . f (4) g (4)  55.





2. Since f  g  log3  x  3 x  3  log3 x 2  9 ,

f

 g  (6)  log3  36  9   3log3 3  3 .

3. f  g  log 2  x  2  x  1  log 2  x  2   log 2

 x  2  x  1  log 2  x  1  x  2

Therefore,  f  g  (9)  log 2  9  1  3log 2 2  3 4. Since f  g ( x )   e 3ln( x  3)  ( x  3) 3 , f  g (5)   (5  3) 3  8 . Answer: 1. 55

30

2. 3

3. 3

4. 8

Tips Periodic Functions

TIP 21

If a function f is periodic if there exists a number p such that

f ( x  p)  f ( x) for all number x. 1.

The smallest period is called the fundamental period of the function.

2.

If a periodic function f has period p , then



1) y  cf ( x) still has period p.

y

Period



x

p 2) y  f  cx  has period . c

 

x

x p

3. The smallest period is simply called the period.

PRACTICE 1. If a function f ( x)  sin x has period 2 , then what is the period of the function

f ( x)  3sin3x ?

   2. What is the period of the function y  2 cos   x    5 ? 12  

3. If a function is defined by f ( x)  f ( x  2 ) , what is the period of the function?

EXPLANATION 1. p 

2 2  c 3

2.

Dr. John Chung's SAT II Math Level 2

p

2



2

3. p  2

31

Tips Inverse Functions

TIP 22

An inverse function is a function that reverses function f . If f is a function mapping x to y, then the inverse function of f maps y back to x.

x

f ( x)  x  5

y

f 1 ( x)  x  5

y x

: (1, 6), (2, 7), (3, 8), (4, 9)

If

f ( x)  x  5

then

f 1 ( x)  x  5 : (6, 1), (7, 2), (8, 3), (9, 4)

In order to form inverse function of f just interchange x and y coordinates and express y in terms of x. If the function g is the inverse function of f , 1.

f  g ( x)   x and g  f ( x)   x.

2.

The domain of f is the range of g and the range of f is the domain of g.

3.

f 1 ( x) is a reflection of the graph of f in the line y  x.

4.

If point (a, b) lies on graph of f , then point (b, a) must lie on the graph of f 1.

PRACTICE 1. What is the inverse function of f ( x )  2. If f (4)  35, then f  1  35  

EXPLANATION 1. x 

3y  5 , 2

f 1 : y 

2x  5 3

2. f :  4, 35   f  1 : (35, 4) Answer: 1. y 

32

2x  5 3

2. 4

3x  5 ? 2

Tips The Existence of an Inverse Function

TIP 23

If a function f is one-to-one, then its inverse is a function. 1. If f is increasing on its entire domain, then f is one-to-one. 2. If f is decreasing on its entire domain, then f is one-to-one. 3. If f is increasing on its entire domain, then f 1 is a function. 4. If f is decreasing on its entire domain, then f 1 is a function. To check one-to-one, the horizontal line test can be used. y

y  f ( x)

By the horizontal line test, f does not have an inverse function.

x

O

PRACTICE 1. Does the function f ( x)  x  2  3 have an inverse function?

2. Does the function g ( x)  x  3 have an inverse function?

EXPLANATION y

y 1.

2.

 (2,3)

O

f is one - to - one. Its inverse is a function.

Dr. John Chung's SAT II Math Level 2

x

(3, 0) O

x

g is not one- to- one Its inverse is not a function.

33

Tips TIP 24

Leading Coefficient Test (Behavior of Graph)

Whether the graph of a polynomial rises or falls can be determined by the Leading Coefficient Test as follows.

P( x)  an x n  an 1 x n 1      a1 x  ao

: an is the leading coefficient.

1. When n is odd and an is positive, the graph falls to the left and rises to the right. 2. When n is odd and an is negative, the graph rises to the left and falls to the right. 3. When n is even and an is positive, the graph rises to the left and right. 4. When n is even and an is negative, the graph falls to the left and right.

PRACTICE 1. What are the right-hand and left-hand behaviors of the graph of f ( x)  x5  2 x3  3x  5 ?

EXPLANATION y

1. Using the test, leading coefficient is positive and n is odd.

lim x5    and lim x5  

x 

x 

x O Answer: The graph falls to the left and rises to the right.

34

Tips TIP 25

Arithmetic Sequences

A sequence is arithmetic if the differences between consecutive terms are the same. If a1  a and d is the common difference, then the nth term of an arithmetic sequence is

an  a  (n  1)d . And the sum of a finite arithmetic sequence with n terms is Sn 

n ( a1  an ) 2

PRACTICE 1. If the first term of a sequence an  is a1  3 and an1  an  4, what is a25 ?

2. If three numbers m, a , and k form an arithmetic sequence in that order, the sum of the numbers is 21, and the product of the numbers is 315, what is the greatest number in the sequence?

3. If the first term of an arithmetic sequence is 5 and the common difference is 3, what is the sum of the first 100 terms?

EXPLANATION 1. Since a1  3, an1  an  4, and d  4, a25  a   n  1 d  3  24  4  99 . 2. Let three numbers be a  d , a, a  d . Then a  d  a  a  d  3a  21  In an arithmetic sequence, the middle number (median) is equal to the average.

a7

(7  d )  7  (7  d )  315  49  d 2  45  d 2  4  d  2 Therefore, three numbers are 5, 7, 9 3. Since a1  5 and d  3 , then a100  5  (100  1)  3  302 and number of terms is 100. Therefore, S100  Answer: 1. 99

100  5  302  2 2. 9

Dr. John Chung's SAT II Math Level 2

 15,350 3. 15,350

35

Tips Geometric Sequences

TIP 26

A sequence is geometric if the ratios of consecutive terms are the same. If a1  a and the common ratio is r , then 1) the n th term of the geometric sequence is an  ar n 1 and

a(1  r n ) . 1 r

2) the sum of the sequence is Sn 

a(1  r n ) . n  1  r

3) The sum of the infinite series is given by S  lim Sn  lim n 

If r  1 , then the sum of the infinite series is as follows. S 

a 1 r

PRACTICE 1. The first term of a geometric sequence an  is 2 and an1  2an . What is the value of a10 ?

2. In a geometric sequence, the second term is 3 and fifth term is 24. What is the 8th term?

3. What is the sum of the infinite series 15  3 

3 3    ? 5 25

EXPLANATION 1. Since a1  2 and r  2, then a10  2  2   210 . 9

2. a2  a1r  3 and a5  a1r 4  24 ,

a5 a1r 4 24   3 a2 a1r

 r3  8 .

3 3 7 . The 8th term a8     2   192 2 2 a 15 1 3. Since r   , S    12.5 1  r 1   1 5  5

Therefore r  2 and a1 

Answer: 1. 1024

36

2.

192

3.

12.5

Tips Exponential Functions

TIP 27

A. Laws of Exponents: If a, b, and c are positive integers. 1. xa  xb  xa b

: Multiplication Law

2. x a  xb  x a b

: Division Law

3.

x 

 x ab

: Power Law

4.

 xy a

 xa  y a

: Power of a Product Law

a b

a

x xa 5.    a y  y

: Power of a Quotient Law

6. xo  1 ( x  0)

: Zero Exponent

a 7. x  a

a

1 1    ( x  0) : Negative Exponent xa  x 

b

8. x b  x a 

 x b

a

: Fractional Exponent

B. Exponential Function f with base a :

f ( x)  a x , where a  0, a  1 , and x is any real number. C. Graphs of Exponential Functions 1) a  1 y

2) 0  a  1

x

y

O

O

x

The graph of y  2  x is the reflection in the y-axis of the graph of y  2 x . Domain:  ,   , Range:  0,   , Horizontal asymptote: y  0 D. The natural base e  e  2.718281828

 1 e  lim 1   x   x

Dr. John Chung's SAT II Math Level 2

x

or

1

e  lim 1  x  x x0

37

Tips PRACTICE 1. If a total of $10,000 is invested at an annual interest rate of 5%, compounded annually, what is the balance in the account after 5 years?

2. If $5,000 is invested at an annual interest rate of 6%, compounded quarterly, what is the amount of the balance after 5 years?

3. If 32 x  3x  72  0 , what is the value of x ?

2

2 x 1  16 , then x  4. If 2 x 1

EXPLANATION 1.

A  10, 000 1  0.05   12762.82

2.

 0.06  A  5,000 1   4  

5





4(5)

nt

 r  6734.28 , because A  P 1   .  n



3. Since 3x  9 3x  8  0 , then 3x  8 and 3x  9 . Therefore, x  2. 2

2 x 1 x2  x  2  24 , then x2  x  2  4  x 2  x  2  0 . 4. Since x 1  2 2  x  2 x  1  0 , x  2 or  1 . Answer: 1. $12762.82

38

2. $6734.28

3. 2

4. x  2 or x  1

Tips TIP 28

Logarithmic Functions

The function is given by

y  loga x

x  a  , y

where x  0, a  0, and a  1 .

Properties of Logarithms: 1.

loga 1  0

2.

loga a  1

3.

log a a x  x log a a  x

4. 5.

aloga x  x If loga x  loga y , then x  y.

6.

log10 x  log x

: (Common logarithms)

7.

loge x  ln x

: (Natural logarithms)

8.

log a x 

9.

loga  xy   log a x  loga y

log b x log b a

: (Change of base)

x  log a x  log a y y

10.

log a

11.

log a x n  n log a x

12.

loga x  log an x

13.

log a x 

: (Product property) : (Quotient property) : (Power property)

n

: (All real n except 0)

1 log x a

: (Reciprocal property)

Graphs of Logarithmic Functions: y  log a x

y

a 1

y  log 2 x

x

O

y

0  a 1

y  log1 2 x

O

x

The graph of y  log 1 x is the reflection in the x -axis of the graph of y  log 2 x . 2

Domain:  0,   , Range:  ,   , Vertical asymptote: x  0

Dr. John Chung's SAT II Math Level 2

39

Tips PRACTICE 1. If ln  x  2   ln  2 x  3  2ln x , then x 

2. $10,000 is invested in an account at an interest rate of 5%, compounded continuously. The value A , of the

investment at any time is given by the equation A  Pert , where t represents the number of years. How long will it take the balance to double?

3. If log3 x  log3 ( x  8)  2 , then x 

4. If y  3e x  2 , what is its inverse function?

EXPLANATION





1. ln 2 x 2  7 x  6  ln x 2  2 x2  7 x  6  x 2   x  6  x  1  0

Since x  2 , the answer is x  6 only. 2. 20, 000  10, 000e 0.05t  0.05t  ln 2  t 





3. log3 x2  8x  2  x2  8x  32



ln 2  13.86 0.05

 x  9 x  1  0

, x  9 or 1

Since domain is x  8 , x  1 cannot be the solution. 4. x  3e y  2  e y 

Answer: 1. x  6

40

x2  x2  f 1 : y  ln   3  3 

2. 13.86

3. 9

 x2 4. f 1 ( x )  ln    3 

Tips TIP 29

Basic Trigonometric Identities

Functions

Domain

Range

Period

sin 

All real numbers

1  sin   1

2 ,360

cos 

All real numbers

1  cos   1

2 ,360

All real numbers

 ,180

tan 

All real numbers except

  180 n  90 csc 

All real numbers except

csc  1

2 ,360

sec  1

2 ,360

All real numbers

 ,180

180 n sec 

All real numbers except 90  180 n

cot 

All real numbers except

180 n Reciprocal Identities sec  

1 cos 

Quotient Identities sin  tan   cos 

csc  

1 sin 

cot  

cos  sin 

cot  

1 tan 

Pythagorean Identities

sin 2   cos2   1

Dr. John Chung's SAT II Math Level 2

1  tan 2   sec2 

1  cot 2   csc2 

41

Tips PRACTICE 1. If sin   cos  

1 , then sin  cos   2

2. If sin   cos  

1 , what is the value of tan   cot  ? 4

3. If cos   

4 and 90o    180o , what is the value of tan  ? 5

4. If the roots of the equation 3x2  kx  1  0 are sin  and cos , what is the positive value of k ?

EXPLANATION 1.

 sin   cos  2 

1 4

 sin 2   cos 2   2 sin  cos  

1 1  1  2 sin  cos   4 4

3 8

Therefore, sin  cos    . 2.

 sin   cos  2  tan   cot  

3. Since sec  

1 16

 1  2 sin  cos  

1 16

 sin  cos   

15 32

sin  cos  sin 2   cos2  1 32     cos  sin  cos  sin  cos  sin  15

1 5  , cos  4 2

9  5 1  tan 2   sec2   tan 2   sec2   1  tan 2       1  . 16  4 9 3 3   . In the second quadrant, tan   0 . Therefore, tan    . 16 4 4 1 k 4. Sum and product of the roots. sin   cos    and sin  cos    . 3 3 tan   

 sin   cos 2  1  2sin  cos  Therefore, 1 

2 k2  3 9

Answer: 1. 

3 8

42



2. 

k2 9

1 k2  3 9

 k2  3 . k  3

32 15

3. 

3 4

4.

3

Tips Circle (Trigonometry)

TIP 30 a 0 c b cos     0 c b tan     0 a

sin  

b 0 a a sin     0 c b cos     0 c tan  

a 0 c b cos    0 c b tan    0 a

sin  

c

a

c

b a

a b

c

a

c

 is the terminal angle from initial side,

b 0 c a sin     0 c b tan     0 a cos  

PRACTICE 1. If cos   

4 and tan   0 , then what is the value of sin  ? 5

2. In the interval 0  x  360 , sin x  cos x where x is

3. If tan x  

3 and cos x  0 , then what is the value of x ? 2

EXPLANATION 1. Since the angle terminates in Quadrant III, sin   

3 . 5

2. They are equal in Quadrant I and III. 3. The angle is in Quadrant II.  3 x  tan 1     56.3  2

Therefore, x  180  56.3  123.7 

Answer: 1. 

3 5



2. 45 , 225

Dr. John Chung's SAT II Math Level 2



3. 123.7

43

Tips Reference Angle and Cofunction

TIP 31

If  is the terminal angle, its reference angle is the acute angle   formed by the terminal side of  and the horizontal axis.





Initial Side

Terminal Side Cofunction: Any trigonometric function of an acute angle is equal to the Cofunction of its complement.

If A  B  90 , then sin A  cos B , sec A  csc B , and

tan A  cot B .

PRACTICE 1. Rewrite cos 258 as a function of a positive acute angle.





2. Write an expression equivalent to cot 118 as a function of an acute angle whose measure is less than 45 .





3. If  is an acute angle and csc   15  sec 45 , then sin  

EXPLANATION 1.    258  180  78 ,

Using reference angle Using cofunction     cos 258   cos 78   sin12

2.    180  118  62 ,

cot(118 )  cot 62  tan 28



3. Cofunction:   15  45  90    30   sin 30 

Answer: 1.  cos 78 or  sin12

44

2. tan 28

1 2

3.

1 2

Tips TIP 32

Trigonometric Graphs

1. y  a sin  bx  c   d and y  a cos  bx  c   d .

Amplitude : a Period : p 

2 b

Middle line : y  d 2. y  tan(bx  c)  d Amplitude : Does not exist.

Period : p 



b Middle line : y  d

PRACTICE 1. Find the period and amplitude of the function f (t )  5 cos

t 4. 12

2. What is the period and amplitude of y  4sin x cos x  1 ? 3. If the height is given by h  30 sin

 15

 t  25   15 , what are the maximum and minimum

values of the height?

EXPLANATION 1. p 

2  24 and amp  5  5 .  12

1.

2. y  4sin x cos x  1  2sin 2 x  1 2 Therefore, amp is 2 and p   .

p  24

0

2

3. Since amplitude is 30, maximum  30  15  45 and minimum  15  30  15.

5

y 1

24 y  4 middle line y  9

Answer: 1. p  24 and amplitude  5

Dr. John Chung's SAT II Math Level 2

2. p   , amplitude  2

3. maximum  45 , minimum  15

45

Tips TIP 33

Inverse Trigonometric Functions y

  1,   2

Inverse Sine Function y

1



    2 , 1

y  sin x

y  arcsin x

x

     ,  1   2 

1

1

y  sin x has an inverse function only on this interval.



   1,   2 

Inverse Cosine Function

y

 1,  

y

 0, 1 

x

1



y  cos x x

y  arccos x

  ,  1 y  cos x has an inverse function only on this interval.



1, 0 

1

Inverse Tangent Function

y

y

 y  tan x



2





2

2

x

y  tan x has an inverse function only on this interval.

46

x

y  arctan x

x



 2

Tips The ranges of the inverse functions   1.   arcsin x  2. 0  arccos x  

3. 

Compositions of Functions: 1. sin(arcsin x)  x

3. tan(arctan x)  x

2

2

2. cos(arccos x)  x

 2

 arctan x 

 2

PRACTICE 1. What is the value of tan  arctan(3)  ?

5  2. What is the value of arcsin  sin 3  3.

 ? 

2  tan  arccos   3 



 3   5 

4. cos  arcsin     



EXPLANATION 1.

Let arctan(3)  x.

2. sin

5 3 ,  3 2

3. arccos

 tan x  3 . Therefore tan  arctan(3)   tan x  3 .    5   3  arcsin  sin     arcsin      3 2 3      2 5   Therefore, tan  arccos   tan x  . 3 2 

2 2  x  cos x  3 3

3  4  3  3   Therefore, cos  arcsin      cos x  . 4. arcsin     x  sin x   5 5 5 5     

Answer: 1. 3

2. 

Dr. John Chung's SAT II Math Level 2

 3

3.

5 2

4.

4 5

47

Tips Sum and Difference of Angles

TIP 34

Functions of the Sum of Two Angles

Functions of the Difference of Two Angles

sin( A  B)  sin A cos B  cos Asin B

sin( A  B)  sin A cos B  cos Asin B

cos( A  B)  cos A cos B  sin Asin B

cos( A  B)  cos A cos B  sin Asin B

tan( A  B ) 

tan A  tan B 1  tan A tan B

tan( A  B ) 

tan A  tan B 1  tan A tan B

PRACTICE

1. If sin x 

1 1 and cos y  , where x and y are positive acute angles, what is the value of cos( x  y) ? 2 3

2. tan(180  y)  

3. If sin A  0.6, sin B  0.8,

2

 A   , and   B 

3 , what is the value of sin( A  B)? 2

EXPLANATION

1. cos( x  y )  cos x cos y  sin x sin y  2

3 1 1 2 2 32 2 .     2 3 2 3 6 3 y

1

x

1

3

2. tan(180  y ) 

tan180  tan y   tan y 1  tan180 tan y

3. sin( A  B)  sin A cos B  cos Asin B  (0.6)(0.6)  (0.8)(0.8)  0.28.

Answer: 1.

48

32 2 6

2.  tan y

3. 0.28

2 2

Tips Double Angle Formulas

TIP 35

Functions of the double angle

cos 2 A  cos2 A  sin 2 A

sin 2 A  2sin A cos A

tan 2 A 

2 tan A 1  tan 2 A

cos 2 A  1  2sin 2 A cos 2 A  2cos2 A  1

PRACTICE 1. If sin  

4  and     , what is the value of sin 2 ? 2 5

2. If sin x  

5 , then cos 2 x  13

3. If cos   

1  and     , then tan 2  2 2

EXPLANATION 24 3  4  3  1. In Quadrant II, cos    . sin 2  2 sin  cos   2        25 5  5  5  2

50 119  5 2  2. cos 2 x  1  2sin x  1  2     1  169 169  13  3. In Quadrant II, tan    3 . Therefore,

tan 2 

 

2  3 2 tan   1  tan 2  1   3

Answer: 1. 

24 25

Dr. John Chung's SAT II Math Level 2

2.

 

2

 3.

119 169

3.

3

49

Tips Half Angle formulas

TIP 36

Half Angle Formulas

sin

A 1  cos A  2 2

cos

A 1  cos A  2 2

tan

A 1  cos A  2 1  cos A

PRACTICE 1. If sin A  

4 3 1  and   A  , what is the value of cos  A  ? 2 5 2 

2. If cos x  

5  x and  x   , then what is the value of cos ? 2 2 13

3. If y is a positive acute angle and sin

y 1  , then what is the value of y ? 2 2

EXPLANATION A 1  cos A  A 3 3  1. Since (Quadrant II), then cos A   and cos     2 2 2 2 4 5



x  x 1  cos x   , cos    2. Since 2 2 4 2 2 3.

y  30 2



y  60  .

Answer: 1. 

50

 5 1    13   2 13 . 2 13

5 5

2.

2 13 13



3. 60

3 5  5 . 2 5

1

Tips TIP 37

Trigonometric Equation

PRACTICE 1. In the interval 0  x  360 , what is the value of x that satisfies the equation 4sin 2 x  4cos x  5  0 ?

2. In the interval 0  x  360 , what is the value of x that satisfies the equation cos2 x  2sin x  1  0 ?

3. In the interval 0  x  2 , what is the value of x that satisfies the equation 2sec2 x  3tan x  1  0 ?

4. What is the measure of the positive acute angle that satisfies the equation 1  sin x  2cos2 x ?

EXPLANATION 1. 4(1  cos2 x)  4cos x  5  0  4cos2 x  4cos x  1  0  (2cos x  1)2  0

1  x  60 , 300 2 2 2. (1  sin x)  2sin x  1  0  sin 2 x  2sin x  0  sin x(sin x  2)  0 cos x 

sin x  0 or sin x  2 (reject)

Therefore, x  0 , 180

3. Since sec2 x  1  tan 2 x , 2(1  tan 2 x)  3tan x  1  0  2 tan 2 x  3tan x  1  0

 2 tan x  1 tan x  1  0

 tan x 

1 or tan x  1 2

If tan x  0.5 , then x  tan 1 (0.5)  0.46 or x  0.46    3.61. 1

If tan x  1 , then x  tan 1  0.79 or x  0.79    3.93. Therefore, x  0.46, 3.60 , 0.79, 3.93 4. 1  sin x  2(1  sin 2 x ) sin x 

 2 sin 2 x  sin x  1  0



Remember:For these questions, You can use a calculator. Graph and find the zeros.

 2 sin x  1 sin x  1  0

1 or sin x  1 . Therefore, the positive acute angle is 30 . 2

Answer: 1. 60 , 300

Dr. John Chung's SAT II Math Level 2

2. 0 , 180

3. 0.46, 0.79, 3.61, 3.93

4. 30

51

Tips The Law of Sines

TIP 38

Area of a Triangle:

If ABC is a triangle with sides a, b, and c, then Area 

bc sin A ab sin C ac sin B .   2 2 2

C

ch c  b sin A bc sin A   2 2 2 h  b sin A

a

b

Area 

h

A

B c

Law of Sines:

If ABC is a triangle with sides a, b, and c, then a b c   sin A sin B sin C

PRACTICE 1. In  ABC , what is the ratio a : b if A  30 and B  45o ?

2. How many possible triangles can be constructed if a  10, b  12, and B  20 ?

52

Tips 3. How many distinct triangles can be constructed if the measures of two sides are 4 and 6 and the measure of the

angle opposite the smaller side is 30 ?

EXPLANATION 1. Law of sines: a b   sin 30 sin 45

2.

10 12  sin A sin 20

1 a b a 2     2  12 b 2 2 2 2 2  10sin 20  sin A   0.28 : A lies in Quadrant I or II. 12

A  sin 1 0.28  16 or 164o

10sin 20  3.4 Or

A

B

C

16

20

164

20

144 No triangle

4 6 6sin 30   B   0.75 : sin 4 sin 30 sin B

3.4 20

B

Therefore, only one triangle can be constructed. 3.

a  10

Since b  12 is greater than 10, one triangle.

B lies in Quadrant I or II.

A  sin 1 0.75  49 or 131

A

B

C

30

49

101





30

131

6sin 30  3 Or 6 4



19

4

3

30

Therefore, two triangles can be constructed.

Since 3  4  6, two triangles.

Answer: 1.

2 2

2. One triangle

Dr. John Chung's SAT II Math Level 2

3. Two triangles

53

Tips The Law of Cosines

TIP 39

Law of Cosines:

a 2  b 2  c 2  2bc cos A

cos A 

b2  c2  a2 2bc

b2  a2  c2  2ac cos B

cos B 

a 2  c 2  b2 2ac

c2  a2  b2  2ab cos C

cos C 

a 2  b2  c 2 2ab

PRACTICE 1. In  ABC , if a  3 and b  1 , and C  30 , then c 

2. If two forces of 30 pounds and 40 pounds act on a body with an angle of 120 between them, what is the magnitude of the resultant?

EXPLANATION 1. The Law of cosines:

c2 

 3

2

 1  2 2

 3  (1) cos 30



 c2  1  c  1

2. In a parallelogram, two consecutive angles are supplementary and opposite sides are  .

30

120

30

R

60 40

R 2  302  402  2(30)(40) cos 60  1300 , Answer: 1. c  1

54

2. 36.06 lb

R  1300  36.06lb

Tips Permutation

TIP 40

A permutation of a set of values is an arrangement where order is important.

The number of ways of obtaining an ordered r elements from n elements is given by n

Pr 

n!  n  r !

PRACTICE

1. In how many ways can a class with 15 students choose a president, a vice-president, and a treasurer?

2. In how many different orders can the program for a music concert be arranged if 6 students are to perform?

3. How many 8-letter arrangements can be made from the letters in the word PARALLEL?

EXPLANATION 1.

15 P3

2.

6 P6



15!  15  14  13  2730 12!

 6!  720

3. Permutations with repetition: This is a permutation of 8 letters taken at a time when 2A and 3L are identical.

Therefore,

8! 8 P8   3360 2!3! 2!3!

Answer: 1. 2730

Dr. John Chung's SAT II Math Level 2

2. 720

3. 3360

55

Tips Combination

TIP 41

A selection in which order is not important is called a combination. The number of combinations of n things taken r at a time is

1.

n Cn

n Cr



n Cr



1

n Pr

r!

or

 n  n Pr    r  r!

n!  n  r ! r !

2.

n Co

1

3.

n Cr

 n Cnr

PRACTICE 1. There are 10 novels and 6 biographies in a book reading list. If a student chose 5 novels and 3 biographies to read, how many different combinations can be chosen?

2. There are 6 boys and 5 girls in a chess club. In how many ways can 2 boys and 3 girls be selected to attend the regional tournament?

 n 3. If n Pr  120 and    20 , then what is the value of r ? r 

EXPLANATION 10  6  1.     5, 040  5  3   6  5  2.     150  2  3  n P 120  20 3.    n r  r!  r  r! Answer: 1. 5040

56

 2.

Therefore, r !  6  r  3 . 150

3. 3

Tips Dividing Group

TIP 42

When we divide group into several groups, 1) Each group has different number of people How many ways to divide group of 10 people into one group of 7 people and the other group f 3 people?

The number of ways  10 C7  3 C3  120 2) Some groups has same number of people How many ways to divide of 10 people into two groups of 3 people and one group of 4 people? The number of ways 

10 C 3

 7 C3  4 C 4  2100 2!

Because two groups has the same number of people,

10 C3  7 C3  4 C4

should be divided by 2!.

If the number of people n  a  a  a  b  b  c , then the number of ways divides into the groups of

a, a, a, b, b and c is  n  n  a  n  2a  n  3a  n  3a  b  n  3a  2b         a  a  a  b  b c    The number of ways  (3!)(2!)

PRACTICE 1. How many ways are there to divide 8 people into two groups of three people and one group of two people?

2. Eight people are divided into one groups of 5 people and the other group of 3 people. How many ways are there?

EXPLANATION  8  5  2  1 (56)(10)(1)  280 1.      2  3  3  2  2!  8  3  2.     56  5  3 

Dr. John Chung's SAT II Math Level 2

57

Tips Binomial Expansion Theorem

TIP 43

Formula for expanding  x  y  for positive integers n is n

 x  y n  n Co  x n  y 0  n C1  x n 1  y 1  n C2  x n  2  y 2      n Cn 1  x 1  y n 1  n Cn  x 0  y n 1. For any binomial expansion of

 x  y n , there are

n  1 terms.

2. The general term of the expansion is n Cr

 x n  r  y r

or

 n  nr r    x   y  , where r  0, 1, 2, 3  n. r  

PRACTICE 1. What is the third term of the expansion of  a  2b  ? 4

2. What is the middle term of the expansion of  x  2 y  ? 6

10

1   3. In the expansion of  x 2  2  , what is the value of the constant term? x  

EXPLANATION  4 2 2 1. The third term is    a   2b   6(2)2 a 2b2  24a 2b2 .  2 6 3 3 2. The middle term is    x   2 y   160 x3 y 3 . 3 10  3.   x 2 r 

  x  10  r

2 r

10  10  r r     1 x 20 2 r  2 r     1 x 20 4 r , 20  4r  0  r  5 r  r 

 10  5 Therefore, the constant term is    1  252 .  5 Answer: 1. 24a 2b2

58

2.

160x3 y 3

3.

252

Tips TIP 44

Sum of Coefficients of a Binomial Expansion

From Binomial Expansion Theorem below

 x  y n  n Co  x n  y 0  n C1  x n 1  y 1  n C2  x n  2  y 2      n Cn 1  x 1  y n 1  n Cn  x 0  y n , Sum of Binomial Coefficients (SBC) is n Co

 n C1  n C2  n C3  n C4    n Cn1  n Cn .

Since Binomial expansion is true for all real values of x and y, when you put x  1 and y  1 in the expansion,

1  1n  n Co 1n 10  n C1 1n 1 11  n C2 1n  2 12      n Cn 1 11 1n 1  n Cn 10 1n or

2n  n Co  n C1  n C2    n Cn 1  n Cn . Therefore, SBC  2n.

Example 1: What is the sum of coefficients in the binomial expansion of

 2 x  3 y 3

?

3 Putting x  1 and y  1 , SBC   2  3   1.

Check:

 2 x  3 y 3  8 x 3  36 x 2 y  54 xy 2  27 y 3

 SBC  8  ( 36)  54  ( 27)  1

PRACTICE 1. What is the sum of coefficients in the binomial expansion of 1  2 x  ? 8

2. What is the value of

n Co

 n C1  n C2  n C3  n C4    n Cn1  n Cn ?

EXPLANATION 1. Putting x  1: SBC  1  2    1  1 8

2.

8

 x  y n  n Co  x n  y 0  n C1  x n 1  y 1  n C2  x n  2  y 2      n Cn 1  x 1  y n 1  n Cn  x 0  y n Putting x  1 and y  1: SBC  1  1  2 n n

Answer: 1. 1

Dr. John Chung's SAT II Math Level 2

2.

2n

59

Tips Binomial Probability

TIP 45

If the probability of success is p and the probability of failure is 1  p  q , then the probability of exactly r successes in n independent trials is:

P  n Cr p r q n  r

PRACTICE 1. If a fair coin is tossed 10 times, what is the probability of rolling a head exactly 7 times?

2. If a fair coin is tosses 5 times, what is the probability of obtaining at most 3 heads?

3. A coin is loaded so that the probability of heads on a single throw is three times the probability of tails. What is the probability of at most 3 heads when the coin is tossed 6 times?

EXPLANATION 7

3

15 1 1  0.12 1. P  10 C7       2   2  128 0

5

1

4

2

3

3

2

26 1 1 1 1 1 1 1 1  0.8125 2. P  5 C0      5 C1      5 C2      5 C3      2 2  2  2 2 2  2   2  32 Or, binomcdf (5, 0.5, 3)  0.8125 (Graphing Calculator) 3. Since P(H )  0.75 and P(T )  0.25 , the probability is: 6 C0

 0.75 0  0.25 6  6 C1  0.75 1  0.25 5  6 C2  0.75 2  0.25 4  6 C3  0.75 3  0.25 3

= binomcdf (6, 0.75, 3) = 0.16943359  0.17 Answer: 1.

60

15 128

2.

13 16

or 0.8125

3. 0.17

Tips Probability with Combinations

TIP 46

If 5 cards are drawn at random from a standard deck, what is the probability that all 5 cards are hearts? Method 1:

Since there are 13 hearts in the 52-card deck, the probability that the first card drawn is a heart is probability that the second card drawn is a heart is P (all hearts) 

13 and the 52

12 . Therefore, continuing in this way, the probability is 51

13 12 11 10 9      0.000495 52 51 50 49 48

Method 2:

Using combinations:

P (all hearts) 

13 C5 52 C5

 0.000495

PRACTICE 1. If three marbles are picked at random from a bag containing 4 red marbles and 5 white marbles, what is the probability that exactly 2 marbles are red?

EXPLANATION

1. Method 1: There are three ways to pick: RRW ,

The probability of RRW :

RWR, WRR

4 3 5 5 5 5 . Each has the same probability. Therefore,    3 9 8 7 42 42 14

Method 2: Using combinations. P

 5 C1 30 5   84 14 9 C3

4 C2

Answer: 1.

5 14

Dr. John Chung's SAT II Math Level 2

61

Tips TIP 47

Heron’s Formula

Given any triangle with sides a, b, and c , the area of the triangle is given by Area 

, where s 

s ( s  a )( s  b )( s  c )

abc . 2

PRACTICE 1. Find the area of a triangle having sides of lengths 5, 12, and 15.

EXPLANATION 1.

Using Heron’s Formula: s

5  12  15  16 , 2

Area  16(16  5)(16  12)(16  15)  26.53

Using the law of cosines: In the figure, find angle  using the law of cosines. cos  

52  152  122 106  2(5)(15) 150

The area of the triangle 

62

 106    45.036  150 

  cos 1 

ab sin  2

(5)(15) sin 45.036   26.53 2

Answer: 1. 26.53





15

5 12

Tips TIP 48

Vectors in the Plane  

For a vector V  AB A



Directed Line Segment

Initial Point

B Terminal Point

1. If the vectors have the same magnitude and direction, then they are equivalent. 2. Magnitude of a vector is the length of the segment: Distance formula.  If A  (0, 0) and B  (3,4) , then AB  (3  0) 2  (4  0) 2  5 . 3. If a vector has an initial point in the origin, then the vector is in standard position.   V  (v1 , v2 ) is the component form in standard position and V  v12  v2 2 .





4. Unit vector u has a magnitude of 1 and the same direction as V .   V u   : Divide a vector by its length. V

PRACTICE  1. Find the unit vector in the direction of V  (3, 4) .

 2. If A is represented by the directed line segment from P  (2,3) to Q  (2, 8) , what is the magnitude of  vector A ?

Dr. John Chung's SAT II Math Level 2

63

Tips     3. If a  (2,3) and b  (4,6) , then what is the value of 2a  b ?

EXPLANATION 

1. V is in standard position and the initial point is at (0, 0).  V  (3)2  42  5   V (  3, 4) Therefore, u       0.6, 0.8  5 V 2.

 PQ  (2, 8)  (2, 3)  ( 4, 5)  PQ  (4)2  52  41 Or, use distance formula:

D  (2   2)2  (3  8)2  41   3. 2a  b  2(2,3)  (4,6)  (0,12)   Therefore, 2a  b  02  122  12

Answer: 1.

64

 0.6, 0.8

2.

41

3. 12

Tips Interchange of Inputs

TIP 49

Example 1. If f ( x)  5x  2, what is f ( x  5)? x

y y  f ( x)

Input

x5

Output

y y  f ( x  5)

Input

Output

Interchange of input means replacement x with x  5. If f ( x)  5x  2, then f ( x  5)  5( x  5)  2  5x  27. The best way to interchange the input is as follows.

Step 1: Change the input variable of original function: f (k )  5k  2 Step 2: Find relationship between two inputs: k  x  5 Step 3: Replace with a new input and simplify: f ( x  5)  5( x  5)  2  5x  27 x3 , then f ( x  1)  5 k 3 Step 1: f ( k  5)  Step 2: k  5  x  1  k  x  4 5 ( x  4)  3 x  1 Step 3: f ( x  1)   5 5

Example 2. If f ( x  5) 

PRACTICE

x 1. If f (2 x)  x2  5, what is f   ? 2

 x 1 2. If f    2 x  3, then f ( x)   4 

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Tips x  3. If f   5   x 2  1, then 2 

f (2 x  1) 

EXPLANATION 1. Step 1: f (2 x)  x 2  5

Step 2: 2 k 

f (2k )  k 2  5



x x  k 2 4 2

 x  x Step 3: f       5 2 4

2  x x f   5  2  16



 k 1 2. Step 1: f    2k  3  4 

Step 2:

k 1  x  k  1  4x  k  4x  1 4

Step 3: f ( x)  2(4x  1)  3



x  3. Step 1: f   5   x 2  1 2  

Step 2:

k  5  2x  1 2

 2

66

k  f   5  k2 1 2  



k  2x  4 2

Step 3: f (2 x  1)   4 x  8   1

Answer: 1.

f ( x)  8x  5

2 x x f   5  2  16



2.



k  4x  8

f (2 x  1)  16 x 2  64 x  63

f ( x)  8x  5

3.

f (2 x  1)  16 x2  64 x  63

Tips Polynomial Inequalities

TIP 50

Graphical Solution: 1. Solve

( x  4)( x  2)( x  2)  0 .

Step 1. Find critical points. ( x  4)( x  2)( x  2)  0  x  4, 2,  2 Step 2. Graph y  ( x  4)( x  2)( x  2) . y0

y0 x

y0

2

2

y0

4

Step 3) From the graph above, find the intervals for y  0 . The solution is 2  x  2 or x  4 . 2 2. Solve ( x  4)( x  2) ( x  2)  0 .

Step 1. Find the critical points. ( x  4)( x  2)2 ( x  2)  0  x  4,

2,  2

2

Step 2. Graph y  ( x  4)( x  2) ( x  2) . Zeros: x  4, x  2(two equal roots), and x  2. y0

y0 2

y0

2

y0

4

Step 3. Find the intervals for y  0 . The solution is 2  x  2 or 2  x  4 .

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Tips PRACTICE 1. Solve the inequality ( x  1)( x  2)2 ( x  1)( x  2)2  0 .

2. What is the solution set of

2 x3  3x2  24 x  12 ?

EXPLANATION 1. Graphical solution: use graphing utility. y

y0



2

y0



1



O

1

From the graph, the solution set is 2. By using a graphing utility,

y0



y0 x

2

 ,  1  1,  

y

y0

y0 3.06

x

O 0.48

4.08

The solution set is  3.06, 0.48    4.08,   . Answer: 1.

68

x  1 or x  1

2.

 3.06,

0.48    4.08,  

Tips TIP 51

Rational Inequalities

Graphic Solution: 1. Solve the rational inequality

( x  1)( x  2)  0. x2

Method 1. Test value using critical points

Critical points from the equation:

( x  1)( x  2)  0  x  1 and x  2 (closed) x2 x  2 (open)

Critical points from undefined: There are four possible solution set

2

1

2

Test value

Results

x  3

( 3  1)( 3  2) 0 3  2

No

x0

(0  1)(0  2) 0 02

Satisfied

x  1.5

(1.5  1)(1.5  2) 0 1.5  2

No

x3

(3  1)(3  2) 0 32

Satisfied

Solution: 2  x  1 or

x2

Method 2. Using the graph

Since

 x  2 2

 x  2 2

 0 , multiply both sides by  x  2  . 2

( x  1)( x  2) 2  0   x  2 x2





 2

1

 x  2 x  1 x  2  0



 2

From the graph above, the solution set is  2, 1   2,   .

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Tips PRACTICE

1. Solve the inequality

( x  2)2 ( x  3) 0. ( x  1)( x  2)

EXPLANATION 1. Method 1) Multiply both sides by ( x  1) 2 ( x  2)2 which is positive.

( x  1)( x  2)( x  2) 2 ( x  3)  0 and x  1 , x  2 y y  ( x  1)( x  2)( x  2)2 ( x  3)

  1  2  3

 2

x

Therefore, the solution set is  x  2  1  x  3 .

Method 2) Use a graphing utility.

y

2 

1





3

From the graph, the solution set is  x  2  1  x  3 .

70

x

Tips Limit

TIP 52

lim f ( x)  L x c

The limit of f ( x) is L as x approaches c . 1. If f ( x) is a polynomial function, then lim f ( x)  f (c) : direct substitution. x c

2. If f ( x) is a rational function given by

lim f ( x)  lim x c

x c

3. If f ( x ) 

N ( x) such that D(c)  0 , then D( x)

N ( x ) N (c )  D( x) D(c)

1 1 , then lim   or  : the limit does not exist. x 0 x x

x3  x x ( x 2  1)  lim  lim x  1 x 1 x 2  1 x 1 x 2  1 x 1

4. lim

1

 1 5. e  lim(1  x ) x or e  lim 1   x   x0 x

x

PRACTICE

1.

lim

x 1

2. lim

x 2

3. lim

x 1

4. lim

x 3

x2  x  6 = x2

x2  x  6  x2

x2  2 x  3 x 1



x3  x2  9

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Tips x4 2  x

5. lim

x 0

EXPLANATION x2  x  6 1  1  6   6 : (direct substitution) x 1 x2 1  2

1. Since D(1)  0 , lim

2. Since D(2)  0 and N (2)  0 , lim

x 2

( x  3)( x  2)  lim( x  3)  5 x 2 ( x  2)

3. Rationalizing Skill: Since D(1)  0 and N (1)  0 , then lim

 x  3 x  1 



x 1



x 1

  lim  x  3 x  1 

x 1



x 1

  lim  x  3

x 1

 x  1

x 1

x 1





x 1  8 .

 x  3 1 1  lim  . x 3  x  3 x  3 x 3 x  3 6

4. Since D(3)  0 and N (3)  0 , then lim 5. Rationalizing Skill: D(0)  0 and N (0)  0

lim



x 0

x4 2 x



x42

x4 2

Answer: 1.  6

72



 2. 5

  lim  x 0

x

x4



3. 8



2

 22

x42

4.

 1 6

 lim

x 0

x



x42

5.

1 4

x



 lim

x 0



1 x42





1 4

Tips TIP 53

Rational Function and Asymptote

A rational function can be written in the form R ( x) 

N ( x) , where N ( x) and D(x) are polynomials D( x)

and D( x)  0. 1. The line x  a is a vertical asymptote of the graph if D(a)  0 . 2. The line y  b is a horizontal asymptote of the graph if R( x)  b as x   or x  . y 1 Example 1: R ( x )  x Domain:  , 0   (0,  ) or x  0

Range:  , 0    0,    or y  0 Vertical asymptote: x  0 Horizontal asymptote: y  0 2x x 1 D( x)  0 , vertical asymptote: x  1

Example 2: f ( x) 

2x  2 , horizontal asymptote: y  2 x 1 Domain: x  1 Range: y  2 lim

x

O

y

2

x 

Example 3: g ( x) 

1

2x x 1

x

2

y

D( x)  0  x 2  1  0 , no vertical asymptote. 2x  0 , horizontal asymptote: y  0 . x  x  1 Domain: all real x Range: 1  y  1 (using graphing utility) lim

1

2

O

x

1

3. Slant asymptote (or, oblique asymptote): If the degree in the numerator is greater than the degree in denominator, the original function should be rearranged by long division. 2 x 2  3x  1 6 f ( x)   2x  5   The slant asymptote is y  2x  5 . x 1 x 1

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Tips y Example 4: h( x) 

h( x ) 

2

x  2x  3 x2  1

 x  3  x  1  x  3   x  1  x  1  x  1

D( x )  0 y  lim

x 

(1, 2) y 1

 x  1:vertical asymptote

O

x

x3  1  y  1:horizontal asymptote x 1

f (1) is still undefined. Domain: (,  1)  (1, 1)  (1, )

x  1

Range : (, 1)  (1, 2)  (2, )

Example 5: f ( x) 

x2  x x 1

f ( x)  x  2 

D( x)  0

2 (Use long division) x 1



y  lim ( x  2)  x 

x  1 : vertical asymptote

2  ( x  2) x 1



y  x  2 : slant asymptote

Domain: (,  1)  (1, )

y

Range: (,  5.82]  [0.17, ) (using graphing utility)

y x2 O

x  1

74

x

Tips PRACTICE 1. Find all asymptotes of the rational function f ( x) 

2. If g ( x ) 

2 x2  1 ? x

x2  x  2 , then its vertical asymptote(s) is(are) x2  2 x  3

3. If the vertical asymptote of the rational function R ( x ) 

x 2  3x  b is x  1 , what is the value of b ? ( x  1)( x  1)

EXPLANATION 1. Since f ( x)  2 x 

2. g ( x) 

1 , vertical asymptote: x  0 , and slant asymptote: y  2x . x

( x  2) ( x  1) ( x  3) ( x  1)



( x  2) ( x  3)



vertical asymptote: D( x)  0  x  3

3. Since x  1 is not asymptote, the numerator must have a factor of ( x  1) . Therefore,

f (1)  0



Answer: 1. x  0 ,

1 3  b  0



y  2x

2.

Dr. John Chung's SAT II Math Level 2

b2 x  3

3.

b2

75

Tips Parametric Equations

TIP 54

Parametric equations define a relation using parameters. Conversion from two parametric equations to a single equation: Eliminating the parameter from the simultaneous equations.

Parametric Solve for t in Substitute in Rectangular    equations one equation second equation equation Example 1: If x  a cos t and y  b sin t , what is the graph of the parametric equations?

cos t 

x a

and sin t 

y b

To eliminate the parameter t ,

cos2 t  sin 2 t  1



x2 y 2   1 : Ellipse a2 b2

PRACTICE 1. What is the curve given by the parametric equations x  t and y  t  2 ?

2. What is the graph given by the parametric equations x  sec and y  tan  ?

EXPLANATION 1. t  x2  y  x2  2, x  0 (Domain is restricted because x  t ) 2. Since 1  tan 2   sec2  , 1  y 2  x 2  x 2  y 2  1: hyperbola y

y  x 2  2, ( x  0)

2 O

76

 x Answer: 1. Parabola

2. Hyperbola

Tips Polar Coordinates

TIP 55 y

(r ,  )

Polar Coordinates and Rectangular Coordinates Conversions x  r cos 

r

 O

y  r sin  y tan   x

y

x

x

   tan 1

y x

r 2  x2  y 2  r  x2  y 2

PRACTICE   1. If the coordinates of a polar point are  5,  , what are its rectangular coordinates?  2

2. If the rectangular coordinates are ( 3, 1) , what are its polar coordinates?

Equation Conversion: 3. What is the graph of the polar equation r  5 ?

4. What is the graph of the polar equation r  sec  ?

5. Convert the polar equation r 

6 to rectangular form. 2  3sin 

6. Convert the rectangular equation 3x  6 y  2  0 to polar form.

Dr. John Chung's SAT II Math Level 2

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Tips 7. Convert the rectangular equation x2  y 2  6 x  0 to polar form.

EXPLANATION

1. x  r cos  5cos

 3

2. r 

2

 2

 0 and y  r sin x  5sin

 2

 5. Therefore, (0,5) .

   1      30   . Therefore,  2,   . 6 6   3

  tan 1 

 12  2 ,

3. r  x 2  y 2  5  x 2  y 2  25 4. r  sec 

Therefore, r 

5. r 

sec 

1 x , cos  cos  r



sec 

r x

r  x 1. x

6 6  r  2r  3 y  6  2r  3 y  6 y 2  3sin  23 r





2 x2  y 2  3 y  6  4 x2  y 2   3 y  6 

2

 4 x 2  4 y 2  9 y 2  36 y  36

Therefore,

4 x 2  5 y 2  36 y  36  0. Hyperbola 6. 3x  6 y  2  0  3(r cos )  6(r sin  )  2  0  r (3cos  6sin  )  2

r

2 3cos   6sin 

7. x2  y 2  r 2 and x  r cos  . Therefore, r 2  6r cos  r  6cos .

Answer: 1. (0, 5)

  2.  2,   or 6 

 11   2,  6  

5. 4 x2  5 y 2  36 y  36  0 , Hyperbola

78

3. x2  y 2  25 , circle 6. r  

2 3cos   6sin 

4. x  1 7. r  6 cos 

Tips Matrix

TIP 56

A. Order of Matrix: An m  n ( m by n ) matrix is a rectangular array of numbers arranged in m rows (horizontal lines) and n columns (vertical lines). 5 2  2 1    1 5 

This matrix has three rows and two columns. The order of the matrix is 3  2 (3 by 2).

B. Addition of Matrices: If two matrices have the same order, then you can add two matrices by adding their corresponding entries.

 1  2 

3 2  2 1  2   4  1  3   2  1

3  2 1  4  3  3

1 1

C. Scalar Multiplication:

 1  3  3 3  4   6  2

9 12 

D. Scalar Multiplication and Matrix Subtraction:

 1 If A    2

2  2 4 and B     , then 2 A  3 B is 3  3 2 

 1 2  2 4  2 2  3    2 3  3 2   4

4  6 12  4  8   6   9 6  13 0

E. Multiplication of Matrices: Row by Column Multiplication The number of columns of the first matrix must equal to the number of rows of the second matrix.

 2  3 1   2  1  (3)  3 7     1 2  3 (1)  1  2  3   5  

Dr. John Chung's SAT II Math Level 2

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Tips 

A m n  Order of A

B



n p Order of B

AB m p  Order of AB

F. Determinant of a 2  2 matrix: a c

b  ad  bc d

PRACTICE  2  3 1. What is the determinant of the matrix A   ? 2  1

2. If matrices A, B, C, and D have orders of 2  3, 2  3, 3  2, and 2  2 , respectively, what are the orders of following operations?

II) D( A  3B)

I) A(2B)

III) ( BC  D) A

EXPLANATION

1.

2 3  2  2  (3)(1)  4  3  1 1 2

2. 1) A(2 B ) 

 2  3 2  3 :

2) D( A  3B) 

Invalid

 2  2 2  3   2  3

3.  BC  D  A  BC : 

 2  3 3  2   2  2 , BC  D : ( BC  D) A :   2  2  (2  3)  (2  3)

80



 2  2

Tips TIP 57

Inclination Angle

Inclination angle is the angle measured counterclockwise from the x-axis to the line.

y y  mx  b m  tan 

  Inclination angle O

   Inclination Angle:    

x

Horizontal line:   0 Vertical line :   90 Positive slope :   Acute angle Negative slope:   Obtuse angle

PRACTICE 1. If a line has an equation of 2 x  3 y  5 , what is the inclination angle of the line?

EXPLANATION 2 5 2 x  m 3 3 3 2 2 tan      tan 1  33.69006753  33.7 3 3

1. 2 x  3 y  5  y 

Answer: 1. 33.7o

Dr. John Chung's SAT II Math Level 2

81

Tips Angle between Two Lines

TIP 58

y

y  m2 x  b2 y  m1 x  b1



2 1 x

O

1.    2  1

2. tan 1  m1

tan   tan  2  1  

3. tan  2  m2

tan  2  tan 1 m  m1  2 1  tan 2 tan 1 1  m2 m1

 m  m1  Therefore,   tan 1  2 .  1  m2 m1 

PRACTICE 1.

Find the angle between the two lines y   x  2 and y  2 x  1 .

EXPLANATION y

y  x  2

y  2x  1

1 O

2

m2  tan 2  1 and m1  tan 1  2

   2  1 x

tan  

m2  m1 1  2  3 1  m1m2 1  (1)(2)

  tan 1 3  71.56505118  71.6 Or, find the angles directly.  2  tan 1 (1)  135 and 1  tan 1 (2)  63.4    135  63.4  71.6

82

Tips TIP 59

Intermediate Value Theorem

If f is continuous on a closed interval [ a, b] and k is any number between f (a) and f (b), then there is at least one number c in  a, b such that f (c)  k. 1. If f (a)  0 and f (b)  0 on a closed interval  a, b , then there are at least one real zeros somewhere between a and b .

PRACTICE 1. If P( x)  x3  4 x2  6 , what are the intervals needed to guarantee real zeros?

(Use the Intermediate Value Theorem to find intervals of length 1.)

PRACTICE 1. Using a graphic utility,

Since f (2)  18  0 and f (1)  1  0 , there is a zero between  2 and 1.

f (1)  3  0 and f (2)  2  0 , there is a zero between 1and 2. f (3)  3  0 and f (4)  6  0 , there is a zero between 3 and 4.

Answer: 1.  2,  1 , 1, 2 ,  3, 4 

Dr. John Chung's SAT II Math Level 2

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Tips TIP 60

Rational Zero Test

If the polynomial f ( x)  an x n  an 1 x n 1     a1 x  ao has integer coefficients, the possible rational zeros of f are

Possible Rational Zeros =

factors of constant term factors of leading coefficient

PRACTICE 1. What is all possible rational zeros of f ( x)  x3  x  1 ?

2. Find the possible rational zeros of f ( x)  2 x3  3x 2  8 x  3 .

PRACTICE

1.

Factors of constant term 1   1 Factors of leading coefficients 1 By testing these zeros, you can see that neither works. f (1)  1 and f (1)  1 Therefore, you can conclude that the polynomial function has no rational zeros. The real zero is x  1.324718    irrational root.

2.

1,  3 1 3  1,  3,  ,  2 2 1,  2 By using a graphing utility, you can find the real rational zeros of f are 1 x  1, x  , and x  3 . 2 Answer: 1.  1

84

1 3 2.  1,  3,  ,  2 2

Tips TIP 61

Descartes Rule of Sign

If f ( x)  an x n  an 1 x n 1     a1 x  ao with real coefficients and a0  0, 1. The number of positive zeros of f is either equal to the number of variations

in sign of f ( x) or less than the number by an even integer. 2. The number of negative zeros of f is either equal to the number of variations

in sign of f ( x) or less than the number by an even integer.

PRACTICE 1. Find the possible real zeros of f ( x)  4 x3  6 x 2  3x  3 ?

PRACTICE 3 2 1. Check the number of variations in sign of f ( x)  4 x  6 x  3x  3

 to   to   to     4  6 6  3 3  3 Three variations in sign.

3 2 Check the number of variations in sign of f ( x)  4 x  6 x  3x  3 No variations in sign.

Conclusion: 1) Three positive zeros and no negative zeros. or 2) One positive real zero and no negative zeros. (One positive real zero and two imaginary roots)

Answer: 1. Three positive real zeros or one positive real zero and has no negative real zeros.

Dr. John Chung's SAT II Math Level 2

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No Test Material on This Page

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Test 1                                               Dr. John Chung’s                SAT II Mathematics Level 2

Dr. John Chung's SAT II Math Level 2 Test 1

87

2

 

2

    

MATHEMATICS LEVEL 2 TEST REFERENCE INFORMATION    

 

THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME   OF THE QUESTIONS IN THIS TEST   1   Volume of a right circular cone with radius r and height h : V   r 2 h 3   1 Lateral Area of a right circular cone with circumference of the base c and slant height  : S  c 2  

  Volume of a sphere with radius r : V  4  r 3 3   Surface Area of a sphere with radius r : S  4 r 2   1   Volume of a pyramid with base area B and height h : V  3 Bh

       

88

2

 

2 Dr. John Chung’s SAT II Math Level 2

Answer Sheet 01 02

A

B

C

D

E

A

B

C

D

E

 

03

A

B

C

D

E

B

C

D

 

04 05

A A

B

C

 

06

A

B

 

07

A

08

26 27

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

E

28 29

A

B

C

D

E

D

E

30

A

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C

D

E

C

D

E

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B

C

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31 32

A

B

C

D

E

A

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C

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33

A

B

C

D

E

09

A

B

C

D

E

34

A

B

C

D

E

10 11

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B

C

D

E

A

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C

D

E

A

B

C

D

E

35 36

A

B

C

D

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12

A

B

C

D

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37

A

B

C

D

E

13 14

A

B

C

D

E

B

C

D

E

A

B

C

D

E

38 39

A

 

A

B

C

D

E

 

15

A

B

C

D

E

40

A

B

C

D

E

 

16 17

A

B

C

D

E

A

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C

D

E

A

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C

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41 42

A

B

C

D

E

 

18

A

B

C

D

E

B

C

D

E

 

19 20

B

C

D

E

43 44

A

A

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

 

21 22

45 46

A

B

C

D

E

47

A

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C

D

E

 

23

A

B

C

D

E

48

A

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C

D

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24

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D

E

49

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25

A

B

C

D

E

50

A

B

C

D

E

              

 

       

 

       

     

           The number of right answers:                              

                

           The number of wrong answers:                             

 

          

 

# of correct

Dr. John Chung's SAT II Math Level 2 Test 1

1    4 # of wrong

Raw score

 

89

2

 

2 Score Conversion Table 6

Scaled Score 480

630

5

470

26

620

4

470

800

25

620

3

460

46

800

24

610

2

460

45

800

23

610

1

450

44

800

22

600

0

450

43

800

21

600

42

800

20

590

41

800

19

590

40

780

18

580

39

760

17

570

38

750

16

560

37

740

15

550

36

720

14

540

35

710

13

530

34

700

12

520

33

690

11

510

32

680

10

500

31

670

9

490

30

660

8

490

29

650

7

480

28

Scaled Score 640

800

27

48

800

47

50

Scaled Score 800

49

Raw Score

90

Raw Score

Raw Score

 

2

2

  

MATHEMATICS LEVEL 2 TEST For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding circle on the answer sheet Note: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions in this test. For each question you will have to decide whether or not you should use a calculator. (2) For some questions in this test you may have to decide whether your calculator should be in the radian mode or the degree mode. (3) Figures that accompany problems in this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers x for which f ( x) is a real number. The range of f is assumed to be the set of all real numbers f ( x) , where x

is in the domain of f . (5) Reference information that may be useful in answering the questions in this test can be found on the page preceding Question 1. USE THIS SPACE FOR SCRATCHWORK 1.

If a( x  2)  b( x  1)  3 for all x , then a  (A) 1

2.

(C) 1

(D) 2

(E) 3

If a  b  2 and ab  1 , then a 2  b 2  (A) 4

3.

(B) 0

(B) 5

(C) 6

(D) 8

(E) 10

If the graph of 3x  4 y  5 is perpendicular to the graph of kx  2 y  5 , then k  (A) 2 (B) 2.67 (C) 2.15 (D) 3.20 (E) 4

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  Dr. John Chung's SAT II Math Level 2 Test 1

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

4.

If K 

(A) (B) (C) (D) (E)

5.

AB , then B  A B

A 1 A AK A K AK KA Ak A A K AK

If log 3  a , then log 90  (A) 1  2a (B) (C) (D) (E)

10a 2 10  2a 30a 10  3a

` 6.

If f ( x)  3ln x and g ( x)  e x , then g  f ( x)   (A) 3x (B) e x (C) e2x (D) x3 (E) x 2  1 20 cm

7.

In Figure 1, the slant height of a regular circular cone is 20 cm and the radius of the base is 10 cm . Find the volume of the cone? (A) 1813.8 cm3

(B) 3000.5 cm3 (C) 4120.4 cm3

(D) 7024.8 cm3

(D) 7046.6 cm3

10 cm Figure 1

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

8.

If 2  i is one of the zeros of the polynomial p ( x ) , then a factor of p ( x) could be (A) x 2  2 (B) x 2  4 (C) x 2  4 x  4 (D) x 2  4 x  5 (E) x 2  4 x  3

9.

When a polynomial function f ( x)  x 2  5 x  k is divided by ( x  2) , the remainder is 5. What is the value of k ? (A) (B) (C) (D) (E)

10.

19 18 16 10 9

Figure 2 shows a cube with edge of length 6, what is

P

the length of diagonal PQ ? (A) 18 (B) 15

Q

(C) 6 6 (D) 6 3 Figure 2

(E) 6 2

y

11.

An equation of line  in figure 3 is (A) 3x  4 y  4  0 (B) 3x  4 y  4  0



4

(C) 4 x  3 y  4  0 (D) 4 x  3 y  12  0 (E) 4 x  3 y  12  0

O

3

x

Figure 3

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

12.

The mean score of 10 students of an algebra class was 85. When two new students enrolled, the mean increased to 86. What was the average of the new students? (A) (B) (C) (D) (E)

13.

88 89 90 91 92

If sin   cos  

1 , then tan   cot   2

(A) 4.12 (B) 2.67 (C) 1.35 (D) 2.67 (E) 4.12 14.

 3x   What is the period of the function f , where f ( x)  5  cos   2  (A) 3 2 (B) 3 4 (C) 3

 ? 

(D) 2 (E)

15.

3 2

Find all of the asymptotes of y 

x 2  3x  2 . x2  1

(A) x  1 (B) x  1, x  1 (C) x  1, y  1 (D) x  1, y  1 (E) x  1, x  1, and y  1

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

16.

If f (2 x)  (A)

17.

2x 3

x , then f ( x)  3 x x (B) (C) 3 6

(D) 3x (E) 6x

If three numbers log a, log b, and log c in that order form an arithmetic progression, which of the following is true? (A) b  ac (B) c  ab (C) b 

ac 2

(D) b 2  ac (E) b 2  a  c 18.

The point of set ( x, y ) such that x 2  y 2  0 is (A) (B) (C) (D) (E)

19.

What is the sum of the geometric series 5 5 5 10  5        ? 2 4 8 (A) (B) (C) (D) (E)

20.

A circle An ellipse A hyperbola A point Two lines

30 25 20 19 18

Which of the following is an equation whose graph is the set of points equidistant from the point (0, 4) and (2, 2) ? (A) y  2 (B) x  2 (C) y  x (D) y  x  2 (E) y   x  2 GO ON TO THE NEXT PAGE

Dr. John Chung's SAT II Math Level 2 Test 1

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

21.

The solution set of 3x  5 10 is (A)

 x  5

(B)

 x  5

 5  (C)   x  5  3  5  (D)  x  5 or x    3  (E)  x  5 or x  15 y

22.

In the graph in Figure 4, which of the following could be the equation of the graph? (A) y  sin x  3 (B) y  sin 2 x

3

(C) y  3sin x  3 (D) y  3sin 2 x  3 1 (E) y  3sin x  3 2

23.

O

4

x

Figure 4

What is the range of f ( x)  x2  5x  6 ? (A) y  0 (B) y  0.25 (C) 2  y  3 (D) y  5 (E) All real numbers

24.

If sin( x  10)  cos(2 x  28) , then which of the following could be the value of x ? (A) (B) (C) (D) (E)

0 10 20 24 30

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

25.

Which of the following is the equation of the inverse of

y  3x 1 ? (A) y  log x 1 3 (B) y  log( x)  1 x (C) y  log3   3 (D) y  log3  x   1

(E) y  log 3 ( x  1)

26.

If f ( x)  x2  kx  9 is always greater than 0 for all real x , which of the following could be the value of k ? (A) 10

27.

(B) 5

(C) 10

(D) 12

(E) 13

Which of the following is the center of a circle

x2  y 2  10 x  6 y  10 ? (A) (B) (C) (D) (E)

28.

 5, 3  5,3 10,6   10, 6  10, 10 

What is the domain of the function defined by 2 x  10 ? f ( x)  2 x  3x  2 (A) (B) (C) (D) (E)

All real numbers except 1 All real number except 2 All real numbers except 1 and 2 x5 x5

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

29.

If x2  y 2  20 and x2  y , then x  (A) (B) (C) (D) (E)

30.

3 2, 2 4 10 4,10

If f ( x)  log 2  x  1  2 , then f 1  x   (A) 2 x1 (B) x 2  1 (C) 2 x 2 (D) 2 x 2  1 (E) 22 x  2 x  1

31.

Figure 5 shows a right triangle. If the length of AD is 8 and

B

the length of BD is 4, then the length of AC is (A) (B) (C) (D) (E)

32.

D

8 A

C

Figure 5

If complex number z  4  6i , then z  2  (A) (B) (C) (D) (E)

33.

15.23 16.42 17.89 18.44 20.25

4

5.87 6.32 6.38 7.21 7.31

The coefficient of the middle term of the expansion of

 x  3 y 6

is

(A) 540 (B) 270 (C) 135 (D) 270 (E) 540

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

34.

If

x2  6 x  9  x , which of the following is the solution?

(A) x   (B) x 

1 2

(C) x   (D) x 

3 2

1 4

1 4

(E) No solution 35.

A sequence is defined as a1  1 and an 1  an  3 . Which of the following is the n th term of this sequence? (A) 3n  2 (B) 4 n  3 (C) 5n  4 (D) 6n  5 (E) n 2  n  1

36.

If f ( x)  x2  2 x  4 for x  0 , what is the value of f 1  4  ?

(A) (B) (C) (D) (E) 37.

2 4 5 8 10

In triangle ABC , a  6, b  6 3 , and A  30 . What is the measure of C ? (A) 30  only (B) 90  only (C) 30  or 90  (D) 120 only (E) 60  or 120

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

38.

If log 2 5  a and log 2 3  b , then log 2 75  (A)

2a  b

(B) 2a  b 1 a  b 2 1 (D)  a  2b  2 1 (E) a  b 2

(C)

39.

Which of the following could be the graph of the parametric equations represented by x  3sin  and y  4cos ? (A)

y

O

(C)

x

y

O

(E)

(B)

O

(D)

x

y

x

y

O

x

y

O

x

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100

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

40.

If the three sides of a triangle are 4, 5, and 7, then the area of the triangle is (A) 9.80

41.

If

6 7 8 10 12

50 68 70 75 80

If the 11th term of an arithmetic sequence is 30 and the 21st term is 0, what is the 10th term of the sequence? (A) (B) (C) (D) (E)

44.

n?

The radius of the base of a right circular cone is 5 and the slant height of the cone is 10. What is the surface area of the cone? (A) (B) (C) (D) (E)

43.

(D) 11.43 (E) 11.56

 n  1!  56 , what is the value of  n  1!

(A) (B) (C) (D) (E)

42.

(B) 9.92 (C) 10.2

lim

30 33 36 39 42

n2  1

n 1

(A) 0

n 1



(B) 1

(C) 2 (D) 4 (E) Limit does not exist.

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

45.

If one of the roots of a polynomial function f ( x)  0 is 1

3 i , which of the following could be f ( x) ? 2

(A) 2 x 2  4 x  6 (B) x 2  2 x  10 (C) 3 x 2  6 x  4 (D) 4 x 2  8 x  7 (E) 5 x 2  10 x  7

46.

If p( x)  x3  2 x2  mx  n is divisible by x 2  3 x  2 , what is the value of m ? (A) (B) (C) (D) (E)

47.

2 1 0 1 2

If f ( x)  cos x and g ( x)  x2  3 , which of the following is not true? (A) f ( x)  g ( x)  2 is an even function. (B) f ( x)  g ( x) is an even function. (C) g  f ( x)  is an even function. (D) f ( x)  g ( x)  2 is an even function. (E) f ( x  1)  g ( x  1) is an even function

48.

Under which of the following conditions is x 0 ?  x  1 x  2  (A) x  0 or x  2 (B) x  2 or x  1 (C) x  1 or x  0 (D) x  2 or 0  x  1 (E) 2  x  0 or x  1

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102

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

49.

What are the asymptotes of the hyperbola whose equation is

16 x2  25 y 2  400  0 ? 16 x 25 25 y x 16 4 y x 5 5 y x 4 16 y x 5

(A) y   (B) (C) (D) (E)

50.

If kx 2  4 x  2  0 have two different real roots, which of the following is the values of k ? (A)

k  2

(B)

k  2

(C) 2  k  0 or k  0 (D) 2  k  2 (E) 1  k  3

STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK.

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103

No Test Material on This Page

104

2

2 S ANSWERS

TEST 1 # 1 2 3 4 5 6 7 8 9 10

answer C C B B A D A D E D

# 11 12 13 14 15 16 17 18 19 20

answer E D B C D C D E C D

# 21 22 23 24 25 26 27 28 29 30

answer C E B D C B A E B D

# 31 32 33 34 35 36 37 38 39 40

answer C B A E A A C E D A

# 41 42 43 44 45 46 47 48 49 50

answer B D B D D B E D C C

Explanations: Test 1 1. (C)

a( x  2)  b( x  1)  3  (a  b) x  2a  b  3 To be equal for all numbers x : a  b  0 and 2a  b  3. Therefore, ab0

2a  b  3 3a  3 2. (C)

 a 1

(a  b) 2  a 2  b 2  2ab  22 and ab  1 . Therefore, a 2  b 2  4  2ab  4  2(1)  6.

3.

(B)

4.

(B)

5. (A)

3 5 k 5 3x  4 y  5  y   x  and kx  2 y  5  y   x  . 4 4 2 2 To be perpendicular, the product of the two slopes should be 1. 8  3  k  3k  2.67.  1 . k  Therefore,       3  4  2  8

AB , KA  KB  AB or KA  AB  KB  ( A  K ) B. A B AK . Therefore, B  A K Since K 

Since log 90  log  3  3  10   log 3  log 3  log10 , log 3  a and log10  1, log 90  a  a  1  2a  1.

6. (D)

3

g  f ( x)   e3ln x  eln x  x3

Dr. John Chung's SAT II Math Level 2 Test 1

105

2 7. (A)

2 202  102  300  10 3.

By Pythagorean Theorem the height of the cone is

Therefore, the volume of the cone is

 r 2h

20cm

20

3

10 3 10

2





 10  10 3 3

  1813.8cm . 3

10 cm

Figure 1

8. (D)

Since 2  i is one of the roots, the other root must be 2  i (conjugate). Therefore, the equation with these two roots is  x  (2  i )  x  (2  i )   ( x  2)  i  ( x  2)  i   0. ( x  2)2  i 2  0  x 2  4 x  4  1  0  x 2  4 x  5  0

9. (E)

Remainder Theorem x 2  5 x  k  ( x  2)Q( x)  R , where Q( x) is quotient and R  5. When x  2 , both sides have the same number. Since 22  5(2)  k  5,  k  9 Or, (1) use long division. x 7

x  2 x2  5x  k

Therefore, k  9.

2

x  2x 7x  k 7 x  14 k  14  5 (2) Synthetic Division.

2

1 5 k 2 14 1 7

Therefore, the remainder is 14  k  5  k  9 .

14  k 62  62  62  108  6 3.

10. (D)

The length of diagonal PQ is

11. (E)

If a and b are x- and y-intercept respectively, the equation of the line is Since

12. (D)

106

x y   1. a b

x y   1, the equation is 4 x  3 y  12. Therefore, 4 x  3 y  12  0. 3 4

The sum of scores of 10 students is 85  10  850. Let the average of score of two student be x .

2

2 Since

13. (B)

14. (C)

850  2 x  86  850  2 x  12(86)  2 x  182 , x  91. 12

sin  cos  sin 2   cos 2  1    ------- (1) cos  sin  sin  cos  sin  cos  1 3  sin  cos   ------- (2)  sin   cos 2  1  2sin  cos   4 8 Substitute (2) into (1) 1 8  3 3  8 tan   cot  

The frequency of the periodic function is

3 , because the function can be expressed as follow. 2

3  f ( x)  5  cos  x   , where the coefficient of x is the frequency. 2 3 2 2 4   Therefore, the period of the function is . f 32 3 15. (D)

16. (C)

( x  2)( x  1) x  2  , there is no asymptote at x  1. ( x  1)( x  1) x  1 Denominator : x  1  0  x  1 (vertical asymptote) x2  1 (horizontal asymptote) For other asymptote: y  lim x  x  1 Since y 

Since f (2 x) 

x (2 x)  , 3 6

 f ( x) 

x . 6

Or, to avoid confusion, change the function to f (2k ) 

x k . Let 2k  x , then k  . 2 3

x x 2  . Substitute the function in terms of x . f ( x)  3 6 17. (D)

Since log a, log b, and log c are arithmetic progression, log c  log b  log b  log a . c b c b    b 2  ac. By log operation, log  log b a b a

18. (E)

x 2  y 2  0   x  y  x  y   0  y   x or y  x . That represent two lines.

19. (C)

a(1  r n ) , where a is the first term and r is the n  1  r a 10 10   20. . Therefore, S  common ratio. When r  1 , the sum is S  1 1 1 r 1 2 2 The sum of geometric series is S  lim

Dr. John Chung's SAT II Math Level 2 Test 1

107

2 20. (D)

2 Line m which is equidistant from two points must be perpendicular bisector of the line segment. 42 02 42 The midpoint is   1 ,   1,3 and the slope of the segment is 02 2   2 y

m

 (0, 4)



22. (E)

(2, 2) x

O

21. (C)

Therefore, the slope of line m is 1. The equation of line m is y  3  1( x  1)  y  x  2 .

3x  5  10

  10  3 x  5  10   5  3x  15 

5  x5 3

From the graph, the middle line is 3, amplitude is 3, and period is 4 . The frequency is 2 1  . 4 2 y  A sin( Bx)  C

A : amplitude

B : frequency

C : middle line

Period 

2 B

1 Therefore, the equation of the trigonometric graph is y  3sin x  3 . 2 23. (B)

The graph of y  x 2  5 x  6 is concave up which has a minimum on axis of symmetry. b 5 The axis of symmetry is x    2.5 . Therefore, the minimum of y is 2a 2 f (2.5)  (2.5) 2  5(2.5)  6  0.25 . The range is y   0.25 . Or, by completing squared form, y  ( x  2.5) 2  0.25 . You can use a graphing calculator.

24. (D)

Cofunction: Since sin( x  10) and cos(2 x  28) are cofunctions, x  10  2 x  28  90. Therefore, 3x  72  x  24.

25. (C)

Switch x and y , then express y in terms of x. Therefore, x  3 y 1

x  y  1  log 3 x  y  log3 x  1  log3 x  log3 3  log3 . 3

26. (B)

Since f ( x) is greater than 0 for all x , f ( x)  0 cannot have x-intercept ( zeros). Therefore, discriminant must be negative (imaginary roots). D  b 2  4ac  k 2  4(1)(9)  0  (k  6)(k  6)  0   6  k  6

27. (A)

Complete squared form: x 2  10 x  25  y 2  6 y  9  10  25  9  ( x  5) 2  ( y  3)2  44 Therefore, the center is (5, 3) .

108

2 28. (E)

2 Domain of the polynomial function is 2 x  10  0 and ( x  2)( x  1)  0 . Therefore,  x  5   x  2,1   x  5 .

29. (B)

Substitute y  x 2 into x 2  y 2  20. Then x 4  x 2  20  0  ( x 2  5)( x 2  4)  0 Since x 2  5  0 , x 2  4  0  x  2 or x  2 .

30. (D)

Switch x and y : x  log 2  y  1  2  x  2  log 2 ( y  1) Therefore, y  1  2 x  2

31. (C)

33. (A)

34. (E)

y  2x2  1 .

By the formula, BD  DC  82  DC  AC 2  CD  CB

32. (B)



64  16 4

 AC 2  16  20  320  AC  320  17.89

z  2  4  6i  2  2  6i ,

2  6i  22  (6)2  40  6.32

The middle term is the 4th term. The 4th term is 6 C3 ( x)3 (3 y )3  540 x3 y 3 .



x2  6 x  9

Or, since



2

 x 2  x  1.5 But x  0 . No solution.

x 2  6 x  9  ( x  3) 2  x  3 , then x  3  x .

If x  3 , x  3  x  3  0 No solution 3 3 But is not less than 3 . No solution If x  3 ,  x  3  x  x  2 2 35. (A)

From the recursive equation an 1  an  3 , it is arithmetic sequence with common difference 3 and the first term 1. Therefore, an  a1  (n  1)d  1  (n  1)(3)  3n  2.

36. (A)

From original function, you can find f 1 (4). The input of f 1 was the output of f . Since, x 2  2 x  4  4  x 2  2 x  8  0  ( x  4)( x  2)  0 , x  2 ( x  0) . This value of x is the output of f 1 . Therefore, f 1 (4)  2.

37. (C)

sin 30 sin B 6 3 sin 30 3   sin B   6 6 2 6 3 Therefore, B  60 or 120. Now find C .

The law of sines:

A : 30

30

B : 60 C : 90

120 30

Dr. John Chung's SAT II Math Level 2 Test 1

C  90 or 30.

109

2

2

38. (E)

Since log 2 5  a and log 2 3  b , 1 1 1 log 2 75  log 2 75   log 2 5  5  3   log 2 5  log 2 5  log 2 3 2 2 2 1 1   2a  b   a  b 2 2

39. (D)

Parametric equation: x  3sin  and y  4 cos  From cos 2   sin 2   1 (Pythagorean identity), sin   x2 y 2  1 32 42

40. (A)

41. (B)



x y , and cos   3 4

It is ellipse and major axis is on the y -axis. Graph (D) is correct.

Three sides of a triangle are given. Heron’s formula: The area of the triangle is abc A  s ( s  a )( s  b)( s  c) , where s  2 457 Therefore, s   8 and the area is 8(8  4)(8  5)(8  7)  9.80 . 2 (n  1)! (n  1)n (n  1) !  (n  1)n ,  (n  1)! (n  1) !

n 2  n  56  (n  8)(n  7)  0

Since n  8 , n  7. 42. (D)

10

 180 o 50

10

5

25  5

Since the circumference of the base is 10 and the circumference of the lateral side is 20 , lateral side is exactly a semicircle which has a central angle 180 . Therefore, the surface area is 25  50  75 . Or, use the formula S f   rs   r 2 , s is a slant height.  (5)(10)   (52 )  75 43. (B)

110

Since a11  a  10d  30 and a21  a  20d  0 , a  10d  30 a  20d  0  d  3 and a  60.  10d  30 Therefore, a10  a  9d  60  27  33. Or, since a10  (3)  a11 , then a11  30  (3)  33 .

2

2 n2  1

44. (D)

lim

45. (D)

Since 1 

x 1





n 1

 lim

x 1

(n  1)(n  1)( n  1) ( n  1)( n  1)

 lim

x 1

(n  1) (n  1)( n  1) (n  1)



2 2 4 1

i 3 i 3 , the other root mist be 1  (conjugate). Therefore, 2 2

  i 3    i 3   i 3 i 3  i 3 2 0   x  1     x  1      x  1   x  1     x  1        2     2    2  2   2    7 When simplified, x 2  2 x   0 is equivalent to 4 x 2  8 x  7  0 . 4

2

Therefore, f ( x)  4 x 2  8 x  7. Or, 1) Use sum and product of two root. The equation is defined by ax 2  bx  c  0  x 2 

b c x 0. a a

b b  i 3   i 3   1    1    2   2 a a  2   2  c  i 3  i 3  7 Product of roots   1  1   a  2  2  4  7 When substitute, x 2  2 x   0  4 x 2  8 x  7  0. 4

Sum of roots 

2) Because the coefficients of the equation is real, let x  1  x 1 

i 3 i 3  ( x  1)2    2  2 

2

Therefore, the equation is x 2  2 x  46. (B)

 x2  2x  1  

i 3 . 2

3 4

7  0  4 x 2  8 x  7  0. 4

Remainder Theorem: Let x3  2 x 2  mx  n  ( x  1)( x  2)Q ( x) , where Q( x) is the quotient. When x  1 , 1  2(1)  m(1)  n  0  m  n  1 ----- (1) When x  2 , 8  2(4)  m(2)  n  0  2m  n  0 ------ (2) (2)  (1) m  1 and n  2 . Or, you can solve it using long-division and synthetic division.

Dr. John Chung's SAT II Math Level 2 Test 1

111

2

2 x 1 2

3

x  3x  2 x  2 x 2  mx  n x3  3x 2  2 x x 2  (m  2) x  n

Therefore, m  1 and n  2 .

2 x 2  3x  (m  1) x  (n  2)

47. (E)

If f ( x) and g ( x) are even, f  g  even, f  g = even, and f  g = even. (E) is not even, because it is translated to the right by 1.

48. (D)

Multiplying both sides by ( x  1) 2 ( x  2)2

b  b 2  4ac , it will be simplified as follows. 2a

x  0  ( x  1) 2 ( x  2) 2 ( x  1)( x  2) y  x( x  1)( x  2)  0 Using graphic solution, ( x  1) 2 ( x  2)2



2



0



1



x

From the graph, the value of y is negative for x  2 or 0  x  1 . 49. (C)

Asymptote of hyperbola: 16 x 2 25 y 2 x2 y 2  1   1 400 400 25 16 4 Since a  5 and b  4, the asymptote is y   x. 5 16 x 2  25 y 2  400 

50. (C)

END

112

In order to have two different real roots, discriminant D  b 2  4ac  0 . Since 16  4k (2)  0 , k  2. But the leading coefficient of quadratic equation cannot be zero. k  0 . Therefore, 2  k  0 or k  0.

Test 2                                               Dr. John Chung’s                SAT II Mathematics Level 2

Dr. John Chung's SAT II Math Level 2 Test 2

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2

 

2

    

MATHEMATICS LEVEL 2 TEST REFERENCE INFORMATION    

 

THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME   OF THE QUESTIONS IN THIS TEST   1   Volume of a right circular cone with radius r and height h : V   r 2 h 3   1 Lateral Area of a right circular cone with circumference of the base c and slant height  : S  c 2  

  Volume of a sphere with radius r : V  4  r 3 3   Surface Area of a sphere with radius r : S  4 r 2   1   Volume of a pyramid with base area B and height h : V  3 Bh

       

114

2

 

2 Dr. John Chung’s SAT II Math Level 2

Answer Sheet 01 02

A

B

C

D

E

A

B

C

D

E

 

03

A

B

C

D

E

B

C

D

 

04 05

A A

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06

A

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07

A

08

26 27

A

B

C

D

E

A

B

C

D

E

A

B

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D

E

E

28 29

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D

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D

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30

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31 32

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33

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09

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34

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10 11

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35 36

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12

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37

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13 14

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D

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B

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38 39

A

 

A

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15

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40

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16 17

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41 42

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18

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19 20

B

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43 44

A

A

A

B

C

D

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D

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A

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21 22

45 46

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47

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23

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48

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24

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49

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25

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50

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D

E

              

 

       

 

       

     

           The number of right answers:                              

                

           The number of wrong answers:                             

 

          

 

# of correct

Dr. John Chung's SAT II Math Level 2 Test 2

1    4 # of wrong

Raw score

 

115

 

2

2 Score Conversion Table 6

Scaled Score 480

630

5

470

26

620

4

470

800

25

620

3

460

46

800

24

610

2

460

45

800

23

610

1

450

44

800

22

600

0

450

43

800

21

600

42

800

20

590

41

800

19

590

40

780

18

580

39

760

17

570

38

750

16

560

37

740

15

550

36

720

14

540

35

710

13

530

34

700

12

520

33

690

11

510

32

680

10

500

31

670

9

490

30

660

8

490

29

650

7

480

28

Scaled Score 640

800

27

48

800

47

50

Scaled Score 800

49

Raw Score

116

Raw Score

Raw Score

2

2

 

  

MATHEMATICS LEVEL 2 TEST For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding circle on the answer sheet Note: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions in this test. For each question you will have to decide whether or not you should use a calculator. (2) For some questions in this test you may have to decide whether your calculator should be in the radian mode or the degree mode. (3) Figures that accompany problems in this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers x for which f ( x) is a real number. The range of f is assumed to be the set of all real numbers f ( x) , where x

is in the domain of f . (5) Reference information that may be useful in answering the questions in this test can be found on the page preceding Question 1. USE THIS SPACE FOR SCRATCHWORK 1.

If

x2  1  3x  5 , then x  3  x 1

(A) 3 (B) 2 (C) 0 (D) 2 (E) 4

2.

The slope of a line which contains the points  a  3,  4  1 and  6a  2, 6  is  . What is the value of a ? 2

(A) 3 (B) 2 (C) 2 (D) 3 (E) 5           

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  Dr. John Chung's SAT II Math Level 2 Test 2

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

3.

What is the equation of a line whose x -intercept is 5 and y -intercept is  4 ? (A) 4 x  5 y  15 (B) 5 x  4 y  20 (C) 4 x  5 y  15 (D) 4 x  5 y  20 (E) 4 x  5 y  20

4.

If c  5  4c  1 , what is the value of c ? (A) 2 (B) 12 (C) 2 or 12 (D) 2 (E) 12

5.

If (a  b)  (a  b)i  1  5i , where a and b are real numbers, then a  (A) (B) (C) (D) (E)

6.

If z1  1  2i and z2  3  5i , then z1  z2  (A) (B) (C) (D) (E)

7.

0 1 2 3 4

10 5 4 3 2

If 90    180 and sin  

1 , then sin  2   2

(A) 0.87 (B) 0.94 (C) 0.60 (D) 0.87 (E) 0.94

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

8.

If figure 1 shows the graph of y  f ( x) , then which of the following is the graph of y  f (  x ) ? y

(A)

y

y  f ( x)

y

(B)

O x

O

x

x

O

Figure 1 (C)

(D)

y

x

O

x

O

y

(E)

x

O

9.

y

If f ( x)  (A) 

1 2

3x 4  3x3  2 x  1000 , what values does f ( x) approach as x gets infinitely larger? 5 x 4  2000 (B) 0

(C)

1 2

(D)

3 5

(E) Infinite

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

10.

In the graph of Figure 2, the equation of the graph is y  A cos  Bx  C   D . What is the value of B ?

y

3

(A) 6 (B) 3 (C) (D)



O

6

6

x

 3

3

(E) 6

11.

3

Figure 2

 3  cos      2  (A) sin  (B) cos (C)  sin  (D)  cos (E) 2sin  cos 

12.

In how many ways can 15 people be divided into two groups, one group with 10 and the other with 5 people? (A) (B) (C) (D) (E)

13.

3003 6006 12000 48600 9018009

If the polar equation is r  sin  , which of the following represents the graph? (A) (B) (C) (D) (E)

An ellipse A circle A line A parabola A hyperbola

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

14.

Figure 3 shows a parallelogram with sides 10 and 15. If

Q

Q  135 , what is the area of the parallelogram? (A) (B) (C) (D) (E)

15.

10 P

15 Figure 3

S

Figure 4 shows a rectangular solid. If the area of face I is 15, the area of face II is 20, and the area of face III is 18, then what is the volume of the solid? (A) (B) (C) (D) (E)

16.

53.03 66.02 83.33 106.07 121.67

R

135

67.48 73.48 88.98 96.76 101.44

Figure 4

If 180    270 and cos   0.707 , what is the value of   tan   ? 2 (A) (B) (C) (D) (E)

17.

2.414 2.414 3.424 3.424 1.414

What is the smallest positive value of x, in radian that satisfies the equation 3sin x  sin 2 x  0 ? (A) (B) (C) (D) (E)

0.01 2.14 3.14 6.28 9.42

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

18.

Figure 5 shows the graph of y  log 2  2 x  . What is the sum

y

of the areas of the three shaded rectangles? (A) (B) (C) (D) (E)

24 32 36 42 48

y  log 2  2 x 

O

2

4

8

12

x

Figure 5 Note: Figure not drawn to scale.

19.

The current population of Lake Pond is 50,000. The population at any time t can be calculated by the function

P(t )  Ae0.025t , where A is initial population and t is the time in years. How many years would it take for the population to reach half the present population? (A) (B) (C) (D) (E)

20.

12.4 15.5 18.4 24.6 27.7

Which of the following are asymptotes of the graph of x2  1 ? y x ( x  1) I. x  0 II. x  1 III. y  1 (A) (B) (C) (D) (E)

I only II only I and II only I and III only I, II, and III

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

21.

What is the range of the function defined by

 x  1 , x  1 f ( x)   2  1  x , x  1 (A) y  1 (B) y  0 (C) y  0 (D) 1  y  1 (E) All real numbers 22.

If x  3  y  2  0 and x  y  2  0 , then x  (A) (B) (C) (D) (E)

23.

0.5 1.0 1.5 2.0 2.5

If the 5th term of a geometric sequence is 24, and the 7th term is 144, what is the first term of the sequence? (A) 2 3 2 2 (C) 3 1 (D) 3 1 (E) 4 (B)

24.

If a cube is inscribed inside a sphere of radius 10, what is the volume of the cube? (A) 1539.60 (B) 1450.56 (C) 1300.48 (D) 1148.04 (E) 1200

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

25.

If sin x  2cos x  0 , where 0  x  2 , then x could equal (A) (B) (C) (D) (E)

26.

1.017 1.107 2.034 2.412 3.003

Figure 6 shows a cube with edge 10. What is the area of triangle ABC ? (A) (B) (C) (D) (E)

129.90 88.83 86.60 82.37 50.00

A

C B

Figure 6 2

27.

 n  1!  2  n  1!

(A) n  1 (B) n 2 (C) n 2  n (D) n2 (n  1)2 (E) (n  1)(n  1)

28.

For all  , sin(90   )  cos(180   )  tan(180   )  (A) sin  (B) cos  (C) tan  (D) sin   cos  (E) cos   sin 

GO ON TO THE NEXT PAGE

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

29.

If f (2 x  3)  4 x  2 , f ( x)  (A) (B) (C) (D) (E)

30.

x4 x4 2x  4 3x  5 3x  5

If a circle is defined by the equation x 2  10 x  y 2  2 y  10 , what are the coordinates of the center of the circle? (A) (B)

 5, 1  5,  1

(C) (5, 1) (D) (E)

31.

If x , 12 , 3x  6 ,… are the first three terms in a geometric progression, then the 5th term could be (A) (B) (C) (D) (E)

32.

 5,  1  10,  2 

1.5 36.5 40.5 60 96

The multiplicative inverse of

1 i is 3i

1  2i 10 (B) 2  i

(A)

(C) 1  2i (D) 2  i 3i (E) 2

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

33.

 1  1   The value of sin  2cos      2   (A)

1 3

(B)

3 2

(C) 

3 2

(D)

2 2 3

(E) 

34.



b0 a0 b0 b0 d 0

( 1, 0)

O



(1, 0)

Figure 7

A solution for the equation cos 2 x  2sin x  7  0 , where 0  x  2 , is (A) 4

36.

y

Figure 7 shows the graph of p( x)  ax3  bx 2  cx  d . Which of the following must be true. (A) (B) (C) (D) (E)

35.

2 2 3

(B) 2

(C)

 4

(D) 

 2

(E) No solution

What is the period of the graph of y  2sin  3 x     1 ? (A)  2 3 2 (C) 3

(B)

(D)

 3

(E) 2 GO ON TO THE NEXT PAGE

126

x

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

37.

Which of the following could be the graph of f ( x)  x   x  ? (A)

(B)

y

y

x

x

(C)

(D)

y

y

x

x

(D) y

x

38.

If a function f has the property of the fundamental period such that f ( x )  f ( x  2) , which of the following could be f ? (A) 2sin x  1 (B) sin 2x (C) cos 2 x  1 (D) 3tan 2x (E) 4 tan

 2

x GO ON TO THE NEXT PAGE

Dr. John Chung's SAT II Math Level 2 Test 2

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

39.

h

lim

h0

4h 2

(A) (B) (C) (D) (E)

0 2 4 Infinite Undefined



A

40.

In Figure 8, if BD  10 , what is the length of AC ? (A) (B) (C) (D) (E)

41.

20 25.12 34.64 36 36.56

B

60 D 10 Figure 8

30

C

Which of the following is the equation of the common chord of the circles with equations x 2  y 2  16 and

x 2  y 2  8 x  8 y  16  0 ? (A) y  x (B) y   x (C) y  x  4 (D) y  x  4 (E) y   x  4 42.

Which of the following functions are odd? I. f ( x)  x 4  5 x 2  3 II. f ( x)  3x3  5 x  1 III. f ( x)  x3  x (A) (B) (C) (D) (E)

I only II only III only II and III I, II, and III

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

43.

Figure 9 is a tetrahedron such that OB  OA  OC and OA  4 , OB  3 , and OC  4 . What is the area of  ABC ? (A) (B) (C) (D) (E)

11.66 16.42 18.44 20.25 21.32

B

O C

A 44.

Figure 9

3log9 18  (A) 3 2 (B) 3 3 (C) 2 5 (D) 2 6 (E) 2 7

45.

If sin x  t for all x in the interval

 2

 x  ,

then sin 2x  (A) (B) (C)

t 1 t2

t2 1  t2 1

t 1 t

(D) 2t 1  t 2 (E)

2t t 2  1

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129

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

46.

 If a vector a  (5,  12) , then which of the following is the unit  vector of a ? (A) 1, 1  1 1  (B)  ,   5 12  (C)  0.38,  0.92  (D)  0.5,  1 (E)  0.25,  0.75  y

47.

In Figure 10, if OH is perpendicular to the line 3x  4 y  28 ,

3x  4 y  28

what is the length of OH ? (A) (B) (C) (D) (E) 48.

49.

5.6 6.5 8.0 8.5 8.7

H

If matrix A has dimension m  n , matrix B has dimension p  m , and matrix C has dimension n  p , which of the following must be true? (A) (B) (C) (D) (E)

The product The product The product The product The product

x

O

Figure 10

AB exists. BC exists. ABC exists. CBA exists. BCA exists.

If the height of a cylinder is increased by 10 percent, by what percent must the radius of the circular base be increased so that the volume of the cylinder is increased by 25 percent? (A) (B) (C) (D) (E)

5.6% 6.2% 6.6% 7.5% 7.7% GO ON TO THE NEXT PAGE

130

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

50.

If the binomial expansion is defined by

 x  y n  n C0 x n y o  n C1 x n 1 y1  n C2 x n 2 y 2  ....  n Cn xo y n , then (A)

n C1

 n C2  n C3  ....  n Cn 

n(n  1) 2

(B) n 2  n (C) 2n1 (D) 2 n  1 (E) 2n  2n1

STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK.

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131

No Test Material on This Page

132

2

2

TEST 2 # 1 2 3 4 5 6 7 8 9 10

ANSWERS

answer C A E B D B A D D D

# 11 12 13 14 15 16 17 18 19 20

Explanations: 1. (C)

Since

answer C A B D B B C B E D

# 21 22 23 24 25 26 27 28 29 30

answer E C C A C C D C C A

# 31 32 33 34 35 36 37 38 39 40

answer C C B C E C B E C C

# 41 42 43 44 45 46 47 48 49 50

answer E C A A D C A D C D

Test 2

x 2  1 ( x  1) ( x  1)   x  1 , then x  1  3 x  5  x  3. x 1  x  1

Therefore, x  3  3  3  0. 2 1 6  4 10 2 , then   . Therefore, a  3.   a 1 2 6a  2  (a  3) 5a  5 a  1

2. (A)

Since

3. (E)

The equation is

x y   1 . Therefore, 4 x  5 y  20. 5  4 

Or, y  mx  4 passes through a point (4,0) . Therefore, by substitution, m  4. (B)

 c  5 2  

4c  1



2

4 . 5

 c 2  14c  24  0  (c  2)(c  12)  0

c  2 , 12. From the equation, c  5 . Hence c  12. 5. (D)

Complex number identity: a  b 1 ab 5 Hence a  3. 2a  6

6. (B)

z1  z2  1  2i  3  5i  4  3i ,

7. (A)

z1  z2  4  3i  42  (3) 2  5

 1  3 2 sin   . 1 in the second quadrant  90    180  , then cos   2 2  3 2  1   3   3 Therefore, sin  2   2sin  cos   2     0.87.   2 2 2   

Dr. John Chung's SAT II Math Level 2 Test 2

133

2

2

8. (D)

y  f ( x) is a reflection with y -axis . (D) is correct.

9. (D)

3 x 4  3 x 3  2 x  1000 3  x  5 5 x 4  2000

10. (D)

From the graph, period is 6. Frequency B 

11. (C)

 3   3 cos      cos  2    2

12. (A)

15 C10

13. (B)

Since sin  

lim



15 C5



  3  cos   sin    2

15

C5 

2 2    . P(period) 6 3

  sin   0  cos    1 sin    sin  

15  14  13  12  11  3003 5!

y y , r  sin   r   r 2  y. r 2  x 2  y 2 in rectangular coordinates. r r

Therefore, x 2  y 2  y  x 2  y 2  y  0 represents a circle. 2

1 1  The equation of the circle is x 2   y    . 2 4   14. (D)

The area  10  15  sin135  106.07

15. (B)

Let the dimensions of the solid be a, b, and c. The volume of the solid is V  abc. Therefore, the areas of each face are as follows. Multiply both sides and find the volume. ab  15 bc  20 ca  18 V  abc  5400  73.48. (abc) 2  5400

16. (B)

  cos 1 (0.707)  135 , but   225 where 180    270 . Therefore, tan

225  2.414. 2

17. (C)

3sin x  sin 2 x  0  3sin x  2sin x cos x  0  sin x(3  2 cos x)  0 3 (  1  cos x  1) . Therefore, sin x  0  x   , 2 ,... and cos x  2 The smallest positive number is   3.14.

18. (B)

Sum of the areas: 2  log 2 4   4  log 2 8   4  log 2 16   2 log 2 22  4 log 2 23  4 log 2 24



 4  12  16  32

134

 

 



2 19. (E)

2 Calculator will be needed. 25,000  50, 000e0.025t 

20. (D)

21. (E)

1 ln 0.5  e0.025t  t   27.7 2 0.025

( x  1)( x  1) x  1 , then  x( x  1) x Denominator: x  0 (Vertical asymptote) x 1 y  lim  1 (Horizontal asymptote). x  x Since y 

The range of f ( x) is all real from the graph below. Ry   ,   y

1

y

x 1 x

O 1

y   x2  1

22. (C)

If x  3 , x  3  y  2  0  x  y 1  0

 x  y  2  0

No solution.

3  0 If x  3, x  3  y  2  0   x  y  5  0   x  y  2  0 Therefore, x  2 x  3  0

3  1.5. 2

23. (C)

Geometric sequence: a5  ar 4  24 a 24 2 ar 6 , 7  4  r 2  6 . Therefore, a(6)2  24  a   . 6 36 3 a7  ar  144 a5 ar

24. (A)

Let the length of a side of the cube be x . Since the length of the diagonal of the cube is equal 20 to the diameter of the sphere, x 2  x 2  x 2  x 3  x 3  20  x  . 3 3

 20  Therefore, the volume of the cube is x     1539.60 .  3 3

25. (C)

sin x  2cos x  0  sin x  2cos x 

sin x  2  tan x  2 cos x

x  tan 1 (2)  1.1071 . Therefore x  2.035 or 5.176 in the interval  0, 2  . Or, use your graphing calculator.

Dr. John Chung's SAT II Math Level 2 Test 2

135

2 26. (C)

2  ABC is equilateral and the length of a side is 10 2 . Therefore, the area of  ABC is 10 2  10 2  sin 60  86.60 . 2 2

27. (D)

28. (C)

 n  1! 2   (n  1)n   (n  1) 2 n 2 2  n  1!

Trigonometry identities: sin(90   )  cos(180   )  tan(180   )  tan180  tan     sin 90 cos   cos 90sin     cos180 cos   sin180sin       1  tan180 tan    cos   ( cos  )  ( tan  )  tan 

29. (C)

Since f (2 x  3)  2(2 x  3)  4 , then f ( x)  2 x  4. Or , change variable : f (2k  3)  4k  2 and let x  2k  3 and k 

x3 . Then replace 2

with x.

 x  3 Therefore, f ( x)  4    2  2x  4 .  2  30. (A)

x 2  10 x  y 2  2 y  10 

 x  5 2   y  12  36

Center is at (5, 1). 31. (C)

The common ratio between two consecutive terms of geometric progress is equal. 12 3x  6 hence . From the equation  x 12 3x 2  6 x  144  0  x 2  2 x  48  0  ( x  8)( x  6)  0



x  8 or 6 .

4

3 3 and a5  8    40.5 . 2 2 12 If x  6 , then r   2 and a5  (6)(2) 4  96. 6

If x  8 , then r 

1 i 3i (3  i )(1  i ) 2  4i is    1  2i . 3i 1 i (1  i)(1  i ) 2

32. (C)

The multiplicative inverse of

33. (B)

Use your calculator. 1 1 Or,   cos 1  cos   , where 0     . 2 2  3  1  3 3 Because sin   , sin 2  2sin  cos   2  .     2 2 2 2    3 . Or,   60 and sin 2  sin120  2

136

2 34. (C)

2 From the graph, a  0 and the sum of the roots is

b  (1  0  1)  0  b  0 . a

d  (1)(0)(1)  0  d  0. a Or, find p (0)  d  0, p (1)  a  b  c  0, and p(1)  a  b  c  0 , and solve it.

And product of the root is

35. (E)

36. (C)

cos 2 x  2sin x  7  0  (1  sin 2 x)  2sin x  7  0  sin 2 x  2sin x  8  0 (sin x  4)(sin x  2)  0  sin x  4 or sin x  2. No solution.

1  y  2sin(3 x   )  1  y  2sin 3  x    1 3  From the equation frequency is 3 and period is P 

37. (B)

2 2  . 3 3

f ( x)  x   x  . For the following intervals,

 0  x  1, f ( x)  x  1  x  2, f ( x)  x  1    2  x  3, f ( x)  x  2  ... (B) is correct.

38. (E)

f ( x)  f ( x  2) is a periodic function with the fundamental period 2. 2 2 (A) period  2 (B) period    (C) period   2 2 (D) period 

39. (C) 40. (C)

lim

h 0



(E) period 

2

h 4h 2

 lim

h 0 (

 2  2

h( 4  h  2) 4  h  2)( 4  h  2)

A

 lim

h 0

h ( 4  h  2) 4 h

In the figure, AD  20 and AB  10 3 . AC  20 3  34.64. 20

10 3

60o B

41. (E)

42. (C)

10

D

30o

C

From the equation x 2  y 2  8 x  8 y  16  0 and x 2  y 2  16 16  8 x  8 y  16  0  8 x  8 y  32  0  y   x  4 f ( x)  x3  x : odd function + odd function = odd function

Dr. John Chung's SAT II Math Level 2 Test 2

137

2 43. (A)

2 By Pythagorean Theorem, AC  5 , BC  5, and 4 2 .

5

5

The height is The area is

52  (2 2)2  17 .

4 2  17  11.66 2

4 2

Or, use Heron’s formula. A  s ( s  a )( s  b)( s  c) , where s  44. (A)

By the logarithmic operation : log a b  log 3log9 18  3log3

45. (D)

18

If sin x  t on

a

abc . 2

b  log a 2 b 2

 18  3 2

 2

 x   , then cos x 

 1  t2 . 1

Therefore, sin 2 x  2sin x cos x  2(t )( 1  t 2 )  2t 1  t 2 .

46. (C)

   a Since unit vector u   and a  52  (12)2  13 , then a  (5, 12)  5 12  u = ,   (0.38,  0.92) . 13  13 13 

47. (A)

The distance from (0,0) to the line 3x  4 y  28  0 is D 

48. (D)

(A) AB   m  n  p  m  doesn’t exist

3(0)  4(0)  28 2

3 4

2



28  5.6. 5

(B) BC   p  m  n  p  doesn’t exist

(C) ABC   m  n  p  m  n  p  doesn’t exist

(D) CBA   n  p  p  m  m  n  exist

(E) BCA   p  m  n  p  m  n  doesn’t exist

49. (C)

138

Let the volume be V   r 2 h and new radius be R . Since h increased by 10% and R 2 1.25 h V increased by 25%, 1.25V   R 2 (1.1h)  1.25 r 2 h   R 2 (1.1h)  2  1.1 h r R 1.25   1.066  1  0.066 Therefore, the radius is increased by 6.6%. r 1.1

2 50. (D)

2 In order to find the sum of the coefficients of the binomial expansion, substitute x  y  1 .

(1  1)n  n C0  n C1  n C2  ......  n Cn 2n

 1  n C1  n C2  ......  n Cn

2n  1  n C1  n C2  ......  n Cn END

Dr. John Chung's SAT II Math Level 2 Test 2

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No Test Material on This Page

140

Test 3                                               Dr. John Chung’s                SAT II Mathematics Level 2

Dr. John Chung's SAT II Math Level 2 Test 3

141

2

 

2

    

MATHEMATICS LEVEL 2 TEST REFERENCE INFORMATION    

 

THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME   OF THE QUESTIONS IN THIS TEST   1   Volume of a right circular cone with radius r and height h : V   r 2 h 3   1 Lateral Area of a right circular cone with circumference of the base c and slant height  : S  c 2  

  Volume of a sphere with radius r : V  4  r 3 3   Surface Area of a sphere with radius r : S  4 r 2   1   Volume of a pyramid with base area B and height h : V  3 Bh

       

142

2

 

2 Dr. John Chung’s SAT II Math Level 2

Answer Sheet 01 02

A

B

C

D

E

A

B

C

D

E

 

03

A

B

C

D

E

B

C

D

 

04 05

A A

B

C

 

06

A

B

 

07

A

08

26 27

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

E

28 29

A

B

C

D

E

D

E

30

A

B

C

D

E

C

D

E

A

B

C

D

E

B

C

D

E

31 32

A

B

C

D

E

A

B

C

D

E

33

A

B

C

D

E

09

A

B

C

D

E

34

A

B

C

D

E

10 11

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

35 36

A

B

C

D

E

12

A

B

C

D

E

37

A

B

C

D

E

13 14

A

B

C

D

E

B

C

D

E

A

B

C

D

E

38 39

A

 

A

B

C

D

E

 

15

A

B

C

D

E

40

A

B

C

D

E

 

16 17

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

41 42

A

B

C

D

E

 

18

A

B

C

D

E

B

C

D

E

 

19 20

B

C

D

E

43 44

A

A

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

 

21 22

45 46

A

B

C

D

E

47

A

B

C

D

E

 

23

A

B

C

D

E

48

A

B

C

D

E

24

A

B

C

D

E

49

A

B

C

D

E

25

A

B

C

D

E

50

A

B

C

D

E

              

 

       

 

       

     

           The number of right answers:                              

                

           The number of wrong answers:                             

 

          

 

# of correct

Dr. John Chung's SAT II Math Level 2 Test 3

1    4 # of wrong

Raw score

 

143

 

2

2 Score Conversion Table 6

Scaled Score 480

630

5

470

26

620

4

470

800

25

620

3

460

46

800

24

610

2

460

45

800

23

610

1

450

44

800

22

600

0

450

43

800

21

600

42

800

20

590

41

800

19

590

40

780

18

580

39

760

17

570

38

750

16

560

37

740

15

550

36

720

14

540

35

710

13

530

34

700

12

520

33

690

11

510

32

680

10

500

31

670

9

490

30

660

8

490

29

650

7

480

28

Scaled Score 640

800

27

48

800

47

50

Scaled Score 800

49

Raw Score

144

Raw Score

Raw Score

2

2

 

  

MATHEMATICS LEVEL 2 TEST For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding circle on the answer sheet Note: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions in this test. For each question you will have to decide whether or not you should use a calculator. (2) For some questions in this test you may have to decide whether your calculator should be in the radian mode or the degree mode. (3) Figures that accompany problems in this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers

x for which f ( x) is a real number. The range of f is assumed to be the set of all real numbers f ( x) , where x is in the domain of f . (5) Reference information that may be useful in answering the questions in this test can be found on the page preceding Question 1. USE THIS SPACE FOR SCRATCHWORK 1.

If a( x  1)  b( x  1)  2  0 for all real x , then a  (A) (B) (C) (D) (E)

2.

2 1 0 1 2

If i  1 , which of the following is a negative integer?

(A) i 24 3.

(B) i 33 (C) i 46

(D) i 55

(E) i 72

If f (2)  0 and f (1)  0 , which of the following must be a factor of f ( x) ?

(C) x 2  x  2

(A) x  2

(B) x  1

(D) x 2  x  2

(E) None of these

GO ON TO THE NEXT PAGE

  Dr. John Chung's SAT II Math Level 2 Test 3

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

4.

5.

If

x 2  x , then the solution consists of

(A) (B) (C) (D) (E)

Zero only Positive real numbers only Negative real numbers only All real numbers No real numbers

If x 

2 , then 9 x 2  4 is equivalent to 3

(A) 9 x 2  4 (B) 9 x 2  4 (C) 4  9x 2 (D) 4  9x 2 (E) 3 x  2 6.

If 1 is a root of the equation kx 2  6 x  4  0 , then the other root is (A) 4

7.

(B) 2

(C) 0.4

(D) 0.4

(E) 2.5

y

Figure 1 shows the graph of the equation 2 x  3 y  9  0 . What is the value of  ? (A) 33.7 

2x  3 y  9  0



(B) 34.2

O



(C) 37.8

x

(D) 38.1 (E) 40.6 8.

If f ( x)  x  1 and

Figure 1

f

 g  ( x)  x 2  3 x  2 , which of the

following is g ( x ) ? (A) x  2 (B) x 2  3 x (C) x 2  3 x  1 (D) x 2  3 x  1 (E) ( x  1)2 ( x  2)

GO ON TO THE NEXT PAGE

146

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

9.

If x  3 is a factor of 2 x 3  kx  3 , then k  (A) (B) (C) (D) (E)

3 7 9 15 17

10. If f ( x)  x

(A) 4



2 3

1 , then f    8

(B)  4

(C)

1 4

(D) 

1 4

(E) 8

11. If the pendulum of a clock swings through an angle of 2.8 radians and the length of the arc that its tip travels is 40, then the length of the pendulum is

(A) (B) (C) (D) (E)

12.54 13.58 14.29 52 112

12. If y  10  5sin  2 x  , what is the minimum value of y ?

(A) 5 (B) 10 (C) 5 (D) 10 (E) 15 13. The period of the graph of y  3cos 2  2 x  is

(A)

 8

(B)

 4

(C)

 2

(D) 

(E) 2

14. If 2  7 is a root of the equation x 2  4 x  3k  0 , then k 

(A) 1

(B) 1

(C) 2

(D) 2

(E) 4 GO ON TO THE NEXT PAGE

Dr. John Chung's SAT II Math Level 2 Test 3

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

15. If ax 2  bx  c  0 for all real numbers x , then which of the following must be true?

(A) a  0 and b 2  4ac  0 (B) a  0 and b 2  4ac  0 (C) a  0 and b 2  4ac  0 (D) a  0 and b 2  4ac  0 (E) a  0 and b 2  4ac  0

16. If the roots of the equation x 2  x  3  0 are  and  ,

then

(A)

1

 1 3



1





(B)

2 3

(C)

2 3

(D)

2 5

(E)

3 5

17. If the points  3, 7  , 1, k  , and  1, 1 are collinear, what

is the value of k ? (A) 2

(B) 4

(C) 5

(D) 7

18. The graph of the function f ( x) 

(E) 8

x2  x  2 has a vertical x 2  3x  2

asymptote at x  (A) (B) (C) (D) (E)

1 only 2 only 2 only 1 and 2 only 1,  2 , and 2

GO ON TO THE NEXT PAGE

148

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

19. Which of the following is the length of the radius of the

sphere x 2  2 x  y 2  2 y  z 2  3 (A)

3

(B) (C) (D)

5 3 5

(E) 2 5

20. Which of the following is the equation whose graph is the set of points equidistant from points  0, 4  and  2, 0  ?

(A) y  x  1 (B) y  2 x  1 1 1 x 2 2 1 3 (D) y  x  2 2 1 3 (E) y  x  4 2

(C) y 

n

21.

1  i n  i 1 2 lim 

(A) (B) (C) (D) (E)

0 1 5 10 Infinite

22. Which of the following is equivalent to

(A) (B) (C) (D) (E)

x 1  0? x

x0 0  x 1 x0 0 x5 x 1

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

23.

A school committee of 5 is to be chosen from a group consisting of 5 boys and 6 girls. How many ways can the committee be made up of 3 boys and 2 girls? (A) (B) (C) (D) (E)

24.

The interquartile range of a data set is 12. If the first quartile is 65, which of the following could be the median? (A) (B) (C) (D) (E)

25.

100 150 600 1200 1800

50 64 70 78 80

What is the range of the function defined by f ( x)  (A) (B) (C) (D) (E)

1 2 ? x

All real numbers All real numbers except 2 All real numbers except 0 All real numbers except 2 All real numbers between 2 and 3 nt

26.

r  The formula A  P 1   gives the amount A in a  n savings account with initial investment P which is compounded monthly at an annual interest rate of 6 percent for t years. How many years will it take the initial investment to double?

(A) (B) (C) (D) (E)

5.3 6.5 8.3 11.6 13.1

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150

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

27.

A line has parametric equations x  5  t and y  7  2t , where t is the parameter. The slope of the line is (A)

28.

5 7

(B) 2

(C)

7  2t 5t

(D)

7 5

(E) 2

 ln (n  1)2  lim    n 0 n   (A) (B) (C) (D) (E)

0 1 2 5 Undefined y

29. If the graph of f in Figure 2 is a polynomial of degree 7,

which of the following could be f ? (A) x3 ( x  3)2 ( x  2)( x  4)



(B) x 2 ( x  3)( x  2)( x  4)( x  1)2 (C) x 2 ( x  3)( x  2)( x  3)( x  1)2 (D) x 2 ( x  3) 2 ( x  3) 2 ( x  4)

3

 O

 2

 4

x

Figure 2

(E) x 2 ( x  3)( x  2)( x  4)( x 2  1) 30. If a1  1 , a2  3 , and an 1 

an  an  2 , then what is the 2

20th term of this sequence? (A) (B) (C) (D) (E)

36 39 41 43 60

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

31. If f ( x)  2 x3  3x  2 , which of the following statements is true?

(A) The function is increasing for all real x. (B) The function is increasing for x  0 (C) The equation f ( x)  0 have three real roots. (D) The equation f ( x)  0 have two imaginary roots. (E) The inverse of the function is also a function.





32. If f ( x)  x2  1 , where x  0 , then f  f 1 ( x) could

equal (B) x (C) x2 (D) x 2  1

(A) 1

(E)

1 x 1 2

33. If log a x  3 and log b x  4 , then log ab x 

(A) 12

(B)

8 3

(C)

12 7

(D)

7 12

(E)

3 8

34. If a cylinder whose height is equal to the diameter of its base is inscribed in a sphere, then the ratio of the volume of the cylinder to the volume of the sphere is

(A)

1 2

(B)

2 3

(C)

2 2 3

(D)

2

(E)

3 2 8

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

35. Which of the following is the graph of

(A)

x 2



y

(B)

y

1?

3 y

2 3

3 O

2 x

2

x

O

2

(C)

(D)

y

y 3

2 3

3

O

2

x

2

O

2

x

3

(E)

y

3 2

2 O

x

3

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

36. Figure 3 shows the part of a circle with a radius r and central angle  in radians. What is the area of the figure?

(A) r (B) r 2 r 2 (C) 2 (D) (E)

 r 2 360

r 

r

 Figure 3

2

37. When a fair coin is tossed 4 times, what is the probability of tossing at least 3 heads?

(A) (B) (C) (D) (E)

5 16 1 2 9 16 3 4 5 8

  38. If a  10 and b  18 , then which of the following could   NOT be a  b ? (A) (B) (C) (D) (E)

7 8 15 22 28

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

1 39. If   2 (A) (B) (C) (D) (E)

2a

b

a 1    , what is the value of ? b  3

0.33 0.67 0.79 0.81 0.87

40. If the equation sin 2   8cos   5  0 , where

0    360 , how many solutions are there in the interval? (A) 1

(B) 2

(C) 3

(D) 4

(E) 0

41. If  n !  118n !  240  0 , what is the value of n ? 2

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6

42. If the equation of an ellipse is 18 x 2  5 y 2  90  0 , what is the length of the major axis of the ellipse?

(A) (B) (C) (D) (E)

2.24 5.00 8.49 9.00 10.0

43. If 2 x  4 x  4  2 x , then x 

(A) 2 (B) 3 (C) 5 (D) log 2 3 (E) log 4 3

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

44. Which of the following intervals contain a root of

x3  3 x 2  3 x  2  0 (A) (B) (C) (D) (E)

2  x  1 1  x  0 0  x 1 2 x3 3 x  4

45. If the statement is “If xy  0 , then x  0 or y  0 ”, an indirect proof of the statement begins with the assumption that

(A) x  0 or y  0 (B) x  0 and y  0 (C) xy  0 (D) x  0 or y  0 (E) x  0 and y  0 46. By the rational zero theorem, which of the following could not be a possible rational zero of the equation

y  2 x 4  3x 2  2 x  10 ? (A) 

1 2

(B)

1 5

(C) 

5 2

(D)

5 2

(E) 5

a  1 2  3 and matrix B    . If AB    , 47. Matrix A    b  3 4  4 then a  (A) 3

(B) 2

(C) 1

(D) 1

(E) 2

48. If a set A  1, 2, 3, 4, 5, 6, 7, 8, 9, 10 , how many

subsets are there containing the elements 3, 4, and 5? (A) (B) (C) (D) (E)

64 128 256 512 1024

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

49. If a cone with a slant height equal to the diameter of the base is inscribed in a sphere with a radius of 10, what is the volume of the cone?

(A) (B) (C) (D) (E)

375 300 250 200 160

50. Given the parametric equations x  sec and y  tan  ,

which of the following is the graph of the points ( x, y ) ? (A) (B) (C) (D) (E)

Circle Ellipse Parabola Hyperbola None of these

STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK.

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No Test Material on This Page

158

2

2

TEST 3 # 1 2 3 4 5 6 7 8 9 10

ANSWERS

answer D C D D C C A D E A

Explanations:

# 11 12 13 14 15 16 17 18 19 20

answer C A C B D A B C B D

# 21 22 23 24 25 26 27 28 29 30

answer B B B C B D E C E B

# 31 32 33 34 35 36 37 38 39 40

answer D B C E D C A A C B

# 41 42 43 44 45 46 47 48 49 50

answer D C D B E B E B A D

Test 3

1. (D)

a ( x  1)  b( x  1)  2  0  (a  b) x  a  b  2  0 must be true for all real x. Therefore, a  b  0 and a  b  2 . By addition a  1 and b  1.

2. (C)

i 46  i 44  i 2  1

3. (D)

Since ( x  2) and ( x  1) are factors of f ( x), f ( x)  ( x  2)( x  1)Q( x) .





Therefore, f ( x)  x 2  x  2 Q( x) . 4. (D) 5. (C)

x 2  x is always true for any value of x. 2 2 2 , then   x  . 3 3 3 4   For this interval: 9 x 2  9    9 x 2  4  9 x 2  4  0 . 9

Since x 

Therefore, 9 x 2  4  (9 x 2  4)  4  9 x 2 . 6. (C)

The product of the roots is

4 6 and the sum of the roots is . Let the other root be r. k k

6 6  1  r  k     (1) k r 1 4 4  1r  k    (2) k r 6 4   4r  4  6r  r  0.4 From (1) and (2), r 1 r 7. (A)

tan   slope of the line.

Dr. John Chung's SAT II Math Level 2 Test 3

3 y  2x  9  y 

2 2 x  3 The slope is . 3 3

159

2

2 Therefore, tan  

2 2    tan 1    33.7o 3 3

8. (D)

Since f  g ( x)   g ( x)  1  x 2  3x  2 , then g ( x)  x 2  3x  1 .

9. (E)

By factor theorem: f (3)  2(3)3  k (3)  3  0  51  3k  k  17

10. (A)

f ( x)  x



2 3

1 1 f    8 8



2 3



 

 23



2 3

 22  4. Or use a calculator.

11. (C)

The length of a arc s  r , where r is a radius(the length of the pendulum) and  is the central angle in radian. 40  14.29. Therefore, 40  r  2.8  r  2.8

12. (A)

Since 5  5sin(2 x)  5 , the minimum will be 10  5  5.

13. (C)

The period of y  cos(2 x) is follows.

2    . Therefore, the period of y  3cos 2 (2 x) is as 2 2

y

y  3cos 2 (2 x)

O

x



y  cos(2 x )

14. (B)

If 2  7 is one of the roots, then the other root is 2  7 . The product of the roots is 3k . Therefore, 3k  (2  7)(2  7)  (2) 2  ( 7)2  3  k  1 .

15. (D)

The graph must be as follows. Therefore, the graph is concave down and the function do not have real roots (Imaginary). x

y  f ( x)

16. (A)

Since sum of the roots     1



160



1





   1 1   .  3 3

1 3  3 ,  1 and product of the roots   1 1

2

2

17. (B)

Because the three point on the same line, the slope between any two points are equal. k 7 k 1 k  7 k 1     k  4. 1  3 1  (1) 2 2

18. (C)

Since f ( x) 

19. (B)

x 2  2 x  y 2  2 y  z 2  3  ( x  1)2  ( y  1) 2  z 2  5

( x  2) ( x  1) ( x  2) ( x  1)



( x  2) , f ( x) has a vertical asymptote at x  2 . ( x  2)

Therefore, the radius is r  5 . 20. (D)

The equation which pass through (0, 4) and (2, 0) : y  2 x  4 40 02 40  2 , midpoint   ,   (1, 2) 2  02  2 Therefore, the line of equidistance is perpendicular to the line and pass through (1, 2). 1 1 3 y  x  b  2  (1)  b  b  2 2 2 1 3 The equation is y  x  . 2 2

slope 

21. (B)

Series: S 

a 1 r

1 1 1 1 1 1 lim  i      ...  2  1. 1 n  2 4 8 16 i 1 2 1 2 n

x 1  0  x 2  x( x  1)  0  0  x  1 x Or, you can use the test value method.

22. (B)

Because x 2  0, x 2 

23. (B)

5 C3

24. (C)

Interquartile range = upper quartile – lower quartile = 12 Since Interquartile has the range  65, 77  , the median must be in this range.

25. (B)

y  2 is an asymptote.

26. (D)

Since the interest is compounded monthly, n  12.

 6 C2  150.

12t

 0.06  2 P  p 1   12   12t  log1.005 2

Dr. John Chung's SAT II Math Level 2 Test 3

2  1.00512t

 

t

log 2  11.6 12log1.005

161

2

2

27. (E)

t  5  x  substitute  y  7  2(5  x)  y  2 x  17 Slope is 2.

28. (C)

Since e  lim 1  n  n , lim

29. (E)

At x  0 , the graph is bounced on x-axis and pass at x  3, 2, and 4. Therefore, the polynomial have factors as follows. P( x)  x m ( x  3)n ( x  2) n ( x  4) n (imaginary roots ) , where m  even and n  odd. (E) could be the polynomial function.

30. (B)

Since 2an 1  an  an  2  an 1  an  an  2  an 1 , the sequence is arithmetic progress.

1

n 0

1 2 ln(n  1)  lim 2 ln 1  n  n  2 ln e  2 . n 0 n 0 n

a1  1 and d  2 . Therefore, a20  a1  (20  1)d  1  19  2  39. Or, find the pattern. You can find a1  1, a2  3, a3  5, a4  7,..... , which is arithmetic sequence. 31. (D)

Use your graphic calculator. The graph will be as follows.

(D) is correct. f ( x) have one real root and two imaginary roots. (E) is incorrect. Use horizontal line test to check whether it’s inverse is a function. When the function is both increasing and decreasing, it’s inverse is not a function.





32. (B)

f f 1 ( x)  f 1  f ( x)   x

33. (C)

log ab x 

log x log x  log ab log a  log b log x log x  3  log a  log a 3

-------- (1) and

log x log x  4  log b  log b 4

Substitute into (1) log x 12 log ab x   log x log x 7  3 4 Or, use the formula: log ab x 

162

1 1 1 1 12     1 1 7 log x ab log x a  log x b 7  3 4 12

2

2 3

2

34. (E)

d3 d  Vc     d  4 2

d 2 4   2  d3 2  , where the diameter of the sphere and Vs   3 3

 d3

Vc 3 3 2 4    3 Vs  d 2 4 2 8 3 Or, you can use a convenient number for d . Use d  2.

is d 2 . d 2

d d

35. (D)

Memorize the graph of x  y  1 is as follows. In first quadrant, x  0 and y  0 , then x  y  1  y  x 1

1 y  x 1 1

1

Therefore, the graph of



y

2 3 When x  0 , y -intercept is 3 .

1

r 2 2

x

because A   r 2 



36. (C)

A

37. (A)

At least 3 heads is greater than or equal to 3 heads. Therefore, 3

2

 1 is (D).

.

4

4 1 5 1 1 1 P  4 C3      4 C4      . 2 2 2 16 16 16       38. (A)

        If a and b are in the same direction, a  b  a  b  10  18  28 . If a and b are in the       opposite direction, then a  b  b  a  18  10  8 . Therefore, 8  a  b  28. It cannot be 7.

2a

39. (C)

b

 1  b  1 b     2  3

Therefore,

2a

1 b 1     3 2



1 log 2a 1 3  log 3  log 1  1 log 2 b 3 2 log 2

log 3 a  0.79.  b 2log 2

Dr. John Chung's SAT II Math Level 2 Test 3

163

2 40. (B)

2 Use graphic utility.

O

2

Or algebraically as follows. Since sin 2   1  cos 2  , the equation is 1  cos 2   8cos   5  0  cos 2   8cos   6  0 . 8  64  24  cos   8.69 or cos   0.69 . cos   8.69 and 2 cos   0.69 . Therefore,   2.3328 or   3.9503 .

Therefore, cos  

41. (D)

42. (C)

The equation can be factored as follow.  Since n !  2 , then n !  120  n ! 120  n ! 2   0 Change the equation into a standard form.

18 x 2 5 y 2 90   90 90 90

 5!  120 

 n  5.

x2 y2  1 90 90 18 5

90 90 90  , major axis is on y -axis. Therefore, a 2   a  18 . 5 18 5 Major axis is 2a  2 18  8.49 .

Since

43. (D)

Since 2 x  22 x  4  2 x  22 x  3  2 x  0  2 x (2 x  3)  0 and 2 x  0 , the solution is 2 x  3  0  2 x  3  x  log 2 3.

44. (B)

If f (a)  f (b)  0 , then f ( x)  0 has at least one solution on interval (a, b) . Since f (1)  5  0 and f (0)  2  0 , then f (1)  f (0)  0 . Therefore, the equation has a solution on interval  1, 0  . Or, use a graph calculator to find the zeros. The graph will be as follow.

0.44

45. (E)

164

Indirect proof is a proof in which a statement to be proved is assumed false and if the assumption leads to impossibility, then the statement assumed false has been proved true. Therefore,  ( x  0 or y  0) is x  0 and y  0. In symbolic notation is as follows.  ( x  0  y  0)   ( x  0)  ( y  0) : De Morgan’s Law

2

2

46. (B)

The possible rational zeros are obtained as follows. factors of 10 1,  2,  5,  10  factors of 2 1,  2 1 Therefore, cannot be the rational root of the equation. 5

47. (E)

From matrix equation 1 2   a   3  3 4   b    4       a  2b  3 and 3a  4b  4 3a  4b  4 2 a  4b  6 Therefore, a  2

48. (B)

Number of subsets of 1, 2,6,7,8,9,10 is 27  128 . Now add elements 3, 4, and 5 to those subsets. Therefore, number of subsets containing 3, 4, and 5 is also 128.

49. (A)

10 O 5 30o 5 3

From the figure above, the radius of the cone is 5 3 and the height is 15.  (5 3)2  15 Therefore, the volume of the cone is V   375 . 3 50. (D)

Parametric equation: eliminate  . Since 1  tan 2   sec2  , substitute x and y. Therefore, 1  y 2  x 2  x 2  y 2  1 , which represents hyperbola.

END

Dr. John Chung's SAT II Math Level 2 Test 3

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No Test Material on This Page

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Test 4                                               Dr. John Chung’s                SAT II Mathematics Level 2

Dr. John Chung's SAT II Math Level 2 Test 4

167

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2

    

MATHEMATICS LEVEL 2 TEST REFERENCE INFORMATION    

 

THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME   OF THE QUESTIONS IN THIS TEST   1   Volume of a right circular cone with radius r and height h : V   r 2 h 3   1 Lateral Area of a right circular cone with circumference of the base c and slant height  : S  c 2  

  Volume of a sphere with radius r : V  4  r 3 3   Surface Area of a sphere with radius r : S  4 r 2   1   Volume of a pyramid with base area B and height h : V  3 Bh

       

168

2

 

2 Dr. John Chung’s SAT II Math Level 2

Answer Sheet 01 02

A

B

C

D

E

A

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03

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04 05

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06

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07

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08

26 27

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E

28 29

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30

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31 32

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33

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09

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34

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10 11

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A

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35 36

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12

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37

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D

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13 14

A

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D

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B

C

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A

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38 39

A

 

A

B

C

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15

A

B

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D

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40

A

B

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D

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16 17

A

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D

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A

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41 42

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18

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B

C

D

E

B

C

D

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19 20

B

C

D

E

43 44

A

A

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

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A

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21 22

45 46

A

B

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47

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23

A

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48

A

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24

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49

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25

A

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D

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50

A

B

C

D

E

              

 

       

 

       

     

           The number of right answers:                              

                

           The number of wrong answers:                             

 

          

 

# of correct

Dr. John Chung's SAT II Math Level 2 Test 4

1    4 # of wrong

Raw score

 

169

 

2

2 Score Conversion Table 6

Scaled Score 480

630

5

470

26

620

4

470

800

25

620

3

460

46

800

24

610

2

460

45

800

23

610

1

450

44

800

22

600

0

450

43

800

21

600

42

800

20

590

41

800

19

590

40

780

18

580

39

760

17

570

38

750

16

560

37

740

15

550

36

720

14

540

35

710

13

530

34

700

12

520

33

690

11

510

32

680

10

500

31

670

9

490

30

660

8

490

29

650

7

480

28

Scaled Score 640

800

27

48

800

47

50

Scaled Score 800

49

Raw Score

170

Raw Score

Raw Score

2

2

 

  

MATHEMATICS LEVEL 2 TEST For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding circle on the answer sheet Note: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions in this test. For each question you will have to decide whether or not you should use a calculator. (2) For some questions in this test you may have to decide whether your calculator should be in the radian mode or the degree mode. (3) Figures that accompany problems in this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers

x for which f ( x) is a real number. The range of f is assumed to be the set of all real numbers f ( x) , where x is in the domain of f . (5) Reference information that may be useful in answering the questions in this test can be found on the page preceding Question 1. USE THIS SPACE FOR SCRATCHWORK 2

1.     

2

1  1  a   a      a  a  (A) 4 (B)  4 (C) 2 (D)  2 (E) 2a

    2.     

1  If   cos x    for   0  x  , then  sin 2x    3 2 (A) (B) (C) (D) (E)

0.25 0.30 0.50 0.63 0.75

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  Dr. John Chung's SAT II Math Level 2 Test 4

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

3.

The inverse of which of the following graphs is also a function? (A)

(B) y

y

O

x

(C)

O

x

(D) y

y

O

x

O

x

(E) y

O

x

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172

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

4.

If

x 2  5 , then x 

(A) 5 only (B) 5 only (C) 5 and 5 (D) 25 and 15 (E) 25 only

5.

2 2 The radius of the circle x  2 x  y  4 y  9 is

(A) (B) (C) (D) (E)

6.

3.00 3.74 4.12 5.43 6.15

If f ( x)  9 x 2  x and g ( x) 

(A) 7.

2 3

(B)

3 3

(C)

x 1 , then f  g (2)   x 1

5 3

(D)

6 3

(E)

2 2 3

If f ( x)  5 for all real numbers x , then f ( x  2)  f ( x  2) 

(A) 0 8.

(B) 2

(C) 5

(D) 10

(E) 20

10!  10!   9!2 100! 81! 100 (B) 81

(A)

(C) 100 1000 81 10000 (E) 9

(D)

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

9.

The set of all real numbers of x such that x 2  1  1 consists of

 x   2 (B)  2  x  0  0  x  2 (C) 0  x  2 (D)  x   2   x  2 (E)  x  2 (A)

10.

If the line y  x  k is tangent to the graph of the circle

x2  y 2  4 , then k  (A) 2 2 only (B) 3 2 only (C) 4 2 only (D)  2 2 (E)  4 2

11.

x If f ( x)  2 ln  x  1 and g ( x)  e , then

 g  f  ( x) 

(A) e x 1 2

(B) e x 2 x 1 (C) x  1 (D) 2( x  1) (E) x 2  2 x  1 12.

3 If f ( x)  x  3x  1 , then f 1  f ( x)  

(A) x (B) x 2 (C) x3  3x  1 3

(D) (E)

x3  3x  1

x

3



 3x  1

3

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174

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

13.

log16 81  log 2 3  (A) 2

14.

(C) 0

(D) 1

(E) 3

Three numbers have a sum of 36, a product of 1680, and form an arithmetic sequence. What is the largest number? (A) (B) (C) (D) (E)

15.

(B) 1

10 12 14 16 18

If a  bi 

3i , which of the following is true? 1 i

(A) a  1, b  2 (B) a  2, b  1 (C) a  2, b  1 (D) a  2, b  1 (E) a  2, b  1

16.

If sin( A  B)  0.25 , sin A 

3 , and 2

90  A  B  180 , then B could be (A) 65 (B) 80.5 (C) 105.5 (D) 120.5 (E) 125.4

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175

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

17.

2 If the equation of a parabola is y  2 x , then the directrix of the graph is

(A) y  2 1 2 1 (C) y  8

(B) y 

1 8 1 (E) y   2 (D) y  

18.

Which of the following is symmetric with respect to the origin? 2 (A) y  x  x

(B) y  x  5 5 3 (C) y  x  3x  x 6 4 2 (D) y  x  x  x 7 5 (E) y  x  x  1

19.

What is the x-intercept of the hyperbola

 x  12  y  2 2 10



4

1?

(A) (4.16, 0) and (4.16, 0) (B) (3.12, 0) and (3.12, 0) (C) (4.16, 0) and (2.16, 0) (D) (5.47, 0) and (3.47, 0) (E) (5.12, 0) and (5.12, 0)

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176

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

20.

  sin      2  (A) sin  (B)  sin  (C) cos  (D)  cos  (E) sin  cos 

21.

Which of the following is the solution set of ( x  2)( x  1)2 0 x2 (A) (B) (C) (D) (E)

22.

There are 4 boys and 5 girls in a chess club. In how many ways could 3 boys and 3 girls be selected to attend the school tournament? (A) (B) (C) (D) (E)

23.

x  2 x0 2  x  1 2  x  2 2  x  1 or x  2

40 80 120 360 720

1 , what 10 is the probability that a package of 10 light bulbs has exactly two defective bulbs?

If the probability that a light bulb is defective is

(A) 0.01

(B) 0.10

(C) 0.19

(D) 0.25

(E) 0.33

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177

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

24.

A polynomial P( x) has remainder of three when divided by ( x  1) and remainder of five when divided by ( x  2) . If P( x) is divided by ( x  1)( x  2) , then the remainder is (A) (B) (C) (D) (E)

25.

8 x8 2x  1 2x  1 3x  1

In Figure 1, a triangle is inscribed in a semicircle. If BC  10 , what is the area of  ABC in terms of  ? (A) 50sin 

B

(B) 50cos  (C) 50sin  cos  (D) 50 tan 





O

C

Figure 1

50 tan 

(E)

26.

A

10

If a rectangular prism has dimensions a, b, and c , which of the following represents the length of its diagonal?

abc

(A) (B)

3

a 2  b2  c 2

(C)

a 2  b2  c 2

(D)

a3  b3  c3

(E)

3

a3  b3  c3

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178

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

27.

    If vectors a  (3,  4) and b  (2, 3) , then a  b  (A) (B) (C) (D) (E)

28.

If 3  4i is a root of 2 x 2  ax  b  0 , then b  (A) (B) (C) (D) (E)

29.

5.68 7.07 8.60 9.13 10.87

25 25 50 50 It cannot be determined from the information given.

If cos   

(A)

3 3 8

(B)  (C)

30.

1 and 90    180 , then sin  2  equals 3

3 2 8

4 2 9

(D) 

4 2 9

(E) 

5 2 11

What is the distance from the plane 3x  4 y  5 z  10  0 to the point (0, 0, 0)? (A)

2

(B) 2 (C) 2 2 (D) 4 (E) 4 2

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179

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

31.

If x0  1 and xn 1  xn  2n , then x10  (A) (B) (C) (D) (E)

32.

20 91 162 268 381

The line ax  by  4  0 forms a triangular region with the x-axis and y -axis . What is the area of the region in terms

of a and b ? 2 ab

(A)

33.

(B)

4 ab

(C)

ab

8

(D)

ab

16

(E)

8 ab

Which of the following is equivalent to the expression tan 70  tan 20 ? 1  tan 70 tan 20 (A) tan 90 (B) tan 50 tan 90 tan 50

(C)

tan 50 1  tan 50 1  tan 500 (E) tan 50o (D)

34.

In  ABC ,  B is an obtuse angle, AB  15 , BC  20 , and the area of the triangle is 90. What is the measure of  B ? (A) (B) (C) (D) (C)

0.36 0.64 2.50 5.48 5.63

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180

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK. 10

35.

1  The constant term of the expansion of  x   is x  (A) 1

36.

(C) 150

(D) 210

(E) 252

The lateral surface area of a right cylinder in Figure 2 is 80. If the height of the cylinder is 10, what is the volume of the cylinder? (A) (B) (C) (D) (E)

37.

(B) 45

10

48.4 50.9 54.8 60.3 61.4

Figure 2

2 Figure 3 shows the graph of y  ax  bx  c . Which of the

y

following could NOT be true? (A) ab  0 (B) bc  0 (C) ac  0

O

(D) b 2  4ac

Figure 3

(E) b 2  4ac 38.

x

Which of the following is an odd function? 2

(A) f ( x)  x  5 (B) f ( x)  x  sin x (C) f ( x)  x  x (D) f ( x)  3 (E) f ( x) 

39.

1 x 1 2

If   Arc cos

(A)

1 2

1  , what is the value of sin ? 2 2

(B) 

1 2

(C)

3 2

(D) 

3 2

(E)

3 3

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181

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

40.

2 If f ( x)  ( x  1)( x  x  1) , which of the following

statements are true? I. The function f is increasing for x  1 II. The function f ( x)  0 has three real solutions. III. The domain of the function f ( x) is all real numbers. (A) (B) (C) (D) (E)

41.

I only II only I and III only II and III only I, II, and III

3 1 If  A is obtuse and cos A   , cos A is 5 2 3 5 1 (B)  5 1 (C) 5 (A) 

42.

(D)

5 5

(E)

2 5 5

What is the length of the major axis of an ellipse whose 2 2 equation is 4 x  16 x  y  4 y  16  0 ?

(A) (B) (C) (D) (E)

1 2 4 6 8

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182

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

43.

Which of the following is an even function? (A) f ( x)  sin x (B) f ( x)  tan x 2x (C) f ( x)  e 2 (D) f ( x)  2x  3

(E) f ( x)  log x 44.

What is the sum of the infinite series 2 4 8 1    ....? 3 9 27 (A) (B) (C) (D) (D)

lim

45.

n 

(A) (B) (C) (D) (E) 46.

0 0.2 0.4 0.6 Infinite 2 n2  n  n



0 2 4 10 Infinite





If y  log5 x 2  6 x  14 , what is the minimum value of the equation ? (A) 2 (B) 1 (C) 1 (D) 2 (E) 5

47.

If

n 1 P2

(A) 5

 n P2  12 , what is the integer value of n ? (B) 6

(C) 7

(D) 8

(E) 10

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183

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

48.

The middle term of the expansion of  x  2 y  is 4

2 2 (A) 12x y 2 2 (B) 12x y

(C) 24x 2 y 2 2 2 (D) 24x y 2 2 (E) 32x y

49.

If f ( x) 

9  x 2 and x  0 , what is the inverse of f ( x) ?

(A) f 1 ( x)  x 2  3 (B) f 1 ( x) 

9  x 2 and x  0

(C) f 1 ( x)   9  x 2 and x  0 (D) f 1 ( x)   9  x 2 and x  0 (E) f 1 ( x)  x 2  9 and x  0

50.

y

 0, 2 

In Figure 4, what is the equation of line  that is tangent to 2

2

the circle x  y  1 and passes through the point (0, 2) ? (A) y   x  2 (B) y   2 x  2

x

O

(C) y   3x  2 (D) y  2 x  2 (E) y  3 x  2

 Figure 4

STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK.

GO ON TO THE NEXT PAGE

184

No Test Material on This Page

Dr. John Chung's SAT II Math Level 2 Test 4

185

2

2

TEST 4 # 1 2 3 4 5 6 7 8 9 10

answer B D C C B D D C B D

ANSWERS # 11 12 13 14 15 16 17 18 19 20

answer E A C C C C D C D D

# 21 22 23 24 25 26 27 28 29 30

answer D A C D E C C C D A

# 31 32 33 34 35 36 37 38 39 40

answer B E B C E B E B A C

Explanations: Test 4 2

1. (B)

1 1  Since  a    a 2  2  2 and a a  2

2

1 1  2 a    a  2  2 , a a 

2

1  1  then  a     a    4 . a a    2. (D)

In Quadrant I:

3 x 1

 2 2  1  sin 2 x  2sin x cos x  2      3  3   0.628539...  0.63

2 2

3. (C)

Only choice (C) passed a horizontal line test. Only one-to-one functions have an inverse function.

4. (C)

Since

5. (B)

x2  2 x  y 2  4 y  9

x 2  x , x  5 . Therefore, x  5 . 

 x  12   y  2 2  14

r 2  14  r  3.74165...  3.74

2 1 1  , 2 1 3

2

2 6 1 1 1 f    9     3 3  3  3 3

6. (D)

g (2) 

7. (D)

f ( x)  5 is a constant function for any real x . Therefore, 5  5  10.

186

# 41 42 43 44 45 46 47 48 49 50

answer D C D D C C B C C C

2 8. (C) 9. (B)

2 Since 10!  10  9! , x2  1  1 

10. (D)



 x  0  

10  9! 10  9!  100 . 9! 9!  1  x2  1  1

2x 2





x

2

 



 0  x2  2  0

x2  ( x  k )2  4

Substitute y  x  k .

x 2  x 2  2kx  k 2  4  2 x 2  2kx  k 2  4  0 Since the line is tangent to the ellipse, its discriminant should be 0. D  4k 2  4(2)(k 2  4)  0  k2  8  k   2 2 y Or, since OM  2 , OM  MP , OPR is an isosceles triangle, and MP  2 ,

OP  22  22  2 2 . Therefore, the values of k are 2 2 or 2 2 .

O

R 2 M

P

x

2

2

2

11. (E)

g  f ( x)   e2ln( x 1)  eln( x 1)  ( x  1)2  x 2  2 x  1

12. (A)

f 1  f ( x)   x and f f 1 ( x)  x.





13. (C)

Since log 2 3  log 24 34  log16 81 , then log16 81  log16 81  0 .

14. (C)

Let three numbers be a  d , a, a  d , where d  0 .

 a  d   a   a  d   3a  36 12  d 12 12  d   1680 

 a  12

144  d 2  140  d 2  4  d  2

Therefore, the largest number is 12  2  14. 15. (C)

3  i  3  i 1  i  4  2i    2i 1  i 1  i 1  i  2 Since a  bi  2  i , a  2 and b  1.

16. (C)

Since sin( A  B)  0.2 , A  B  165.5224...  165.5 in Quadrant II. sin A 

3 2



A  60 or 120

Therefore, 165.5  60  105.5 or 165.5  120  45.5 .

Dr. John Chung's SAT II Math Level 2 Test 4

187

2 17. (D)

2 The standard form of the parabola is x 2 

1 1 y  4  y . 2 8

1  1 Therefore, the focus is at  0,  and the directrix is y   . 8  8 18. (C)

Odd functions are symmetric with respect to the origin. (C) is odd function.

19. (D)

Let y  0 .

20. (D)

    sin      sin  cos  cos  sin   cos  2 2 2 

21. (D)

Method 1) Graphic Solution: multiply by ( x  2) 2  0 .

 x  12

 x  12  20

1  1 



10 x  1  2 5  x  5.47 or  3.47

( x  2) 2

( x  2)( x  1) 2  0  ( x  2)2 ( x  2)



x

  2 1

Therefore, y  0 in the interval 2  x  2 . Method 2) Test value: Test value

3

0 2

 1

1.5

()()  0  F () ()()  0 (T) At x  0  () ()()  0 (T) At x  1.5  () ()()  0 (F) At x  3  ( )

 2

3

At x  2 

Method 3) Use graphic utility directly.

188

20

 ( x  2)( x  2)( x  1)2  0 and x  2

y  ( x  2)( x  2)( x  1) 2

y

2

 x  1  

2  x  2

2

2  4    4 C3 3

22. (A)

 4 5       4  10  40 3  3

23. (C)

10   1   9        0.1937...  0.19  2   10   10 

2

24. (D)

25. (E)

8

P( x)  ( x  1)Q1 ( x)  3 -------(1) P( x)  ( x  2)Q2 ( x)  5 ------(2) P( x)  ( x  1)( x  2)Q( x)  ax  b -------(3) From equations (1) and (3) P(1)  3  a  b --------(4) From equations (2) and (3) P(2)  5  2a  b ------(5) From (4) and (5) a  2 and b  1 . Therefore, the remainder is 2 x  1.

10 10 , AB  . tan  AB 1  10 The area of  ABC  10   2  tan  Since tan  

50  .   tan 

26. (C)

The length of the diagonal  a 2  b 2  c 2

27. (C)

   z  a  b   3  (2),  4  3  (5,  7) ,

28. (C)

 z  52  (7) 2  8.602325  8.60

Since 3  4i is a root of the equation, then its conjugate 3  4i is also the root of the equation. b The product of the roots is . 2 b (3  4i )(3  4i )  25   b  50 2

29. (D) 22

3

 2 2  1  4 2 sin 2  2sin  cos   2        9  3  3 

II

1 III

30. (A)

D

IV

3(0)  4(0)  5(0)  10 32   4    5  2

Dr. John Chung's SAT II Math Level 2 Test 4

2



10 5 2



2 2

 2

189

2

2

31. (B) n  1,

x2  x1  2

x2  x1  2  1  2

n  2,

x3  x2  4

x3  x2  4  1  2  4

n  3,

x4  x3  6

x4  x3  6  1  2  4  6

...... n  9,

x10  x9  18  1  2  4  6      18 (2  18)  9 Therefore, x10  1   2  4  6      18   1   91 2

32. (E)

33. (B)

34. (C)

35. (E)

x10  x9  18

4 , a 1 4 The area of the triangle A   2 a Since tan( A  B) 

4 b

 4  8   b   ab : area cannot be negative.  

tan A  tan B tan 70  tan 20 . , tan(70  20 )  1  tan A tan B 1  tan 70 tan 20

15  20  sin B 3  3  90  sin B   B  sin 1    0.6435    2 5 5 Since B is obtuse, B    0.6435  2.50 . Area of  ABC 

Since the general term is the constant term is

36. (B)

y-intercept: by  4  y 

x-intercept: ax  4  x 

10 Cr

10 C5 x



 x r 

10  r

1   x

when r  5.

The lateral area: 2 rh  80  r 

 10 Cr x r  x 10  r  10 Cr x 2 r 10 , That is

10 C5

 252 .

80 80 4   2 h 20r 

2

160 4  50.9295    50.9 V   r 2 h      10     37. (E)

From the graph, (1) Concave down ----- a  0 b 0  b0 2a (3) y-intercept ----- f (0)  c  0 (2) Axis of symmetry ----- 

(4) Two unequal roots ---- D  b 2  4ac  0  b 2  4ac (E) is not true because y  f ( x) has two real roots. 38. (B)

Since y  x and y  sin x are odd functions, then f ( x)  x  sin x is an odd function.

39. (A)

 60  1 1  Since   arccos    60 , sin    sin 30  . 2 2  2 

190

2 40. (C)

2 Graphic utility: The graph of f ( x)  ( x  1)( x 2  x  1) is as follows. y

i) The function is increasing for x  1. ii) The domain of the function is all real x .

 1

x

Algebraically: x 2  x  1 is positive for all real x , because its discriminant D  0 . Therefore, for the interval x  1 , f ( x) is always positive. 41. (D)

Since cos A  

A  63.4349  acute angle 2

3 and A is obtuse, A  126.8698976 . 5

 3 1   A 1  cos A  5  1  5 . Therefore, cos    2 2 2 5 5 Or, cos (63.4349)  0.447214  42. (C)

5 . 5

The standard expression of the ellipse: 4( x 2  4 x  4)  ( y 2  4 y  4)  16  16  4 ( x  2)2 ( y  2) 2   1  a2  4  a  2 1 4 Therefore, the length of the major axis  2a  4 4( x  2)2  ( y  2) 2  4 

43. (D) 44. (D)

f ( x)  2 x 2  3 , because f ( x)  f ( x) . Since r  

2 3 a and   1 , the sum of the series   3 4 1 r

2

45. (C)

lim

x 





n2  n  n

n2  n  n





n2  n  n



 lim

x 

2



n2  n  n 2

n nn

2

1 3  .  2 5 1    3

  lim 2  x 

n2  n  n



n

 1  lim 2  1   1  4 x  n  

Dr. John Chung's SAT II Math Level 2 Test 4

191

2 46. (C)

2 The graph of f ( x)  x 2  6 x  14 has a minimum of 23 at x  3 (axis of symmetry). Therefore, the minimum of y  log 5 5  1. Or using graphic utility: Trace the minimum. y



x

O

Minimum x  2.999999982  y  1

47. (B)

n 1 P2

 n P2  12  (n  1)n  n(n  1)  1  n 2  n  n 2  n  12

n6 48. (C)

49. (C)

 4 2 2 The middle term is the third term:    x   2 y   24 x 2 y 2 2   4   4 0    x   2 y  is the first term. 0   The domain of function

f :x0



The range of f 1 ( x) : y  0

The range of function f : y  0  The domain of f 1 ( x) : x  0 Therefore, the inverse can be obtained as follows,

y  9  x2

 switch x and y 

 

 x  9  f 1

2



f 

1 2

 9  x2

f 1   9  x 2

Since the range of the inverse is y  0 and its domain is x  0 , f 1 ( x)   9  x 2 and x  0 . y

y  9  x2

O

x

f 1   9  x 2

192

2 50. (C)

2 Let the equation of line  be y  mx  2 . Since the line is tangent to the circle, the discriminant of the equation x 2  (mx  2)2  1 must be 0.

x 2  m 2 x 2  4mx  4  1  0  (1  m 2 ) x 2  4mx  3  0 D   4m   4(1  m 2 )(3)  0  16m 2  12  12m 2  0  4m 2  12 2

m 2  3  m   3 , Since the line  has a negative slope, the equation of line  is y   3x  2 . END

Dr. John Chung's SAT II Math Level 2 Test 4

193

No Test Material on This Page

194

Test 5                                               Dr. John Chung’s                SAT II Mathematics Level 2

Dr. John Chung's SAT II Math Level 2 Test 5

195

2

 

2

    

MATHEMATICS LEVEL 2 TEST REFERENCE INFORMATION    

 

THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME   OF THE QUESTIONS IN THIS TEST   1   Volume of a right circular cone with radius r and height h : V   r 2 h 3   1 Lateral Area of a right circular cone with circumference of the base c and slant height  : S  c 2  

  Volume of a sphere with radius r : V  4  r 3 3   Surface Area of a sphere with radius r : S  4 r 2   1   Volume of a pyramid with base area B and height h : V  3 Bh

       

196

2

 

2 Dr. John Chung’s SAT II Math Level 2

Answer Sheet 01 02

A

B

C

D

E

A

B

C

D

E

 

03

A

B

C

D

E

B

C

D

 

04 05

A A

B

C

 

06

A

B

 

07

A

08

26 27

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

E

28 29

A

B

C

D

E

D

E

30

A

B

C

D

E

C

D

E

A

B

C

D

E

B

C

D

E

31 32

A

B

C

D

E

A

B

C

D

E

33

A

B

C

D

E

09

A

B

C

D

E

34

A

B

C

D

E

10 11

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

35 36

A

B

C

D

E

12

A

B

C

D

E

37

A

B

C

D

E

13 14

A

B

C

D

E

B

C

D

E

A

B

C

D

E

38 39

A

 

A

B

C

D

E

 

15

A

B

C

D

E

40

A

B

C

D

E

 

16 17

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

41 42

A

B

C

D

E

 

18

A

B

C

D

E

B

C

D

E

 

19 20

B

C

D

E

43 44

A

A

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

 

21 22

45 46

A

B

C

D

E

47

A

B

C

D

E

 

23

A

B

C

D

E

48

A

B

C

D

E

24

A

B

C

D

E

49

A

B

C

D

E

25

A

B

C

D

E

50

A

B

C

D

E

              

 

       

 

       

     

           The number of right answers:                              

                

           The number of wrong answers:                             

 

          

 

# of correct

Dr. John Chung's SAT II Math Level 2 Test 5

1    4 # of wrong

Raw score

 

197

 

2

2 Score Conversion Table 6

Scaled Score 480

630

5

470

26

620

4

470

800

25

620

3

460

46

800

24

610

2

460

45

800

23

610

1

450

44

800

22

600

0

450

43

800

21

600

42

800

20

590

41

800

19

590

40

780

18

580

39

760

17

570

38

750

16

560

37

740

15

550

36

720

14

540

35

710

13

530

34

700

12

520

33

690

11

510

32

680

10

500

31

670

9

490

30

660

8

490

29

650

7

480

28

Scaled Score 640

800

27

48

800

47

50

Scaled Score 800

49

Raw Score

198

Raw Score

Raw Score

2

2

 

  

MATHEMATICS LEVEL 2 TEST For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding circle on the answer sheet Note: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions in this test. For each question you will have to decide whether or not you should use a calculator. (2) For some questions in this test you may have to decide whether your calculator should be in the radian mode or the degree mode. (3) Figures that accompany problems in this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers

x for which f ( x) is a real number. The range of f is assumed to be the set of all real numbers f ( x) , where x is in the domain of f . (5) Reference information that may be useful in answering the questions in this test can be found on the page preceding Question 1. USE THIS SPACE FOR SCRATCHWORK 1.

Which is the negation of the statement “Some numbers are even”? (A) (B) (C) (D) (E)

2.

All numbers are even. Some numbers are not even. All numbers are not even. All numbers are odd. Some numbers are not odd.

In how many ways can 2 juniors and 2 seniors be selected from a group of 8 juniors and 6 seniors? (A) (B) (C) (D) (E)

4 48 420 480 840

GO ON TO THE NEXT PAGE

  Dr. John Chung's SAT II Math level 2 Test 5

199

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

3.

Find the largest integral value of k such that the roots of

x 2  5 x  k  0 are real? (A) 4

4.

(E) 10

B

10

8 A

D

C

Figure1

If a and b are positive numbers, and a 2  b 2  29 and ab  10 , then a  b 

7

(B) 7

(C) 8

(D) 5 2

(E) 7 2

If x  2i is a solution to the equation x 3  kx  0 , what is the value of k ? (A) (B) (C) (D) (E)

7.

(D) 8

66.67 112.45 125.36 133.33 150

(A)

6.

(C) 7

In Figure 1, BD is the altitude to the hypotenuse AC . If BD  8 and BC  10 , which is the area of  ABC ? (A) (B) (C) (D) (E)

5.

(B) 6

8 6 4 2 1

1 and g ( x)  x  5 , what is the domain of x f  g ( x)  ?

If f ( x) 

(A) (B) (C) (D) (E)

All All All All All

x such that x such that x such that x such that x such that

x  5 x0 x0 x  0 and x  5 x  0 and x  5

GO ON TO THE NEXT PAGE

200

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

8.

Which of the following graphs best describes the set of points a  a , b  for which  b  2 in the xy -plane ? 2 (A) (B) y y

x

O

(C)

x

O

(E)

y

O

x

y

x

O

9.

(D)

y

x

O

Which of the following is symmetric with respect to the y -axis ? (A) y   x 1

2

(B) x  y 2 (C) x 2  4 y 2  4 (D) x 2  2 x  y 2  3 (E) y  x  2

GO ON TO THE NEXT PAGE Dr. John Chung's SAT II Math Level 2 Test 5

201

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

10.

If x 0.4  10 , then what is the value of x 0.6 ? (A) 10 10 (B) 15 5 (C)

100

(D) 15 10 (E) 11.

10000

1 If e x  3 , what is the value of  3  e 

(A) 729

(B) 64

(C) 27

(D)

2 x

? 1 64

(E)

1 729

1 x  5 , what is the value of x ? If 1 1 2 x x

12.

(A) 5

13.

(C) 1.25

(D) 1.25

(E) 5

Figure 2 shows a hemisphere with a radius of 4. Find the surface area of that figure. (A) (B) (C) (D) (E)

14.

(B) 3

20 32 36 42 48

4 

Figure 2

If two forces of 10 pounds and 15 pounds act on a body with an angle of 60  between them, what is the magnitude of the resultant? (A) 16.80 (B) 18.21 (C) 20.42 (D) 21.80 (E) 24.92

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202

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

15.

 2  3  x   2  If the matrix equation       , what is the 5  y   3  2 value of y ? (A) (B) (C) (D) (E)

16.

0.625 0.505 0.125 3.500 4.254

In the arithmetic progression an  , a2  50 and a4  44 . Which is the first term that is a negative number? (A) (B) (C) (D) (E)

17.

If log 2 x  log 2 ( x  1)  1 , which of the following is the solution set of the inequality? (A) (B) (C) (D) (E)

18.

17th 18th 19th 20th 21th

 x x  1  x  1  x  2  x 0  x  2  x 1  x  2  x x  2

In a game, the probability of winning is

1 and the 4

3 . If 3 games are played, what is 4 the probability of winning at least 2 games?

probability of losing is

(A)

3 64

(B)

5 64

(C)

5 32

(D)

10 27

(E)

5 16

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

19.

The graph of a polynomial function is shown in Figure 3. Which of the following could be the equation of the polynomial function?

y

(A) P( x)  x( x  1)( x  3)( x2  1) (B) P( x)  x( x  1)( x2  9)

x

O

(C) P( x)  x( x  1)( x  3)( x2  5) (D) P( x)  x( x  1)( x  3)( x2  5x  10) Figure 3

(E) P( x)  x( x  1)( x  3)2 ( x2  1)

20.

If sin  A  60   cos 40 , the measure of  A is (A) 110

21.

(C) 80

(D) 45

(E) 20

If tan A  4 and tan B  3 , what is the value of tan( A  B) ? (A) (B) (C) (D) (E)

22.

(B) 90

0.065 0.077 0.126 0.245 0.333

x If

2

 3x  4 

 0 , which of the following is the solution x2 of the inequality?

(A) (B) (C) (D) (E)

x  1 x  1 1  x  4 1  x  0 or 0  x  4 All real x

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204

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

23. In Figure 4, line  is the perpendicular bisector of AB at point E . What is the area of  ADE ? (A) (B) (C) (D) (E)

y 

 A(0, 2)

2 3 4 5 10

 E

x  B (4, 0)

D

Note: Figure not drawn to scale. 24.

In Figure 5, if the radius of the semicircle is 5, what is the area of the inscribed square? (A) 9

25.

(C) 20

(D) 25

(E) 36

1   What is the value of sec  arctan ? 3  (A)

26.

(B) 16

3 2

(B)

2 3 3

(C)

2 2 3

Figure 4

(D)



3 4 (E) 5 7

5

Figure 5

If two lines y  2 x  4 and y  mx  5 are parallel, where m is a constant, then the distance between the two lines is

(A) (B) (C) (D) (E)

27.

4.02 5 5.4 6.25 8

If the parametric equations are x  4sin 2 and y  2 cos 2 , which of the following represents the graph of point  x, y  ? (A) (B) (C) (D) (E)

Line Parabola Hyperbola Ellipse Circle

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

28.

If the equation of a circle is x2  2 x  y 2  4 y  1 , then the area of the circle is (A) (B) (C) (D) (E)

29.

5 6 25 36 42

z

In Figure 6, the graph of plane 2 x  3 y  4 z  12 in three

B

dimensions forms a triangular pyramid with base  AOC . What is the volume of the pyramid? (A) (B) (C) (D) (E)

12 18 36 48 72

2 x  3 y  4 z  12

C

O

x

A y

Note: Figure not drawn to scale. Figure 6

30.

sin   sin   tan   2

2

2

(A) sin 2  (B) cos 2  (C) tan 2  (D) cot 2  (E) sec 2 

31.

If the value of f ( x)  x2  3x  k is always positive for any x , which of the following could be the value of k ? (A) 3 (B) 2 (C) 0 (D) 2 (E) 3

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206

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

32.

In Figure 7, AB  5 and BC  10 . What is the area of the quadrilateral? (A) (B) (C) (D) (E)

33.

25.6 28.4 32.5 42.6 62.9

A

5 130

B 10

If a( x  1)  b( x  1)  ( x  2)  0 for all real x , where

C D Note: Figure not drawn to scale. Figure 7

a and b are constants, what is the value of a ? (A) 1.5

34.

(B) 2.0

(C) 2.5

(D) 3.5

(E) 4.5

2 When polynomial f ( x)  2 x  5x  k is divided by 2 x  3 ,

the remainder is 5. What is the value of constant k ? (A) (B) (C) (D) (E)

35.

1 3 5 7 8

What is the value of  sin x  cos x   sin 2 x ? 2

(A) 1

36.

(B) 0

(C) 1

(D) 2

(E) 3

What is the value of  in the interval 0   

 2

that satisfies

the equation 6 cos   1  5sec  ? (A) (B) (C) (D) (E)

0.45 0.59 0.62 0.78 0.82

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

37.

If the ratio of the two roots of the equation x 2  kx  18  0 is 1: 2 , which of the following is all the values of constant k? (A) 3,6 (B) (C) (D) (E)

38.

9, 9 9 10,12 6, 9

If f ( x) 

2x  1 and f 1 ( x) is the inverse of f ( x) , then x 1

f 1 (3)  (A) 3

39.

(C) 4

(D) 6

(E) 10

Figure 8 shows a triangle in a circle with center O . If the radius of circle O is 2, what is the area of the triangle in terms of  ? (A) (B) (C) (D) (E)

40.

(B) 2

sin 2 2sin  2sin 2 2 cos  2 cos 2

O

2



Figure 8

In five years, the population of Spring Lake decreased steadily from 50,000 to 45,000. Find the rate of decrease per year? (A) (B) (C) (D) (E)

1.2% 2.1% 2.4% 2.5% 3.0%

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

41.

Find the measure of the angle between two forces of 10 pounds and 20 pounds if the magnitude of their resultant is 25 pounds. (A) 45.3 (B) 71.8 (C) 108.2 (D) 123.5 (E) 135.7

42.

Find the asymptotes of

x2 y 2   1. 8 18

3 (A) y   x 2 2 (B) y   x 3 2 (C) y   x 9 9 (D) y   x 2 9 (E) y   x 4

43.

If x   log 27 3 (A) 3

44.

log3 27

(B) 3

, then log 3 x  (C)

1 3

(D)

1 9

(E)

1 27

In Figure 9, the volume of the right circular cone is 12 and the radius of the base is 3. What is the lateral area of the cone? (A) (B) (C) (D) (E)

4 6 15 18 36



3



Figure 9

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

45.

The area of a triangle whose sides are of lengths 7, 20, and 23 is (A) 20 5 (B) 25 (C) 25 5 (D) 30 (E) 30 5

46.

What is the value of 1  i  ? 10

(A) 8i (B) 16i (C) 32i (D) 32 (E) 64 47.

In how many ways can 10 people be divided into three groups, one group with 4 people and the other two groups with 3 people each? (A) (B) (C) (D) (E)

48.

210 420 2100 4200 326000

Which of the following equations could be the graph shown in Figure 10? y (A)

x2 y 2  1 16 6

(B)

( x  8) 2 ( y  3) 2  1 8 3

(0,3)

( x  8) 2 ( y  3) 2  1 (C) 16 9

O

(D)

( x  8) 2 ( y  3) 2  1 64 9

(E)

( x  8) 2 ( y  3) 2  1 64 9

(8,0)

x

Figure 10

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

49.

Day 1

Day 2

Day 3

16GB

10

15

20

64GB

5

9

13

128GB

11

17

18

The table above shows the number of smart phones that were sold during a three-day sale. The prices of models 16GB, 64GB, and 128GB were $300, $400, and $500, respectively. Which of the following matrix representations gives the total daily income, in dollars, received from the sale of the smart phones for each of the three days?

10 15 20  (A)  5 9 13  300 400 500 11 17 18  10 15 20  300  (B)  5 9 13   400  11 17 18  500  10 5 11  300  (C) 15 9 17   400   20 13 18  500  300  10 5 11  (D)  400  15 9 17  500   20 13 18 

(E) 300 10 5 11  400 15 9 17   500  20 13 18

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

50.

From the binomial expansion of  2 x  3 , what is the 6

coefficient of x 4 ? (A) (B) (C) (D) (E)

60 68 720 2160 4320

STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK.

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212

No Test Material on This Page

Dr. John Chung's SAT II Math Level 2 Test 5

213

No Test Material on This Page

214

2

2

TEST 5 # 1 2 3 4 5 6 7 8 9 10

ANSWERS

answer C C B A B C A E C A

# 11 12 13 14 15 16 17 18 19 20

answer A E E D C C D C D A

# 21 22 23 24 25 26 27 28 29 30

answer B D D C B A D B A C

# 31 32 33 34 35 36 37 38 39 40

answer A E A E C B B C C B

# 41 42 43 44 45 46 47 48 49 50

answer B A B C E C C D C D

Explanations: Test 5 1. (C)

2. (C)

Remember the negation of the word “some” is “all.” Therefore, the negation is “All numbers are not even.” 8   6        28  15  420  2  2

3. (B)

Discriminant: D  25  4k  0  4k  25   k  6.25 The largest integer value of k is 6.

4. (A)

DC  6

B

10

8 8

A

D

6

C

BD 2  AD  DC  64  AD  6  AD 

32 3

1 32  The area of  ABC   6    8   66.6666    66.7. 2 3  5. (B)

(a  b)2  a 2  b 2  2ab  29  2(10)  49  a  b  7  a and b are positive.

6. (C)

Substitution:  2i   k (2i)  0   8i  2ki  0   2k  8   k  4

7. (A)

Since f  g ( x)  

3

Dr. John Chung's SAT II Math Level 2 Test 5

1 x5

, the domain is x  5  0  x  5.

215

2 8. (E)

2 If y  0 , then y  

a a  2 , and if y  0 , then y   2 . The graph is as follows. 2 2

y

y

a  2, where y  0 2 x

O

a  2, where y  0 2

y

9. (C)

(A) is symmetric with respect to x  1. (B) is symmetric with respect to the x-axis. (C) is symmetric with respect to the x-axis or y -axis.

 x  12  y 2  4

(D) is symmetric with respect to x  1 or the y -axis. (E) is symmetric with respect to x  2.

 

10. (A)

x 0.6  x 0.4

11. (A)

1  3 e 

2 x

3 2

3

 10 2  10 10

 

 e3

2 x

 

 e6 x  e x

6

 36  729 6

Or, since e x  3  x  ln 3 , then e6 x  e6ln 3  eln 3  36. 12. (E)

Compound fraction: multiply common denominator by x 2 . 1 2  2 x x x3  x x x  1 x    x 1  2 x2  1  x2  1  x 1   x2   Therefore, x  5.





13. (E)

Since the surface area of a sphere is 4 r 2  4 (4) 2  64 , the surface area of a hemisphere is 32 . The area of the circular base is 16 . Therefore, the entire surface area is 48 .

14. (D)

In a parallelogram, two consecutive angles are supplementary. R

10 60

o

10 120

o

15

Law of cosine: R  102  152  2(10)(15) cos120  21.79441    21.8

216

2 15. (C)

2 From the matrix equation: 2x  3y  2  2x  5 y  3

y  0.125

,

 8 y  1 16. (C)

Since a2  a1  d  50 and a4  a1  3d  44 , then d  3 and a1  53 . an  a1  (n  1)d  an  53  (n  1)(3)  0  3n  56 Therefore, n  18.666    and the first negative term is the 19th term.

17. (D)









log 2 x  log 2 ( x  1)  1  log 2 x 2  x  1  log 2 x 2  x  log 2 2 Since base 2 is greater than 1, x 2  x  2  x 2  x  2  0  ( x  2)( x  1)  0 . The solution of the inequality is 1  x  2 , but x  1 from the logarithmic equation. Therefore, 1  x  2 . 2

18. (C) 19. (D)

1

3

0

 3   1   3   3  1   3  9 1 5               64 64 32  2   4   4   3  4   4 

The function has one zero at x  0 , one negative zero, and one positive zero. Choice (D) has one zero at x  0 , one negative zero, and one positive zero. x 2  5 x  10 has imaginary roots.

20. (A)

Cofunction: A  60  40  90  A  110

21. (B)

tan( A  B) 

22. (D)

Test value: Graphing utility: x 2  3x  4  0  x 2  3x  4  0  ( x  4)( x  1)  0   1  x  4 and x  0. 2 x Therefore, the solution set is 1  x  0  0  x  4

23. (D)

The slope of AB  

tan A  tan B 43 1    0.07692    0.077 1  tan A tan B 1  4  3 13

2 1 04 20   and the midpoint E is  ,    2, 1 . 4 2 2   2 The slope of the perpendicular line is 2 and passes through (2,1). The equation of line  is y  2 x  3 . y

h

(2,1)

 D(0, 3)

Dr. John Chung's SAT II Math Level 2 Test 5

Since AD  5 and the height h  2 , 52  5. the area of  ADE is 2



A(0, 2)

x B (4, 0)

217

2 24. (C)

2 From the figure below: 52  x 2  (2 x)2 .  25  5 x 2  x  5 The length of an edge is 2 5 . 5

x



25. (B)

1  1  cos  tan    Or , algebraically 1 Let X  tan 1 . tan X  3

Use calculator:

28. (B)

2

 20.

1   3  1 3

 1.1547   

2 3 3

and 90  X  90 .

Therefore, sec X  1

1 2 2 3 .   cos X 3 3

Choose one point on y  2 x  4 : that is  (0,  4) . Two lines are parallel: m  2. The distance between a point (0, 4) and the line 2 x  y  5  0 is D

27. (D)



5

2 X 3

26. (A)



The area of the square is 2 5

2x

2(0)  (4)  5 2

2  (1)

2



9 5

 4.02492    4.02

x y x2 y 2 and cos 2  , sin 2  2   cos 2  2   1   1. 4 2 16 4 The graph of the parametric equations is an ellipse.

Since sin 2 

x 2  2 x  y 2  4 y  1  ( x  1) 2  ( y  2)2  6  r  6

Therefore, the area of the circle are,  r 2  6 . 29. (A)

The coordinates of each intercept is: C ( x,0, 0), When y  0 and z  0 , 2 x  3(0)  4(0)  12 x  0 and z  0 , 2(0)  3 y  4(0)  12 x  0 and y  0 , 2(0)  3(0)  4 z  12

A(0, y ,0), and B (0, 0, z )  x6.  y4  z 3

 6 4  3 Bh  2    12 The volume of the cone: 3 3

30. (C)

218





sin 2   sin 2  tan 2   sin 2  1  tan 2   sin 2   sec 2   sin 2  

1  tan 2  cos 2 

2

2

31. (A)

Since f ( x) is always positive, f ( x)  0 must have imaginary roots. 9 2 D  b 2  4ac   3  4(1)(k )  0  k   4 9 Choice (A) : 3   4

32. (E)

EC  10sin 40 , BE  10cos 40 , and DC  5  EC sin 40 . A

5

B 130o 40

EC  10sin 40

10

BE  10 cos 40 is the height of the trapezoid. D

E

C

The area of the trapezoid: h(b1  b2 ) 10cos 40  (5  5  10sin 40)   62.922415    62.9 A 2 2 33. (A)

a ( x  1)  b( x  1)  ( x  2)  0  (a  b  1) x  (a  b  2)  0 To be identical, a  b  1  0 and a  b  2  0 . 3 Therefore, 2a  3  a  . 2

34. (E)

Remainder theorem: (or, long division)

f ( x)  2 x 2  5 x  k   2 x  3 Q ( x )  5 35. (C)

 sin x  cos x 2  sin 2 x

36. (B)

Use a graphic utility. Graph y 



 3  9 15 f      k  5  k  8  2 2 2

 sin 2 x  2sin x cos x  cos 2 x  2sin x cos x  1

5  6cos x  1 and find the zero. cos x

Or algebraically, 6cos   1  5sec  6cos   1 

 6 cos   5 cos 1  0

5 cos 

 6cos 2   cos   5  0

5  and cos   1 . Since 0    , (calculator must be in radian mode) 2 6 5 5    cos 1  0.58568    0.58 cos   6 6

Therefore, cos  

37. (B)

Define the two roots as n and 2n. The product of the two roots: 2n 2  18  n  3 The sum of the roots: n  2n  3n  9

Dr. John Chung's SAT II Math Level 2 Test 5

219

2

2 2y 1  y 1

x 1 4  f 1 (3)   4 x2 1 2y 1  y4 Or, 3  y 1

f 1 : y 

38. (C)

From the inverse: x 

39. (C)

Since AB  2 cos  and OB  2sin  , 4cos   2sin   4sin  cos   2sin  2  the area of OAC  2

2

OB  OB  2sin  2 AB cos    AB  2 cos  2

sin  

O 

C



B

A

40. (B)

45000  50000(1  r )

r 41. (B)

1  1  0.9 5

 1 r 

1 0.9 5

The Law of Cosines: ABC and  are supplementary. 102  202  252 A cos    0.3125 2(10)(20) B

43. (B)

 0.9  (1  r )

5

 0.0208516376  2.1%

25

10

42. (A)

5

10

 20

C

  cos 1 0.3125  108.2099569o mABC  180  108.2099569  71.79004  71.8

x2 y 2   1  a  8 and b  18 8 18 b 18 3 Therefore, the asymptotes are : y   x   x x a 2 8

Since log 27 3 

 

log 3 1 log 27  and log3 27  3 , then x   log 27 3 3  31 3log 3 3

3

 33 .

Therefore, log3 33  3 . 44. (C)

A   rs , where r is a radius and s is a slant height.  r 2h   9  h V   3 h  3 h  12  h  4 3 3 Therefore, s  5 . s h A    3  5  15 3

220

2 45. (E)

2 Heron’s Formula: s 

abc 2

7  20  23 50   25 2 2 The area is s ( s  a)( s  b)( s  c)  25(25  7)(25  20)(25  23)  30 5 s

46. (C) 47. (C) 48. (D)

Since 1  i   1  2i  i 2  2i , (1  i )10   2i   25 i 5  32i . 2

10 C4

5

 6 C3  3 C3  2100 2!

From the graph: center  8,3 , a  8 , and b  3. y

The equation of the ellipse is

 x  8 2  y  32 (3, 0) O

49. (C)

50. (D)

82 (8, 0)



32

 1.

x

Total income each day: Day 1: 10  (300)  5  (400)  11  (500) Day 2: 15  (300)  9  (400)  17  (500) Day 3: 20  (300)  13  (400)  18  (500) Its matrix form is choice (C). 6  6  6 r nr r r 6r    2 x   3     2   3 x r  r  Since x6 r  x 4 , r  2 . 6 2 The coefficient of x 4 is    24   3  2160. 2  

END

Dr. John Chung's SAT II Math Level 2 Test 5

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No Test Material on This Page

222

Test 6                                               Dr. John Chung’s                SAT II Mathematics Level 2

Dr. John Chung's SAT II Math Level 2 Test 6

223

2

 

2

    

MATHEMATICS LEVEL 2 TEST REFERENCE INFORMATION    

 

THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME   OF THE QUESTIONS IN THIS TEST   1   Volume of a right circular cone with radius r and height h : V   r 2 h 3   1 Lateral Area of a right circular cone with circumference of the base c and slant height  : S  c 2  

  Volume of a sphere with radius r : V  4  r 3 3   Surface Area of a sphere with radius r : S  4 r 2   1   Volume of a pyramid with base area B and height h : V  3 Bh

       

224

2

 

2 Dr. John Chung’s SAT II Math Level 2

Answer Sheet 01 02

A

B

C

D

E

A

B

C

D

E

 

03

A

B

C

D

E

B

C

D

 

04 05

A A

B

C

 

06

A

B

 

07

A

08

26 27

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

E

28 29

A

B

C

D

E

D

E

30

A

B

C

D

E

C

D

E

A

B

C

D

E

B

C

D

E

31 32

A

B

C

D

E

A

B

C

D

E

33

A

B

C

D

E

09

A

B

C

D

E

34

A

B

C

D

E

10 11

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

35 36

A

B

C

D

E

12

A

B

C

D

E

37

A

B

C

D

E

13 14

A

B

C

D

E

B

C

D

E

A

B

C

D

E

38 39

A

 

A

B

C

D

E

 

15

A

B

C

D

E

40

A

B

C

D

E

 

16 17

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

41 42

A

B

C

D

E

 

18

A

B

C

D

E

B

C

D

E

 

19 20

B

C

D

E

43 44

A

A

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

 

21 22

45 46

A

B

C

D

E

47

A

B

C

D

E

 

23

A

B

C

D

E

48

A

B

C

D

E

24

A

B

C

D

E

49

A

B

C

D

E

25

A

B

C

D

E

50

A

B

C

D

E

              

 

       

 

       

     

           The number of right answers:                              

                

           The number of wrong answers:                             

 

          

 

# of correct

Dr. John Chung's SAT II Math Level 2 Test 6

1    4 # of wrong

Raw score

 

225

 

2

2 Score Conversion Table 6

Scaled Score 480

630

5

470

26

620

4

470

800

25

620

3

460

46

800

24

610

2

460

45

800

23

610

1

450

44

800

22

600

0

450

43

800

21

600

42

800

20

590

41

800

19

590

40

780

18

580

39

760

17

570

38

750

16

560

37

740

15

550

36

720

14

540

35

710

13

530

34

700

12

520

33

690

11

510

32

680

10

500

31

670

9

490

30

660

8

490

29

650

7

480

28

Scaled Score 640

800

27

48

800

47

50

Scaled Score 800

49

Raw Score

226

Raw Score

Raw Score

2

2

 

  

MATHEMATICS LEVEL 2 TEST For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding circle on the answer sheet Note: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions in this test. For each question you will have to decide whether or not you should use a calculator. (2) For some questions in this test you may have to decide whether your calculator should be in the radian mode or the degree mode. (3) Figures that accompany problems in this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers

x for which f ( x) is a real number. The range of f is assumed to be the set of all real numbers f ( x) , where x is in the domain of f . (5) Reference information that may be useful in answering the questions in this test can be found on the page preceding Question 1. USE THIS SPACE FOR SCRATCHWORK 1.

If

1 1  x   x , then x  x x

(A) 1

2.

If

1 1 1 x

(A) 2

3.

(B) 0

3 5

(D) 2

(E) undefined

 2 , what is the value of x ?

(C) 

(B) 2

If 3x  5 y , then

(A)

(C) 1

(B)

5 3

1 2

(D)

1 2

(E)

1 4

x  y

(C) log3 5

(D) log5 3

(E) 35

GO ON TO THE NEXT PAGE

  Dr. John Chung's SAT II Math Level 2 Test 6

227

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

4.

What is the number of arrangements of letters that can be formed from the letters of the word “abscissa”? (A) (B) (C) (D) (E)

5.

40320 20160 6720 3360 1680

If  log x   log x 2  3 , then which of the following could 2

be the value of x ? (A) 10 6.

8.

(C) 5

(D) 8

(E) 10

What is the minimum value of y  sin x  3? (A) 0

7.

(B) 3

(B) 1

(C) 2

(D) 3

(E) 4

1 A B for all real x , what is the value of   x 1 x 1 x 1 constant B ?

If

2

(A) 1

(B) 

If x 

1  3 

1 2 2

(C)

1 2

(D) 1

(E) 2

 3 , then x 

(A) 1  3

(B) 4  3 (C) 2  3

9.

(D)

32

(E)

34

y

In Figure 1, if  ABC is equilateral, what is the slope of BC ? (A)  3

(B)  2

(C) 1

(D) 1

(E)

3

O

C

A (2, 0)

B (6, 0)

x

Figure 1

GO ON TO THE NEXT PAGE

228

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

10. If tan  

(A) 1.8

1 2 , then  sin   cos    2 (B) 2.0

(C) 2.2 (D) 2.4

(E) 2.6

11. The graph of y  f ( x) is shown in Figure 2. Which of the following is the equation of the graph? y (3,3)

(A) y  x  3

y  f ( x)

(B) y  x  3  3 (C) y   x  3  3

O

(D) y   x  3  3

x

Figure 2

(E) y   x  3  3



(6, 0)



12. If f ( x)  log 2 x 2  7 and f  g (1)   4, which of the

following could be g ( x)? (A) g ( x)  x 2  x  2 (B) g ( x)  2 x 2  x  1 (C) g ( x)  cos  x   4 (D) g ( x)  sin  x   2 (E) g ( x)  3x  1 13. If the roots of 2 x 2  kx  14  0 are integers, then which of the following could be the value of constant k ?

(A) 6

(B) 8

(C) 10

(D) 12

(E) 16

14. Which of the following is an equation with roots 0 and

2  3? (A) 0  x3  3x 2  x (B) 0  x3  4 x 2  x (C) 0  x3  4 x 2  x (D) 0  x 2  2 x  2 (E) 0  x3  2 x 2  2 x

GO ON TO THE NEXT PAGE Dr. John Chung's SAT II Math Level 2 Test 6

229

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

15. How far is the point  2,1 from the line 3x  y  4? (A) (B) (C) (D) (E)

0.316 0.542 1.358 2.855 3.282

2    , which of 3   the following are the rectangular coordinates of point A ?

16. If the polar coordinates of point A are 10,

(A) (B) (C) (D) (E)

 5 3, 5 5,5 3  5,  5 3   5, 5 3   5,  5 3  B

17. In Figure 3, AD  20 and BD is perpendicular to AC . What is the length of CD ? (A) (B) (C) (D) (E)

10.35 12.07 13.06 14.85 15.50

A

32

46 20

D

C

Note: Figure not drawn to scale. Figure 3

18. If

 3

(A)

1 3

46 x

 27 x , then x 

(B)

1 2

(C)

2 3

(D)

3 4

(E)

3 2

GO ON TO THE NEXT PAGE

230

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

19. Which of the following is the solution of

x  0? ( x  1)( x  2)

(A)    x  1 (B) 5  x  2 (C)    x  2 (D)    x  0 or 1  x  2 (E)    x  1 or x  2

20. What are the asymptotes of f ( x) 

x ? x x 3

(A) x  0 (B) x  0 and x  1 (C) x  1 and y  0 (D) y  0 (E) x  0 and y  0

21. If f ( x)  log3

 x   3 and g ( x) is the inverse of

f ( x),

what is the value of g (2)? (A)

1 27

(B)

1 9

(C)

1 3

(D) 3

1 22. If sin 2  , what is the value of 4 (A) (B) (C) (D) (E)

(E) 9

 cos   sin  2 ?

1.25 0.75 0.50 0.25 0.15

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231

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

23. Which of the following is a horizontal tangent to the ellipse

 x  32  y  2 2 49



25

1?

(A) y  2 (B) y  3 (C) y  5 (D) y  7 (E) y  10

24. Which of the following is true of the graph of the function

xy  x 2  1 ?

(A) (B) (C) (D) (E)

Even function Odd function Symmetric with respect to x-axis Symmetric with respect to y -axis Symmetric with respect to y  x

25. What is the value of

(A) 1

(B)

2

2i ? i2 (C)

3

(D) 2

(E) 2 3

B

26. In  ABC , if A  30 , a  5 and b  10, then  ABC in Figure 5 is

(A) (B) (C) (D) (E)

A

An acute triangle A right triangle An obtuse triangle An acute or an obtuse triangle An isosceles triangle

a

30 b

C

Note: Figure not drawn to scale. Figure 5

27. If cos 2   cos   1 , then sin 4   sin 2  

(A) 1

(B)

2

(C)

3

(D)

1 2

(E)

3 2 GO ON TO THE NEXT PAGE

232

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

28.

x  What is the period of the function y  2 tan   1  4? 3  (A)

29.

1 3

(B) 2

(C) 3

(D) 6

(E) 8 y

In Figure 6, point P is on the x-axis. What is the

(A) (B) (C) (D) (E) 30.

7.48 8.60 9.25 9.75 13.75

O

P ( x, 0)

x

Note:Figure not drawn to scale. Figure 6

The roots of x 2  kx  1  0 are p and q, where k is a constant. If (A) (B) (C) (D) (E)

31.

B (3, 4) 

A(2, 3) 

minimum length of AP  PB ?

1 1   10, what is the value of k ? p q

10 5 5 10 15

If  x  1 is a factor of x 6  5 x 4  4 x3  x  k , then what is the value of k ? (A) 1

32.

(B) 2

(C) 3

(D) 4

(E) 5

If the quadratic equation x 2  2ax  2a 2  2a  3  0 has real roots, then which of the following could NOT be the value of a ? (A) 2 (B) 1 (C) 0 (D) 1 (E) 2 GO ON TO THE NEXT PAGE

Dr. John Chung's SAT II Math Level 2 Test 6

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

33.

What is the distance from the plane x  2 y  3z  5 to the point  2, 2, 0  ? (A) (B) (C) (D) (E)

34.

1.871 2.225 2.786 3.125 4.750

If the line through (5, 4) and (2, k ) is perpendicular to the line with equation 3x  4 y  4, what is the value of k ? (A) (B) (C) (D) (E)

35.

If the radius of a right circular cone is 6 and the height of the cone is 8, what is the lateral surface area of the cone? (A) (B) (C) (D) (E)

36.

20 40 60 96 120

If y  3log ( 10 x  x 2 ), what is the maximum value of y ? (A) (B) (C) (D) (E)

37.

2 4 6 8 10

3.56 4.19 5.25 6.32 7.41

What is the length of the major axis of the ellipse whose equation is 5 x 2  18 y 2  90  0 ? (A) 6.25

(B) 7.25

(C) 8.49

(D) 9.34

(E) 10.25

GO ON TO THE NEXT PAGE

234

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

38.

In Figure 7, A, B, and C , the vertices of the squares, are

C

collinear. What is the value of k ? (A) (B) (C) (D) (E)

39.

k 7 Figure 7

4

5 6

(B) 

 6

(C)

 6

(D)

 2

(E)

5 6

In a box there are 4 red marbles and 5 white marbles. If marbles are drawn one at a time and replaced after each drawing, what is the probability of drawing exactly 2 red marbles when 3 marbles are drawn? (A) (B) (C) (D) (E)

41.

A

 3 If   Arc cos    , then    2 

(A) 

40.

B

8.45 10.38 12.25 13.12 13.74

0.329 0.235 0.198 0.110 0.102

In Figure 8, P is a point in the square of side-length 10 such that it is equally distant from two consecutive vertices

B

and from the opposite side AD. What is the length of BP ? (A) (B) (C) (D) (E)

5 5.25 5.78 6.25 7.07

C P

A

D Figure 8

GO ON TO THE NEXT PAGE Dr. John Chung's SAT II Math Level 2 Test 6

235

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

42.

If log 2  log 3  log 2 x    1 , what is the value of x ?

(A) (B) (C) (D) (E)

43.

1 a   2  1 If matrix A    , B    , C    , and AB  C ,  2 1 b  3 where a and b are constants, what is the value of a ? (A) (B) (C) (D) (E)

44.

126 256 512 1024 2048

1 2 3 4 5

In the arithmetic progression, the first term is 5 and the common difference is 3. What is the sum of the first 20 terms? (A) (B) (C) (D) (E)

300 475 670 850 925



45.

The sum of the infinite series (A) (B) (C) (D) (E)

k

k

 1 1    2    3  is k 1 k 1

1 1.5 2 3 4.5

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236

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

46.

1 2  3    n  n  n2 lim

(A) (B) (C) (D) (E)

47.

A committee of 5 is to be chosen from 8 men and 5 women. What is the probability that the committee consists of 2 men and 3 women? (A) (B) (C) (D) (E)

48.

0.5 1 2 3 4

0.185 0.218 0.302 0.387 0.425

From the expansion of the binomial  ax  y  , where a is a 6

positive constant, the coefficient of x 2 y 4 is 60. What is the value of a ? (A) 2

49.

(B) 3

(C) 4

(D) 6

(E) 8

If f ( x)  e 2 x and g ( x)  ln( x 2  1) , then  f  g  x   (A) x 2  1 (B) x 3  x (C) 2 x( x 2  1) (D) x 4  2 x 2  1





(E) 2 x ln x 2  1

GO ON TO THE NEXT PAGE Dr. John Chung's SAT II Math Level 2 Test 6

237

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

50.

If

ab  0 , then which of the following could be true? ab

I. 0  b  a II. b  a  0 III. a  b  0 (A) (B) (C) (D) (E)

I only I and II only II and III only I and III only I, II, and III

STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK.

GO ON TO THE NEXT PAGE

238

No Test Material on This Page

Dr. John Chung's SAT II Math Level 2 Test 6

239

2

2 S ANSWERS

TEST 6 # 1 2 3 4 5 6 7 8 9 10

answer E B C D E C C B A A

# 11 12 13 14 15 16 17 18 19 20

answer D C E C A D B A D D

# 21 22 23 24 25 26 27 28 29 30

answer B B D B A B A C B A

# 31 32 33 34 35 36 37 38 39 40

answer A E A D C B C C E A

# 41 42 43 44 45 46 47 48 49 50

answer D C C C B A B A D B

Explanations: Test 6 1. (E)

2. (B)

1 1 x x  x0 x x But x  0 is extraneous. The solution is undefined. 1( x) x x    2  x  2x  2  x  2 1  x 1 x 1  1  ( x )   x   x

y



1 3x y

1 5y y

   

3. (C)

3 5

4. (D)

Permutation with repetition:

5. (E)

 log x 2  log x 2  3



x y

 3 5 

x  log3 5 y

8!  3360 2!3!

 log x 2  2 log x  3  0



 log x  3 log x  1  0

3

Since log x  3 and log x  1 , x  10 or 10. 6. (C)

The graph is symmetric with respect to y-axis. Therefore, the minimum of the function is 1  3  2.

2

240

2 7. (C)

8. (B)

2 1 A B 1 A( x  1) B( x  1) 1 ( A  B) x  B  A    2  2  2  2  x 1 x 1 x 1 x 1 x 1 x 1 x 1 x2  1 1 To be identical, A  B  0 and  A  B  1 . Therefore, B  . 2 2

x

1  3 

2

1  3 

3 

2

 3  x  1 3  3  x 

3 1  3  x

Therefore, x  4  3 . 9. (A)

 CM 2 3   3. Since BM  2 , A  60 , and CM  2 3 , then the slope of BC   MB 2 y

O

10. (A)

C

A (2, 0) M

Since tan  

B (6, 0)

x

1 , then 2 1 2  (1) sin   5 and cos   5  or  1 2 (2) sin    and cos     5 5

1

2 2

1

2

(1)

2  9  1  sin   cos 2       1.8 5 5  5

(2)

 sin   cos 2   



1



5

2



2  9    1.8 5 5

11. (D)

Translations: y   x

T3,3  y   x  3  3 

12. (C)

Let g (1) be k , then f  k   log 2 k 2  7  4 .



2

4



2

k  7  2  16  k  9  k  3 (B) g (1)  2 (C) g (1)  3 (A) g (1)  4 (E) g (1)  2 (D) g (1)  2

Dr. John Chung's SAT II Math Level 2 Test 6

241

2 13. (E)

2 k 14 , Product of the roots: r1  r2  7 2 2 Therefore, the roots are 7 and 1, or 7 and  1 .  k  16 and  16. Sum of the roots: r1  r2 

(Check) Choice (E): 2 x 2  16 x  14  0  x 2  8 x  7  0  ( x  1)( x  7)  0 The roots are 1 and 7.

14. (C)









The polynomial equation is x  x  2  3   x  2  3   0 .    x( x  2  3)( x  2  3)  0  x ( x  2) 2  ( 3)2   0

Therefore, x( x 2  4 x  1)  0  0  x3  4 x 2  x Or, use sum and product of the roots: b c Let the quadratic equation be x 2  x   0 . a a b c SUM  2  3  2  3  4    4 , PRODUCT  2  3 (2  3)  1  1 a a The equation is x 2  4 x  1  0 . Because of zero at x  0 , the equation is x( x 2  4 x  1)  0 .



15. (A)

16. (D)

17. (B)

Distance from a point (2, 1) to a line 3x  y  4  0 : D 

BD  20 tan 32 and CD 

A

32  (1) 2

BD 20 tan 32o   12.06858    12.07 tan 46 tan 46

 3

32o

46o 20

46 x

 27

x

D

 1   32     

C

46 x

 33 x  32 3 x  33 x

Exponents: 2  3x  3x  6 x  2  x 

242

3(2)  1  4

 2  x  r cos   x  10cos    5  3   2  y  r sin   y  10sin  5 3  3 

B

18. (A)



1 3

 0.316227766  0.316

2 19. (D)

2 Test value:

x 0 ( x  1)( x  2) 





0

1

2

( ) ( ) ( )  0 (Ok). At x  0.5,  0 . At x  1.5, 0 ()() ()() ()() ( )  0 . Therefore, the solution set is   x  0  1  x  2 . At x  3, ()() At x  1,

Or, multiply by ( x  1)2 ( x  1) 2







1

0

x( x  1)( x  2)  0

2

The solution set is x  0 or 1  x  2. 20. (D)

f ( x) 

x 1  2 x  x x 1 3

x 2  1  0  No vertical asymptote 1 y  lim 2  0  y  0 : Horizontal asymptote x  x  1 21. (B)

Since g ( x)  f 1 ( x)  32( x 3) , g (2) 

22. (B)

 cos  sin  

23. (D)

The graph of

2

1 . 9

 cos 2   sin 2   2 cos  sin   1  sin 2  1 

 x  32  y  2 2

a  7 and b  5.

49



25

1 3  4 4

 1 is as follows.

y

y7 5 (3, 2)  5

x y  3

There are two horizontal tangent lines: y  7 and y  3

Dr. John Chung's SAT II Math Level 2 Test 6

243

2 24. (B)

25. (A)

2 x2  1 x Choice (A): f ( x)  f ( x) : not even function Choice (B): f ( x)   f ( x) : odd function Choice (C): f ( x)   f ( x) : not symmetric with respect to x-axis Choice (D): f ( x)  f ( x) : not symmetric with respect to y-axis Choice (E): not symmetric with respect to y  x xy  x 2  1  y 

2i i2



2i 2  i



22  12 2

2

(2)  1



5 5

1

sin 30 sin B 10sin 30   sin B   1  B  90 5 10 5

26. (B)

The law of sines:

27. (A)

cos 2   cos   1  cos   1  cos 2   cos   sin 2  Since sin 4   cos 2  , then sin 4   sin 2   cos 2   sin 2   1 . 1  , the period is  3 . 1 3 3

28. (C)

Since the frequency is

29. (B)

To have a minimum length point P( x,0) should be on the segment AB . Since AP  AP , the minimum of APB is equal to the length of AB . y

AB  (3   2)2  (4   3)2  74  8.602325    8.60 A( 2, 3) 

O

B (3, 4) 

P ( x, 0)

x

 A( 2, 3)

1 1 p  q k    . p q pq 1

30. (A)

Since p  q  k and pq  1 ,

31. (A)

Factor Theorem: f (1)  1  5  4  1  k  0  k  1

32. (E)

Discriminant: D   2a   4(1)(2a 2  2a  3)  0  (a  3)(a  1)  0 2

The solution is: 3  a  1

244

k  10  k  10 .

2

2

33. (A)

Distance from a point to a line: D 

34. (D)

Since the slope of 3x  4 y  4 is

2  2(2)  3(0)  5 2

2

2

1  (2)  3

7



14

 1.870828    1.871

3 4 , the slope perpendicular to the line is  . 4 3

k 4 4 k 4 4     k 8 25 3 3 3 62  82  10 . The lateral surface area is  rs   (6)(10)  60 .

35. (C)

The slant height is

36. (B)

y  3log(10 x  x 2 ) has a maximum at x  5 [Axis of symmetric of ( f ( x)  10 x  x 2 ) ] Therefore, the minimum of y is 3log(10  5  52 )  3log 25  4.193820026    4.19

37. (C)

38. (C)

x2 y 2  1 18 5 The length of the major axis is 2 18  8.485281374  8.49 5 x 2  18 y 2  90  0 

The slopes of AB and BC are equal. C B A

3

7

4

4

39. (E)

3 k 7   k  12.25 4 7

k 7

7

k

1) Using calculator: 2) Algebraic solution:  3 3 Let   Arccos    , where 0     . cos    2  2 

 

5 6

2

40. (A)

4 5      0.329218107  0.329 9 9 Or, the ways to draw two red marbles: 3 C2

For WRR, the probability is

Dr. John Chung's SAT II Math Level 2 Test 6

WRR, RWR, RRW

5 4 4 80  80     . Therefore, 3     0.329. 9 9 9 729  729 

245

2 41. (D)

2 If PM  x , then BN  10  x . 10

B 10  x N

P

5

x

A

42. (C)

43. (C)

44. (C)

Pythagorean Theorem: x 2  (10  x) 2  52  x  6.25

C

x

D

M

log 2  log 3  log 2 x    1  log 3  log 2 x   2  log 2 x  9  x  29  512

1 a  2   1        2  ab  1 and 4  b  3  b  1  2 1  b   3  Therefore, 2  a (1)  1  a  3 . a20  5  (20  1)3  62 , S 20 

n(a1  a20 ) 20(5  62)   670 2 2

1 1 k k  1 1       2   2 1  1 and   3   3 1  0.5 k 1 k 1 1 1 2 3 Therefore, the sum is 1  0.5  1.5 . 

45. (B)

46. (A)

1 2  3    n lim  lim n  n  n2

47. (B)

P

48. (A)

Since 6 Cr  ax 

n(1  n) n2  n 1 2  lim  n  2n 2 2 n2

 5 C3  0.2175602176  0.218 13 C5

8 C2

 y

6 r

r

 60 x 2 y 4 , 6 Cr a 6  r x 6 r  1 y r  6 Cr a 6 r  1 x 6  r y r . r

r must be 4. Therefore, 6 C4 a 2  1  60  15a 2  60  a  2 4

49. (D)

246

f

 g  x   e



  eln  x2 1

2ln x 2 1

2





 x2  1

2

 x4  2 x2  1

r

2 50. (B)

2 ab 0 ab Method 1: ab  0 a  b and a  0, b  0  a  b  0    a  b a  b and a  0, b  0  0  a  b

ab  0  a  b and a  0, b  0  a  b  0 (Not working)    a  b  a  b and a  0, b  0  b  0  b Method 2: Plug-in test number 2 1 I. a  2, b  1   2  0 (OK) 2 1 (1)(2)  1  0 (OK) II. b  2, a  1  (1)  (2) (2)(1)  2  0 (NO) III. a  2, b  1  (2)  (1) END

Dr. John Chung's SAT II Math Level 2 Test 6

247

No Test Material on This Page

248

Test 7                                               Dr. John Chung’s                SAT II Mathematics Level 2

Dr. John Chung's SAT II Math Level 2 Test 7

249

2

 

2

    

MATHEMATICS LEVEL 2 TEST REFERENCE INFORMATION    

 

THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME   OF THE QUESTIONS IN THIS TEST   1   Volume of a right circular cone with radius r and height h : V   r 2 h 3   1 Lateral Area of a right circular cone with circumference of the base c and slant height  : S  c 2  

  Volume of a sphere with radius r : V  4  r 3 3   Surface Area of a sphere with radius r : S  4 r 2   1   Volume of a pyramid with base area B and height h : V  3 Bh

       

250

2

 

2 Dr. John Chung’s SAT II Math Level 2

Answer Sheet 01 02

A

B

C

D

E

A

B

C

D

E

 

03

A

B

C

D

E

B

C

D

 

04 05

A A

B

C

 

06

A

B

 

07

A

08

26 27

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

E

28 29

A

B

C

D

E

D

E

30

A

B

C

D

E

C

D

E

A

B

C

D

E

B

C

D

E

31 32

A

B

C

D

E

A

B

C

D

E

33

A

B

C

D

E

09

A

B

C

D

E

34

A

B

C

D

E

10 11

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

35 36

A

B

C

D

E

12

A

B

C

D

E

37

A

B

C

D

E

13 14

A

B

C

D

E

B

C

D

E

A

B

C

D

E

38 39

A

 

A

B

C

D

E

 

15

A

B

C

D

E

40

A

B

C

D

E

 

16 17

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

41 42

A

B

C

D

E

 

18

A

B

C

D

E

B

C

D

E

 

19 20

B

C

D

E

43 44

A

A

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

 

21 22

45 46

A

B

C

D

E

47

A

B

C

D

E

 

23

A

B

C

D

E

48

A

B

C

D

E

24

A

B

C

D

E

49

A

B

C

D

E

25

A

B

C

D

E

50

A

B

C

D

E

              

 

       

 

       

     

           The number of right answers:                              

                

           The number of wrong answers:                             

 

          

 

# of correct

Dr. John Chung's SAT II Math Level 2 Test 7

1    4 # of wrong

Raw score

 

251

 

2

2 Score Conversion Table 6

Scaled Score 480

630

5

470

26

620

4

470

800

25

620

3

460

46

800

24

610

2

460

45

800

23

610

1

450

44

800

22

600

0

450

43

800

21

600

42

800

20

590

41

800

19

590

40

780

18

580

39

760

17

570

38

750

16

560

37

740

15

550

36

720

14

540

35

710

13

530

34

700

12

520

33

690

11

510

32

680

10

500

31

670

9

490

30

660

8

490

29

650

7

480

28

Scaled Score 640

800

27

48

800

47

50

Scaled Score 800

49

Raw Score

252

Raw Score

Raw Score

2

2

 

  

MATHEMATICS LEVEL 2 TEST For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding circle on the answer sheet Note: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions in this test. For each question you will have to decide whether or not you should use a calculator. (2) For some questions in this test you may have to decide whether your calculator should be in the radian mode or the degree mode. (3) Figures that accompany problems in this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers

x for which f ( x) is a real number. The range of f is assumed to be the set of all real numbers f ( x) , where x is in the domain of f . (5) Reference information that may be useful in answering the questions in this test can be found on the page preceding Question 1. USE THIS SPACE FOR SCRATCHWORK 1.

If x  10 , then

(A) 5

2.

(B) 9

x x 1  x x  xx

(C) 10

(D) 100

(E) 1000

If x and y are positive integers and x 2  y 2  21 , then which of the following could be the value of x ? (A) (B) (C) (D) (E)

13 11 10 9 4

 

 

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

3.

If f ( x)  4 x  2 , then the inverse function f 1 ( x) is x8 4 2x  1 2 x2 4 2x  1 2 x2 4

(A) (B) (C) (D) (E)

4.

In how many points do the graphs of x 2  y 2  1 and y 2  x 2  1 intersect? (A) (B) (C) (D) (E)

5.

If f ( x)  3x and g ( x)   2 x  , then f  g (5.4)   (A) (B) (C) (D) (E)

6.

0 1 2 4 8

33 30 25 20 3

A

In Figure 1, AD  CD and AB  5 . Which of the following is the value of tan ACD ? (A) (B) (C) (D) (E)

0.201 0.309 0.407 0.414 0.500

C

45

D

B

Note: Figure not drawn to scale. Figure 1

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

7.

If sin  cos  

1 , then which of the following is the value(s) of 4

sin   cos  ?

1  (A)   2  1 (B)     2

 2    2 

(C) 



(D) 



2   2 

 2 2  ,  2   2

(E) 

8.

If x increases from  (A) (B) (C) (D) (E)

9.

 2

to

 2

, then the value of sec x

decreases, then increases. increases, then decreases. increases. decreases. none of these

 3x  2 If f    x  1 , what is the value of f (6) ?  1 x   (A) (B) (C) (D) (E)

3 5 18 37 42

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

10.

x 1 1 and f  g ( x)   , then which of the x x 1 following could be g ( x ) ?

If f ( x) 

(A) x  1 1 x 1

(B)

(C) x  1 1 x 1 x 1 (E) 1 x

(D)

11.

What is the range of the function defined by

 x2 , x  0  f ( x)   1 , x0   x (A) y  0 (B) y  0 (C) y  0 (D) y  0 (E) All real numbers

12.

If 2log x  log  2 x  3 , then which of the following is the solution set of x ? (A) (B) (C) (D)

1 1, 3 3 3, 4

(E) All real numbers

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256

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

13.

Which of the following could be the graph of y   (A)

(B)

y

O

x x

?

y

x

x

O

x

(C)

(D)

y

y

 x

O

(E)

x

y

O

14.

O 

x

Which of the following is the equation of the polynomial with roots 1  2 and i ? (A) x3  x  1 (B) x 3  2 (C) x 4  2 (D) x 4  3x 2  2 (E) x 4  2 x 3  2 x  1

15.

If a  log 5 9 , then 252 a is (A) 81

(B) 729

(C) 2187

(D) 6561 (E) 13122

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

16.

If the volume of a cube is 63a , then the surface area of the cube is (A) 62 a (B) 6(6a ) (C) 6(32 a ) (D) 62 a1 (E) 62 a 3

17.

If cos 2 x 

1 , then sin x  2

3 2 1 1 (B)  or 2 2 (A) 

(C)

3 2

(D)

3

(E) 

18.

 3 If Arc sin     k , then tan k   2  (A) 1

19.

2 2 or 2 2

(B)  3

(C)

3

(D)

1 2

(E) 2

Which of the following is the domain of f ( x)  (A) (B) (C) (D) (E)

x 1 ? x2

All real numbers All real numbers except 2 All real numbers greater than or equal to 1 All real numbers greater than or equal to 2 All real numbers greater than or equal to 1 except 2

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

20.

If x C x  2  21 , then x  (A) (B) (C) (D) (E)

21.

On a multiple- choice test, there are 5 choices for each question. What is the probability that a student who guesses every answer will have exactly 10 correct answers on a test that consists of 20 questions? (A) (B) (C) (D) (E)

22.

4 5 6 7 8

0.002 0.02 0.2 0.25 0.5

What is the sum of the numerical coefficients of  x  2 y  ? 4

(A) (B) (C) (D) (E)

0 1 16 32 48

100

23.

What is the value of

 in ? n 1

(A) 0 (B) 1 (C) i (D) i (E) 1

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

24.

If 3  i 2 is the root of the quadratic equation ax 2  12 x  c  0 , what is the value of c ?

(A) (B) (C) (D) (E) 25.

11 22 25 33 44

In Figure 2, which of the following is the equation of the graph?

y

(A) y  3sin 2

3

(B) y  3sin 4

O

(C) y  3sin 8 (D) y  3sin (E) y  3sin 26.

2

4



3

 2

Figure 2

 4

What is the length of the major axis of an ellipse whose equation is x 2  4 y 2  4 x  8 y  8 ? (A) 4

27.

(B) 8

(C) 12

(D) 16

(E) 20

What is the slope of the line tangent to the circle

 x  12   y  12  25

at the point  4,5  ?

4 3 3  4 5  4 4  5 4 5

(A)  (B) (C) (D) (E)

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260

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

28.

  If f ( x)  f  x   , then which of the following could be 2  f ( x) ? (A) f ( x)  sin x (B) f ( x )  2sin 2 x (C) f ( x)  cos x (D) f ( x)  2cos 2 x (E) f ( x)  2 tan 2 x

29.

In Figure 3, ABCD is a rectangle and tan CBE  tan EAD 

(A) (B) (C) (D) (E) 30.

1 and 7

1 . What is the value of tan BDA ? 3

C E

0.488 0.476 0.434 0.421 0.306

If f ( x) 

B

A

D

Note:Figure not drawn to scale. Figure 3 1  2 , what is the range of the function? x3

(A) y  2 (B) y  2 (C) y  3 (D) y  2 (E) All real 31.

y

y  f ( x)

Figure 4 shows the graph of f ( x) . Which of the following could be the function f ( x) ? (A) y  x  x  1

1

O

x 1

(B) y  x  1  x  1 (C) y  x  1  x  1 (D) y  x  1  x  1

Note: Figure not drawn to scale. Figure 4

(E) y  x  1  x

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

32.

The right circular cone is sliced horizontally forming two pieces, each of which has the same height. What is the ratio of the volume of the smaller piece to the volume of the larger piece? 1 (A) 2 1 (B) 4 1 (C) 5 1 (D) 7 1 (E) 8

33.

How many possible rational zeros does

f ( x)  2 x3  3x2  8x  4 have? (A) (B) (C) (D) (E) 34.

6 8 10 12 14

What is the polar form of the rectangular equation

x2  y 2  4 x  0 ? (A) r  sin 

35.

(B) (C) (D) (E)

r 2  4sin  r  4 cos  r  4sin  r  2 cos 

lim

x 1  x  x2  x  1

(A)

1 4

x 1

3

(B)

1 2

(C)

2 3

(D) 2

(E) 5

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

36.

If the demand equation for a graphic utility is given by





D  x   200  0.4 e0.005 x , where D is in dollars, which of the following is the demand x for a price D of $150? (A) 966

37.

(D) 1450

(E) 2048

Parabola Circle Ellipse Hyperbola Two perpendicular lines

If g ( x)  3 (A) (B) (C) (D) (E)

39.

(C) 1368

What is the graph of the parametrically defined equations x  4  2 cos  and y  sin   1 ? (A) (B) (C) (D) (E)

38.

(B) 1024

x 1 , then g 1 (1.5)  2

4.25 5.75 6.52 7.12 8.45

On a math exam, the scores of ten students were 66, 81, 85, 97, 86, 58, 76, 73, 88, and 80. What is the standard deviation of the scores? (A) (B) (C) (D) (E)

10.72 12.29 12.88 13.16 13.58

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

40.

Which of the following is the function of the graph in Figure 5? y

f ( x)   x3  ax 2  bx  c

I. II.

f ( x)  x5  ax 4  bx3  cx 2  dx  e

III.

f ( x)  x 7  ax 6  bx5  cx 4  dx3  ex 2  fx  g

(A) (B) (C) (D) (E)

41.

I only II only II and III only I, II, and III None of those

y  f ( x)

O

x

Figure 5

What is the smallest positive value of  , in radians, which satisfies the equation 2sin 2   2 cos   1  0 ? (A) (B) (C) (D) (E)

42.

0.56 0.86 1.24 1.56 1.95

Which of the following is the solution set of the equation log 3 ( x  5)  log 9 (2 x  5) ? (A) 2 (B) 5 (C) 10 (D) 2,10 (E) 2, 5, 10

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264

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

43.

What is the distance between two points of 1, 2, 3 and

 0,  4,  2 ? (A) (B) (C) (D) (E)

44.

7.87 8.24 8.48 10.25 11.24

Which of the following could be the solution to

 n 2  2  n   3 , where  n (A) (B) (C) (D) (E)

is the greatest integer function?

2n3 0  n 1 1  n  0 1  n  0 1  n  0

y C

45. In Figure 6, ABCDE is a regular pentagon with side of length 6. What is the x -coordinate of D ?

(A) (B) (C) (D) (E)

D

B

10.3 10.5 10.7 10.9 11.9 O

A (4, 0)

E

x

FIgure 6 46.

If the line y  3x  k is tangent to the hyperbola whose equation is 4 x 2  y 2  16 , which of the following could be the value of k ? (A) 2 (B) 1 (C)

5

(D) 2 5 (E) 3 5

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

47.

What is the area of the polygon formed by the points ( x, y ) which satisfy the inequality x  (A) (B) (C) (D) (E)

48.

49.

2

1?

2 3 4 8 10

If lim

x 1

(A) (B) (C) (D) (E)

y

x2  2 x  k  4 , then what is the value of k ? x 1

4 3 2 3 4

What is the radius of the sphere whose equation is

x2  y2  z 2  2x  4 y  6z  0 ? (A) (B) (C) (D) (E)

50.

3 3.74 8.56 12.45 14

    If a  (2, 1) and b  (1,  2) , then 3a  2b  (A) (B) (C) (D) (E)

7.44 8.06 8.45 9.12 10.14

STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK.

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266

No Test Material on This Page

Dr. John Chung's SAT II Math Level 2 Test 7

267

2

2

TEST 7 # 1 2 3 4 5 6 7 8 9 10

ANSWERS

answer B B C A A D E A B E

# 11 12 13 14 15 16 17 18 19 20

answer E C E E D D B B E D

# 21 22 23 24 25 26 27 28 29 30

answer A B A B E B B E B D

Explanations: Test 7 1. (B)

2. (B)

x x 1  x x x x ( x  1)   x 1 xx xx Since x  10, x  1  10  1  9 .

 x  y  x  y   21  x and y are positive integers   x  y  21  x  y  1

or

x  y  7  x  y  3

2 x  22

2 x  10

x  11 x5 Or, substitute choices and check.



x  4y  2 

f 1 : y 

3. (C)

y  4x  2

4. (A)

The asymptotes of both graphs are y   x . The graphs are as follows.

x2 4

y

x

5. (A)

268

g (5.4)   10.8  11



f (11)  33  33

# 31 32 33 34 35 36 37 38 39 40

answer D D B C B A C B A C

# 41 42 43 44 45 46 47 48 49 50

answer E C A D E D C B B B

2 6. (D)

2 tan C 

5

 0.414

55 2

A

5

5 2 45o

C

7. (E)

D

5 2

B

5

1 1 (sin   cos  ) 2  1  2sin  cos   1  2    4 2 2 2 1 1 1  sin 2   2  30 ,150 ,390 ,510... Or, sin  cos   sin 2  2 4 2   15 , 75 ,195 , 255... Therefore, 2 sin15  cos15  0.707   2 2 sin 75  cos 75  0.707  2 sin 75  cos 75  0.707 ….. The answer is (E). sin   cos   

8. (A)

sec x 

1 cos x

y  sec x y  cos x



9. (B)

10. (E) 11. (E)

Since





2

2

3x  6  x  2 , then f (6)  22  1  5. x 1

f (g) 

g 1 1 1 x   xg  x  g  1  g ( x)  g 1 x 1 x

The graph is as follows. 1 lim  0 , f (0)  0 x  x 1 Since lim  f (0) , the range is all real numbers x 0 x

Dr. John Chung's SAT II Math Level 2 Test 7

y



x

269

2

2

12. (C)

2log x  log(2 x  3)  x 2  2 x  3  x 2  2 x  3  0 ( x  3)( x  1)  0  x  3 or x  1 (rejected)

13. (E)

Piecewise-defined function: x  x  0,  1 y  x  x  0, 1

14. (E)

Reconstruct equation: conjugate roots [ x  (1  2)][ x  (1  2)][ x  i ][ x  i]  0 [( x  1) 2 

 2  ][ x 2

2

 1]  0  ( x 2  2 x  1)( x 2  1)  0

 x 4  2 x3  2 x  1  0 15. (D)

Since a  log5 9  9  5a ,

 

252 a  54 a  5a 16. (D)

4

 94  6561 .

x : the length of an edge

 

Volume  x3  63a  x  63a

 

Surface area  6 x 2  6 6a 17. (B)

18. (B)

2

1 1 1  sin 2 x   sin x   2 4 2 Or, 2 x  60 , 300  x  30 , 150 1 1 Therefore, sin 30  or sin 300   2 2 cos 2 x  1  2sin 2 x 

Use a calculator.  3 3 , where 90  k  90 . Or, Arcsin     k  sin k   2  2 

2

N ( x)  x  1  x  1 D ( x)  x  2  x  2 Domain:  x x  1 but x  2

270

 6a

 62 a 1

tan k   3

1

19. (E)

1 3

 3

2

2 x! x( x  1) x( x  1)    21  ( x  7)( x  6)  0 2!( x  2)! 2 2 x  7 ( x must be a positive integer and greater than 2) 

20. (D)

x Cx 2

21. (A)

P

22. (B)

Let x  1 and y  1 .

20 C10

 0.2 10 (0.8)10  0.0020314137  0.002

 x  2 y 4  4 C0  x 4  2 y 0  4 C1  x 3  2 y 1  4 C2  x 2  2 y 2  4 C3  x 1  2 y 3 0 4  4 C4  x   2 y  When you substitute x  1 and y  1 , you can get the sum of the coefficients. Both sides are equal for any x . Therefore, the sum of all coefficients is 1  2  1  1 . 4

100

23. (A)

 i n  i  i 2  i3  i 4  ...  i100 : Geometric sequence n 1

a (1  r n ) i (1  i100 ) i (1  1)   0 1 r 1 i 1 i Or, i  i 2  i 3  i 4  0 : The sum of every four terms is 0. There are 25 of the terms. 0  25  0 . S100 

24. (B)

Since 3  i 2 is the root, the other root is 3  i 2 . 12 Sum of the roots: S   6  a  2 a c Product of the roots: P   3  i 2 3  i 2  11  c  22 2



25. (E)



From the graph: Period is 8 and frequency is b  Amplitude : 3

26. (B)





2   . 8 4

y  3sin  4

x 2  4 y 2  4 x  8 y  8  ( x  2)2  4( y  1)2  16  a  4  Major axis  2a  8

( x  2) 2 ( y  1) 2  1 16 4

m

27. (B)

 (4,5) (1,1) 



Dr. John Chung's SAT II Math Level 2 Test 7

5 1 4  4 1 3 3 The slope of line    4

The slope of line m 

271

2 28. (E)

2 f ( x) is periodic with period of The period of f ( x)  2 tan 2 x is

29. (B)

30. (D)

 2



2

. .

CE a ED b  and tan EAD   . BC 7a AD 3b AB a  b 7a tan BDA   B AD 3b 3b Since 7 a  3b  a  , ab 7 3b b ab 7 10b 10 A     0.476 . 3b 3b 3b 21b 21 tan CBE 

C a E b

D

1 : shift to the right by 3 and up by 2 x The range: y  2

The graph of f : Transformation of y  y

2 O

3

x

31. (D)

From the graph: y2  If x  1,   If  1  x  1, y  2 x  If x  1, y  2  Choice (D): If x  1, then y  ( x  1)  ( x  1)  2 If 1  x  1, then y  ( x  1)  ( x  1)  2 x If x  1, then y  ( x  1)  ( x  1)  2

32. (D)

Since the ratio of the lengths is 1:2, the ratio of their volumes is 13 : 23  1: 8 . Therefore, the ratio of the volumes is 1:7.

h

2h

 

33. (B)

272

The rational zero test: 1,  2,  4 1  1,  2,  4,  1,  2 2

: 8 possible rational zeros

2 34. (C)

2 Since r 2  x 2  y 2 and r cos   x , x 2  y 2  4 x  0  r 2  4r cos   0  r  4 cos  ( x  1) 1 x 1 ( x  1)   lim 2  lim 2 2 1 1 x  x  x  x  x 1 x ( x  1)  ( x  1) ( x  1) ( x  1) 2

35. (B)

lim

36. (A)

150  200  0.4 e0.005 x

37. (C)

38. (B)

x 1

3





 x   ln125   0.005  966

 125  e0.005 x

x4 , sin   y  1 2 ( x  4)2 Since cos 2   sin 2   1,  ( y  1) 2  1 : Ellipse 4 cos  

g 1 (1.5)  k  (1.5, k ) Since g 1 ( x) is the inverse of g ( x) , point (k , 1.5) must be on g ( x). Therefore,

3

k 1 k 1  1.5   1.53  k  2  1.53  1  5.75 . 2 2

Or, you can find g 1 ( x) . 39. (A)

Use your calculator : Statistics.  x  10.72

40. (C)

Check the end of the graph. f ( x) have three real roots. Therefore, II and III could be f ( x). I. lim f ( x)   ( False) x 

II. lim f ( x)   and lim f ( x)   . Three real and two imaginary roots. x 

x 

III. lim f ( x)   and lim f ( x)   . Three real and four imaginary roots. x 

41. (E)

x 

Use a graphic utility:



Zero x  1.9455308

42. (C)

y0

Theorem: log a b  log a 2 b 2 log3 ( x  5)  log9 (2 x  5)  log9 ( x  5)2  log9 (2 x  5)  ( x  5) 2  2 x  5 ( x  2)( x  10)  0  x  2 or x  10 But x  5  0 , x  2 rejected.

Dr. John Chung's SAT II Math Level 2 Test 7

273

2

2

43. (A)

Distance  (1  0) 2  (2   4) 2  (3   2)2  62  7.87

44. (D)

Let  n   x . x 2  2 x  3  0  ( x  3)( x  1)  0  x  3 or x  1

If  n   3, then 3  n  4 .

If  n   1, then 1  n  0 .

45. (E)

y

( x, y ) 6

x  10  6 cos 72  11.854    11.9 72

x

10

46. (D)

Discriminant: Substitute y  3x  k into the equation of the hyperbola. 4 x 2  (3x  k ) 2  16  5 x 2  6kx  k 2  16  0 D  (6k )2  4(5)(k 2  16)  0  k 2  20

y  3x  2 5

Therefore, k   2 5 . y 2

2

x y  1 4 16 Since a  2 and b  4, Asymptotes: y   2 x . 4 x 2  y 2  16 

 x



y  3x  2 5

274

2 47. (C)

2 Remember: The graph of x  y  1 is as follows. y 1

1

1

x

1

Therefore, the graph of x 

y

2

 1 is

y 2

1

1

x

2

The area 

48. (B)

2 1 4 4 2

x2  2 x  k  4 : Numerator must have a factor of ( x  1) . x 1 x 1 N (1)  1  2  k  0  k  3

lim

( x  3) ( x  1) x2  2 x  k x2  2 x  3 4  4  lim  4  lim x 1 x 1 x 1 x 1 x 1 ( x  1)

lim

49. (B)

Since ( x  1)2  ( y  2)2  ( z  3)2  14 , r  14  3.74 .

50. (B)

  3a  2b  3(2,1)  2(1, 2)  (4, 7)   3a  2b  42  7 2  65  8.06

END

Dr. John Chung's SAT II Math Level 2 Test 7

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No Test Material on This Page

276

Test 8                                               Dr. John Chung’s                SAT II Mathematics Level 2

Dr. John Chung's SAT II Math Level 2 Test 8

277

2

 

2

    

MATHEMATICS LEVEL 2 TEST REFERENCE INFORMATION    

 

THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME   OF THE QUESTIONS IN THIS TEST   1   Volume of a right circular cone with radius r and height h : V   r 2 h 3   1 Lateral Area of a right circular cone with circumference of the base c and slant height  : S  c 2  

  Volume of a sphere with radius r : V  4  r 3 3   Surface Area of a sphere with radius r : S  4 r 2   1   Volume of a pyramid with base area B and height h : V  3 Bh

       

278

2

 

2 Dr. John Chung’s SAT II Math Level 2

Answer Sheet 01 02

A

B

C

D

E

A

B

C

D

E

 

03

A

B

C

D

E

B

C

D

 

04 05

A A

B

C

 

06

A

B

 

07

A

08

26 27

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

E

28 29

A

B

C

D

E

D

E

30

A

B

C

D

E

C

D

E

A

B

C

D

E

B

C

D

E

31 32

A

B

C

D

E

A

B

C

D

E

33

A

B

C

D

E

09

A

B

C

D

E

34

A

B

C

D

E

10 11

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

35 36

A

B

C

D

E

12

A

B

C

D

E

37

A

B

C

D

E

13 14

A

B

C

D

E

B

C

D

E

A

B

C

D

E

38 39

A

 

A

B

C

D

E

 

15

A

B

C

D

E

40

A

B

C

D

E

 

16 17

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

41 42

A

B

C

D

E

 

18

A

B

C

D

E

B

C

D

E

 

19 20

B

C

D

E

43 44

A

A

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

 

21 22

45 46

A

B

C

D

E

47

A

B

C

D

E

 

23

A

B

C

D

E

48

A

B

C

D

E

24

A

B

C

D

E

49

A

B

C

D

E

25

A

B

C

D

E

50

A

B

C

D

E

              

 

       

 

       

     

           The number of right answers:                              

                

           The number of wrong answers:                             

 

          

 

# of correct

Dr. John Chung's SAT II Math Level 2 Test 8

1    4 # of wrong

Raw score

 

279

 

2

2 Score Conversion Table 6

Scaled Score 480

630

5

470

26

620

4

470

800

25

620

3

460

46

800

24

610

2

460

45

800

23

610

1

450

44

800

22

600

0

450

43

800

21

600

42

800

20

590

41

800

19

590

40

780

18

580

39

760

17

570

38

750

16

560

37

740

15

550

36

720

14

540

35

710

13

530

34

700

12

520

33

690

11

510

32

680

10

500

31

670

9

490

30

660

8

490

29

650

7

480

28

Scaled Score 640

800

27

48

800

47

50

Scaled Score 800

49

Raw Score

280

Raw Score

Raw Score

2

2

 

  

MATHEMATICS LEVEL 2 TEST For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding circle on the answer sheet Note: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions in this test. For each question you will have to decide whether or not you should use a calculator. (2) For some questions in this test you may have to decide whether your calculator should be in the radian mode or the degree mode. (3) Figures that accompany problems in this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers x for which f ( x) is a real number. The range of f is assumed to be the set of all real numbers f ( x) , where x

is in the domain of f . (5) Reference information that may be useful in answering the questions in this test can be found on the page preceding Question 1. USE THIS SPACE FOR SCRATCHWORK 1.                 

If  e x  32 , then  x      (A)  1.4  (B)  1.7  (C)  1.9  (D)  2.2  (E)  2.4 

2. 

If  f

           

  (A)  21  (B)  46  (C)  84  (D)  117  (E)  256 

 x  x

2

 4 x , then  f (3)   

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  Dr. John Chung's SAT II Math Level 2 Test 8

281

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

3.

If 3 

1 3 1  9  , then 3   x x x

(B) 1

(A) 3

4.

ab 18

(B)

(E) 9

2a b

(C)

2b a

b 2a

(D)

(E)

a 2b

If 32 y1  5 , then y  (A) 0.23

6.

(D) 3

3 ax  6 bx (A)

5.

(C) 0

(B) 0.32

(C) 2.75

(D) 3.12

(E) 3.44

Which of the following could be the graph of the polar equation r  2cos  ? y

(A)

y

(B)

x

x

O

y

y

(C)

(D) x

O

x

y

(E) x

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282

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

7.

Which of the following ordered pairs is not an element of x the greatest- integer function y    ? 2

(D)

8, 4  2.78, 1  4.8,  3  5.64,  2

(E)



(A) (B) (C)

8.



10, 1

What is the range of the function f ( x)  x  2 for the domain 3  x  3 ? (A) 3  y  5 (B) 1  y  5 (C) 0  y  5 (D) 1  y  5 (E) 3  y  5

9.







If 2  5 3  5  a  b 5 , where a and b are rational numbers, then a  b  (A) 1

10.

(B) 2

(C) 3

(D) 4

(E) 5

a a 4a  3

(A) a 16 (B)

1 a8 3

(C) a 4 (D) a (E) a 2

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283

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

11.

Figure 1 shows the prism with dimensions 3, 5, and 7. What is the perimeter of the triangle ABC ? (A) (B) (C) (D) (E)

A

B

15 17.45 20.71 22.38 23.12

7

5 3

C

Figure 1 12.

cos 4   sin 4  

(A) 1 (B) sin  2  (C) 2 cos  (D) cos  2  (E) sin   cos 

13.

If 2  3 is a root of the polynomial P( x) , then a factor of P( x) is (A) x 2  3 (B) x 2  4 (C) x 2  4 x  1 (D) x 2  4 x  1 (E) x 2  4 x  1

B 14.



The angle of elevation from point A to point B is 40 and the angle of elevation from D to B is 60 . If the length of AD is 100, then what is the length of BC ? (A) (B) (C) (D) (E)

80.50 93.97 98.56 120.45 162.76

A

40

60

D

C

Note: Figure not drawn to scale. Figure 2

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284

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

15.

1 and f  g ( x)   x , then which of the x 1 following is g ( x) ?

If f ( x) 

(A) x  1 x x 1 x 1 (C) x x 1 (D) x x (E) 1 x

(B)

16.

In  ABC , if A  30 , b  1 , and c  3 , then B  (A) 30 (B) 45 (C) 52 (D) 60 (E) 75

17.

If sin   0.4 , then which of the following could be cos(90   ) ? (A) (B) (C) (D) (E)

0.2 0.4 0.5 0.6 0.8

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285

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

18.

The set of points ( x, y, z ) such that ( x  1) 2  ( y  1) 2  z 2  0 is (A) (B) (C) (D) (E)

19.

A point A circle A plane A sphere Empty

If the graph of the rational function R( x) 

x 2  ax  b x  x  1

does not have vertical asymptotes, then what is the value of a? (A) (B) (C) (D) (E)

20.

0 1 2 3 4

If the four numbers log 3, x, log81, and y form a geometric sequence in that order, which of the following could be the value of y ? (A) log126 (B) log162 (C) log 324 (D) log 1296 (E) log 6561

21.

If 52 x 3  7 x 1 , then x  (A) (B) (C) (D) (E)

4.35 2.26 1.32 2.26 4.35

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286

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

22.

If lim

x 

(A) (B) (C) (D) (E)

ax 2  bx  1  4 , then what is the value of a  b ? 2x  5

4 6 8 10 14

n

23.

 n  lim

i 2

(A) (B) (C) (D) (E)

24.

1  3i

1 6 1 3 2 3 3 2 4 3

If a and b are values in the domain of f ( x) and f (a)  f (b) , where b  a , then which of the following must be true? (A) f ( x) is an odd function. (B) f ( x) is an even function. (C) f ( x) increases as x increases. (D) f ( x) decreases as x increases. (E) f ( x) is a linear function.

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

25.

What is the measure of the largest angle in a triangle with sides of lengths 3, 4, and 6? (A) 62 (B) 84 (C) 98 (D) 117 (E) 128

x 3

26.

1 If f ( x)     2 and g ( x) is the inverse of f , then 2 what is the value of a which satisfies g (a )  5 ?

(A) (B) (C) (D) (E)

27.

2 2.25 2.5 4.45 5.25

If f ( x )  3 x , then what is the range of the function? (A) y  0 (B) y  0 (C) y  1 (D) y  1 (E) 0  y  1 y

y  3 x

28.

Figure 3 shows the graph of y  3 x and three inscribed rectangles. What is the sum of the areas of the rectangles? (A) 0.25 (B) 0.50 (C) 2.45 (D) 6.26 (E) 12.68

O

2

4

8

x

Note: Figure not drawn to scale. Figure 3

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288

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK. 2

29.

What is the domain of f ( x)  ( x 2  4) 3 ? (A) (B) (C) (D) (E)

30.

3 5 , A is acute, sec B   , and B is obtuse, 5 3 then what is the value of tan( A  B ) ?

If sin A 

(A) (B) (C) (D) (E) 31.

0.58 0.29 0.29 0.58 0.86

If f (3x  4)  6 x  5 for all real numbers x , then f ( x )  (A) (B) (C) (D) (E)

32.

x0 x  2 or x  2 2  x  2 2  x  2 All real numbers

3x  4 2 x  15 2 x  13 2 x  10 3 x  13

Which of the following is true for the graph of the equation x 2  y 2  kx , where k is a positive constant? (A) A circle with center on the y -axis is tangent to the x-axis. (B) A circle with center on the x-axis is not tangent to the y -axis. (C) A circle with center on the x-axis is tangent to the y -axis. (D) An ellipse with center on the x-axis is tangent to the y -axis. (E) None of these

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

33.

What is the area of an equilateral triangle that is inscribed in a circle of radius 2? (A) 3 3 (B) 2 3 (C) 3 2 (D) 2 2 (E) 4 2

34.

What is the remainder when the polynomial x 4  3 x 2  10 is divided by x 2  1 ? (A) (B) (C) (D) (E)

35.

In Figure 4, if the radius of the base is 6 and the height of the cone is 8, what is the surface area of the cone? (A) (B) (C) (D) (E)

36.

10 14 x  10 x  14 2x  5

36 64 96 136 148



6

8

FIgure 4

A cylinder with height 10 is inscribed in a sphere with radius 8. What is the volume of the cylinder? (A) (B) (C) (D) (E)

201 652 844 1225 1412

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290

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

37.

If one zero of f ( x)  x 2  kx  k  7 is 3, where k is a constant, then what is the other zero of f ? (A) (B) (C) (D) (E)

38.

If $5,000 is invested in a bank at a rate of 5% annual interest compounded monthly for 3 years, what is the amount of the balance after 3 years? (A) (B) (C) (D) (E)

39.

1 2 3 4 5

$5300.45 $5800.12 $5807.36 $6000.50 $6900.56

 2x  3  If f    4 x  8 , then f (2)   3 

(A) 8 (B) 10 (C) 12 (D) 14 (E) 16

40.

In Figure 5, if the volume of the inscribed circular cone is 100, then what is the volume of the square pyramid? (A) (B) (C) (D) (E)

112.43 127.32 135.25 151.38 167.12

h r

2r

2r Figure 5

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291

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

41.

Which of the following is the point(s) of intersection of x 2  y 2  16 and  y 2  x  4 ? (A)  4, 0  (B)  5, 0  (C)  4, 0  and  5,  3 (D)  4, 0  and  4, 0  (E)  5, 0  and  5, 0 

42.

What is the coefficient of x3 in the expansion of  5 x  1 ? 9

(A) (B) (C) (D) (E)

43.

The standard deviation of a data set is 8.5. If a new data set is created by subtracting 5.5 from each data value, what is the standard deviation of the new data set? (A) (B) (C) (D) (E)

44.

5 25 125 9500 10500

3.0 5.5 7.0 8.5 8.7

 Which of the following is the unit vector of a  (3, 4, 12) ?

(A) (1, 0, 0) (B) (1, 1, 1) (C) (0.31, 0.23, 0.92) (D) (0.23, 0.31, 0.92) (E) (0.23, 0.92,  0.31)

GO ON TO THE NEXT PAGE

292

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

45.

The polynomial equation xy  x  2  0 can be expressed by a set of parametric equations as a function of t . If y (t )  2t  3 , then x(t )  (A) 1  3t 1 2t 1 (C) t2 1  3t (D) 2

(B)

(E) 3t  1

46.

If i  1 , what is the third term in the binomial expansion of (2  3i )6 ? (A) 1120 (B) 1120i (C) 2160 (D) 2160 (E) 2160i

47.

If there are 10 points on a circle, how many line segments can be made by connecting any two given points? (A) (B) (C) (D) (E)

48.

100 90 80 72 45

In  ABC , if A  45 , a  7 , and b  10 , how many distinct triangles can be formed? (A) (B) (C) (D) (E)

0 1 2 3 4 GO ON TO THE NEXT PAGE

Dr. John Chung's SAT II Math Level 2 Test 8

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

49.

Which of the following could be the curve represented by 1 2t  1 and y  ? the parametric equations x  t t y

(A)

x

O

y

(C)

(E)

x

O

y

(D)

x

O

y

(B)

O

x

y

O

x

GO ON TO THE NEXT PAGE

294

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

50.

Figure 6 shows the rectangular solid with dimensions 3, 4, and 5. If PQ is the diagonal of the solid, what is the value of  formed by the diagonals and the side QR ?

4

3

(A) 30

P

(B) 32.07

5

(C) 34.45 (D) 46.28



Q

(E) 55.55

R Note: Figure not drawn to scale. Figure 6

STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK.

GO ON TO THE NEXT PAGE Dr. John Chung's SAT II Math Level 2 Test 8

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2

2

TEST 8 # 1 2 3 4 5 6 7 8 9 10

ANSWERS

answer D D C D A D D C B D

# 11 12 13 14 15 16 17 18 19 20

answer C D E E C A B A B E

# 21 22 23 24 25 26 27 28 29 30

answer D C A D D B E A E C

# 31 32 33 34 35 36 37 38 39 40

answer C C A B C D E C B B

# 41 42 43 44 45 46 47 48 49 50

answer A E D D B D E A C E

Explanations: Test 8 1. (D)

x  ln 9  2.2

2. (D)

Since

3. (C)

3x  1  9 x  3  6 x  2  x 

4.

 3     abx  3b b  ax    6a 2a  6     abx   bx 

(D)

x 3 

x  9.

log 5 log 3

f ( 9)  92  4(9)  117

1  log 5  y   1  0.23 2  log 3 

5. (A)

2 y  1  log3 5 

6. (D)

Since r  x 2  y 2 and x  r cos  , r  2 cos  



1  33 0 3

x2  y 2  2

x 2

x y

2

7. (D)

 5.64   2    2.82  3  

8. (C)

From the graph of f ( x)  x  2 , the range is 0 y5

296

 x  12  y 2  1 .

 x2  y 2  2 x  0 

y

 (3,5)

3



O

3

x

2 9. (B)

10. (D)

11. (C)

2 The equation can be simplified as follows. 1  5  a  b 5 Since a and b are rational numbers, a  1 and b  1. a  b  2 . 1 a  a2

3 a2



1

3  3 2   a2   a4 .    

AC  32  52  7 2  83

3

1

4

Therefore, a 4  a 4  a 4  a.

AB  3

BC  5  7  74 2

2

Perimeter  83  74  3  20.71







12. (D)

cos 4   sin 2   cos 2   sin 2  cos 2   sin 2   cos 2   sin 2   cos 2

13. (E)

 x   2  3   x   2  3   0 ,  x  2  3  x  2  3    x  2

14. (E)

15. (C)

sin 20 sin 40 100sin 40   y 100 y sin 20 100sin 40  sin 60  z  y sin 60   162.76 sin 20

Law of sines:

2

 3  x2  4 x  1  0 B

20o

z

y 40o 100

A

If f  g ( x)   x , then g ( x)  f 1 .

60o

C

D

1 1 1  y 1   y  1 y 1 x x 1 1 1 x 1  x  g 1   g  1  Or, f ( g )  g 1 x x x For f 1 , x 

B

16. (A)

Law of cosines: a2 

 3

2

 12  2

 3  1 cos30



 4  3 1

a  1. Since AC  BC , A  B . B  30

a

3 A

30o

C

1

17. (B)

cos(90   )  cos 90 cos  sin 90sin   sin   0.4

18. (A)

It is not a sphere because the radius is 0. It indicates the point (1, 1, 0).

19. (B)

In order not to have any vertical asymptotes, x 2  ax  b must have factors of x( x  1) . Therefore, N (0)  b  0 and N (1)  1  a  0 . a  1.

20. (E)

r

x log81   x   2 log 3 . log 3 x y  log  81  2  log 6561

Dr. John Chung's SAT II Math Level 2 Test 8

Therefore, r  2 or  2 .

297

2

2

21. (D)

52 x 3  7 x 1  (2 x  3) log 5  ( x  1) log 7  x(2log 5  log 7)  log 7  3log 5 log 7  3log 5 x  2.26 2log 5  log 7

22. (C)

To have a finite limit value, degree of numerator and degree of denominator should be same. Therefore, a must be 0. To have limit value of 4, b must be 8. Therefore, a  b  8 .

23. (A)

Sum of geometric infinite series: S 

a , where a is the first term and r  1 . 1 r

1 1 1 1 1 1     9  lim  t   1 n  9 27 81 6 t 2 3 1 3 n

24. (D)

Since f (b)  f (a ) where b  a, f ( x) is decreasing for all real number x .

25. (D)

The longest side is opposite the greatest angle. a 2  b 2  c 2 32  42  62 11 cos C    2ab 2(3)(4) 24

 11  C  cos 1     117  24  26. (B)

1 If g (a )  5 , then f (5)  a . Therefore,   2

27. (E)

f ( x) has a maximum at x  0 and lim 3

53

x

x 

2a  a

9 . 4

 0.

Therefore, 0  y  1 . Or use graphic utility. y

1 O

28. (A) 29. (E)

x

1 1  1  The areas  2 f (2)  2 f (4)  4 f (8)  2    2    4    0.25 9  81   6561  y

Using a graphic utility: Domain: All real numbers. O

298

x

2

2 3 4 and tan B   , 4 3 3  4    tan A  tan B 4  3 tan( A  B )    0.29 . 1  tan A tan B  3  4  1       4  3 

30. (C)

Since tan A 

31. (C)

Let f (3k  4)  6k  5 . Then x  3k  4 or k 

x4 . 3

 x4 Therefore, f ( x)  6    5  2 x  13.  3  2

32. (C)

k  k x 2  y 2  kx   x    y 2    2  2 k   , 0 2 

y

O

33. (A)

2



The area of triangle is A 

x

(2)(2)sin120 3  3 3 . 2

2 120o 2

34. (B)

Use long division. x2  4

x 2  1 x 4  3x 2  10 x4  x2  4 x 2  10  4x2  4 14 35. (C)

Lateral area   r  where  is a slant height.   10. Therefore, the lateral area is  (6)(10)  60 and the base area is 36 . The surface area is 96 .

Dr. John Chung's SAT II Math Level 2 Test 8

299

2 36. (D)

2 The radius of the base is

V   r 2h  





156  39 . Therefore, the volume of the cylinder is 2

2

39 10  390  1225 .

16

10

d  162  102  156

d

37. (E)

Because 3 is a zero of f ( x) , f (3)  0 . f (3)  9  3k  k  7  0  k  8.

f ( x)  x 2  8 x  15  ( x  3)( x  5)  0 Therefore, the other root is 5. 12(3)

38. (C)

 0.05  A  5, 000 1   12  

39. (B)

2x  3 2 3

40. (B)

Since

41. (A)

Substitute y 2  x  4 in the other equation. That is,

 r 2h 3

 x

 100 

 5807.36 9 2

9 Therefore, f (2)  4    8  10. 2

 r 2h  r 2 h 100  100   . The volume of a pyramid : V  4    4   127.32 .  3      3 

x 2  x  4  16  x 2  x  20  0 

 x  5 x  4   0 ,

x  4,  5

2

But when x  5, y  5  4  9 (rejected). When x  4, y 2  4  4  0  y  0 . Therefore, the intersection is (4, 0) . y

O

42. (E)

43. (D)

300

y2  x  4 x

9 9 r r For the coefficient of each term    5 x   1 , r  6 to have x3 . r   9   3 6 Therefore, the coefficient is    5   1  10500 . 6   Because all data values have been decreased by 5, the standard deviation has not been changed.

2 44. (D)

45. (B)

2

  a u  , a

  1 a  32  42  122  13 . Therefore u   3, 4,12    0.23, 0.31, 0.92  . 13

Since y  2t  3 , the polynomial equation will be as follows. 2 1 x(2t  3)  x  2  0  x(2t  4)  2  x   x 2t  4 2t

46. (D)

6 4 2 The third term is    2   3i   2160 . 2  

47. (E)

10     45 2

48. (A)

Law of sines: 7 10 10sin 45   sin B   1.01 , sin B  1.01 (No solution) sin 45 sin B 7

49. (C)

x

50. (E)

PQ  32  42  52  50 and QR  4 .

1 t

cos  

 0 and t 

1 2t  1 1 . y  2   y  2  x 2 (Domain : x  0) . 2 t t x

 4  o    cos 1    55.55 50  50 

4

END

Dr. John Chung's SAT II Math Level 2 Test 8

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No Test Material on This Page

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Test 9                                               Dr. John Chung’s                SAT II Mathematics Level 2

Dr. John Chung's SAT II Math Level 2 Test 9

303

2

 

2

    

MATHEMATICS LEVEL 2 TEST REFERENCE INFORMATION    

 

THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME   OF THE QUESTIONS IN THIS TEST   1   Volume of a right circular cone with radius r and height h : V   r 2 h 3   1 Lateral Area of a right circular cone with circumference of the base c and slant height  : S  c 2  

  Volume of a sphere with radius r : V  4  r 3 3   Surface Area of a sphere with radius r : S  4 r 2   1   Volume of a pyramid with base area B and height h : V  3 Bh

       

304

2

 

2 Dr. John Chung’s SAT II Math Level 2

Answer Sheet 01 02

A

B

C

D

E

A

B

C

D

E

 

03

A

B

C

D

E

B

C

D

 

04 05

A A

B

C

 

06

A

B

 

07

A

08

26 27

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

E

28 29

A

B

C

D

E

D

E

30

A

B

C

D

E

C

D

E

A

B

C

D

E

B

C

D

E

31 32

A

B

C

D

E

A

B

C

D

E

33

A

B

C

D

E

09

A

B

C

D

E

34

A

B

C

D

E

10 11

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

35 36

A

B

C

D

E

12

A

B

C

D

E

37

A

B

C

D

E

13 14

A

B

C

D

E

B

C

D

E

A

B

C

D

E

38 39

A

 

A

B

C

D

E

 

15

A

B

C

D

E

40

A

B

C

D

E

 

16 17

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

41 42

A

B

C

D

E

 

18

A

B

C

D

E

B

C

D

E

 

19 20

B

C

D

E

43 44

A

A

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

 

21 22

45 46

A

B

C

D

E

47

A

B

C

D

E

 

23

A

B

C

D

E

48

A

B

C

D

E

24

A

B

C

D

E

49

A

B

C

D

E

25

A

B

C

D

E

50

A

B

C

D

E

              

 

       

 

       

     

           The number of right answers:                              

                

           The number of wrong answers:                             

 

          

 

# of correct

Dr. John Chung's SAT II Math Level 2 Test 9

1    4 # of wrong

Raw score

 

305

 

2

2 Score Conversion Table 6

Scaled Score 480

630

5

470

26

620

4

470

800

25

620

3

460

46

800

24

610

2

460

45

800

23

610

1

450

44

800

22

600

0

450

43

800

21

600

42

800

20

590

41

800

19

590

40

780

18

580

39

760

17

570

38

750

16

560

37

740

15

550

36

720

14

540

35

710

13

530

34

700

12

520

33

690

11

510

32

680

10

500

31

670

9

490

30

660

8

490

29

650

7

480

28

Scaled Score 640

800

27

48

800

47

50

Scaled Score 800

49

Raw Score

306

Raw Score

Raw Score

2

2

 

  

MATHEMATICS LEVEL 2 TEST For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding circle on the answer sheet Note: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions in this test. For each question you will have to decide whether or not you should use a calculator. (2) For some questions in this test you may have to decide whether your calculator should be in the radian mode or the degree mode. (3) Figures that accompany problems in this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers x for which f ( x) is a real number. The range of f is assumed to be the set of all real numbers f ( x) , where x

is in the domain of f . (5) Reference information that may be useful in answering the questions in this test can be found on the page preceding Question 1. USE THIS SPACE FOR SCRATCHWORK 1.

If 2(a  b)  5(a  b) , then which of the following must be true? (A) (B) (C) (D) (E)

2.

a0 b0 ab a  b ab0

If f ( x)   f ( x) for all real x and a point (3, 5) is on the line, then which of the following points is also on the line? (A) (3, 5) (B) (3, 5) (C) (5,  3) (D) (3,  5) (E) (5, 3)                                                      

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  Dr. John Chung's SAT II Math Level 2 Test 9

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

3.

If

10!  n ! , then n  90  56

(A) 3

4.

(B) 4

(C) 5

(D) 6

(E) 7

Figure 1 shows the graph of the linear function whose 3 equation is defined by f ( x)   x  3 . What is the value of 4 x -axis ?  formed by the line and (A) (B) (C) (D) (E)

36.9 36.9 45.5 45.5 55.8

y

y  f ( x)

 O

x

Note: Figure not drawn to scale. FIgure 1

5.

If f ( x)  3 x  1 and g ( x)  5 , then g  f (7.82)   (A) (B) (C) (D) (E)

6.

10.3 5 5 10.3 15.6

Two circles are symmetric with respect to y  x . If the equation of a circle is x2  y 2  2 x  4 y  1  0 , then which of the following is the equation of the other circle ? 2 2 (A) ( x  2)  ( y  1)  4

(B) ( x  1)2  ( y  2)2  2 2 2 (C) ( x  1)  ( y  2)  4

(D) ( x  2)2  ( y  2)2  4 (E) ( x  1)2  ( y  2)2  4

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

7.

If 6  (A) (B) (C) (D) (E)

8.

cos  ?

0.116 0.201 0.217 0.328 0.466

If x  3, then (A) (B) (C) (D) (E)

10.

1  n  13 2  n  13 2  n  26 n  1 or n  26 n  2 or n  13

If tan   4.5 , what is the value of (A) (B) (C) (D) (E)

9.

n  3  4 , which of the following is the solution set? 2

 x  10 2



10  x 10  x x  10  x  10

( x  10)

Which of the following is true? (A) sin     sin  (B) cos      cos  (C) tan     tan  (D) sec     sec (E) csc     csc

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

11.

When a polynomial P( x)  x2  ax  b is divided by ( x  1) , the remainder is 3, and when the polynomial is divided by ( x  2) , the remainder is 3 . What are the values of a and b? (A) a  9, b  11 (B) a  9, b  11 (C) a  5, b  3 (D) a  5, b  3 (E) a  3, b  5

12.

If sin a  cos(2a  30) , then what is the value of tan a  ? (A) (B) (C) (D) (E)

13.

0.21 0.36 0.42 0.60 0.75

What is the range of the function f ( x)   3x  9  4 ? (A) y  3 (B) y  3 (C) y  4 (D) y  4 (E) y  4 y

14.

If line  is perpendicular to the line y  3x , then what is the

y  3x



area of  ABO ?

A (A) (B) (C) (D) (E)

1 1.5 2 2.5 3

O

B (3, 0)

x

Note: Figure not drawn to scale.

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

15.

If xy  1 , then (A) 1

16.

(B) 2

If log

3

(E) 5

x  10 , then log 3 x3  (B) 15

(C) 30

If tan   3 and     (A) (B) (C) (D) (E)

19.

(D) 4

2 2 5 5 10

(A) 10

18.

(C) 3

If one of the roots of 2 x 2  ax  b  0 is 1  2i , what is the value of b ? (A) (B) (C) (D) (E)

17.

x y   x 1 y 1

(D) 45

(E) 60

3 , what is the value of cos  2  ? 2

0.2 0.4 0.8 0.8 0.4

If the surface area of a cylinder, whose height is twice the radius, is 50, then what is the value of the radius? (A) (B) (C) (D) (E)

 r

1.63 1.84 2.45 3.87 4.56

h

FIgure 3

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

20.

If f ( x)  log( x  1)  log( x  1) , then f 1 ( x)  2

1

(A)

10 x

(B)

x 2  10

10 x  1

(C)

(D)  10 x  1 (E)  10 x  1 y

21.

In Figure 2, f ( x)  x  b is tangent to the graph of a circle

f ( x)  x  b

whose equation is x 2  y 2  4 . What is the value of b ? (A) (B) (C) (D) (E)

22.

5.45 3.48 2.14 2.21 2.83

O

x

Figure 4

What is the distance between the two points of intersection of the circles whose equations are x 2  y 2  16 and

 x  4 2   y  4 2  16 ? (A) (B) (C) (D) (E)

23.

1.12 2.73 3.35 4.87 5.66

  If a   3, 2 , 3 and b  1, 5, 2  , which of the following is   the value of a  b ?

(A) 2.52 (B) 3.74 (C) 4.25 (D) 7.58 (E) 8.02

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

24.

A pencil holder contains only five black pencils and three white pencils. If three pencils are drawn at random, what is the probability to have two black pencils and one white pencil? (A)

25.

3 5

(B)

3 8

(C)

15 28

(D)

5 7

(E)

2 3

( x  1) 2  0 , then which of the following is the complete x solution set of the inequality?

If

(A)  x  0 (B) 0  x  1 (C)  x  0 (D)  x  0 or x  1 (E)  x  0 or x  1 26.

Which of the following includes all asymptotes of the x3 rational function f ( x)  2 ? x 1 (A) x  1, x  1 (B) x  1, y  0 (C) x  1, x  1, and y  0 (D) x  1, x  1, and y  x (E) x  1, x  1, and y  1

27.

Which of the following is the distance from the origin to the plane x  y  z  3  0 ? (A)

2

(B) 3 (C) 2 (D)

5

(E)

6

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

28.

If cos 2   3cos   1  0 , then what is the smallest positive value of  ? (A) (B) (C) (D) (E)

29.

What is the interquartile range of the following set of data 10, 13, 15, 18, 25, 30, 40, 60, 75, 80, 80 ? (A) (B) (C) (D) (E)

30.

20 40 50 60 70

3 If angle A is obtuse and tan A   , which of the 2 following is the value of cos 2A ?

(A) (B) (C) (D) (E)

31.

8.16 4.40 1.88 0.92 0.46

0.38 0.30 0.15 1.5 3.6

 3  3 What is the value of arccos    arcsin    ?  2   2  (A) 0 (B) 30 (C) 45 (D) 30 (E) 45

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

32.

If f ( x)  3 2 x  3 , then f 1  3  (A) (B) (C) (D) (E)

33.

The function f ( x)  x 2  4 x  9 is a shift of f ( x)  x 2 (A) (B) (C) (D) (E)

34.

3.6 5 8 10 12

4 units to the right and 9 units up 2 units to the right and 5 units down 2 unit to the left and 5 units up 4 units to the left and 9 units up 2 units to the right and 5 units up

If  cos  i sin   (cos  i sin  )  a  1  bi , where a and b are real numbers, which of the following is true? (A) a  1, b  1 (B) a  1, b  0 (C) a  2, b  0 (D) a  2, b  2 (E) a  2, b  2

35.

If the difference of the roots of x 2  2mx  7 is 8, then what is the positive integer value of m ? (A) 0

(B) 1 (C) 2

(D) 3 (E) 4

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

36.

What is the measure of one of the larger angles of a parallelogram in the xy -plane that has vertices with coordinates  3, 2  ,  6, 2  ,  4, 6  and  7, 6  ? (A) 76.0 (B) 98.2 (C) 104.0     

 

(D) 103.5 (E) 108.6

37.

A used car was purchased for $20,000 and the car loses k % of its value each year. If the car is worth $10,000 after 5 years, what is the value of k ? (A) (B) (C) (D) (E)

38.

Which of the following is the polar form of the rectangular equation y  4 ? (A) (B) (C) (D) (E)

39.

10.5 11.6 12.9 13.6 14.8

r 3 r4 r  4 sin  r  4 cos  r  4 csc 

If function f ( x ) 

1 x  3 , and f 1 ( x) is the inverse 4





function of f ( x) , then f 1 f 1 (3)  (A) 0 (B) 3 (C) 6 (D) 12 (E) 18

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

40.



(A) (B) (C) (D) (E) 41.

42.



If 2 32 x5  5  10 , what is the value of x ? 1.28 2.46 3.42 3.68 4.12

What is the domain of f  x   (A) (B) (C) (D) (E)

x  3 x 1 1 x  3 x3 1 x  5

lim

x2  x  x2  x  2

x 1

x 1 x2  x  6

?

1 3 1 (B) 2

(A)

(C) 1 (D) 2 (E) Undefined 43.

How many integer values of x satisfy the inequality x( x  6)( x  8)( x  2)  0 ? (A) 4

(B) 6

(C) 8

(D) 10

(E) 11

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

44.

What is the range of the following function?

 1  x, f ( x)    x  1  1, (A) (B) (C) (D) (E) 45.

 , 0 0, 1  0, 1 0,     0,   

10 100 1024 2048 4096

What is the coefficient of x in the binomial expansion of





4

x 5 ?

(A) (B) (C) (D) (E)

47.

if x  1

In how many ways can a 10 question true-false math exam be answered? (Assume that no questions are omitted.) (A) (B) (C) (D) (E)

46.

if x  1

20 150 500 625 875

If 14  n P3  value of n ? (A) (B) (C) (D) (E)

n  2 P4

, then which of the following could be the

6 7 8 9 15

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48.

Assume that the probability of having a boy is 40%. In a family with three children, what is the probability that there is at least one boy? (A) (B) (C) (D) (E)

49.

0.40 0.486 0.562 0.765 0.784

Figure 5 shows the graph of an ellipse whose equation is

 x  h

2

y k

2

  1 . If the area A of an ellipse is given a2 b2 by A   ab , then what is the area of the ellipse in Figure 5?

y

(A) 9.42 (B) 15.71 (C) 28.42

(5, 0) 

O

x  (10,  3)



(D) 47.12 (E) 52.25

Figure 5

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2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

50.

Which of the following could be the graph of the parametric equations x  t  1 and y  t  2 ? (A)

(B)

y

x

O

(C)

x

x

O

(D)

y

O

y

y

O

x

(E) y

O

x

STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK.

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320

No Test Material on This Page

Dr. John Chung's SAT II Math Level 2 Test 9

321

2

2

TEST 9 # 1 2 3 4 5 6 7 8 9 10

ANSWERS

answer C D D A C A C E A D

# 11 12 13 14 15 16 17 18 19 20

answer B B D B A E B D A C

# 21 22 23 24 25 26 27 28 29 30

answer E E B C C D B C D A

# 31 32 33 34 35 36 37 38 39 40

answer B E E C D C C E D C

# 41 42 43 44 45 46 47 48 49 50

Explanations: Test 9 1. (C)

ab 0  a b

2. (D)

Since f ( x) is an odd function, (3,  5) must be on the line.

3. (D)

10! 10!   6! , n  6 90  56 10  9  8  7

4. (A)

Since tan  

5. (C)

g ( x)  5 is a constant function for any value of x .

6. (A)

Switch x and y. y 2  x 2  2 y  4 x  1  0

7. (C) 8. (E)

6

( x  2)2  ( y  1)2  4

Since tan   4.5 and cos   0 ,  lies in Quadrant I. y 1 21.25 Therefore, cos    0.466 4.5 21.25 1

x

( x  10) 2  x  10 If x  3 , then x  10   x  10 .

322



n n n n   6  7   7   6  7   1   13   2  n  26 2 2 2 2

O

9. (A)

3 3 ,   tan 1    36.9 . 4 4

answer D A D E C B A E D D

2

2

10. (D)

sec  is an even function.

11. (B)

P(1)  1  a  b  3 and P (2)  4  2a  b  3 . ab2  2a  b  7 a

a  9 and b  11

 9

12. (B)

Cofunction: a  2a  30  90  a  20  tan 20  0.36

13. (D)

Check with a graphing calculator.

f ( x)   3x  9  4   3( x  3)  4

y

 (3, 4)

x

O

14. (B)

Range: y  4

1 The equation of line  is y   x  b . The line passes through a point (3,0). Therefore, the 3 1 3 1  1.5. equation of line  is y   x  1 . AO  1 and OB  3 . The area is 3 2 x y x( y  1)  y ( x  1) x y2 2 xy  x  y      1. x 1 y 1 xy  x  y  1 xy  x  y  1 x  y  2 1 1 x 1 x 1 1 y Or, since y  , . Therefore,    1.  x  1 x 1 x 1 x 1 x y 1 1 x 1 x

15. (A)

Since xy  1,

16. (E)

Product of the roots:

17. (B)

log

3

 1  2i  1  2i  

b b  5  b  10 2 2

x  log 3 x 2  10  2 log 3 x  10  log3 x  5

log3 x3  3log3 x  3(5)  15

Dr. John Chung's SAT II Math Level 2 Test 9

323

2 18. (D)

2 Since  lies in Quadrant III, cos   

2

4  1  . cos 2  2cos 2   1  2   1   5 10  10 

1

1 3

10

19. (A)

 Surface area = 2 r 2  2 r (2r )  50  6 r 2  50  r 

20. (C)

f ( x)  log( x  1)  log( x  1)  log( x 2  1) (Domain: x  1 )

50  1.63 6

For the inverse f 1 :





x  log y 2  1

21. (E)



y 2  1  10 x  y  10 x  1 (Range: y  1 )

y  x  b and x 2  y 2  4 intersect at one point. Substitute y  x  b into a circle equation. x 2  ( x  b) 2  4  2 x 2  2bx  b2  4  0 To have two equal roots, its discriminant should be zero. D   2b   4(2)(b 2  4)  4b 2  32  0  b 2  8  b  2 2 or 2 2 2

In Figure 2, the y-intercept b is negative. Therefore, b  2 2 .

Or, the distance from the origin to x  y  b  0 is d

00b 12  (1) 2

 2  b  2 2  b  2 2 : the y -intercept is negative.

y

22. (E)

The length is 4 2  5.66.

( x  4)2  ( y  4)2  16

(0, 4)

(4, 4)  (4,0)

x 2  y 2  16

23. (B)

324

x

The graphs intersect at points  0, 4  and  4,0  .

  a  b  (3, 2, 3)  (1, 5, 2)  (2,  3, 1)   a  b  (2,  3, 1)  22  (3) 2  12  14  3.74

2 24. (C)

25. (C)

2  3 C1 15  28 8 C3 Or, find the probability to have BBW , BWB, and WRR. 5 4 3 5 P( BBW )     . P( BBW )  P( BWB)  P(WBB) . 8 7 6 28 5 15 Therefore, 3  . 28 28 P

5 C2

( x  1)2 ( x  1) 2  0  multiply by x 2  x 2  0 x 2  x( x  1) 2  0  x  0  x x Using graphic solution: The complete solution set is  x  0 .

 

 0

Or, use test value. 26. (D)

 1

 



x3 x  x 2 2 x 1 x 1 Vertical asymptotes : D ( x)   x  1 x  1  0  x  1, x  1 f ( x) 

Slant asymptote: y  x as x   . 0003

D

28. (C)

Use a graphing calculator: Radian mode.

2

2

2

1 1 1



3

27. (B)

3

 3

 x  1.8784

y0

Algebraic Solution: Quadratic formula. 3  13 cos    cos   3.30277 (rejected) or  0.3027756377 2 cos 1 (0.3027756377)  1.8783999  1.88 29. (D)

Interquartile range  upper quattile  lower quartile = 75  15 = 60 10, 13, (15), 18, 25, (30), 40, 60, (75), 80, 80  10

15

Dr. John Chung's SAT II Math Level 2 Test 9

30

75

 80

Box and Whisker Plot

325

2 30. (A)

2 Method 1: Find the angle using a calculator. 3  3 tan A    A  tan 1     56.30993247  180  123.6900675  A is obtuse  2  2 cos 2 A  cos(2  123.6900675)  0.3846153846  0.38 Method 2: Using diagram A

2

 2  cos 2 A  2cos 2 A  1  2   1  13   0.3846153846  0.38

13

3 2

31. (B)

Method 1: Using a calculator (Degree mode)  3  3  arccos    arcsin     30 2 2     Method 2: Algebraic solution 3 3 , where 0    180 . cos      30 Let   arccos 2 2  3 3      60 Let   arcsin    , where  90    90 . sin    2 2   Therefore, 30  60  30

32. (E)

33. (E)

f 1 : y 

x3  3  2

f 1  3  12

Method1:

x  3 2y  3 

Method 2:

3  3 2 y  3  27  2 y  3  y  12

f ( x)   x  2   5 : Two units to the right and 5 units up 2

34. (C)

cos 2   sin 2   a  1  bi  1  a  1  bi  a  1  1 and b  0 Therefore, a  2 and b  0.

35. (D)

If r1  r2 , then r1  r2  8 , r1  r2  2m, and r1  r2  7 .

 r1  r2 2   r1  r2 2  4r1r2

 4m 2  64  28  m 2  9  m  3

Or, substitute the choices into the equation. For m  3, x 2  6 x  7  ( x  1)( x  7)  0  x  1,  7

326

(Difference is 8.)

2

2

36. (C)

(4,6)

(7,6)

x (6, 2) 1

(3, 2)

tan x  4  37. (C)

4



x  tan 1 4  75.96

Therefore,   180  75.96  104.0 .

Compound decay yearly: A  P(1  r )t 10,000  20,000 1  0.01k 

5

1   (1  0.01k )5 2

1

 1 5     1  0.01k 2

1

k

1   0.5  5 0.01

 12.94404367  12.9

38. (E)

y  4  r sin   4  r 

39. (D)

f 1 ( x)  4 x  12 

40. (C)

41. (D)

4  4csc sin 

f 1 (3)  0 

f 1 (0)  12

15  7.5  2 x  5  log3 7.5 2  1  log 7.5 Therefore, x    5   3.417021884  3.42 2  log 3  2(32 x 5 )  15  32 x 5 

N ( x)  x  1  x  1 D( x)  x 2  x  6  x 2  x  6  0  ( x  3)( x  2)  0  x  3 or x  2

 x  1   x  3   x  3

or  x  1   x  2  

Therefore, the solution set is  x  3 . x( x  1) x 1  lim  x 1 ( x  2)( x  1) x 1 x  2 3

42. (A)

lim

43. (D)

The graph of y  x( x  6)( x  8)( x  2)  0 is as follows.

8



0

2



6

x

Therefore, the intervals for y  0 is 8  x  0 or 2  x  6 . The integers x are 7,  6,  5,  4,  3,  2,  1 and 3, 4, 5  10 integers

Dr. John Chung's SAT II Math Level 2 Test 9

327

2 44. (E)

2 The graph of the piecewise function is as follows. y

y  x 1 1

y  1 x

 O

x

1

Therefore, the range is  0,   . 45. (C)

210  1024

46. (B)

 4   r 

 x

4r

 5 r

 4 Therefore,    2 47. (A)

14  n P3 

 4  4r     x  2 (5)r r 

 x

n  2 P4

2

 r must be 2

 52  150 x

 14  n(n  1) (n  2)  (n  2)(n  1) n(n  1)

14n  28  n 2  3n  2  n 2  11n  30  0  (n  6)(n  5)  0 Therefore, n  5 or 6. 48. (E)

Since P( B) 

2 3 and P(G )  , 5 5

1 2 2 1 3 0  3  2   3   3   2   3   3  2   3  98  0.784 The probability is                      1   5   5   2   5   5   3   5   5  125  3 2 Or, 1  P(all three girs)  1     0.6  (0.4)0  0.784. 3  

49. (D)

Since a  5 and b  3 , A   ab  15  47.12 .

50. (D)

Since x  t  1  0 and x 2  t  1  t  x 2  1 , Therefore, y  x 2  1 and x  0.

END

328

y  t  2  x2  1  2  y  x2  1 .

Test 10                                               Dr. John Chung’s                SAT II Mathematics Level 2

Dr. John Chung's SAT II Math Level 2 Test 10

329

2

 

2

    

MATHEMATICS LEVEL 2 TEST REFERENCE INFORMATION    

 

THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME   OF THE QUESTIONS IN THIS TEST   1   Volume of a right circular cone with radius r and height h : V   r 2 h 3   1 Lateral Area of a right circular cone with circumference of the base c and slant height  : S  c 2  

  Volume of a sphere with radius r : V  4  r 3 3   Surface Area of a sphere with radius r : S  4 r 2   1   Volume of a pyramid with base area B and height h : V  3 Bh

       

330

2

 

2 Dr. John Chung’s SAT II Math Level 2

Answer Sheet 01 02

A

B

C

D

E

A

B

C

D

E

 

03

A

B

C

D

E

B

C

D

 

04 05

A A

B

C

 

06

A

B

 

07

A

08

26 27

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

E

28 29

A

B

C

D

E

D

E

30

A

B

C

D

E

C

D

E

A

B

C

D

E

B

C

D

E

31 32

A

B

C

D

E

A

B

C

D

E

33

A

B

C

D

E

09

A

B

C

D

E

34

A

B

C

D

E

10 11

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

35 36

A

B

C

D

E

12

A

B

C

D

E

37

A

B

C

D

E

13 14

A

B

C

D

E

B

C

D

E

A

B

C

D

E

38 39

A

 

A

B

C

D

E

 

15

A

B

C

D

E

40

A

B

C

D

E

 

16 17

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

41 42

A

B

C

D

E

 

18

A

B

C

D

E

B

C

D

E

 

19 20

B

C

D

E

43 44

A

A

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

 

21 22

45 46

A

B

C

D

E

47

A

B

C

D

E

 

23

A

B

C

D

E

48

A

B

C

D

E

24

A

B

C

D

E

49

A

B

C

D

E

25

A

B

C

D

E

50

A

B

C

D

E

              

 

       

 

       

     

           The number of right answers:                              

                

           The number of wrong answers:                             

 

          

 

# of correct

Dr. John Chung's SAT II Math Level 2 Test 10

1    4 # of wrong

Raw score

 

331

 

2

2 Score Conversion Table 6

Scaled Score 480

630

5

470

26

620

4

470

800

25

620

3

460

46

800

24

610

2

460

45

800

23

610

1

450

44

800

22

600

0

450

43

800

21

600

42

800

20

590

41

800

19

590

40

780

18

580

39

760

17

570

38

750

16

560

37

740

15

550

36

720

14

540

35

710

13

530

34

700

12

520

33

690

11

510

32

680

10

500

31

670

9

490

30

660

8

490

29

650

7

480

28

Scaled Score 640

800

27

48

800

47

50

Scaled Score 800

49

Raw Score

332

Raw Score

Raw Score

2

2

 

  

MATHEMATICS LEVEL 2 TEST For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding circle on the answer sheet Note: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions in this test. For each question you will have to decide whether or not you should use a calculator. (2) For some questions in this test you may have to decide whether your calculator should be in the radian mode or the degree mode. (3) Figures that accompany problems in this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers

x for which f ( x) is a real number. The range of f is assumed to be the set of all real numbers f ( x) , where x is in the domain of f . (5) Reference information that may be useful in answering the questions in this test can be found on the page preceding Question 1. USE THIS SPACE FOR SCRATCHWORK 1.

If a  0 , then

 2a 2 

9a 2 

(A) a (B) 2a (C) 5a (D) 2a (E) 5a

2.

If 28k is an integer, which of the following is the smallest integer value of k ? (A) (B) (C) (D) (E)

2 4 6 7 28

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  Dr. John Chung's SAT II Math Level 2 Test 10

333

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

3.

If a3  64 and

a2  b , then b 

(A) 4 (B)  4 (C) 8 (D)  8 (E) 16 4.

How many positive values of a are there to satisfy that 40  a is an integer? (A) 3

5.

(C) 5

(D) 6

(E) 7

Which of the following sets of data has a standard deviation of 0 ? (A) (B) (C) (D) (E)

6.

(B) 4

3,  2,  1, 0, 1, 2, 3  3,  3,  3, 1, 3, 3, 3  2,  2,  2, 0, 2, 2, 2  0, 0, 0, 3, 5, 5, 5  5, 5, 5, 5, 5, 5, 5 

How many integers satisfy with the inequality

5  x2  10 ? (A) 4 (B) 6 (C) 8 (D) 10 (E) Infinitely many

7.

If a  3 , then

(A) (B) (C) (D) (E)

a2 

 a  2 2



8 5 3 5 8

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334

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

8.

Which of the following is true? (A)

 5

25   5

(C)

( 11) 2  11

(D)

( 6) 2  6



16



1 2

4

If x 2  5 , then ( x  3)( x  2)  ( x  1)( x  6)  (A)  4

10.

2

(B)

(E)

9.

 5

(B)  2

(C) 0

(D) 5

(E) 10

If the polynomial x 2  kx  1 has a factor of  x  2  ,

then k  (A)

11.

2 3

(B)

3 2

(C) 

2 3

(D) 

3 2

(E) 

5 2

If x 2  10 x  a   x  b  for all real x , where a and b are 2

constants, what is the value of a ? (A) 30

12.

(B) 20

(C) 0

(D) 25

(E) 30

If one root of the equation x 2  10 x  a  2  0 is 5  10 , what is the value of the constant a ? (A) (B) (C) (D) (E)

7 13 15 18 23

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

13.

Which of the following equations has no solution? (A) x3  8

x2  4

(B)

(C) x 2  2 x  1  0 (D) e x  0.5 (E) x  5  2 14.

If f ( x)  8sin x  2 , which of the following includes all values of x in the interval 0  x   , where f ( x)  6 ? (A) (B)

 3



6 5 (C) 6

 3

5 6 6  2 and (E) 3 3

(D)

15.



and

and

2 If the coordinates of the vertex of f ( x)  x  4 x  k are

(2, 5) , what is the value of k ? (A) 5

16.

(B) 7

(C) 9

(D) 11

(E) 14

What is the range of the quadratic function

f ( x)   x2  7 x  12 ? (A)  y y  4.25 (B)  y y  3.50

(C)  y y  2.36 (D)  y y  1.25 (E)  y y  0.25

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336

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

17.

2 If the minimum value of y  x  2kx  k is 12 , then what

is all value of k ? (A) (B) (C) (D) (E)

3 3 3,  4 3, 4 4, 5 A

18.

In Figure 1, a circular cone is inscribed in the sphere at center O with radius 10. If AH  16 , what is the volume of the cone? (A) (B) (C) (D) (E)

4188.8 3217.0 2495.2 1072.3 1010.6

O

B

H

C

Note: Figure not drawn to scale. Figure 1

19.

3 If f ( x)  2 x  1 , then the inverse of f ( x) is

(A) 2 x 2  1 (B)

3

x 1

3

x 1 2

(C) (D)

3

x 1 2

(E)

3

x 1 2

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337

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

20.

If the period of the function y  2sec(2kx  0.3)  5 is 2, then what is the value of k ? (A) (B) (C) (D) (E)

21.

If a sphere has a volume of 64 , then what is its surface area? (A) (B) (C) (D) (E)

22.

6.28 4.77 3.14 1.57 0.79

166 175 184 225 289

Which of the following functions has an inverse function? 2 (A) f ( x)  x

(B) f ( x)  x  5 (C) f ( x)  x3  x  1 (D) f ( x )  x 3 2 (E) f ( x)   x  10

23.

In Figure 2, the surface area of the cylinder is 24 and the volume of the cylinder is 12 . Which of the following is the value of the radius? (A) (B) (C) (D) (E)



r

0.52 1.33 1.59 2.77 3.78

h

Figure 2

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338

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

24.

If two six-sided dice are tossed, what is the probability that a total of 7 is rolled? (A) (B) (C) (D) (E)

25.

1 18 1 12 3 6 2 6 1 6



If a vector P has the magnitude of 10 and the same



direction as the vector Q  (3,  4) , which of the following



is the vector P ? (A) (6, 8) (B) (10, 1) (C) (3, 4) (D) (6,  8) (E) (9,  12)

26.

If sin t 

(A) (B) (C) (D) (E)

4 , then sin(  t )  5

3 5 3  5 4 5 4  5 3 4

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339

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

27.

If f (t )  sin t , g (t )  cos t , and h(t )  f (t ) g (t ) , which of the following is not true? (A) f (t ) is an odd function. (B) g(t) is an even function (C) h(t ) is an odd function (D) h(t ) is an even function

f (t ) is an odd function g (t )

(E)

28.

y  f ( x)

Which of the following is the equation of the graph in Figure 3 ? 1 (A) y  2sin x  2 2 (B) y  2cos2x  2

4 2 4

1 (C) y  2cos x  2 2 1 (D) y  2cos x  2 2 1 (E) y  2cos x  2 4

29.

4

O

2 4 Figure 3

In the xy-coordinate plane, which of the following is the set of points whose distance from the origin is two times the distance from the point (6, 0) ? (A) (B) (C) (D) (E)

a line a parabola a hyperbola an ellipse a circle

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340

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

30.

If f ( x)  2 x  1 and g ( x)  x 2  2 x  1 , then ( f  g )(4)  (A)

19

(B) 10 (C) 3 (D) 9 (E) 27

31.

If g ( x) is the inverse function of f ( x) , which of the following is not true? I. The graphs of g ( x) and f ( x) are reflections of each other in the line y  x . II. f  g ( x)   x III. g  f ( x)    x (A) (B) (C) (D) (E)

32.

I only II only III only I and II only I and III only

If a soft-drink manufacturer has daily production costs of

C (n)  80, 000  120n  0.05n 2 , where C is the total cost in dollars and n is the number of units produced, what is the minimum cost each day? (A) (B) (C) (D) (E)

8,000 7,500 7,000 6,700 6,000

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

33.

In Figure 4, if a line is drawn around the box from vertex P to Q passing through AB , what is the length of the shortest distance of the line? (A) (B) (C) (D) (E)

34.

6 B A 7

Q

Note: Figure not drawn to scale. Figure 4

If a polynomial function y   x7  100 x2  5x  4 , what is the right-end and left-end behavior of the graph of the function? (A) (B) (C) (D) (E)

35.

12.6 13.9 14.6 15.1 15.8

5 P

The graph falls to the left and rises to the right. The graph falls to the left and falls to the right. The graph rises to the left and rises to the right. The graph rises to the left and falls to the right. The graph approaches to the x  axis.

( x  2)( x  5) 2  0 , which of the following is the ( x  3) solution set of the inequality?

If

(A)  x x  3 (B)  x  3  x  2 (C)  x  3  x  2 (D)  x 2  x  5 (E)  x  3  x  2   x x  5

36.

a If log 3 (ab)  10 , log3    2 , and b  0, what is the b value of a ? (A) (B) (C) (D) (E)

9 81 243 729 2187 GO ON TO THE NEXT PAGE

342

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

37.

If the ellipse x 2  2 y 2  2 x  4 y  k has a major axis of 10, where k is a constant, then what is the value of k ? (A) (B) (C) (D) (E)

5 10 15 22 25

2

38.

If x 3  4 , which of the following is the complete solution set ? (A)  x x  8 (B)  x x  8 (C)  x x  4 (D)  x x  8 or  8 (E)  x x  4 or  4

39.

What is the amplitude of y  3sin   4cos  ? (A) (B) (C) (D) (E)

3.5 4 5 5.8 7

n

40.

lim  5(0.2)k 

n 

(A) (B) (C) (D) (E)

k 1

1 1.25 5 6.26 Infinite

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

41.

What is the coefficient of the term x6 y 5 in the expansion of

 2 x  5 y 11 ? (A) (B) (C) (D) (E)

42.

Matrices A, B, C, and D are of orders 2  3 , 2  3, 3  2, and 2  2 , respectively. Which of the following matrices are of proper order to perform the operation? (A) (B) (C) (D) (E)

43.

A  3C B  2C AB BC  3D CB  2D

If the sum of the first n terms of a series is S n  n 2  4n , then what is the 10th term? (A) (B) (C) (D) (E)

44.

3,125 29,568 324,567 90,400,000 92, 400, 000

23 40 85 125 140

In how many ways can 5 different prizes be given to any 5 of 12 people, if no person receives more than one prize? (A) (B) (C) (D) (E)

124020 95040 7650 792 60

GO ON TO THE NEXT PAGE

344

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

45.

If n C2  (A) (B) (C) (D) (E)

46.

7 6 5 4 3

0.084 0.096 0.125 0.358 0.612

If three marbles are chosen at random from a bag containing 4 red marbles and 5 white marbles, what is the probability that exactly two marbles are red? (A) (B) (C) (D) (E)

48.

and n  3, then n 

If three people are randomly chosen, what is the probability that all were born on different days of the week? (A) (B) (C) (D) (E)

47.

n 1 P2

0.36 0.42 0.52 0.63 0.81

1 1 x  x 0 x

lim

(A) 1.5 (B) 0.5 (C) 1.5 (D) 2.0 (E) Undefined

GO ON TO THE NEXT PAGE Dr. John Chung's SAT II Math Level 2 Test 10

345

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

49.

The frequency table in Figure 5 shows the test score for students sampled in a statistics class. What is the standard deviation of the data?

(A) 5.9 (B) 6.1

Test score

70 78 82 92

Frequency

3

(C) 6.6

8

5

2

Figure 5

(D) 7.2 (E) 7.5

50.

What is the surface area of the cast iron solid in Figure 6?

(A) 100 (B) 120 (C) 132

15

(D) 165

12

(E) 190

8 Figure 6

STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK.

GO ON TO THE NEXT PAGE

346

No Test Material on This Page

Dr. John Chung's SAT II Math Level 2 Test 10

347

2

2

TEST 10 # 1 2 3 4 5 6 7 8 9 10

ANSWERS

answer A D A E E C E C E D

# 11 12 13 14 15 16 17 18 19 20

answer D B B D C E D D D D

# 21 22 23 24 25 26 27 28 29 30

answer A D D E D C D D E E

# 31 32 33 34 35 36 37 38 39 40

answer C A B D E D D D C B

# 41 42 43 44 45 46 47 48 49 50

answer E D A B D E A B B C

Explanations: Test 10  2a 2 

1. (A)

Since

x2  x ,

2. (D)

Since

28k  2  2  7  k , k should be 7. (Prime factorization)

3. (A)

a3  64  a  4 and b  a 2  (4)2  16  4 .

4. (E)

9a 2  4a 2  9a 2  2a  3a  2a  (3a)  a .

40  a  0,1, 2,3, 4,5, 6  40  a  0,1, 4,9,16, 25,36 

5. (E)

If all data are equal, the standard deviation is 0.

6. (C)

Since

a  40,39,36,31, 24,15, 4

x 2  x , 5  x  10 . Therefore,

x  6,7,8,9  x  6,  7,  8,  9 (8 integers) .

7. (E)

a  a  2  3  3  2  8

  5

2

 5 (B)

25  5 (C)

 112

 11 (D)

 6 2

6

8. (C)

(A)

9. (E)

( x  3)( x  2)  ( x  1)( x  6)  x 2  5 x  6  x 2  5 x  6  2 x 2



2

 

(E)



2

Since x  5,  2 x  10. 10. (D)

Factor Theorem:

3 Let x 2  kx  1  ( x  2)Q( x) . When x  2, 4  2k  1  0  k   . 2

348

  16

1 2

2

2 11. (D)

2 x 2  10 x  a   x  b 

2

 x 2  10 x  a  x 2  2bx  b 2

Therefore, b  5 and a  b 2 . a  25. 12. (B)



Substitution: 5  10





2



 10 5  10  a  2  0  35  10 10  50  10 10  a  2  0

  15  a  2  0  a  13 Or,







Product of the roots: 5  10 5  10  a  2  15  a  2  a  13 13. (B)

(A) x3  8  x  2

(B)

x 2  4 

x  4

(C) x 2  2 x  1  0  Discriminant  8  0 :two real roots (D) e x  0.5  x  ln 0.5

(E) x  5  2  x  7 or 3

1  5  x  or 2 6 6

14. (D)

8sin x  2  6  sin x 

15. (C)

f (2)  4  8  k  5  k  9

16. (E)

Axis of symmetry: x  The range: y  0.25

7  3.5 , 2

f (3.5)  (3.5)2  7(3.5)  12  0.25

Graphing utility:

Maximum x  3.4999

y  0.25

2k k f (k )  k 2  2k 2  k  12  k 2  k  12  0 2 (k  4)(k  3)  0  k  4 or  3 A

17. (D)

Axis of symmetry: x 

18. (D)

The radius of the circular base is 8 and the height of the cone is 16.  r 2 h 64  16 V   1072.330292  1072.3 3 3 B

19. (D)

f ( x )  2 x3  1 

Dr. John Chung's SAT II Math Level 2 Test 10

f 1 ( x) : x  2 y 3  1  y  3

O  10 6 H 8

C

x 1 2

349

2 20. (D)

2 Since the frequency is 2k , then the period is

 k

2  k 

 2

2   . 2k k

 1.570796327  1.57 1

21. (A)

4 V   r 3  64  r  48 3 3 2

2  1 The surface area: 4 r  4  48 3   4  48 3  165.9729662  166     2

22. (D)

Since f ( x)  x3 is a one- to- one function, its inverse is also a function. (C) y  x3  x  1 is not one-to-one. Use a graphing utility to show the graph.

23. (D)

Since 2 r 2  2 rh  24 and  r 2 h  12 , then r 2  rh  12 and r 2 h  12. h

12 12  12   r 2  r  2   12  r 2   12  r 3  12r  12  0 2 r r r   x  1.1157494   x  2.7687343

 

Zero x  2.7687343

y0 y0

y0

Therefore, r  1.12 or 2.77 . 24. (E)

The total of 7: P

 1, 6  2,5  3, 4  4,3 5, 2  6,1

6 36

25. (D)

Choice (D): Magnitude 62   8   10  6, 8  2  3, 4  : Same direction      Because P  10 and Q  32  (4) 2  5 , P  2Q .

26. (C)

sin(  t )  sin  cos t  cos  sin t  sin t

27. (D)

1 h(t )  sin t cos t  sin 2t : odd function 2

28. (D)

period  4 , middle line  2 , frequency 

2

1  Therefore, y  2cos  x   2 2 

350

 sin t 

2 1  4 2

4 5

2

2

29. (E)

y 

2

(0, 0)

:

 ( x, y )

x2  y 2 2

(6  x)  (0  y )

2



1

 (6, 0)

x

2  x 2  y 2  4(36  12 x  x 2  y 2 )  3 x 2  3 y 2  48 x  144  0 1

x 2  y 2  16 x  48  0  ( x  8)2  y 2  42 : a circle 30. (E)

Multiplication :  f  g  ( x) 







2 x  1  x2  2x  1



 f  g  (4)  3  9  27

cf. ( f  g )( x) ; composition 31. (C) 32. (A)





f f 1 ( x)  x , y  f ( x) and y  f 1 ( x) is symmetric with respect to y  x. Axis of symmetry: x 

120  1200 2  (0.05)

f (1, 200)  80,000  120(1, 200)  0.05(1, 200) 2  8, 000 33. (B)

Since PQ is the shortest distance, PQ  52  132  13.92838828  13.9 6

7

Q

5 P

34. (D)

lim f ( x)   : rises to the left

x 

lim f ( x)   : falls to the right

x 

Dr. John Chung's SAT II Math Level 2 Test 10

351

2 35. (E)

2 By multiplying ( x  3)2 to both sides: ( x  2)( x  5)2  0  ( x  3)( x  2)( x  5) 2  0 , where x  3. ( x  3)  3





 2

 5



x5

3  x  2

The solution set: 3  x  2 or x  5 . At x  5 , the inequality is also true. Or, you can use test value method. 36. (D)

Since ab  310 and

a a a  32  9 , then b  and a    310. 9 b 9

a 2  312  a  36  729 37. (D)

x 2  2 y 2  2 x  4 y  k  ( x  1) 2  2( y  1) 2  k  3 

( x  1)2 ( y  1) 2  1 k 3  k  3 2

a2  k  3  a  k  3 Major axis : 2a  2 k  3  10  k  3  25  k  22

38. (D)

Algebraic solution:

2 x3

2

3

 1  2  2   x 3   4   x 3   43  64  x 2  64  x  8        

Graphing calculator: 

Intersection x  8 x  8

39. (C)



2   y1  x 3   y2  4

y  4 y  4

If y  a sin  and y  b cos  , then the amplitude of y  a sin   b cos  is

a 2  b2 .

The amplitude is 32  42  5 . 4 4 3 3  Because y  5  sin   cos    5sin(  1 ) , where sin 1  and cos 1  . 5 5 5 5  

352

2 40. (B)

2 Infinite series: n n  0.2  lim  5(0.2)k  5 lim  0.2k  5    1.25 n  n   1  0.2  k 1 k 1

41. (E)

r 42. (D)

 2 11 r  5r  x 11r  y r 6 5 must be 5. Therefore, 11C5  2   5  x 6 y 5  92, 400,000

11 Cr

 2 x 11 r  5 y r



11 Cr

Dimension operation: (A) A  3C   2  3   3  2  : wrong

 2  3   3  2  : wrong AB   2  3 2  2  :wrong BC  3D   2  3   3  2    2  2    2  2    2  2    2  2  : true

(B) B  2C  (C) (D)

 3  2   2  3   2  2   3  3   2  2 : wrong

(E) CB  2 D 



 



43. (A)

a10  S10  S9  102  4(10)  92  4(9)  140  117  23

44. (B)

Choose 5 people and assign them to the 5 prizes:

45. (D)

n C2



n 1 P2



n (n  1) 2!

Therefore, n  4. 46. (E)

P

  n  1  n  2  

12 C5

 5!  95040

n  (n  2) 2

 n 1

7  6  5 210   0.612244898  0.612 343 73

people #1 M T W TH F SA S

people #2 M T W TH F SA S

people #3 M T W TH F SA S

All possible outcomes: 7  7  7  343 47. (A)

Probability: P 

 5 C1 30   0.3571428571  0.36 84 9 C3

4 C2

Dr. John Chung's SAT II Math Level 2 Test 10

353

2

2 1 



1 x 1 1 x

48. (B)

lim

49. (B)

Calculator: statistics S x  6.097251068  6.1

50. (C)

END

354

x 0



x 1 1 x



  lim

x 0



x

x 1 1 x

Lateral area of the cone   rs    4  5  20 Lateral area of the cylinder  2 rh  96 The area of the base  16 Therefore, 20  96  16  132



 lim

x 0

1 

1 1 x





1 2

5

3 4 12

8

15

Test 11                                               Dr. John Chung’s                SAT II Mathematics Level 2

Dr. John Chung's SAT II Math Level 2 Test 11

355

2

 

2

    

MATHEMATICS LEVEL 2 TEST REFERENCE INFORMATION    

 

THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME   OF THE QUESTIONS IN THIS TEST   1   Volume of a right circular cone with radius r and height h : V   r 2 h 3   1 Lateral Area of a right circular cone with circumference of the base c and slant height  : S  c 2  

  Volume of a sphere with radius r : V  4  r 3 3   Surface Area of a sphere with radius r : S  4 r 2   1   Volume of a pyramid with base area B and height h : V  3 Bh

       

356

2

 

2 Dr. John Chung’s SAT II Math Level 2

Answer Sheet 01 02

A

B

C

D

E

A

B

C

D

E

 

03

A

B

C

D

E

B

C

D

 

04 05

A A

B

C

 

06

A

B

 

07

A

08

26 27

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

E

28 29

A

B

C

D

E

D

E

30

A

B

C

D

E

C

D

E

A

B

C

D

E

B

C

D

E

31 32

A

B

C

D

E

A

B

C

D

E

33

A

B

C

D

E

09

A

B

C

D

E

34

A

B

C

D

E

10 11

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

35 36

A

B

C

D

E

12

A

B

C

D

E

37

A

B

C

D

E

13 14

A

B

C

D

E

B

C

D

E

A

B

C

D

E

38 39

A

 

A

B

C

D

E

 

15

A

B

C

D

E

40

A

B

C

D

E

 

16 17

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

41 42

A

B

C

D

E

 

18

A

B

C

D

E

B

C

D

E

 

19 20

B

C

D

E

43 44

A

A

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

A

B

C

D

E

 

21 22

45 46

A

B

C

D

E

47

A

B

C

D

E

 

23

A

B

C

D

E

48

A

B

C

D

E

24

A

B

C

D

E

49

A

B

C

D

E

25

A

B

C

D

E

50

A

B

C

D

E

              

 

       

 

       

     

           The number of right answers:                              

                

           The number of wrong answers:                             

 

          

 

# of correct

Dr. John Chung's SAT II Math Level 2 Test 11

1    4 # of wrong

Raw score

 

357

 

2

2 Score Conversion Table 6

Scaled Score 480

630

5

470

26

620

4

470

800

25

620

3

460

46

800

24

610

2

460

45

800

23

610

1

450

44

800

22

600

0

450

43

800

21

600

42

800

20

590

41

800

19

590

40

780

18

580

39

760

17

570

38

750

16

560

37

740

15

550

36

720

14

540

35

710

13

530

34

700

12

520

33

690

11

510

32

680

10

500

31

670

9

490

30

660

8

490

29

650

7

480

28

Scaled Score 640

800

27

48

800

47

50

Scaled Score 800

49

Raw Score

358

Raw Score

Raw Score

2

2

 

  

MATHEMATICS LEVEL 2 TEST For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding circle on the answer sheet Note: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions in this test. For each question you will have to decide whether or not you should use a calculator. (2) For some questions in this test you may have to decide whether your calculator should be in the radian mode or the degree mode. (3) Figures that accompany problems in this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers x for which f ( x) is a real number. The range of f is assumed to be the set of all real numbers f ( x) , where x

is in the domain of f . (5) Reference information that may be useful in answering the questions in this test can be found on the page preceding Question 1. USE THIS SPACE FOR SCRATCHWORK

1.

 3x 3   2   5y  (A)

2.

2



3x 4 9 x6 25 x 6 9x4 25 y 4 (D) (B) (C) (E) 5 y5 25 y 4 9 y4 25 y 6 9 x6

What is the value of k for which x 2  2kx  k 2 is a perfect square? (A) (B) (C) (D) (E)

1 4 9 16 Any real number

GO ON TO THE NEXT PAGE

  Dr. John Chung's SAT II Math Level 2 Test 11

359

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

3.

3 If x 2  x  3 is a factor of 2x  px  q , what are the values

of p and q ? (A) p  4 and q  6 (B) p  8 and q  4 (C) p  8 and q  6 (D) p  8 and q  6 (E) p  8 and q  6

4.

What is the sum of the roots of the equation x3  3x  52  0 ?

(A) (B) (C) (D) (E)

5.

3 0 2 3 52

If a polynomial P ( x) has a remainder of three when divided by ( x  1) and remainder of one when divided by ( x  1) , then the remainder when divided by ( x2  1) is (A) (B) (C) (D) (E)

6.

4 x2 x4 x2 x4

If x  3, then 1  x  2  (A) (B) (C) (D) (E)

x3 x3 x  3 x  3 x  1

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360

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

7.

Which of the following values of x is true for the inequality sin x  cos x ?



(A)

8.

4

 4

(C)

3 4

(B)

 2

(C)

4



7 7 25 25 6

If

(D)

5 4

(E)

3 2

(D)

2



(E)

If

3 4

10

10

10

Figure 1

a 2 4 , then a  a2 a2 2 2 4 6 No solution

(A) (B) (C) (D) (E)

11.

2

What is the product of 3  4i and its conjugate? (A) (B) (C) (D) (E)

10.



In Figure 1, a right cylinder is inscribed in a cube. If the cube has an edge of 10, what is the ratio of the volume of the cylinder to the volume of the cube? (A)

9.

(B)

3

27  x3  3 , then x 

(A)  3 (B) 3 3 2 (C)  33 2 (D) 9 (E)  9

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

12.

What is the domain of f ( x)  (A) (B) (C) (D) (E)

13.

x ? 10  x

x0 0  x  10 0  x  10 0  x  10 0  x  10

Which of the following is the slant asymptote(s) of 2 x 2  3x  1 f ( x)  ? x 1 (A) x  1 (B) y  2 (C) y  2 x (D) y  2 x  5 (E) y  2 x  10

14.

2 If f    3x  5 , then f ( x)   x (A) (B) (C) (D) (E)

15.

3x 5  2 2 3x 5  2 2 3 1  x 5 6 5 x 6 1  x 5

If f ( x)  x  1 , what value does the inverse function

f 1 ( x) have at the point x  9 ? (A) 8

(B) 10

(C) 100

(D) 120

(E) 125

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

16.

The symbol  x  is the greatest integer which is less than or equal to x. If

 x 2  6  x   9 , which of the following

could be the value of x ? (A) (B) (C) (D) (E) 17.

3.3 1.3 2.5 3.9 4.01

How many real numbers are equal to their multiplicative inverses? (A) 4

18.

(D) 1

(E) 0

24 120 360 2880 40320

Find all values of k which satisfy the determinant k 3  27 ? 2k k (A) (B) (C) (D) (E)

20.

(C) 2

How many ways can 8 books be arranged on a shelf if 5 of them are math books and must be kept together? (A) (B) (C) (D) (E)

19.

(B) 3

9 3 6 6 or 3 9 or 3

If y varies directly as x and inversely as the square of z , and y  15 when x  3 and z  5 , what is the value of y when x  5 and z  2.5 ? (A) 1

(B) 25

(C) 50

(D) 100

(E) 125

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

21.

What is the number of points of intersection of the graphs of

x 2  y 2  1 and x 2  4 y 2  4 ? (A) 0

22.

(B) 1

(C) 2

(D) 3

(E) 4

If the equation of the graph in Figure 2 is y   x  10  k ,

y

A

where k is a constant, what is the area of  ABO ? (A) (B) (C) (D) (E) 23.

x

Figure 2

100 192.5 225 475.8 1000

If cos x  x , then how many real solutions are there? (A) (B) (C) (D) (E)

25.

B O

If the length of a diagonal of a cube is 10, then what is the volume of the cube? (A) (B) (C) (D) (E)

24.

100 75 50 25 12.5

0 1 2 3 4

Which of the following is the graph of the polar equation 1 r ? cos   sin  (A) (B) (C) (D) (E)

A line A circle An ellipse A parabola An hyperbola

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

26.

What is the distance between the line y  2 x  5 and the point (1, 4)? (A) (B) (C) (D) (E)

27.

3.25 1.75 1.34 1.19 0.75

If the area of the triangle bounded by the lines y  2 x , x  k , and y  6 is 16, what is the positive value of k ?

(A) (B) (C) (D) (E) 28.

What is the sum of all factors of 210? (A) (B) (C) (D) (E)

29.

556 576 584 616 625

If log 2 ( x 2  3x  2)  log 2 ( x  1)  5 , what is the value of x? (A) (B) (C) (D) (E)

30.

4 5 6 7 8

10 30 34 40 48

If 103 x  27 , then 102 x  (A)

1 3

(B)

1 6

(C)

1 9

(D)

1 18

(E)

1 27

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

31.

If log 2  0.3010 , what is the number of digits in 2100 ? (A) (B) (C) (D) (E)

32.

50 46 34 31 30

If 2i is a root of x3  2 x 2  4 x  8  0 , which of the following are the other roots? (A) 2i (B) 2i and 3 (C) 2i and 2 (D) 2i and  1 (E) 2i and  2

33.

2 3 4 n   1 lim  2  2  2  2      2   n n n n n 

n  

(A) 0

34.

1 2

(C) 1

(D)

3 2

(E) 2

Three numbers have a sum of 30 and a product of 640. If these three numbers form an arithmetic sequence, what is the smallest number? (A) (B) (C) (D) (E)

35.

(B)

1 2 4 10 16

If f ( x)  22 x and g ( x)  log 4 x , then  f  g  ( x)  (A) 0.5 x (B) x (C) 1.5x (D) 2 x (E) 2.5x GO ON TO THE NEXT PAGE

366

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

36.

Which of the following is not a function? 1 (A) y    2

x

(B) y  x5 (C) xy  9 (D) y  10  x 2 (E) y 2  4 x

37.

In Figure 3, what is the value of the angle  between the lines y  x  1 and y  3x  5 ? (A) (B) (C) (D) (E)

63.4 65.8 69.2 72.5 75.6

y

y  x 1

y  3x  5  O

x

Note: Figure not drawn to scale. 38.

sin 2  1  cos 2

(A) (B) (C) (D) (E)

39.

Figure 3

sin  cos  tan  cot  sec 

If cos   0.61, then cos(   )  (A) (B) (C) (D) (E)

0.61 0.39 0.39 0.61 0.93

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

40.

x2 y2   1 , then which of the following is the foci of the 9 16 hyperbola?

If

(A) (B) (C) (D) (E)

41.

 0,  3  0,  4    3, 0    4, 0    5, 0 

Which of the following is the graph of the curve represented by the parametric equations x  t  2 and y  t  5 ? (A)

(B)

y

x

O

(C)

x

(E)

x

O

(D)

y

O

y

y

O

x

y

O

x

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

42.

In Figure 4, if AB  3 , BC  5 , and CD  7 , what is the degree measure of angle ADF ? (A) (B) (C) (D) (E)

50.2 48.4 46.8 45.6 43.1

A

F

B

E C

D

Figure 4 43.

If the rectangular equation is x 2  y 2  4 x  0 , which of the following is an equivalent equation in polar form? (A) (B) (C) (D) (E)

44.

 2 0  1 2  If the matrices A   and B    , what is the   3 1  1 4 determinant of A+B ? (A) (B) (C) (D) (E)

45.

r2  r  0 r  2cos   0 r  4cos   0 r  2sin   0 r  4cos   0

12 12 23 23 28

If the equation of a sphere is given by x 2  y 2  z 2  4 x  6 z  0 , what is the surface area of the sphere? (A) (B) (C) (D) (E)

13 52 84 128 169

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

46.

If the plane passes through the point  3, 2, 5  and is parallel to the xy -plane , then what is the equation of the plane? (A) x  3 (B) y  5 (C) x  5 (D) z  5 (E) y  z

47.

x 2  px  q  5 , where p and q are constants, what x 1 x 1 is the value of p ? If lim

(A) (B) (C) (D) (E)

48.

What is the minimum value of the function defined as x 2  x , f ( x)   ? x 2  2, (A) (B) (C) (D) (E)

49.

2 1 1 3 4

4 2 0 1 2

How many positive integers are there in the solution set of ( x  5)( x  2) 0? x 1 (A) (B) (C) (D) (E)

2 3 4 5 6

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2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

50.

What is the constant term in the binomial expansion of 12

 2 1 x   ? x  (A) (B) (C) (D) (E)

924 792 495 220 66

STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK.

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No Test Material on This Page

372

2

2

TEST 11 # 1 2 3 4 5 6 7 8 9 10

ANSWERS

answer D E A B B D E A D E

# 11 12 13 14 15 16 17 18 19 20

answer B E D D C D C D E D

# 21 22 23 24 25 26 27 28 29 30

answer C A B B A C D B C C

# 31 32 33 34 35 36 37 38 39 40

answer D C B C B E A D A E

# 41 42 43 44 45 46 47 48 49 50

answer D A C D B D D C C C

Explanations: Test 11 2

1. (D)

 3x 3   5 y 2   

2. (E)

 2k  Since    k 2 , then k 2  k 2 . It is always true for all real number k .  2 

3. (A)

Long division:



32 x 6 25 x 6  52 y 4 9 y 4 2

2x  2 2

x  x  3 2 x3 

px 

q

2 x3  2 x 2  6 x  2 x 2  ( p  6) x  q

remainder

 2 x2  2 x 6 ( p  4) x  q  6 Therefore, p  4 and q  6 . 4. (B)

Sum of the roots of polynomial equation an x n  an 1 x n 1  an  2 x n  2      a1 x  a0 a 0 is  n 1 . Therefore, S    0. an 1

5. (B)

P ( x) can be expressed in three different ways as follows.

( x  1)Q1 ( x)  3  P( x)  ( x  1)Q2 ( x)  1 ( x  1)( x  1)Q( x)  ax  b 

Dr. John Chung's SAT II Math Level 2 Test 11

(1) (2) (3)

373

2

2 When x  1 3 P(1)   a  b

(1) (3)

When x  1 1 P(1)   a  b

(2) (3)



ab 3



 a  b 1

Therefore, a  1 and b  2  remainder  x  2 6. (D)

Since x  3 , 1  x  2  1  ( x  2)  x  3  ( x  3)   x  3 .

7. (E)

Graphing calculator: Check each choice:  If  If    sin x  cos x  If   If  If 

8. (A)

x x

 4



, then sin , then sin





 cos . 4 4



 cos



2 2 2 3 3 3  cos , then sin x 4 4 4 5 5 5  cos x , then sin 4 4 4 3 3 3 x , then sin  cos (True) 2 2 2

Volume of the cylinder:  r 2 h  250

Volume of the cube: 103  1000 250  Therefore, the ratio:  . 1000 4 9. (D)

(3  4i )(3  4i )   3   4i   25

10. (E)

Multiply by the common denominator (a  2) : a  4(a  2)  2  a  2. (Extraneous root)

11. (B) 12. (E)

2

3

2

27  x3  3  27  x3  27  x3  54  x  3 3 2

x x is  0 , where x  10 . 10  x 10  x (10  x)  x  0  x( x  10)  0  0  x  10

Domain of f ( x) 

Or, find the critical points; From

374

x  0, critical points are x  0 and x  10. 10  x

2

2 Therefore, for 0  x  10,

13. (D)

Since f ( x) 

14. (D)

Method 1:

x is positive. 10  x

2 x 2  3x  1 6 6    2x  5  , y  lim  2 x  5    2x  5 . x  x 1 x 1 x 1 

   2  2 2 If x is replaced with , f   is changed to f    f ( x) x  x  2   x  6 2 Therefore, f ( x)  3    5   5 x x   Method 2: 2 2 Let f    3k  5 and x  . k k 2 2 k  . Replace k with . x x 6 2 2 f    3k  5  f ( x)  3    5  f ( x)   5 x k  x 15. (C)

Method 1: The inverse function: f 1 ( x)   x  1  Method 2: From f ( x) : f 1 (9)  y is equivalent to 9  2

16. (D)

 x 2  6  x   9   x 2  6  x   9  0  x  3  3  x  4

17. (C)

k

y  1.



y  100

 x   3

2

0

1  k 2  1  k  1 k

18. (D)

4!

4! 5!  2880 19. (E)

f 1 (9)  100

5!

Determinant: k 3  k 2  6k  k 2  6k  27  k 2  6k  27  0 2k k

Dr. John Chung's SAT II Math Level 2 Test 11

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2

2 (k  9)(k  3)  0  k  9 or k  3

20. (D)

Compound proportion: y  k 15  k

21. (C)

3 25

 k  125

x z2 y  125 

,

5  100 2.52

x2 y 2  1 4 1

x 2  y 2  1 and x 2  4 y 2  4  y

1  2

2 

22. (A)

1

f (0)  0   10  k  0  k  10 y

10

A

B

O

10

23. (B)

x

Area A  x

20  10  100 2

Let the length of a edge be x. The length of the diagonal:

x 2  x 2  x 2  x 3  10

3

 10  Volume V  x3     192.4500897  192.5  3 24. (B)

Graphic utility: One point of intersection. 

25. (A)

376

1 cos   sin   x  y 1 r



r cos   r sin   1  r

x y  r 1 r r

2 26. (C)

2 Distance between 1, 4  and 2 x  y  5  0 . D

2(1)  4  5 22   1



2

3

27. (D)

5

 1.341640786  1.34

y

y  2x

( k , 2k ) (3, 6) 



2k  6 y6

x

O

k 3 xk

The area of  because k  0 .

(k  3)(2k  6)  16. (k  3)2  16  k  3  4,  4 2

28. (B)

Prime factorization: 210  21  31  51  71 Therefore, the sum of all factors  (1  2)(1  3)(1  5)(1  7)  576

29. (C)

log 2 ( x 2  3x  2)  log 2 ( x  1)  5 x  2  25

30. (C)

31. (D)

( x  2)( x  1)  log 2 ( x  2)  5 ( x  1)

 x  34

103 x  27 

32 

 log 2

 k  7, 1



103 x





2 3

 27



2 3

 

 10 x  33



2 3

 32

1 9

Using a calculator: 2100  1.2676506  1030  Algebraically: Let X  2100

The number of digits  30  1  31

 log X  log 2100  100log 2  30.10

X  1030.10  1030  X  1031 Therefore, the number of digits is 31. Ex. 210  1024 (4 digits number) log 210  10log 2  10  0.3010  3.010 . 3  log 210  4 . Therefore, it is 4digits number. 32. (C)

b 2   2. a 1 Since 2i is a root, then 2i is also a root. Let the third root be w . Sum of the roots: 2i  (2i )  w  2  w  2 Therefore, the roots are 2i, 2i, and 2 . Sum of the roots of x3  2 x 2  4 x  8  0 is 

Dr. John Chung's SAT II Math Level 2 Test 11

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2 33. (B)

2 2 3 4 n  1  2  3    n  1 lim  2  2  2  2      2   lim  lim n   n n  n n n n  n  n2

n(1  n) 2 n2

n(1  n) n  n2 1  lim  n  2n 2 n  2n 2 2 lim

34. (C)

Let three number be a  d , a, and a  d , where d  0. (a  d )  a  (a  d )  30  a  10 (a  d )a(a  d )  (10  d )(10)(10  d )  640  100  d 2  64  d 2  36 Since d  6 , the smallest number is 10  6  4. 2

 g  ( x)  22log 4 x  2log4 x  2

log 4 x 2

35. (B)

f

36. (E)

Choice (E): If x  2, then y  2. (Vertical line test fails.)

37. (A)

tan  

 2log2 x  x

 x  0

m2  m1 1  m1m2

If m1  1 and m2  3 , then tan  



Or,

tan 1  1  1  45

2

1

3  1  2    tan 1 (2)  63.43494882  63.4 1  (3)(1) tan  2  3   2  tan 1 3  71.56505118  71.565

  180  (45  71.565)  63.435  63.4

38. (D)

Because cos 2  1  2sin 2  and sin 2  2sin  cos  , sin 2 2sin  cos  2sin  cos  cos      cot  2 1  cos 2 1  (1  2sin  ) sin  2sin 2 

39. (E)

cos(   )  cos  cos   sin  sin    cos   0.61

40. (E)

x2 y 2  1 a 2 b2



x2 y 2  1 9 16

 a  3 and b  4

Since c   32  42  5 , the foci are at ( 5, 0) and (5, 0) . 41. (D)

x t2  x0 x  t  2  t  x 2  2  y  ( x 2  2)  5  y  x 2  7

42. (A)

378

Diagonal AD  32  52  7 2  83 and DE  32  52  34 .  34  34   ADF  cos 1  cos ADF    50.2 83  83 

2 43. (C)

44. (D)

45. (B)

2 Since x  r cos  and y  r sin  , x 2  y 2  4 x  0  r 2  4(r cos  )  0 . Therefore, r  4cos   0 .  2 0   1 2   3 2   1 4   3 1    4 5       3 2 Determinant :  (3)(5)  (2)(4)  23 4 5 x 2  y 2  z 2  4 x  6 z  0  ( x  2)2  y 2  ( z  3)2  13 Since r  13 , the surface area is 4 r 2  4 (13)  52 .

46. (D)

The plane is parallel to xy-plane: z  5 z 5 O x y

47. (D)

x 2  px  q  5 , x 2  px  q must have a factor of ( x  1) to have a finite limit x 1 x 1

Since lim

value. f (1)  1  p  q  0  q  1  p x 2  px  q  x 2  px  1  p  x 2  px  1  p  x 2  1  p( x  1)  ( x  1)( x  1)  p ( x  1)  ( x  1)( x  1  p) x 2  px  q ( x  1)( x  1  p ) x 1 p  lim  lim  2 p. x 1 x 1 x 1 x 1 ( x  1) 1 2 p  5  p 3.

Therefore, lim

48. (C)

The graph of f is as follows.  x , f ( x)    2,

x 2 x 2 y

2

O

2

x

The minimum of y  0 .

Dr. John Chung's SAT II Math Level 2 Test 11

379

2 49. (C)

2 ( x  5)( x  2)  0  ( x  1)( x  5)( x  2)  0 and x  1 x 1 Method 1: Graphic solution  2

1

 5

The positive integers are 2, 3, 4, and 5.

( x  5)( x  2) (3)(4) 0   0 (ok) x 1 1 Therefore, there are 4 positive integers, 2, 3, 4, and 5 in 1, 5 . Method 2: Test sign: Test at x  2 in 1, 5

50. (C)

General term:

12 Cr

x 

2 12  r

r

1     x

12 Cr x

For r  8 , it is constant. Therefore,

END

380



24  2 r  r

12 C8 x

x

0

 12 Cr x 24 3r

 495 .

Test 12                                               Dr. John Chung’s                SAT II Mathematics Level 2

Dr. John Chung's SAT II Math Level 2 Test 12

381

2

 

2

    

MATHEMATICS LEVEL 2 TEST REFERENCE INFORMATION    

 

THE FOLLOWING INFORMATION IS FOR YOUR REFERENCE IN ANSWERING SOME   OF THE QUESTIONS IN THIS TEST   1   Volume of a right circular cone with radius r and height h : V   r 2 h 3   1 Lateral Area of a right circular cone with circumference of the base c and slant height  : S  c 2  

  Volume of a sphere with radius r : V  4  r 3 3   Surface Area of a sphere with radius r : S  4 r 2   1   Volume of a pyramid with base area B and height h : V  3 Bh

       

382

2

 

2 Dr. John Chung’s SAT II Math Level 2

Answer Sheet 01 02

A

B

C

D

E

A

B

C

D

E

 

03

A

B

C

D

E

B

C

D

 

04 05

A A

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06

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07

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08

26 27

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E

28 29

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30

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B

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31 32

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33

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09

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C

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34

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C

D

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10 11

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C

D

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E

A

B

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35 36

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12

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37

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13 14

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B

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B

C

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38 39

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A

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15

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D

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40

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16 17

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D

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41 42

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18

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B

C

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19 20

B

C

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43 44

A

A

A

B

C

D

E

A

B

C

D

E

A

B

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D

E

A

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C

D

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A

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21 22

45 46

A

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47

A

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E

 

23

A

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C

D

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48

A

B

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24

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49

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25

A

B

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D

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50

A

B

C

D

E

              

 

       

 

       

     

           The number of right answers:                              

                

           The number of wrong answers:                             

 

          

 

# of correct

Dr. John Chung's SAT II Math Level 2 Test 12

1    4 # of wrong

Raw score

 

383

 

2

2 Score Conversion Table 6

Scaled Score 480

630

5

470

26

620

4

470

800

25

620

3

460

46

800

24

610

2

460

45

800

23

610

1

450

44

800

22

600

0

450

43

800

21

600

42

800

20

590

41

800

19

590

40

780

18

580

39

760

17

570

38

750

16

560

37

740

15

550

36

720

14

540

35

710

13

530

34

700

12

520

33

690

11

510

32

680

10

500

31

670

9

490

30

660

8

490

29

650

7

480

28

Scaled Score 640

800

27

48

800

47

50

Scaled Score 800

49

Raw Score

384

Raw Score

Raw Score

383

2

2

 

  

MATHEMATICS LEVEL 2 TEST For each of the following problems, decide which is the BEST of the choices given. If the exact numerical value is not one of the choices, select the choice that best approximates this value. Then fill in the corresponding circle on the answer sheet Note: (1) A scientific or graphing calculator will be necessary for answering some (but not all) of the questions in this test. For each question you will have to decide whether or not you should use a calculator. (2) For some questions in this test you may have to decide whether your calculator should be in the radian mode or the degree mode. (3) Figures that accompany problems in this test are intended to provide information useful in solving the problems. They are drawn as accurately as possible EXCEPT when it is stated in a specific problem that its figure is not drawn to scale. All figures lie in a plane unless otherwise indicated. (4) Unless otherwise specified, the domain of any function f is assumed to be the set of all real numbers x for which f ( x) is a real number. The range of f is assumed to be the set of all real numbers f ( x) , where x

is in the domain of f . (5) Reference information that may be useful in answering the questions in this test can be found on the page preceding Question 1. USE THIS SPACE FOR SCRATCHWORK 1.

 

If 3 x a (A) 4

2.

 81x12 , then a  b 

(B) 5

(C) 6

If 3a  b  2 a  b , then (A)  3

3.

b

(B) 2

If 2 x1  8 , then (A) (B) (C) (D) (E)

(D) 7

(E) 8

ab  ab

(C) 1

(D) 2

(E) 3

23 x  22 x  2

20 40 48 64 80

GO ON TO THE NEXT PAGE

  Dr. John Chung's SAT II Math Level 2 Test 12

385

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

4.

What is the smallest integer value of x to satisfy the x 2x  1 inequality  1 ? 4 2

(A) 1

(B) 0

(C) 1

(D) 2

(E) 3 m

5.

In Figure 1, when the rectangle with dimensions 4 and 10 is rotated about line m , what is the volume of the resulting solid? (A) (B) (C) (D) (E)

6.

10

100 120 160 320 640

2

Figure 1

 log3 16   log 2 9   (A) (B) (C) (D) (E)

7.

4

8 log 6 144 24 log5 25

36

If sin  

1 and  is acute angle, what is the value of 2

sin 2 ? (A) (B) (C) (D) (E)

8.

0.20 0.42 0.50 0.87 0.95

Which of the following is an asymptote of f ( x)  tan 2 ? (A)  

 8

(B)  

 4

(C)  

 2

(D)  

2 5 (E)   3 3

GO ON TO THE NEXT PAGE

386

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

9.

In Figure 2, if triangle ABC has BC  10 , A  30 , and

B

C  70 , then what is the area of the triangle? (A) (B) (C) (D) (E)

10.

92.5 97.3 112.5 125.8 135.1

10

A

30

70

C

Figure 2

Which of the following is true for the graph of f ( x)  sin 2 x  3 ? (A) (B) (C) (D)

symmetric with respect to the y-axis symmetric with respect to the x-axis symmetric with respect to the origin symmetric with respect to the point (3, 0)

(E) symmetric with respect to the point (0, 3)

11.

2 2 The graph of x  y  2 x  4 y  5  0 is which of the following?

(A) (B) (C) (D) (E)

12.

A point A circle An ellipse A hyperbola A parabola

If  is a positive acute angle and sin 2 

1 , then 2

 sin   cos  2  (A) (B) (C) (D) (E)

1 1.5 2.2 2.5 2.8

GO ON TO THE NEXT PAGE Dr. John Chung's SAT II Math Level 2 Test 12

387

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

13.

In which quadrant is the graph of an ellipse represented by 2 2 the equation x  6x  4 y  16 y  17  0 located?

(A) (B) (C) (D) (E) 14.

I II III IV III and IV

A cannonball is launched from a height of 80 feet. If the height of the cannonball in feet, h , is defined by the 2

equation h(t )  18t  72t  80 , where t is time in seconds , how long does the rocket remain at or above 134 feet from the ground , in seconds? (A) 1

(B) 1.2

(C) 1.5

(D) 2

(E) 2.3 5

15.

1   What is the fourth term in the expansion of  x 2  2  ? x   (A) 5x 4 (B) 10x 2 (C) 15x 3 (D) 10x 2 (E) 15x 3

16.

What is the solution of the equation 2log 9 (5 x)  3 ? (A) (B) (C) (D) (E)

17.

3 5 5.4 6 7.2

Find the sum of the first 30 terms of the recursive sequence defined as a1  3 and an  an 1  5 . (A) (B) (C) (D) (E)

148 1680 2265 2340 3120 GO ON TO THE NEXT PAGE

388

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

18.

If two forces of 30 newton and 40 newton acting on a body form an angle of 50  , what is the magnitude of the resultant force? (A) (B) (C) (D) (E)

19.

2

2

If a  b  (a  b)i  10  5i , where a and b are real numbers, what is the value of a ? (A) (B) (C) (D) (E)

20.

50.25 newton 54.78 newton 63.58 newton 76.45 newton 81.68 newton

2 2.5 3 3.5 4

 1 lim 1   n   n

2n



(A) 1 (B) e (C) e 2 (D) e 2 (E) 0

21.

What is the period of the function defined by x f ( x)  5  2cos 2  ?  3 

(A) (B) (C) (D) (E)

1 2 3 6 8

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389

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

22.

Which of the following is not an odd function? (A) y  sin 2 x (B) y  x5  3 x 3 (C) y   x3  2 x 1

(D) y  x 3 (E) y 

23.

1 1 x

What is the value of sin  Arc tan a  , where a  0 ?

(A) (B) (C) (D)

1  a2 a a2  1 a a

1  a2 a

1  a2 a (E) 1  a2

24.

x 1 1 If f ( x)  e , what is f ( x) ?

(A) ln ( x  1) (B) ln ( x  1)

 x (C) ln   2  x (D) ln   e (E) ln  ex 

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390

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

25.

f  If f ( x)  2 x  1 and g ( x)  x  3 , then   (5)  g (A) (B) (C) (D) (E)

26.

5.5 10 17 22 25

What is the domain of the function f ( x)  8  2 x  x 2 ? (A)  ,  4 (B)  , 5 (C)  4, 2 (D)  1, 4 (E) All real

27.

What is the sum of the infinite geometric series 0.5  0.25  0.125     ? (A) (B) (C) (D) (E)

28.

0 1 2 3 4

Six students are to be seated in a row of 6 chairs. If three of these students must be seated together, how many ways could this be accomplished? (A) (B) (C) (D) (E)

24 48 120 144 210

GO ON TO THE NEXT PAGE Dr. John Chung's SAT II Math Level 2 Test 12

391

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

29.

Suppose the graph of f ( x)  ( x  1) 2  2 is translated 2 units left and 3 units up. if the resulting graph represents g ( x), what is the value of g (2.5)? (A) (B) (C) (D) (E)

30.

29.25 14.62 1.25 14.65 18.75

If sin 2 x  sin x  cos 2 x , then what is the smallest positive value of x ?



(A) (B)

3



2 7 (C) 6 4 (D) 3 11 (E) 6

31.

2 If f ( x)  2 x  x 2 and g ( x)  x  2 , then what is the

domain of ( g  f )( x) ? (A) (B) (C) (D)

x0 x0 0 x2 x  2 and x  0

(E) All real x

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392

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

32.

If the function f ( x) is continuous, what is the value of k ?  x2  4  f ( x)   x  2  k 

(A) (B) (C) (D) (E)

33.

if x  2

1 2 3 4 Undefined

A radioactive nuclide has a half-life of 10 days. At what rate does the substance decay each day? (A) (B) (C) (D) (E)

34.

if x  2

5% 6.7% 8.5% 10% 12.5%

The graph of f ( x)  x  5  2 is translated 6 units left and 3 units down. If the resulting graph is g ( x) , then g (2) is (A) (B) (C) (D) (E)

35.

0 1 2 3 4

1  cot 2   2 sin 

(A) (B) (C) (D) (E)

1 sin  cos  tan  cot 

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393

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

36.

37.

If

x  10 a b   , then a  2 x 4 x2 x2

(A) (B) (C) (D) (E)

3 2 1 1 2

If a die is rolled three times, what is the probability that all three numbers are different? (A) (B) (C) (D) (E)

1 3 1 2 4 9 5 9 2 3 y

38.

m

In Figure 3, if line m is tangent to the circle at point P , which of the following is the equation of line m ? (A) x  2 y  26

P(8, 9) 5

(B) 2 x  2 y  34 (C) 3x  4 y  60 (D) 4 x  3 y  59 (E) 5 x  2 y  58

O

x

5

Note:Figure not drawn to scale. Figure 3

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394

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

39.

The graph in Figure 4 shows a portion of a function. Which of the following could be the function of f ? 3 2 I. f ( x)  x  ax  bx  3

y

5 4 3 2 II. f ( x)  x  ax  bx  cx  dx  5

III. f ( x)   x7  ax6  cx5  dx4  cx3  dx2  ex  5 (A) (B) (C) (D) (E)

I only II only I and II only II and III only I, II, and III

x

O

Figure 4 40.

5log x If f ( x)  2 2 , then what is the smallest integral value of

x such that f ( x)  100 ? (A) (B) (C) (D) (E)

41.

2 3 50 100 1000

If log

3

k  log3 2  log3 (k  4) , then what is the value of

k?

(A) (B) (C) (D) (E)

42.

2 or 4 2 or 5 4 only 4 or 8 Undefined

If triangle ABC with dimensions 10 and 16 is rotated about

A

16

AB , what is the surface area of the resulting solid? (A) (B) (C) (D) (E)

314.2 592.8 634.7 906.9 5026.6

B

C 10 Note: Figure not drawn to scale. Figure 5

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395

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

43.

 What is the measure of the angle between a  (1, 3) and  b  (2, 1) ?

(A) 22.5 (B) 30 (C) 45 (D) 60 (E) 75 44.

Which of the following is the graph of the polar equation r  2csc ? (A) (B) (C) (D) (E)

45.

A point A line A circle A parabola An ellipse

If sin    sin  2  ?

(A) (B) (C) (D) (E) 46.

0.71 0.50 0.31 0.31 0.71

Figure 6 shows an isosceles trapezoid. What is the length of its diagonal? (A) (B) (C) (D) (E)

47.

5 3 and     , then what is the value of 2 13

3 5

6.5 7.4 8.1 8.7 9.3

5 10 Figure 6

If 5  6i  a  2i , then a could be (A) 7.9

(B) 8.6

(C) 11.3

(D) 8.1 (E) 7.5

GO ON TO THE NEXT PAGE

396

2

2 MATHEMATICS LEVEL 2 TEST - Continued USE THIS SPACE FOR SCRATCHWORK.

48.

1 , then what is 3 the probability of at most 2 days of rain during the next 4

If the probability of rain on any given day is

days? (A) (B) (C) (D) (E) 49.

0.5 0.54 0.71 0.89 0.95

If the equation of a parabola is given by y  1 

1  x  2 2 , 4

then what is its focus? (A) (0, 2) (B) (1, 2) (C) (2, 2) (D) (2, 3) (E) (2,  2) 50.

If a hyperbola has the equation y 2  25 x 2  25 , what are the equations of its asymptotes? 1 x 5 2 (B) y   x 5

(A) y  

(C) y   5 x (D) y   10 x (E) y   25 x

STOP IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS TEST ONLY. DO NOT TURN TO ANY OTHER TEST IN THIS BOOK.

GO ON TO THE NEXT PAGE Dr. John Chung's SAT II Math Level 2 Test 12

397

No Test Material on This Page

398

2

2

TEST 12 # 1 2 3 4 5 6 7 8 9 10

answer D E B B D A D B A E

ANSWERS # 11 12 13 14 15 16 17 18 19 20

answer A B D D B C C C D C

# 21 22 23 24 25 26 27 28 29 30

answer C E C D A C B D C B

# 31 32 33 34 35 36 37 38 39 40

answer C D B A A E D C D B

# 41 42 43 44 45 46 47 48 49 50

answer C D C B A B E D C C

Explanations: Test 12 1. (D)

 3x 

a b

 81x12  3b x ab  34 x12

Therefore, b  4 , ab  12 , and a  3. The answer is a  b  7 . 2. (E)

3a  b  2a  b  a  2b a  b 2b  b 3b   3 a  b 2b  b b

3. (B)

2 x 1  8  2 x 1  23  x  2 23 x  2 2 x 2 6  2 4   25  23  32  8  40 2 2

4. (B)

5. (D)

 x 2x  1  4    1(4)  2  4 The smallest integer is 0.

x  2(2 x  1)  4   3x  2  x  

2 3

V   (6) 210   (2) 210  320 m

6 

2

Dr. John Chung's SAT II Math Level 2 Test 12

10

399

2 6. (A)

2 log a b 

log b log a  log16  log 9   4log 2  2log 3     8  log 3  log 2   log 3  log 2 

 log3 16   log 2 9    7. (D)

Method 1: 1 3    30  sin 2  sin 60   0.8660254038  0.87 sin   2 2 Method 2: 3 In Quadrant I : cos   2 3  1  3  sin 2  2sin  cos   2       2  2  2

8. (B)

Period of tan 2 is

 2

.

Therefore, the asymptotes are at x  

9. (A)

Area of a triangle 

80o

10. (E)

30o

3 5 , ,   4 4

The law of sines: sin 80 sin 30 10sin 80   b b 10 sin 30

10

1 10sin 80  Area  10   sin 70  92.54165784  92.5 2 sin 30 

70o C

b

Point of symmetry:

4

, 

ab sin C : 2 B

A



y

 (0, 3) O



x

2

11. (A)

x 2  y 2  2 x  4 y  5  0  ( x  1) 2  ( y  2) 2  0  x  1 and y  2. Since the length of the radius is zero, it represents a point (1, 2) .

12. (B)

 sin   cos 2  1  2sin  cos   1  sin 2 1  sin 2  1 

400

1 3  2 2

2 13. (D)

2 x 2  6 x  4 y 2  16 y  17  0 

x 2  6 x  4( y 2  4 y )  17 ( x  3) 2 ( y  2) 2  1 8 2

 ( x  3)2  4( y  2) 2  8 

a 2  8  a  2 2 , b 2  2  b  2 , and center at (3,  2) . The graph is as follows. y

3

x

O

2

14. (D)

2.82

 1.41

h(t )  18t 2  72t  80  134  18t 2  72t  54  0  t 2  4t  3  0 (t  3)(t  1)  0  t  3, 1 Therefore, 3  1  2 seconds.

feet

h





1

3

y  134

80 O

seconds

5

t

  x 

15. (B)

 2 1  x  2  x  

16. (C)

2 log 9 (5 x)  3  log 9 (5 x) 

 The fourth term = 5 C3 x 2

2

2 3

3

x 17. (C)

27 5





 10 x 4  x 6   10x 2

 

3  5 x  9 2  5 x  32 2

3 2

 27

(Or, use a calculator: 9  (1.5)  2 )

an  an 1  5  an  an 1  5 : Common difference is 5 and the first term is 3. a30  a1  (n  1)d  3  29  5  148 Sum of the arithmetic sequence: S30 

Dr. John Chung's SAT II Math Level 2 Test 12

30(3  148)  2265 2

401

2 18. (C)

2 The law of cosines: The magnitude of the resultant  302  402  2(30)(40) cos130  63.58215365  63.58

30

R 50

o

130 o

30

40

19. (D)

a 2  b 2  (a  b)i  10  5i  a 2  b 2  10, a  b  5 (a  b)(a  b)  10 and a  b  5  a  b  2 . Therefore, a  3.5. n

20. (C)

 1 Since lim 1    e , n   n

 1 lim 1   n   n 21. (C)

22. (E)

23. (C)

2n

 1  lim 1   n   n

 Since the period of cos  3

2n

 e2

2    6 x  is  6 , the period of cos 2  x  is  3 .   3  2 3

1 1  1   1 and  f ( x)      1   1 . x x  x  Since f ( x)   f ( x) , f ( x) is not an odd function. f ( x) 

y  sin  Arc tan a  : Let X  Arctan a .  a  tan X and y  sin X

sin X 

1

X

a

a 1  a2

1  a2

24. (D)

y  e x 1  f 1 ( x) : x  e y 1  y  ln( x)  1  y  ln

25. (A)

f  f (5) 11 Quotient of functions:   (5)    5.5 g (5) 2 g

402

x e

2

2

26. (C)

Domain: 8  2 x  x 2  0  x 2  2 x  8  0  ( x  4)( x  2)  0 Therefore, the domain is 4  x  2 .

27. (B)

The first term is 0.5 and the common ratio is 0.5. 0.5 a S  1 1  r 1  0.5 Permutation:

28. (D)





4

P4  4!

3 P3  3!

Therefore, 3!  4!  144 . 29. (C)

g ( x)    ( x  2)  1  2  3  ( x  1) 2  1  g (2.5)  (2.5  1) 2  1  1.25

30. (B)

sin 2 x  sin x  cos 2 x  sin 2 x  sin x  1  sin 2 x  2sin 2 x  sin x  1  0 (2sin x  1)(sin x  1)  0 1  Therefore, sin x   or sin x  1 . The smallest positive value of x is . 2 2

31. (C)

 g  f  ( x)   x 2  2 x  2

2

The domain of ( g  f )( x) is 0  x  2 . From above, it might appear that the domain of the composition is all real numbers. However, this is not true because the domain of f is 0  x  2 , the domain of  g  f  is 0  x  2 . 32. (D)

x2  4 ( x  2)( x  2) ( x  2)  lim  lim 4 x2 x  2 x2 x2 ( x  2) 1 In order to be continuous, k should be equal to 4.

lim

t

33. (B)

 1 10 Radioactive decay: P  P   2 1

 1 10 The rate each day: P  P    0.9330329915 P 2 Therefore, 1  0.9330329915  0.0669670085  6.7% decay.

34. (A)

Shift 6 left and down 3 f ( x)  x  5  2   g ( x )  x  6  5  2  3  g ( x)  x  1  1

g (2)  2  1  1  0

35. (A)

1 1 cos 2  1  cos 2  sin 2  2   cot     1 sin 2  sin 2  sin 2  sin 2  sin 2 

Dr. John Chung's SAT II Math Level 2 Test 12

403

2 36. (E)

2 x  10 a b x  10 a( x  2)  b( x  2)     2  x  2  x  2  x 4 x2 x2 x2  4 x  10 (a  b) x  2a  2b   x  2  x  2  x2  4 Therefore, a  b  1 and 2a  2b  10 . a  2

37. (D)

The number of all possible outcomes (sample space) is 6  6  6  216 . If the first roll is a 1, the second and third time cannot rolls cannot be 1. If the second roll is a 2, the third cannot be a 2. Therefore, 6  5  4  120 . 120 5  . Probability P  216 9

38. (C)

The tangent is perpendicular to the diameter of the circle. y

95 4  85 3 3 The slope of line m :  4 The equation of line m is 3 x y  9    x  8 4 or 3 x  4 y  60



m

The slope of line  :

 P(8, 9) (5,5)

5 O

39. (D)

5

The function has one positive real root, a positive y-intercept, rises to the left, and falls to the right. I. f ( x)   x3  ax 2  bx  3  Negative y-intercept II. f ( x)   x5  ax 4  bx3  cx 2  dx  5 7

6

5

4

3

 2

Four imaginary roots and one real root

III. f ( x)   x  ax  cx  dx  cx  dx  ex  5

 Six imaginary roots and one real

root 40. (B)

5

f ( x)  25log2 x  100  2log2 x  x5 1

x 5  100  x  100 5  2.5118864 The smallest integer value of x is 3.

41. (C)

log a b  log a n b n : log

3

k  log 3 2  log 3 (k  4)  log 3 k 2  log 3 2  log 3 (k  4)

log3 k 2  log3  2k  8   k 2  2k  8  k 2  2k  8  0

Therefore,  k  4  k  2   0 . k  4 or k  2. From log

404

3

k , k cannot be less than or equal to 0. The value of k is only 4.

2 42. (D)

2 Lateral surface area is  rs . s  356

The Surface area:  rs   r 2

16

 (10)( 356)   (10)2  906.9137817  906.9

43. (C)

3  1  tan 1 3 1 1 1   2  tan 1   tan  2  2 2 1   1   2  tan 1 3  tan 1  45 2 tan 1 

2 sin 

44. (B)

r  2 csc  

45. (A)

 is in Quadrant III.

 r

2 y r

r  10

 (1,3)

1



 (1, 2)

2

O

 y2

5 13 12 cos    13

sin    12 5

13

 5  12  120  0.7100591716  0.71 sin  2   2sin  cos   2        13  13  169 46. (B)

From the figure below: 3 5

A

C 52  3.52

5 6.5

3.5





The length of the diagonal AC  6.52  52  3.52  7.416198487  7.4

Dr. John Chung's SAT II Math Level 2 Test 12

405

2 47. (E)

2 5  6i  52  (6) 2  61 a  2i  a 2  4 a 2  4  61  a 2  4  61  a 2  57  a  7.549834435  7.5

48. (D)

At most two rainy days: 0 days, I day, and 2 days 0

4

1

3

2

2

1  2 1  2 1  2 P  4 C0      4 C1      4 C2      0.88888  3  3  3  3  3  3 Or  1  binomcdf  4, , 2   0.888888  0.89  3  49. (C)

1 1  x  2 2  y   x  2 2  1  p  1 4 4p Vertex  2,1 , p  1 and directrix y  0 . y 1 

The coordinates of the focus is (2, 1  1)   2, 2  . y

 (2, 2) 

x

50. (C)

y 2  25 x 2  25 

y 2 x2  1 1 52

a  1 and b  5 . 5 b Asymptotes: y   x   x  5 x 1 a y

y  5 x

y  5x

5

O

5

END

406

x