Revise Edexcel A level Mathematics Revision Guide 9781292325835, 1292325836

Table of contents :
Cover
Contents
Part 1: Pure mathematics
1: Index laws
2: Expanding and factorising
3: Surds
4: Quadratic equations
5: Functions and roots
6: Sketching quadratics
7: The discriminant
8: Modelling with quadratics
9: Simultaneous equations
10: Inequalities
11: Inequalities on graphs
12: Cubic and quartic graphs
13: Transformations 1
14: Transformations 2
15: Reciprocal graphs
16: Points of intersection
17: Equations of lines
18: Parallel and perpendicular
19: Lengths and areas
20: Equation of a circle
21: Circle properties
22: Solving circle problems
23: The factor theorem
24: Binomial expansion 1
25: Solving binomial problems
26: Proof
27: Cosine rule
28: Sine rule
29: Trigonometric graphs
30: Trigonometric equations 1
31: Trigonometric identities 1
32: Trigonometric equations 2
33: Vectors
34: Solving vector problems
35: Differentiating from first principles
36: Differentiation 1
37: Differentiation 2
38: Tangents and normals
39: Stationary points 1
40: Stationary points 2
41: Modelling with calculus
42: Integration
43: Finding the constant
44: Definite integration
45: Area under a curve
46: More areas
47: Exponential functions
48: Logarithms
49: Equations with logs
50: Exponential equations
51: Natural logarithms
52: Exponential modelling
53: Modelling with logs
54: You are the examiner!
55: You are still the examiner!
56: Proof by contradiction
57: Algebraic fractions
58: Partial fractions
59: Algebraic division
60: Domain and range
61: Graphs and range
62: Inverse functions
63: Inverse graphs
64: Modulus
65: Modulus transformations
66: Modulus equations
67: Arithmetic sequences
68: Arithmetic series
69: Geometric sequences
70: Geometric series
71: Infinite series
72: Recurrence relations
73: Modelling with series
74: Series and logs
75: Binomial expansion 2
76: Radians, arcs and sectors
77: Areas of triangles
78: Sec, cosec and cot
79: Trigonometric equations 3
80: Trigonometric identities 2
81: Arcsin, arccos and arctan
82: Addition formulae
83: Double angle formulae
84: a cos θ ± b sin θ
85: Trig modelling
86: Parametric equations
87: Differentiating sin x and cos x
88: The chain rule
89: Differentiating standard functions
90: The product rule
91: The quotient rule
92: Differentiation and graphs
93: Parametric differentiation
94: Implicit differentiation
95: Differentiating a^x
96: Points of inflexion
97: Rates of change
98: Iteration
99: The Newton-Raphson method
100: Integrating standard functions
101: Reverse chain rule
102: Identities in integration
103: Integration by substitution
104: Integration by parts
105: Integrating partial fractions
106: Area between two curves
107: Areas and parametric curves
108: The trapezium rule
109: Solving differential equations
110: You are the examiner!
111: You are still the examiner!
Part 2: Statistics
112: Sampling
113: Mean
114: Median and quartiles
115: Linear interpolation
116: Standard deviation 1
117: Standard deviation 2
118: Coding
119: Box plots and outliers
120: Cumulative frequency diagrams
121: Histograms
122: Comparing distributions
123: Correlation and cleaning data
124: Regression
125: Using regression lines
126: Drawing Venn diagrams
127: Using Venn diagrams
128: Independent events
129: Tree diagrams
130: Random variables
131: The binomial distribution
132: Hypothesis testing
133: Finding critical regions
134: You are the examiner!
135: You are still the examiner!
136: Measuring correlation
137: Hypothesis testing for 0 correlation
138: Conditional probability
139: Probability formulae
140: The normal distribution 1
141: The normal distribution 2
142: The inverse normal function
143: Finding μ and σ
144: Normal approximations
145: Normal hypothesis testing
146: You are the examiner!
147: You are still the examiner!
Part 3: Mechanics
148: Modelling in mechanics
149: Motion graphs
150: Constant acceleration 1
151: Constant acceleration 2
152: Motion under gravity
153: Forces
154: Forces as vectors
155: Motion in 2D
156: Pulleys
157: Connected particles 1
158: Combining techniques
159: Variable acceleration 1
160: Variable acceleration 2
161: Deriving suvat formulae
162: You are the examiner!
163: You are still the examiner!
164: Moments 1
165: Moments 2
166: Centres of mass
167: Resolving forces
168: Friction
169: Sloping planes
170: Projectiles
171: Projectile formulae
172: Static particles
173: Limiting equilibrium
174: Static rigid bodies
175: Connected particles 2
176: Vectors in kinematics
177: Vectors and bearings
178: Variable acceleration 3
179: Calculus with vectors
180: You are the examiner!
181: You are still the examiner!
Worked solutions
Pure mathematics
Statistics
Mechanics

Citation preview

REVISE PEARSON EDEXCEL A LEVEL

Mathematics

REVISION GUIDE

Notes from the publisher While the publishers have made every attempt to ensure that advice on the qualification and its assessment is accurate, the official specification and associated assessment guidance materials are the only authoritative source of information and should always be referred to for definitive guidance. Pearson examiners have not contributed to any sections in this resource relevant to examination papers for which they have responsibility.

Series Consultant: Harry Smith Author: Harry Smith

REVISE PEARSON EDEXCEL A LEVEL

Mathematics

REVISION GUIDE

Notes from the publisher While the publishers have made every attempt to ensure that advice on the qualification and its assessment is accurate, the official specification and associated assessment guidance materials are the only authoritative source of information and should always be referred to for definitive guidance. Pearson examiners have not contributed to any sections in this resource relevant to examination papers for which they have responsibility.

Series Consultant: Harry Smith Author: Harry Smith

Contents

1-to-1

page

ma

tch w ith th A Lev e el M athem Revis atics ion W orkb ISBN ook 97812 92190

600

Pure mathematics

ii

1

Index laws

34 Solving vector problems

2

Expanding and factorising

35 Differentiating from first principles

3

Surds

36 Differentiation 1

4

Quadratic equations

37 Differentiation 2

5

Functions and roots

38 Tangents and normals

6

Sketching quadratics

39 Stationary points 1

7

The discriminant

40 Stationary points 2

8

Modelling with quadratics

41

9

Simultaneous equations

42 Integration

10

Inequalities

43 Finding the constant

11

Inequalities on graphs

44 Definite integration

12

Cubic and quartic graphs

45 Area under a curve

13

Transformations 1

46 More areas

14

Transformations 2

47 Exponential functions

15

Reciprocal graphs

48 Logarithms

16

Points of intersection

49 Equations with logs

17

Equations of lines

50 Exponential equations

18

Parallel and perpendicular

51

19

Lengths and areas

52 Exponential modelling

Modelling with calculus

Natural logarithms

20 Equation of a circle

53 Modelling with logs

21

54 You are the examiner!

Circle properties

22 Solving circle problems

55 You are still the examiner!

23 The factor theorem

56 Proof by contradiction

24 Binomial expansion 1

57 Algebraic fractions

25 Solving binomial problems

58 Partial fractions

26 Proof

59 Algebraic division

27 Cosine rule

60 Domain and range

28 Sine rule

61

29 Trigonometric graphs

62 Inverse functions

30 Trigonometric equations 1

63 Inverse graphs

31

64 Modulus

Trigonometric identities 1

Graphs and range

32 Trigonometric equations 2

65 Modulus transformations

33 Vectors

66 Modulus equations

67 Arithmetic sequences

90 The product rule

68 Arithmetic series

91

69 Geometric sequences

92 Differentiation and graphs

70 Geometric series

93 Parametric differentiation

71

94 Implicit differentiation

Infinite series

The quotient rule

72 Recurrence relations

95 Differentiating ax

73 Modelling with series

96 Points of inflexion

74 Series and logs

97 Rates of change

75 Binomial expansion 2

98 Iteration

76 Radians, arcs and sectors

99 The Newton-Raphson method

77 Areas of triangles

100 Integrating standard functions

78 Sec, cosec and cot

101 Reverse chain rule

79 Trigonometric equations 3

102 Identities in integration

80 Trigonometric identities 2

103 Integration by substitution

81

104 Integration by parts

Arcsin, arccos and arctan

82 Addition formulae

105 Integrating partial fractions

83 Double angle formulae

106 Area between two curves

84 a cos 𝜃 ± b sin 𝜃

107 Areas and parametric curves

85 Trig modelling

108 The trapezium rule

86 Parametric equations

109 Solving differential equations

87 Differentiating sin x and cos x

110 You are the examiner!

88 The chain rule

111

You are still the examiner!

89 Differentiating standard functions

iii

Statistics

iv

112 Sampling

130 Random variables

113 Mean

131 The binomial distribution

114 Median and quartiles

132 Hypothesis testing

115 Linear interpolation

133 Finding critical regions

116 Standard deviation 1

134 You are the examiner!

117 Standard deviation 2

135 You are still the examiner!

118 Coding

136 Measuring correlation

119 Box plots and outliers

137 Hypothesis testing for 0 correlation

120 Cumulative frequency diagrams

138 Conditional probability

121 Histograms

139 Probability formulae

122 Comparing distributions

140 The normal distribution 1

123 Correlation and cleaning data

141 The normal distribution 2

124 Regression

142 The inverse normal function

125 Using regression lines

143 Finding 𝜇 and 𝜎

126 Drawing Venn diagrams

144 Normal approximations

127 Using Venn diagrams

145 Normal hypothesis testing

128 Independent events

146 You are the examiner!

129 Tree diagrams

147 You are still the examiner!

Mechanics 148 Modelling in mechanics

171 Projectile formulae

149 Motion graphs

172 Static particles

150 Constant acceleration 1

173 Limiting equilibrium

151 Constant acceleration 2

174 Static rigid bodies

152 Motion under gravity

175 Connected particles 2

153 Forces

176 Vectors in kinematics

154 Forces as vectors

177 Vectors and bearings

155 Motion in 2D

178 Variable acceleration 3

156 Pulleys

179 Calculus with vectors

157 Connected particles 1

180 You are the examiner!

158 Combining techniques

181 You are still the examiner!

159 Variable acceleration 1 160 Variable acceleration 2

Worked solutions

161 Deriving suvat formulae

182 Pure mathematics

162 You are the examiner!

200 Statistics

163 You are still the examiner!

204 Mechanics

164 Moments 1 165 Moments 2 166 Centres of mass 167 Resolving forces 168 Friction 169 Sloping planes 170 Projectiles

A small bit of small print: Edexcel publishes Sample Assessment Material and the Specification on its website. This is the official content and this book should be used in conjunction with it. The questions have been written to help you practise every topic in the book.

v

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Index laws You need to be able to work with algebraic expressions confidently for all of your AS maths topics. Make sure you know how to use these six index laws:

1 3 5

am × an = a m + n

x 5 × x −2 = x 3

(am)n = a mn

(x 4)3 = x 12

_1_

__

1 __

na an = √

49 2 = 7 1 __

27 3 = 3

1 __

x2

−

1 __

x2

= 10x

1 1 − __ 2 1 __

1 1 5−2 = ___2 = ___ 5 25

_m_

_7_

− 5x 2

8

x

__

_ −_2 3

n a )m a n = (√

27

= ( 27 3 )

1 −2 __

___

= ( √27 ) 3

−2

1 1 = 3−2 = ___2 = __ 3 9

Do these calculations one step at a time.

10x –__5x2 Given that ________ can be written in the form √x 10xp – 5xq, write down the value of p and the value of q. (2 marks) _7_

1 a−n = __ an

1 __

_7

5x 2 ____

a

x = x2 __ 6

m

16 4 = 2

2 × 2 × 2 ×___ 2 = 16 4 so √16 = 2

10x ____

2 4 6

a = am − n __ n

The fraction acts like a bracket, so you need to divide _both terms in the numerator by √x . Write down the values of p and q when you’ve finished.

Golden rule

1 − __ 2

Convert all roots into fractional powers before applying the other index laws.

= 10x 2 − 5x 3 1 p = __ and q = 3 2

_

√x =

You will be allowed to use your calculator in both of your AS exams, but be careful. __1 If you enter 125 – 3 into your calculator it will give you the answer __1 or 0.2. You 5 need to be able to write it in the form given in the question.

1 __

x2

3

_

√x =

1 __

x3

_1

n

Express 125 −3 in the form 5 where n is an integer. (1 mark) 1 1 __ −__ 3 − 125 3 = (5 ) 3 = 5 3×(−3) = 5 −1 So n = −1 1 __

Check that your answer is in the correct form. n is an integer so it is a positive or negative whole number, or 0.

1 Simplify x( 4x ) −_2 1

2 Simplify ( 9y10 )2

_3

3

__

(2 marks) (2 marks)

There are full worked solutions to all the Now try the questions on page 182.

1

3 Write

5 + 2√x _______

p

q

in the form 5x + 2x , where x2 p and q are constants. (2 marks) Remember that a constant doesn’t have to be an integer.

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Expanding and factorising In your AS exam, you might have to multiply out (or expand) a product of three brackets, or factorise a cubic expression.

Expanding brackets To expand the product of two factors, you have to multiply every term in the first factor by every term in the second factor.

There are 2 terms in the first factor and 3 terms in the second factor, so there will be 2 × 3 = 6 terms in the expanded expression before you collect like terms.

2x ×

(2x + 3)(5x 2 – x + 4) = 10x 3 – 2x 2 + 8x + 15x 2 – 3x + 12 = 10x 3 + 13x 2 + 5x + 12 3×

Simplify your expression by collecting like terms: –2x2 + 15x2 = 13x2 _

_

√ x × √ x = x. ber that Remem _ _ _ _ _ 2 √ √ x √x = 4x √ 4 = x √ 2 × x 2 = ) x (2 __

Show that (3 + 2√x )2 can be written as __ 9 + k√x + 4x, where k is a constant to be found. (2 marks) _

_

_

_

(3 + 2√x )(3 + 2√x ) = 9 + 6√x_+ 6√x + 4x = 9 + 12√x + 4x k = 12

! If you have to find a constant, it’s a good idea to write down the value of the constant when you have finished your working. You will need to use problem-solving skills throughout your exam – be prepared!

Factorising Factorising is the opposite of expanding brackets.

Factorise completely x 3 + x 2 − 6x (3 marks)

ng brac ndi ke pa ts

Ex

x 3 + x 2 − 6x = x(x 2 + x − 6) = x(x + 3)(x − 2)

(x + 4)2 = x 2 + 8x + 16 (2x + 5)(x – 6) = 2 x2 – 7 x – 30 Fa

ctor isi ng

Start by taking the common factor, x, out of every term. You are left with a quadratic expression, which you can factorise into two linear factors.

1 Given that (x + 2)(x + 1)2 = x 3 + bx 2 + cx + d, where b, c and d are constants, find the values of b, c and d. (3 marks) 3 2 2 Factorise completely 3x − 2x − x (3 marks) (2 marks) 3 Factorise completely 25x 2 − 16

Multiply out (x + 1)2 first to get (x + 2)(x 2 + 2x + 1). This is a difference of two squares: a2 – b 2 = (a + b)(a – b)

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Surds You can simplify surds quickly using the fraction and square root functions on your calculator. However, you need to understand how to manipulate surds in order to work with algebraic expressions confidently.

Golden rules

These are the two golden rules for simplifying surds:

1

___

√ab __

__

__

___

= √a × √b __

___

__

2

√b

_a_

__

√

√a __ = ___ √b

__

__

____

__

√3 √3 ___ ___ = = _____ 5 25 √ 25

3 ___

__

__

______

+ √b is not equal to √a + b .

___

___

Simplify √27__ + √48 , giving your answer in the form a√3 , where a is an integer. (2 marks) ___

__

__

__

___

√27 + √48 = √9 × √3 + √16 × √3 __

__

= 3√3 + 4√3 __ = 7√3

a=7

__

= √9 × √5 __ = 3√5 k=3

√a

___

______

√45 = √9 × 5

__

√8 = √4 × √2 = 2√2 __

__

Write √45 in the form k√5 , where k is an integer. (2 marks)

write the answer in the __ You know __ you need to form a√3 so write each surd in the form k√3 . You need to take a factor of 3 out of each number: ______ ___ ×3 _ = √9______ 27 = 9 × 3 so √27 ___ √ 48 = 16 × 3 so √48 = 16 × 3

Rationalising the denominator You can rationalise the denominator of a fraction by removing any surds in the denominator. __

× (4 + √11 )

__

__

4__ + √11 __ 4 + √11 1 __ ________ = __________________ = ________ 5 4 − √11 (4 − √11 )(4 + √11 ) __

× (4 + √11 ) __

__

__

__

(4 − √11 )(4 + √11 ) = 16 − 4√11 + 4√11 − 11 =5

To work out what to multiply the top and bottom by, look at the denominator of the original fraction. Swap a plus for a minus, or swap a minus for a plus.

__

__

1 Expand and simplify (x + √ 3 )(x − √ 3 ). (2 marks) ___ __ 2 Write √98 in the form a√2 , where a is an integer. (1 mark)

3

__ 14 __ Express ______ in the form a + b√2 , 3 + √2 where a and b are integers.__ (2 marks) 14(3__ − √2 ) __ 14 __ _________________ ________ = √ √2 ) 3+ 2 (3 + √2 )(3 − __ 14(3 − √2__ ) _____________________ __ = 9 + 3√2__− 3√2 − 2 2 14(3 − √2 ) ___________ = 71 __ = 6 − 2√2 a = 6 and b = −2

__

If the denominator is in the form p + √q __ then multiply the numerator and denominator of the fraction by p – √q .

3 Simplify the following, giving your answers in __ √ the form a + b 5 , where a and b__are integers. 4 + √5__ 8 __ (2 marks) (b) ______ (4 marks) (a) ______ 3 + √5 2 − √5

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Quadratic equations Quadratic equations can be written in the form ax 2 + bx + c = 0, where a, b and c are constants. The solutions of a quadratic equation are sometimes called the roots of the equation.

Solution by factorising You can follow these steps to solve some quadratic equations: 1. Rearrange the equation into the form ax 2 + bx + c = 0 2. Factorise the left-hand side. 3. Set each factor equal to zero and solve to find two values of x.

(4 marks)

Solve 2(x + 1)2 = 3x + 5 2(x2 + 2x 2x2 + 4x 2x2 + x (2x + 3)(x

The first solution is the value of x which makes the (2x + 3) factor equal to 0.

+ + − −

2x + 3 = 0 3 x = −__ 2

1) = 2= 3= 1) = or

3x + 5 3x + 5 0 0

x−1=0 x=1

Completing the square x2 + 6x − 2 = (x + a)2 + b, where a and b are constants. (a) Find the values of a and b. (3 marks) 2 2 2 x + 6x − 2 = (x + 3) − 3 − 2 = (x + 3)2 − 9 − 2 = (x + 3)2 − 11 a = 3 and b = −11 (b) Hence, or otherwise, show that the roots of x2 + 6x − 2 = 0 ___ are c ± √11 , where c is an integer to be found. (2 marks) 2 (x + 3) − 11 = 0 (+ 11) 2 (√–) (x + 3) = 11 __ x + 3 = ±√11 (−3) __ x = −3 ± √11 c = −3

1 Solve the equation 2(x − 3)2 + 3x = 14

You can write a quadratic expression in the form (x + p)2 + q using these two identities:

1 2

x2 + 2bx + c ; (x + b)2 − b2 + c x2 − 2bx + c ; (x − b)2 − b2 + c

You can use this method to solve a quadratic equation without using a calculator.

Write the left-hand side in completed square form, and use inverse operations to solve the equation. Remember that any positive number has two square roots: one positive and one negative. You need to use the ± symbol when you take square roots of both sides of the equation.

(3 marks)

2 (a) Show that x2 − 10x + 7 can be written as (x + p)2 + q, where p and q are integers to be found. (2 marks) 2 (b) Hence solve the equation x − 10x + __ 7 = 0, giving your answer in the form x = a ± b√2 , where a and b are integers to be found. (3 marks)

2 2 x2 − 10x + 7 = (x − 5) − 5 + 7

You could also solve this by substituting a = 1, b = −10 and c = 7 into the quadra _________ tic formula, 2 √b − ____ −b ± ______ 4ac x = ______ , then 2a simplifying.

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Functions and roots You need to be able to use function notation and to be able to solve quadratic equations in a function of the unknown. Here is an example: You don’t know how to solve a quartic equation like this, but if you write u = x 2 then u2 = (x2)2 = x 4.

x 4 – 7x 2 + 12 = 0 u 2 – 7u + 12 = 0 (u – 3)(u – 4) = 0

Use the substitution to convert back to x. The equation has four solutions.

u=3 or 2 x = 3 __ x = ±√3

Substitute u = x2 to form a quadratic equation in u.

u=4 x =4 x = ±2

You could also write (x2)2 – 7(x2) + 12 = 0 (x2 – 3)(x2 – 4) = 0

2

! __

Solve the equation 3x + √ x – 2 = 0 (4 marks) __ Let u = √__x 3x + √ x − 2 = 0 3u2 + u − 2 = 0 (3u − 2)(u + 1) = 0 3u − 2 = 0 or u + 1 = 0 u = _23_ or u = −1 _

Using u = √x x = u2

x = _49_

__

x = (√ x )2 so you could write the equation as: __ __ 3(√ x )2 + √ x − 2 = 0 __ This is a quadratic equation in √ x . The safest way to solve equations like this is to use the __ substitution u = √x . __ √x is the positive square root of x, so it can only take positive values. You need to ignore the solution u = −1. You will need to use problem-solving skills throughout your exam – be prepared!

Domain Functions will usually be defined for a given domain. This is the set of values that can be used as the input to the function. The domain of this function is all the positive real numbers (R). g(x) = 2x2 – 5x – 3, x ∈ R, x > 0 The roots of a function, g(x), are the values of x for which g(x) = 0. You might need to consider the domain when finding the roots of a function.

1 Solve (4 marks) (a) x4 – 3x2 − 4 = 0 6 3 (4 marks) (b) 8x + 7x −__1 = 0 (4 marks) (c) x + 10 = 7√x __ x = 3, giving your 2 Solve the equation 4√x + __ answer in the form a − b√ 7 , where a and b are integers to be found. (5 marks)

5

g(x) = 2x2 − 5x − 3, x ∈ R, x > 0 Show that g(x) has exactly one root and find its value. (3 marks) Set g(x) = 0 to find roots: 2x2 – 5x – 3 = 0 (2x + 1)(x – 3) = 0 1 x = − __ 2

or

x=3

1 The domain is x > 0, so x = − __ 2 is not a root. The only root is x = 3.

3 f(x) = x4 − 4x2 − 5, x ∈ R, x < 0 Show that f(x) has only one root and determine its exact value. (4 marks)

Look at the domain of the function carefully. x4 – 4x2 – 5 = 0 has two real sol utions but only one of them is a root of f(x).

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Sketching quadratics When you sketch a graph you need to show its key features. You don’t need to use graph paper for a sketch, but you should still draw your axes and any straight lines using a ruler.

Factorised quadratics On a sketch you usually show the points where the graph crosses the axes. y

y = (x + 2)(x – 5)

–2 0

5

–10 When x = 0, y = (0 + 2)(0 – 5) = –10

Negative coefficients If the coefficient of x 2 is negative the graph will be an ‘upside down’ U-shape. You can check the coefficient of x 2 by multiplying out the brackets.

x

Sketch the curve with equation y = x(4 − x) (3 marks)

When y = 0, (x + 2)(x – 5) = 0 x = –2 or x = 5

y

2 y = 4x − x 2, so the coefficient of x is negative.

y = x ( x – 4)

0

Sketch the curve with equation y = (x + 2)2 + 1 Label the minimum point and any points where the curve crosses the coordinate axes. (3 marks) y

4

Sketching y = (x

+ a) + b 2

If a quadratic is written in completed square form then you can find the position of its vertex easily. y

y = (x + 2)2 + 1

y = (x + 2)2 – 5

5 (–2, 1)

0

x

0 –1

x

To work out the point where the curve crosses the y-axis, substitute x = 0 into the equation.

1 On separate diagrams, sketch the graphs of (3 marks) (a) y = (x − 2)2 2 (b) y = (x − 2) + k, where k is a positive constant. (2 marks) Show on each sketch the turning point and the coordinates of any points where the graph meets the axes.

(–2, –5)

x

The curve with equation y = (x + a )2 + b has a vertex at (–a , b ).

2 x 2 + 6x + 15 = (x + a)2 + b (a) Find the values of a and b. (2 marks) 2 (b) Sketch the graph of y = x + 6x + 15, labelling the minimum point and any points of intersection with the axes. (3 marks) (c) Use your graph to explain why the equation x2 + 6x + 15 = 0 has no real solutions. (1 mark)

You can also use the discriminant to determine the number of real solutions. There is more about this on the next page.

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The discriminant The discriminant of a quadratic expression ax 2 + bx + c is the value b 2 − 4ac. You can use the discriminant to work out whether a quadratic equation has any real roots or real solutions. There are three possible conditions for the discriminant:

1

b 2 _ 4ac > 0 y

2

y = x 2 + 4x + 2

O

b 2 _ 4ac = 0

y = x 2 – 6x + 9 y

x

x

Discriminant = (–6)2 – 4 × 1 × 9 =0

Two distinct real roots

Two equal real roots

The equation x2 + 4qx + 2q = 0, where q is a non-zero constant, has equal roots. Find the value of q. (4 marks) b2 − 4ac = 0 (4q)2 − 4 × 1 × (2q) = 0 16q2 − 8q = 0 q(16q − 8) = 0 1 16q − 8 = 0 so q = __ 2

b 2 _ 4ac < 0 y

O

Discriminant = 42 – 4 × 1 × 2 =8>0

q=0

3

y = 2x 2 – x + 3

x

O

Discriminant = (–1)2 – 4 × 2 × 3 = –23 < 0 No real roots

Follow these steps: 1. Work out the values of a, b and c: a = 1, b = 4q and c = 2q. 2. Find an expression for the discriminant (b 2 − 4ac) in terms of q. 3. Set the discriminant equal to 0, because there are two equal roots. 4. Solve this new equation to work out two possible values for q. You are told that q is non-zero, so the _1 correct solution is q = 2.

! The equation must be in the form ax2 + bx + c = 0 before you work out the values of a, b and c. Always write down the condition for the discriminant that you are using, and use brackets when you substitute. You will need to use problem-solving skills throughout your exam – be prepared!

1 Find the value of the discriminant of (1 mark) 3x2 − 2x − 5 2 2 The equation px + 2x − 3 = 0, where p is a constant, has equal roots. Find the value of p. (3 marks) 2 The expression (k + p) must always be greater than or equal to zero.

7

The equation 2x2 − kx + 6 = k has no real solutions for x. Show that k2 + 8k − 48 < 0 (3 marks) 2x 2 − kx + 6 − k = b2 − 4ac < 2 (−k) − 4 × 2 × (6 − k) < k 2 + 8k − 48
0

A car manufacturer currently sells 50 000 units of a popular model each year, at a recommended retail price of £17 000 each. The manufacturer determines that for every £200 they increase the price, they will sell 400 fewer cars each year. The number of cars sold each year, N, for a given price £P, is modelled as N = a + bP (a) Determine the values of the constants a and b in this model. (2 marks) N = a + bP 1 N − 400 = a + b(P + 200) 2 1 − 2 : 400 = −200b so b = −2 50 000 = a − 2(17 000) a = 84 000 The model is N = 84 000 − 2P The total revenue generated, £X, can be modelled as X = P(a + bP), where a and b are the same constants. (b) Rearrange X into the form c – d(P – e)2, where c, d and e are positive constants to be found. (3 marks) X = P (84 000 − 2P ) = −2P 2 + 84 000P = −2(P 2 − 42 000P ) = −2[(P − 21 000)2 − 21 0002] = 441 000 000 − 2(P − 21 000)2 (c) State, with reasons, the amount of money the manufacturer should charge to maximise their revenue, and write down the maximum revenue. (2 marks) The revenue will be maximised when 2(P − 21 000)2 takes its minimum value (zero). This occurs when P = 21 000, so the manufacturer should charge £21 000. This will generate a revenue of £441 million.

(a) Write down the height of the clifftop. (1 mark) (b) Determine the horizontal distance travelled by the stone at the time it hits the water. (3 marks) (c) By completing the square, or otherwise, determine the maximum height of the stone above the water during its flight. (4 marks)

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Simultaneous equations If a pair of simultaneous equations involves an x2 or a y 2 term, you need to solve them using substitution. Remember to number the equations to keep track of your working. Rearrange the linear equation to make y the subject.

y 5 x 2 2 2x 2 7 x 2 y 5 23

y5 x 1 35 Each solution for x has a corresponding value of y. 05 Substitute to find the 05 values of y. x5 The solutions are x 5 5, From

2:

Solve the simultaneous equations x − 2y = 1 1 2 2 x + y = 13 2 From 1 : x = 1 + 2y 3 Substitute 1 + 2y for x in 2 : (1 + 2y)2 + y 2 = 13 1 + 4y + 4y 2 + y 2 = 13 5y 2 + 4y − 12 = 0 (5y − 6)(y + 2) = 0 6_ y = _5

( )

or

1 2

x13 3 2 Substitute x + 3 for y x 2 2x 2 7 in equation 1 . 2 x 2 3x 2 10 (x 2 5)(x 1 2) 5 or x 5 22 y 5 8 and x 5 22, y 5 1.

(6 marks)

y = −2

17 6 x = 1 + 2 __ = ___ x = 1 + 2(−2) = −3 5 5 17 6 Solutions: x = ___, y = __ and x = −3, y = −2 5 5

easier You can substitute for x or y. It is will be to substitute for x because there no fractions. Use brackets to make sure that the whole expression is squared. for Rearrange the quadratic equation 2 + by + c = 0 y into the form ay two Factorise the left-hand side to find solutions for y. Remember that there will be two pairs of solutions. Each value of y will produce a corresponding value of x. You need to find four different valu es in total and pair them up correctly.

y = (1 + x )2

Thinking graphically The solutions to a pair of simultaneous equations correspond to the points where the graphs of the equations intersect. Because an equation involving x 2 or y 2 represents a curve, there can be more than one point of intersection. Each point has an x-value and a y-value. You can write the solutions using coordinates.

y

y =x +1

(0, 1)

(–1, 0)

0

x

There is more on intersections of graphs on page 16.

1 Solve the simultaneous equations x+y=5 (6 marks) x2 + 2y2 = 22 2 You will need to write x − 6x + 7 in completed square form. Have a look at page 4 for a reminder.

9

2 (a) By eliminating y from the simultaneous equations y=x+6 xy − 2x2 = 7 (2 marks) show that x2 − 6x + 7 = 0 (b) Hence solve the simultaneous equations in __ part (a), giving your answers in the form a ± √2 , where a is an integer. (4 marks)

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Inequalities You might need to find a set of values which satisfy an inequality. If the inequality involves a quadratic expression you should always sketch a graph to help you answer the question. This graph shows a sketch of y = (x − 1)(x + 5). For these values of x, the curve is below the x-axis, so y < 0. The solution to (x − 1)(x + 5) < 0 is −5 < x < 1 For these values of x, the curve is above the x-axis, so y > 0. The solution to (x − 1)(x + 5) > 0 is x < −5 or x > 1

y

y = (x – 1)(x + 5)

–5

0

x

1

There are two separate sets of values which satisfy this inequality. You need to give both sets of values, and write OR between them.

Quadratic inequalities Find the set of values for which (a) 8x − 7 < 5x + 5 8x < 5x + 12 3x < 12 x 0 (2x + 1)(x − 3) > 0

(2 marks)

(4 marks)

y

Follow these steps to solve any quadratic inequality: 1. Rearrange so one side is 0. 2. Factorise the other side. 3. Sketch the graph. 4. Write the solutions using ,, ., < or >.

Make sure you write two separate inequalities for your answer. You can’t write _1 3 < x < − 2. It’s not true, because 3 is _1 larger than − 2.

2x 2 – 5x – 3 > 0 – 21

!

x

3

1 x < −__ 2 or x > 3 (c) both 8x − 7 < 5x + 5 and 2x 2 − 5x − 3 > 0

–4 –3 –2 –1 0 1 __

1

2

(3 marks)

3

4

5

6

7

x < − 2 or 3 < x < 4

1 Find the set of values of x for which x(x − 5) < 14 (4 marks) You need to expand the brackets and rearrange the inequality into the form ax 2 + bx + c < 0 first.

You need both inequalities to be true at the same time. Draw them both on a number line and look for the values where they overlap. You will need to use problem-solving skills throughout your exam – be prepared!

You can also write this as __1 {x : x < − 2 } ∪ {x : 3 < x < 4}

2 The equation x 2 + (k − 3)x − 4k has two distinct real roots. (a) Show that k satisfies k 2 + 10k + 9 > 0 (3 marks) (b) Hence find the set of possible values for k. (4 marks)

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Inequalities on graphs You can interpret inequalities graphically by considering curves and regions. This diagram shows the graphs of y = x 2 – 4x and y = –2x + 3: The graphs intersect when x 2 – 4x = –2x + 3 2 x – 2x – 3 = 0 (x – 3)(x + 1) = 0 x = 3 or x = –1 There is more on intersections of graphs on page 16.

y

y = x 2 – 4x

3 3

–1 0 The shaded region satisfies the inequalities y < –2x + 3, y > x2 – 4x and y < 0 simultaneously.

Outside the points of intersection the line is below the curve. So the solution to –2x + 3 < x 2 – 4x is x < –1 or x > 3.

x

4

y = –2x + 3

y

The graph shows the line with equation y = x 2 6 and the curve with equation y = 5x 2 2x2. (a) Determine the coordinates of the points of intersection of the line and the curve. (4 marks) 2 x − 6 = 5x – 2x 2x 2 − 4x − 6 = 0 x 2 − 2x − 3 = 0 (x + 1)(x − 3) = 0 x = −1 or x = 3 When x = −1, y = −1 − 6 = −7. P 1 is (−1, −7) When x = 3, y = 3 – 6 = −3. P2 is (3, −3) (b) Hence solve the inequality x 2 6 > 5x 2 2x2.

y 5 5x 2 2x2 O y5x26

(1 mark)

y

x=2

1

f(x) = 6 – x2 g(x) = _2 x + 1 (a) On the same axes, sketch the graphs of y = f(x) and y = g(x). Indicate any points of intersection with the coordinate axes. (3 marks) (b) Determine the x-coordinates of any points of intersection of the two graphs. (3 marks) 1

11

5

6

x

26

y = f(x)

5

y = g(x )

5 2

The solutions are the x-values when the line is higher than the curve. You can use set notation or give your answer as ‘x < −1 or x > 3’.

{x : x < − 1} ∪ {x : x > 3}

f(x) = x2 2 6x + 5 g(x) = x On a graph, show the region that satisfies the inequalities: y . f(x) y < g(x) x.2 (5 marks)

Between the points of intersection the line is above the curve. So the solution to –2x + 3 > x 2 – 4x is –1 < x < 3.

x

Use dotted lines to show a strict inequality (< or >) and solid lines to show a non-strict inequality (< or >).

(c) Hence solve the inequalities (i) f(x) , g(x) (ii) f(x) > g(x) (2 marks) (d) Shade the region on the graph that satisfies all of the following inequalities: y < f(x), y > g(x), x > 0 (1 mark)

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Cubic and quartic graphs In a cubic function, the highest power of x is x 3. In a quartic function, it is x 4. You need to know the shapes of graphs of cubic and quartic functions and be able to sketch them.

Factorise then sketch

Shapes and roots

You can sketch the graphs of cubic and quartic functions by factorising them to find their roots.

Cubic functions can have 1, 2 or 3 real roots:

y

y = x (x – 2)2(x – 5)

2

y

x

0

One factor is x so the curve passes through the origin.

0

y

Positive x3 term and 3 real roots

5

x

x

0

Negative x3 term and 1 repeated real root

Quartic functions can have 0, 1, 2, 3 or 4 real roots: y

y

(x – 2) is a factor so x = 2 is a repeated root. The curve just touches the x-axis at this point. 2

The curve touches or crosses the x-axis at each root of the function.

0

x

x

0

Negative x4 term and 2 real roots

Positive x4 term and 0 real roots

Considering infinity In the example on the right, as x gets large, the x3 term gets large more quickly than the x2 term. So for large positive x, y gets very large. You can write ‘as x → ∞, y → ∞’. Similarly, as x → − ∞, y → − ∞. This tells you how the curve will behave at either end of the x-axis.

The factorised equation has a factor of x so the curve will pass through the origin. It also has a repeated factor of (x − 4) so the curve will just touch the x-axis at the point x = 4.

1 (a) Factorise completely x3 − 9x (3 marks) (b) Hence sketch the curve y = x3 − 9x (3 marks) 2 Sketch the graph of y = (2x − 1)(x − 3)2, showing clearly the coordinates of the points where the curve meets the coordinate axes. (4 marks)

(a) Factorise completely x 3 − 8x 2 + 16x (3 marks) 2 2 x (x − 8x + 16) = x (x − 4) (b) Hence sketch the curve with equation y = x 3 − 8x 2 + 16x, showing the points where the curve meets the coordinate axes. (3 marks) y

y = x(x − 4)2 0

4

x

3 Sketch the graph of y = x(5 – x)(2x2 + 9x + 4). Show clearly the coordinates of any points where the curve meets or crosses the coordinate axes. (4 marks) You need to show the coordinates of the point where the graph meets the y-axis as well.

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Transformations 1 You can change the equation of a graph to translate it, stretch it or reflect it. These tables show you how you can use functions to transform the graph of y = f(x ). Function

y = f(x ) + a

y = f(x + a)

Transformation of graph

Translation

Translation

( Oa )

y = a f(x )

( −0a )

Stretch in the vertical direction, scale factor a

Useful to know f(x ) + a → move UP a units f(x + a) → move LEFT a units f(x ) − a → move DOWN a units f(x − a) → move RIGHT a units Example

y

y = f(x ) + 3

y

3

y = f(x )

0

x

x-values stay the same

–5 y = f(x + 5)

y = 3f(x )

y y = f(x )

y = f(x )

x

0

x

0

Function

y = f(ax )

y = −f(x )

y = f(−x )

Transformation of graph

Stretch in the horizontal _1_ direction, scale factor a

Reflection in the x-axis

Reflection in the y-axis

‘−’ outside the bracket

‘−’ inside the bracket

Useful to know y-values stay the same Example

y

y

y = f(2x )

y y = f(x )

0

x

x

0

y = f( x )

y = f(– x )

y = f(x )

x

0

y = – f(x )

The diagram shows a sketch of a curve with equation y = f(x).

y

y = f(x + 3)

On the same diagram sketch the curves with equation (a) y = f(x + 3) (3 marks) (b) y = −f(x). (3 marks) Show clearly the coordinates of any Everything in blue is maximum or minimum points, and part of the answer. any points of intersection with the axes.

y = −f(x )

−3

The diagram shows a sketch of a curve C with equation y = f(x). On separate diagrams sketch the curves with equation (a) y = 2f(x) (3 marks) (b) y = f(−x) (3 marks) (c) y = f(x + k), where k is a constant and 0 < k < 4 (4 marks) On each diagram show the coordinates of any maximum or minimum points, and any points of intersection with the x-axis.

13

(3, 2)

2

O

2

5

x

(3, −2)

y 5 f(x)

y

O

(4, 3) 8

x C

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Transformations 2 You need to be able to spot transformed functions from their equations, and sketch transformations involving asymptotes.

Functions and equations Curve C below has equation y = x 3 − 3x 2 − 2. You can sketch the curves of other equations by transforming curve C. y = x 3 – 3x 2 – 2 y C 0 –1 2 x (1, – 4)

y = f(x )

y = ( x + 2)3 – 3( x + 2)2 – 2 y

–3

0

The diagram shows a sketch of the curve with equation y = f(x) where 2x f(x) = _____, x ≠ −1 x+1 y

x

y52

(–1, – 4)

O

y = f(x + 2) y = (x + 2)3 − 3(x + 2)2 − 2

x

x 5 21

Asymptotes An asymptote is a line which a curve approaches, but never reaches. You draw asymptotes on graphs with dotted lines. y

The curve has asymptotes with equations y = 2 and x = −1 (a) Sketch the curve with equation y = f(x + 2) and state the equations of its asymptotes. (3 marks) y

y =5

y =2

y = f(x )

0

x

This curve has an asymptote at y = 5. When you transform a graph, its asymptotes are transformed as well. Transformation y = f(x ) − 1 y = 2f(x ) y = f(x + 4)

New asymptote y=4 y = 10 y=5

The graph is translated 4 units to the left, so the horizontal asymptote does not change.

–2 0

x

x = –3

(b) Find the coordinates of the points where the curve in part (a) crosses the coordinate axes. (3 marks) 2(x + 2) 2x + 4 f(x + 2) = ___________ = _______ (x + 2) + 1 x+3 _4_ When x = 0, y = 3 _4_ (−2, 0) and (0, 3)

y

The diagram shows a curve C with equation y = f(x), where (x + 2)2 f(x) = _______, x ≠ −1 x+1 (a) Sketch the curve with equation y = f(x + 1) and state the new equation of the asymptote x = −1 (3 marks) (b) Write down the coordinates of the points where the curve meets the coordinate axes. (3 marks)

C

22

O

x

x 5 21

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Reciprocal graphs

k k You need to know how to sketch the graphs of y = _x_ and y = ___2 , and transformations of these x graphs.

Shapes and asymptotes The shapes of the reciprocal graphs are different for positive and negative values of k:

1

2

k>0 y

6 The figure shows a sketch of the curve y = _x_ y

k 0 and the angle PRQ = 90°. Find the coordinates of R. (4 marks) R lies on the circle because ∠PRQ = 90°. y

x =6

P (2, 1) 0

R x

It can help to draw a sketch.

Q

x = 6 so (6 – 2)2 + (y – 1)2 = 25 (y – 1)2 = 9 y = –2 or 4 a > 0 so R has coordinates (6, 4).

! A tangent is perpendicular to a radius, so the straight line through A and B is perpendicular to L. Remember that if a line has gradient m, then any line perpendicular to it will have 1 gradient − __ m. Look at page 18 for a recap.

You will need to use problem-solving skills throughout your exam – be prepared!

21

The circle C has centre (5, –2) and radius 10 (a) Write down an equation for C. (2 marks) (b) Verify that the point (–1, 6) lies on C. (1 mark) (c) Find an equation of the tangent to C at the point (–1, 6), giving your answer in the form ax + by + c = 0, where a, b and c are integers. (4 marks)

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Solving circle problems You can solve the equations of a circle and a straight line simultaneously to find the points of intersection. A straight line will intersect a circle once, twice or not at all. If a straight line just There is more about solving touches a circle then it is a tangent to the circle. simultaneous equations on page 9.

! The circle C has centre the origin and radius 3. The straight line with equation 2x + y = k, where k is a positive constant, is a tangent to C. Find the exact value of k. (6 marks)

You can use the discriminant to solve this problem. You need to: 1. Write out the equation of the circle. 2. Substitute y = k 2 2x into this to solve both equations simultaneously. y 3. Find the discriminant of the resulting k quadratic and 3 set it equal to 0. 4. Solve to find 0 the value of k. You are told that 2x + y k is positive so –k ignore the negative square root of 45.

y = k – 2x 1 2 2 x +y =9 2 Substitute 1 into 2 : x2 + (k – 2x)2 = 9 x2 + 4x2 – 4kx + k2 = 9 5x2 – 4kx + k2 – 9 = 0 Using the discriminant: b2 – 4ac = 0 (– 4k)2 – 4(5)(k2 – 9) = 0 16k2 – 20k2 + 180 = 0 k2 = 45 __ k = 3√ 5

x

=k

You will need to use problem-solving skills throughout your exam – be prepared!

Circumcircles For any triangle, you can draw a unique circle which passes through all three vertices. This is called the circumcircle of the triangle. Q

B

C A

O

ABC is a right-angled triangle, so AC is a diameter of the circumcircle. Find the midpoint of the hypotenuse to find the centre of the circle.

P

For any triangle, PQR, the perpendicular bisectors of the sides will intersect at the centre of the circumcircle.

O

R

There is more about properties of chords on page 21.

y

1 Find the coordinates of the points where the line with equation y = x + 7 intersects the circle with equation x2 + (y + 2)2 = 45. (5 marks) 2 The line with equation 2x 2 y + 2 = 0 intersects the circle with centre (k, 0) and radius 2 at two distinct points. Find the range of possible values of k, giving your answer in surd form. (7 marks)

3 A circle passes through the points (0, 0), (0, 10) and (8, –6), as shown in the diagram. Find an equation for the circle. (7 marks)

10 x

O (8, 26)

The perpendicular bisectors of two chords will intersect at the centre of the circle.

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The factor theorem In your exam, you might need to use the factor theorem to help you factorise polynomials like these: f(x) = 2x3 + 5x2 – 15x + 10

The factor theorem If f(x) is a polynomial and f(p) = 0, then (x – p) is a factor of f(x). Only use this theorem with polynomials. Watch out for the sign. If f(–1) = 0 then the factor would be (x + 1). Learn this rule – it’s not in the formulae booklet.

This is a cubic because the highest power of x is 3.

f(x) = x 4 – 4x3 + 2x2 – 3 This is a quartic because the highest power of x is 4.

Synthetic division with polynomials

To completely factorise f(x) = 3x3 – 17x2 + 2x + 40: 1. Use the factor theorem to find one factor: f(5) = 0, so (x – 5) is a factor. 2. Use synthetic division to divide 3x3 – 17x2 + 2x + 40 by (x – 5):

If you have to completely factorise a cubic polynomial, you will usually need to find three linear factors. You can find the first one using the factor theorem. When you take this factor out, your other factor will be a quadratic expression. One quick way to find this is by using synthetic division.

To divide by (x – p), write p here.

+

5

3

–17 2 40 + 15 –10 –40 = = 3 –2 –8 0

Write the coefficient of x3 on the bottom row, then multiply by 5.

f(–3) = 2(–3)3 – (–3)2 – 15(–3) + 18 = –54 – 9 + 45 + 18 = 0 So (x + 3) is a factor.

(4 marks)

(b) Factorise f(x) completely.

Using synthetic division to divide f(x) by (x + 3):

2

–1

–15

18

–6

21

–18

–7

6

0

f(x) = (x + 3)(2x 2 – 7x + 6) = (x + 3)(2x – 3)(x – 2)

1 (a) Use the factor theorem to show that (x – 2) is a factor of x3 – 7x2 – 14x + 48 (2 marks) 3 2 (b) Factorise x – 7x – 14x + 48 completely. (4 marks)

Be careful with the sign (+ or –). The factor theorem says that (x – p) is a factor if f(p) = 0, so you need to evaluate f(–3). You need to write down that (x + 3) is a factor at the end of your working. You know that (x + 3) is a factor, so divide f(x) by (x + 3) using synthetic division. You could check your answer by expanding the brackets: (x + 3)(2x 2 − 7x + 6)

2x 3 – x 2 – 15x + 18 So ___________________ = 2x 2 – 7x + 6 x+3

23

Add the result to the coefficient of x 2, and so on …

These are the coefficients in the quadratic factor. If (x – p) is a factor of f(x ), this number will be zero.

3. The quadratic factor is 3x 2 – 2x – 8. Factorise this to complete the factorisation. 4. So 3x3 – 17x2 + 2x + 40 = (x – 5)(3x2 – 2x – 8) = (x – 5)(3x + 4)(x – 2)

f(x) = 2x3 – x2 – 15x + 18 (a) Use the factor theorem to show that (x + 3) is a factor of f(x). (2 marks)

–3 2

Write the coefficients of f(x ) on the top row.

= 2x 3 − 7x 2 + 6x + 6x 2 − 21x + 18 = 2x 3 − x 2 − 15x + 18 ✓ You need to factorise (2x 2 – 7x + 6) in the normal way. Look at page 2 for a reminder.

2 f(x) = 2x3 – 3x2 – 65x – a (a) Given that (x + 4) is a factor of f(x), find the value of a. (3 marks) (b) Factorise f(x) completely. (4 marks)

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Binomial expansion 1 The binomial expansion is a formula that lets you expand brackets easily. This is how the binomial expansion will appear in the formulae booklet in your AS exam:

n a n − 1b + n a n − 2b 2 + … + ( n ) a n − rb r + …+ b n (n [ ℕ) (a + b)n = a n + r 2 1 ! n __ n ______ The expansion is valid as long where ( r ) = n Cr = r !(n – r)! This means n factorial. as n is a positive integer. n ! = n × (n – 1) × … × 3 × 2 × 1

( )

( )

( nr ) or nCr means n choose r. Use the

function on your calculator to work it out.

Find the first 3 terms, in ascending powers of x, of the binomial expansion of (2 – 3x)6. Give each term in its simplest form. (4 marks) a = 2, b = –3x 6 5 6 4 2 (a + b)6 = a6 + a b+ a b +… 1 2 6 6 (2 – 3x)6 = 26 + × 25 × (–3x) + × 24 × (–3x)2 + … 1 2 = 64 – 576x + 2160x 2 + …

( ) ( )

( )

( )

! b = px, so in the third term you need to square all of px. (px)2 = p 2x 2. 10 or 10C2 on your To work out 2 calculator,

( )

type 10

2

Be careful if either a or b is negative. Always use brackets if you are substituting anything more complicated than a positive whole number.

.

Make sure you simplify each term as much as possible. Don’t leave any powers of numbers or multiplication signs in your final answer. You will need to use problem-solving skills throughout your exam – be prepared!

(a) Find the first 3 terms, in ascending powers of x, of the binomial expansion of (1 + px)10, where p is a non-zero constant. Give each term in its simplest form. (2 marks) a = 1, b = px 10 9 10 8 2 (a + b)10 = a 10 + a b+ a b +… 1 2 10 × 19 × px (1 + px)10 = 110 + 1 10 + × 18 × (px)2 + … 2 = 1 + 10px + 45p 2x 2 + …

( ) ( ) ( )

( )

(b) Given that the coefficient of x2 is 9 times the coefficient of x, find the value of p. (2 marks) 45p 2 = 9(10p) so 45p 2 – 90p = 0 45p (p – 2) = 0 p=0 p=2

1 Find the first 4 terms, in ascending powers of x, of each of these binomial expansions, giving each term in its simplest form. (3 marks) (a) (1 + 3x)9 4 (4 marks) (b) (2 + 5x) 12 (4 marks) (c) (3 – x)

2 (a) Find the first 3 terms, in ascending powers of x, of the binomial expansion of (2 + kx)5, where k is a constant. (4 marks) (b) Given that the coefficient of x is 48, find the value of k. (2 marks) (c) Write down the coefficient of x2

(1 mark)

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Solving binomial problems You can use the binomial expansion to make approximations or to solve harder problems.

You can use this part of the binomial expansion of (a + b)n in the formulae booklet to find the x r term without finding every term up to it:

Find the coefficient of x5 in the binomial expansion x 9 of 6 2 __ (2 marks) 3 x a = 6, b = – __, n = 9 so x r term = ( n )a n – rb r r 3 5 x 9 __ x 5 term = × 64 × – 5 3 1 = 126 × 1296 × – _____ x 5 243 = – 672x 5 The coefficient of x 5 is – 672.

(

)

( )

( )

(

n n–r r … + ( r )a b + … The x part of each term comes from 5 x __ b = – , so use r = 5 to get the x 3 term. Be careful with the fraction part: 1 5 __ x 5 – __ = – 3 x 5 3

)

( )

( )

Binomial approximations You can use a binomial expansion to estimate values. This is especially useful if x is small. If x is less than 1, then larger powers of x get smaller. By ignoring large powers of x you can find a simple approximation. For example: (1 + x)100 < 1 + 100x + 4950x2 This means ‘is approximately equal to’.

d to Write down the value of x you nee wer substitute. You can check your ans 6 = 1.340 095… ✓ with a calculator. 1.05

The first 4 terms of the binomial expansion of x 6 1 + __ are given below. 2 5 x 6 15 1 1 __ = 1 + 3x + ___x2 + __x3 + … 2 4 2 Use the expansion to estimate the value of (1.05)6 (3 marks) x x 6 If x = 0.1, then __ = 0.05 and 1 + __ = (1.05)6 2 2 15 5 (1.05)6 = 1 + 3 × (0.1) + ___ × (0.1)2 + __ × (0.1)3 + … 4 2 = 1 + 0.3 + 0.0375 + 0.0025 + … 6 (1.05) < 1.34

( (

) )

The first 4 terms of the binomial expansion of (2 – x)7 are given below. (2 – x)7 = 128 – 448x + 672x2 – 560x3 + … If x is small, so that x2 and higher powers can be ignored, show that (1 + x)(2 – x)7 < 128 – 320x (2 marks) (1 + x)(128 – 448x + …) = 128 + 128x – 448x – 448x2 + … = 128 – 320x ignoring x2 and higher powers

2 You are going to ignore x and higher powers, so you only need to consider the first two terms of the expansion of (2 – x)7. All the other terms would give you x 2 or higher terms when the brackets are multiplied out.

25

(

)

1 In the binomial expansion of (1 + 2x)30, the coefficients of x3 and x 4 are p and q respectively. (a) Show that p = 32 480 (1 mark) q (b) Find the value of __ (2 marks) p 2 (a) Find the first 4 terms, in ascending powers of x, of the binomial expansion x 12 (4 marks) of 1 1 __ 4 (b) Use your expansion to estimate the value of (1.025)12, giving your answer to 4 decimal places. (3 marks)

()

(

)

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Proof If you need to prove a statement in your exam, you need to construct a logical argument to show that it is always true. You need to know these two methods of proof:

1

Proof by deduction

Also called direct proof. You use known facts and follow logical steps to reach a conclusion.

2

Proof by exhaustion

You consider each of the possible cases separately, in order to show that something is true in every case.

Proof checklist x and y are rational numbers. Prove that the mean of x and y is also a rational number. (3 marks) c a Let x = __ and y = __, where a, b, c and d d b are integers. The mean of x and y is: x+y ad + bc ad ___ bc 1 _a_ 1 ___ ________ ______ _c_ __ = __ 2 ( b + d ) = 2 ( bd + bd ) = 2bd 2 Since a, b, c and d are integers, (ad + bc) x+y and 2bd must also be integers, so ______ 2 is a rational number.

Write down any rules, information or assumptions you need to use. Show each step of your working clearly. Make sure your steps follow logically from each other. Write down what you have proved at the end of your working.

Rational numbers are numbers tha t can be p written in the form __, where p and q are q integers.

! You can prove this result by considering the cases where x and y are positive or negative separately. You know you have covered all possible cases because at least one of the following statements must be true: • x is negative • y is negative • both x and y are non-negative. Remember to write down what you have proved at the end of your working. You will need to use problem-solving skills throughout your exam – be prepared!

Prove that (xy + 1)2 + (x 2 1)2 + (y 2 1)2 > 1 for all real values of x and y. (4 marks) Each of (xy + 1)2, (x – 1)2 and (y – 1)2 is > 0 for all real values of x and y. If x < 0 then (x – 1)2 > 1 If y < 0 then (y – 1)2 > 1 If x and y are both > 0 then (xy + 1)2 > 1 All three terms are > 0 and at least one must be > 1, so the whole expression must be > 1, as required.

1 f(x) = 2x3 + 5x2 2 4 A student is attempting to use the factor theorem to prove that (x + 2) is a factor of f(x). The student writes the following working: f(22) = 2(22)3 + 5(22)2 2 4 = 216 + 20 2 4 =0 (a) Explain why this proof is incomplete. (1 mark) (b) Write an additional line of working to complete the proof. (1 mark)

2 (a) Prove by exhaustion that 2n2 + 11 is a prime number for all n ∈ ℕ, n < 10. (2 marks) (b) By means of a counterexample disprove the statement: n2 + n + 17 is a prime number for all n ∈ ℕ. (2 marks) 2 3 Prove that (4 2 x) > 7 2 2x for all real values of x. (3 marks)

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Cosine rule The cosine rule applies to any triangle. You usually use the cosine rule when you know two sides and the angle between them (SAS) or when you are given three sides and you want to work out an angle (SSS).

1

a2 = b2 + c2 – 2bc cos A Use this version to find a missing side.

2

sin21(0.5) SHIFT ALPHA

1 6

π

SETUP ON

Your calculator might give angles in radians in terms of p.

On some calculators you need to press SHIFT and SETUP to change between degrees and radians mode.

In the triangle ABC, AB = 15 cm, BC = 9 cm and CA = 10 cm. Find the size of angle C, giving your answer to the nearest degree. (3 marks) B c 15 cm A

A c

B

c 2.1 m

Everything in blue is part of the answer.

a

p 4

A

1.7 m b

D

C

The diagram shows a triangle ABC and a sector BCD of a circle with centre C. (3 marks) Find the length of BC. 2 2 2 a = b + c – 2bc cos A p BC2 = 1.72 + 2.12 – 2 × 1.7 × 2.1 × cos __ 4 = 2.2512… BC = 1.50 m (3 s.f.)

You know two sides and the angle between them (SAS) so you can use the cosine rule to find the opposite side. If the angle on the diagram is given in terms of π then it is in radians. Make sure your calculator is set π to radians before working out cos 4.

a 9 cm

10 cm C b

a 2 + b 2 – c2 cos C = ____________ 2ab 2 9 + 102 – 152 = ______________ = –0.2444… 2 x 9 x 10 –1 C = cos (–0.2444…) = 104°

If no diagram is given in the question you can sketch one. You know three sides (SSS) so you can use the cosine rule to find any angle in the triangle. Be careful with the order. You add the squares of the sides adjacent to the angle, and subtract the square of the opposite side.

The diagram shows two triangles PQR and PRS. /RSP = 0.8 radians. Find (a) the length of PR (3 marks) (b) the size of /PQR, giving your answer in radians to 3 significant figures. (3 marks)

27

a

B

In AS exams you only need to be able to work in degrees, but in A-level exams you need to be able to work in degrees or radians. This calculator is in radians mode.

2

This version is useful for finding a missing angle.

Degrees or radians?

You can revise radian measure on page 76.

2

C

b

b +c –a cos A = ____________ 2bc 2

Q

20 cm

R 15 cm

13 cm

0.8 rad P

26 cm

S

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Sine rule You need to learn the sine rule for your exam. It applies to any triangle. The sine rule is useful when you know two angles, or when you know a side and the opposite angle.

1

2

b c = _____ = _____ sin A sin B sin C

a _____

This version is useful for finding a missing side.

sin C sin B = _____ = _____ b c

b

Use this version to find a missing angle.

A

sin A _____

a

C a

c

B

Using radians In the triangle PQR, Q PQ = 4 cm, /PQR = 1.5 radians and 1.5 rad /QPR = 0.8 radians. 4 cm Find the length of the side PR. 0.8 rad (3 marks)

Most length and area problems in your exam will use angles measured in radians. Remember these key facts: Angles in a triangle add up to p radians. p A right angle is __ radians. 2

R

There is more about measuring angles in radians on page 76.

P

/QRP = p – 1.5 – 0.8 = 0.8415… 4 PR _______ = ______________ sin (1.5) sin (0.8415…) 4 × sin (1.5) PR = ______________ = 5.35 cm (3 s.f.) sin (0.8415…)

The sine rule uses opposite sides and angles, so use the fact that the angles in a triangle add up to π radians (or 180°) to work out /QRP first.

Two values If you know sin x, you might be asked to find two possible values for x. Use this rule for angles measured in radians: sin x = sin (p – x) There is more about this on pages 30 and 33.

This sketch shows you why there are two possible values of x : B

10 cm 0.2

A

x2 C2

6 cm

6 cm

x1 C1

In the triangle ABC, AB = 10 cm, BC = 6 cm, /BAC = 0.2 radians and /ACB = x radians. (a) Find the value of sin x, giving your answer to 3 decimal places. (3 marks) sin (0.2) = ________ 10 6 10 × sin (0.2) _____________ = 0.331 (3 d.p.) sin x = 6 (b) Given that there are two possible values of x, find these values of x, correct to 2 decimal places. (3 marks) sin x ____

x1 = sin–1 (0.331) = 0.34 radians (2 d.p.) x2 = p – x1 = p – 0.34 = 2.80 radians (2 d.p.)

A

p In the triangle ABC, BC = 13 cm, /ABC = __ radians, and 3 p /ACB = __ radians. Find the length of AC. (3 marks) 5

Start by finding the size of /CAB.

π 3

π 5

C

13 cm

B

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Trigonometric graphs You need to be able to sketch the graphs of sin, cos and tan, and transformations of them. If you want to recap transformations of graphs, have a look at pages 13 and 14.

y = sin x and y = cos x

y = tan x

y 1

–180 –π

–90 π –2

0

y 3 2 1

y = cos x

90

180 π

π 2

360 x 2π Degrees Radians

270 3π 2

–1

y = tan x

−180 −90 0 −1 π −π − 2 −2 −3

y = sin x

90

180 π

π 2

270 3π 2

360 x 2π

Sketching trig graphs If you have to sketch a trigonometric graph in your exam, make sure you: pay attention to the range of values for x in the question use radians if the range is given in radians p label multiples of __ or 90° on the 2 x-axis put a scale on the y-axis to show the max and min for sin x and cos x draw the asymptotes for tan x.

(

(

y O

P

Q

x

Given that the curve cuts the 6p 16p x-axis at the points P ___, 0 and Q ____, 0 , 5 5 find a and b. (4 marks) 6p 16 p ___ ____ sin a – b = 0 and sin a –b =0 5 5 6p 1 a ___ – b = 0 5 16p a ____ – b = p 2 5 1 2 – 1 : 2pa = p so a = __ 2 3p 1 6p Substituting into 1 : __ ___ – b = 0 so b = ___ 2 5 5

(( ) ) ( ) ( )

(

)

((

( )

29

(

π 2

π

3π 2

π 6

)

2π x

b) Write down the exact coordinates of the points where the graph meets the coordinate axes. (3 __ marks) __ √3 √3 p When x = 0, y = cos __ = ____, so 0, ____ 6 2 2 p When y = 0, 0 = cos x + __ : 6 p p _p_ _p_ – = __, so __, 0 2 6 3 3 4p 4p 3p p and ___ – __ = ___, so ___, 0 2 6 3 3

)

The diagram shows a sketch of y = sin (ax – b), where a > 0 and 0 < b < 2p.

)

–1

π You can write cos 6 exactly as a surd. π The graph of y = cos x + 6 is a translation of y = cos x. The graph units to the left. moves π 6

(

y = cos ( x +

(a) Sketch, for 0 < x < 2p, y 1 the graph of _p_ y = cos x 1 6 0 (2 marks)

(

(

)

(

)

)

)

If sin (ax – b) = 0, then ax – b = 0, or π, or 2π and so on. You can use these facts to write two equations and solve them simultaneously to find a and b.

)

) )

(a) On separate diagrams, sketch, for 0 < x < 2p, the graphs of (i) y = sin (2x) p (ii) y = tan x + __ (4 marks) 2 (b) Write down the coordinates of any points where the curves meet the coordinate axes and the equations of any asymptotes. (6 marks)

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Trigonometric equations 1 You can solve an equation involving sin, cos or tan. You need to be really careful because these equations can have multiple solutions. You will be given a range (or interval) of values for x. This could be in degrees or radians. You need to find values of x that are in that range.

Using graphs to find solutions

1 This graph shows the solutions to the equation cos x = – __ in the range –180° < x < 360°. 2 y y = cos x is symmetrical about the y -axis, so this solution is minus the principal value.

( )

1

120°

–120° –180°

–90°

Subtract the principal value from 360° to get this solution: 360° – 120° = 240° y = cos x

Your calculator will give you this principal value when you work out cos–1 – 21 .

0

90°

240° 180°

270°

x

360°

y = – 0.5

–1

Using a CAST diagram Solve 3 tan x = 5 in the interval 0 < x , 360°. Give your answers to 1 decimal place. (3 marks) 5 tan x = __ Work to 3 d.p. 3 then round your 5 tan–1 __ = 59.036…° final answer. 3

A CAST diagram tells you which trigonometric ratios are positive in which quadrant.

( )

y 3

y=

2 1

0 –1

5 3

y = tan x

90°

180°

270°

360° x

–2 –3

180° + 59.036…° = 239.036…° x = 59.0°, 239.0° (1 d.p.)

1 (a) Sketch the graph of y = sin x in the interval 0 < x < 360°. (2 marks) (b) Find the values of x in the interval 0 < x < 360° for which sin x = 20.3 Give your answers correct to 1 decimal place. (3 marks)

Only sin positive

180°

Only tan positive

All three positive

90°

S

A

T

C 270°

0

Only cos positive

Use your calculator to find the principal value of x. You can find the other solution by sketching a graph or by drawing straight lines like this on a CAST diagram. You know that tan x is positive, so the other solution must be in the third quadrant. 180° – 59.0°✗

Tan x is only positive for angles in the first and third quadrants. So you can reject the angles in the second and 360° − 59.0°✗ fourth quadrants.

59.0°✓

S

A

T

C

180° + 59.0°✓

2 Solve, for 2p < u , p, the equation (3 marks) (a) 3 cos u = 1 (3 marks) (b) tan u + 2 = 0

There will be two solutions to each equation. Find one using your calculator, then sketch the graph to find the other.

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Trigonometric identities 1 You might need to use one of these two trigonometric identities to simplify a trig equation before solving it. They are true for all values of x or u.

1

sin u tan u ≡ _____ cos u

2

sin2 u means (sin u)2.

sin2 u + cos2 u ≡ 1

Quadratic equations If an equation involves sin2 u and sin u (or cos2 u and cos u) then it is a quadratic. You might be able to solve it by factorising: 2 sin2 u – 3 sin u – 2 = 0 (2 sin u + 1)(sin u – 2) = 0 1 sin u = – __ 2

sin u = 2

No solutions exist to sin u = 2, so you would 1 only solve sin u = – __ 2.

(a) Show that the equation 5 cos x = 1 + 2 sin2 x can be written in the form (2 marks) 2 cos2 x + 5 cos x – 3 = 0 5 cos x = 1 + 2 sin2 x 5 cos x = 1 + 2(1 – cos2 x) 5 cos x = 1 + 2 – 2 cos2 x 2 cos2 x + 5 cos x – 3 = 0 (b) Solve this equation for 0 < x < 360°. (3 marks)

Golden rules

When finding solutions to quadratic trigonometric equations, remember these golden rules:

1 2

Write everything in terms of sin2 u and sin u (or cos2 u and cos u). Solutions to sin x = k and cos x = k only exist if –1 < k < 1. Solutions to tan x = k exist for any value of k.

2 You can use sin2 x + cos x = 1 to rewrite 2 sin2 x in terms of cos x : sin2 x = 1 – cos2 x You can then rearrange to get a quadratic equation in cos x. You might find it easier to factorise if you write it as: 2C2 + 5C – 3 = 0 ⇒ (2C – 1)(C + 3) = 0 If cos x + 3 = 0 then cos x = –3, which has no solutions, so you can ignore the second factor.

(2 cos x – 1)(cos x + 3) = 0 1 cos x = __ 2

cos x = –3

( )

1 cos–1 __ = 60° 2 –60° + 360° = 300° x = 60°, 300°

Given that sin u = 4 cos u, find the value of tan u. (1 mark) sin u _____ = 4 so tan u = 4 cos u

sin x _____ Start by writing tan x as cos x

1 Find all the solutions, in the interval 0 < x , 2p, of the equation 3 cos2 x – 9 = 11 sin x giving each solution correct to 3 decimal places. (6 marks)

Use cos2 x = 1 – sin2 x to get a quadratic equation in sin x.

31

2 (a) Show that the equation 5 sin x = 2 tan x can be written in the form sin x(5 cos x – 2) = 0 (2 marks) (b) Solve, for 0 < x , 360°, 5 sin x = 2 tan x (4 marks)

Either sin x = 0, or 5 cos x – 2 = 0. Both these factors will give you solutions.

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Trigonometric equations 2 You need to be careful if a trigonometric equation involves a function of x or u.

Find the exact solutions of the equation cos (u − 50°) = 0.5 (3 marks) in the interval 0 < u < 180°. 0 < u < 180° so –50° < u – 50° < 130° Let Z = u – 50° cos Z = 0.5, –50° < Z < 130° in range ✓ So Z = cos–1 (0.5) = 60° not in range ✗ or Z = –60° not in range ✗ or Z = 360° – 60° = 300° So u – 50° = 60° So u = 110°

The graph of y = sin (3x) is a horizontal stretch of the graph of y = sin x with scale __ 1 factor 3. This sketch shows the solutions of the equation. y 1

y = π 12

0

π 4

3π 4

11π 12

π x

Make sure you remember to transform your solutions for 3x back into solutions for x at the end, and double-check that they all lie within 0 < x < π.

1 Solve, for 0 < x , 360° 1 (a) sin (x − 40°)__= − __ 2 √3 ___ (b) cos (2x) = 2 2 Find all the solutions of the equation p 1 cos2 x + __ = __ 4 6 in the range −p < x < p.

(

)

The safest way to solve an equation involving sin, cos or tan of (x + b) or (ax) is to transform the range. –50°

0 < θ < 180°

–50°

–50° < θ – 50° < 130° If Z = u – 50°, you need to find all the values of Z such that cos Z = 0.5 in the range –50° < Z < 130°. You can then find the corresponding value of u for each solution.

1__ Solve, for 0 < x < p, the equation sin (3x) = ___ , √2 giving your answers in terms of p. (6 marks) 0 < x < p so 0 < 3x < 3p Let Z = 3x 1__ , 0 < Z < 3p sin Z = ___ √2 1__ p in range ✓ So Z = sin–1 ___ = __ 4 √2 9p p in range ✓ or Z = __ + 2p = ___ 4 4 17p p not in range ✗ or Z = __ + 4p = ____ 4 4 3p p or Z = p – __ = ___ in range ✓ 4 4 11p p or Z = p – __ + 2p = ___ in range ✓ 4 4 19 p _p_ ____ or Z = p – + 4p = not in range ✗ 4 4 3p 9p 11p p So 3x = __ or ___ or ___ or ___ 4 4 4 4 3p 11 p p ___ ___ _p_ ___ So x = or or or 12 4 4 12

( )

1 2

y = sin 3 x

–1

Transforming the range

(4 marks) (4 marks)

(6 marks)

You need to consider the positive and negative square roots separately. This is like solving two separate equations: 1 __ _π_ 1 __ _π_ cos(x + ) = and cos(x + 6 ) = – 2 2 6

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Vectors In AS exams you only need to be able to work with 2-dimensional vectors, but in A-level exams you might need to answer questions involving 3-dimensional vectors. These can be written as column vectors, or using i, j, k notation: 2 ⟶ X Y = −1 = 2i − j + 6k ( 6)

Position or direction? z

It is useful to distinguish between position vectors and direction vectors.

B

0

y AB

x

OA A

A position vector starts at the origin. OA tells you the position of point A .

The direction vector AB tells you the direction and distance from A to B.

The points P and Q have position vectors 3i + 4j and −i + 5j respectively. ⟶ (2 marks) (a) Find the vector PQ . ⟶ ⟶ ⟶ PQ = OQ − OP = (−1 − 3)i + (5 − 4)j ~ ~ = −4i~ + j ~ (b) Find the distance PQ. (1 mark) ________ ⟶ |PQ| = √42 + 12 ___ = √ 17 (c) Find a unit vector in the direction of PQ. (1 mark) ⟶ 1 1 ____ ____ ___ PQ = ___ (−4i + j ) ~ ~ √ 17 √ 17 4 1 ____ ___ i + ___ j = −____ √ 17 ~ √ 17 ~

Given that the point P has position vector 2i − 5j + k and the point Q has position vector 6i − 2k, ⟶ (2 marks) (a) find the vector PQ

33

()

0 k= 0 1

O i=

j=

()

() 0 1 0

1 0 0

i, j and k are perpendicular unit vectors. In 2D, you just use i and j.

Magnitude You can find the magnitude of a vector using Pythagoras’ theorem.

| |

2 −4 = |2i − 4j + 5k| ( 5 ) __________ __ = √2 2 + 4 2 + 5 2 = 3 √5

|⟶ AB| =

Ignore minus signs when calculating the magnitude of a vector.

Unit vectors have magnitude 1. The distance between two points A and B ⟶ is the magnitude of the vector AB.

do These are 2D vectors. You would e way sam the the calculations in exactly ak be for 3D vectors, but there would component as well.

You can’t write in bold in your exam! You can underline vectors to make them clearer. If you’re writing the vector between two points, you should ⟶ draw an arrow over the top. PQ is the direction vector from P to Q, whereas PQ is the line segment between P and Q.

⟶ (b) find the exact value of |PQ |, giving your answer as a simplified surd. (2 marks) (c) write down a unit vector in the direction ⟶ (1 mark) of |PQ |.

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Solving vector problems You can use these formulae to find areas of triangles and parallelograms in vector questions:

1 2

The area of the parallelogram is twice the area of the triangle.

1 Area of triangle ABC = __ |a||b|sin θ 2

b

A

D

a

B

Area of parallelogram ABCD = |a||b|sin θ

θ

a

C

b

! ⟶ 3 Three points are such that AB = ( ) and 6 ⟶ −4 AC = ( ). Find the area of the triangle ABC. 1 (6 marks) ⟶ ⟶ ⟶ CB = CA + AB ⟶ ⟶ = − AC + AB 7 3 4 =( )+( ) = ( ) 6 5 − 1 ________ ___ 2 2 √ AB = 3 + 6 = √ 45 _______ 2 √ 2

B

|CB| |AB | C

|AC |

θ A

You will need to use problem-solving skills throughout your exam – be prepared!

___

AC = 4 + 1 = √ 17 ________

Draw a sketch to help you visualise the problem. This will help you plan your strategy before you start. ⟶ ⟶ 1. Find CB (or BC) then find the lengths of all three sides of the triangle. 2. Use the cosine rule to find the size of one angle. 1 3. Use Area = _2 ab sin θ to find the area of the triangle.

___

CB = √ 7 2 + 5 2 = √ 74 b2 + c2 – a2 cos A = ____________ 2bc 45 + 17 – 74 2 ___ ___ ___ = − _____ cos θ = ______________ √ 85 2√ 45 √ 17 θ = 102.528…°

Parallel vectors

If one vector can be written as a multiple of the other then the vectors are parallel.

___ ___ 1 1 __ √ 45 √ 17 sin 102.528… ° ab sin θ = Area = __ 2 2

= 13.5

1 The points A, B and C are such that 5 7 ⟶ ⟶ AB 5 6 , and AC 5 0 (4) (− 2) (a) Show that ABC is an isosceles triangle. (3 marks) (b) Find the area of triangle ABC. (4 marks)

ns for each of the vectors expressio Find→ → ⟶ and HA, JS and KT in terms of a and b, tiples mul show that they can be written as of each other.

p

A

D

1p 2

B

C

M

In this parallelogram M is the midpoint of DC. ⟶ 1 ⟶ AB is parallel to DM so DM = __ 2 AB

2 In triangle OAB, M is the midpoint of OA. OH 5 HJ 5 JK 5 KB B S and T divide the line AB K into three equal segments. T ⟶ ⟶ J OM 5 a and OH 5 b b O

S

H a

M

A

⟶→ ⟶ (a) Prove that HA, JS and KT are all parallel. (6 marks) (b) State the ratio KT : JS : HA (1 mark)

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Differentiating from first principles You can find the gradient of a curve at a point A by considering the gradient of the chord AB as B gets closer to A. y

y = f(x ) B

The gradient of AB is f(x0 + h ) − f(x0) _______________ h The gradient of the curve at A is the same as the gradient of the tangent to the curve at this point.

As point B gets closer to point A, the gradient of the chord gets closer and closer to the gradient of the curve at point A.

A

0

x0

x0 + h

x

As B gets closer to A, h gets closer to 0. You write h → 0.

First principles formula Prove from first principles that the derivative (4 marks) of 4x3 is 12x2. f(x) = 4x3 f(x + h) − f(x) _____________ f9(x ) = hlim →0 h 4(x + h)3 − 4x3 ______________ = hlim →0 h 3 4( x + 3x 2h + 3xh 2 + h 3) − 4x 3 _______________________________ = hlim →0 h 2 2 12 x h + 12 xh + 4h 3 ___________________ = hlim →0 h 2 = hlim (12 x + 12 xh + 4h 2 ) →0 As h → 0, then 12xh → 0 and 4h 2 → 0. So f9(x) = 12x 2, as required.

1 Given that f(x) = x2, use differentiation from first principles to find f 9(7). (5 marks)

The question states ‘first princip les’ so you need to use the formula, or writ e an expression for the chord AB. Make sure you write out which terms in the expres sion → 0 as h → 0.

35

You can solve first principles problems using this formula, which is found in the formulae booklet: First Principles f(x + h) – f(x) ____________ f9(x) = lim h→0 h

! Substitute into the first principles formula. If f(x) = 4x3 then f(x + h) = 4(x + h)3. Multiply out the brackets and simplify the fraction. Any terms with a positive power of h will tend towards 0 as h tends towards 0. You will need to use problem-solving skills throughout your exam – be prepared!

2 The points A and B with x-coordinates 5 and 5 + h respectively lie on the curve with equation y = 2x2 + 3x (a) Show that the gradient of AB is 23 + 2h. (3 marks) (b) Deduce the gradient of the tangent to the curve at A. (1 mark)

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Differentiation 1 You can differentiate a function to find its derivative or gradient function. dy The derivative is written as f9(x) or ___ dx

Differentiating xn y = f(x )

y =x

Differentiation dy = f9(x ) dx

Golden rules

n

1

Multiply by the power …

2

dy … then reduce = nx n − 1 the power by 1 dx

y = 3x − 8 + x d y ___ = 18x 5 − 3x −4 dx 6

Differentiate with respect to x __ x (a) x3 − 3√x + __ 7 1 __ 1 3 2 f(x) = x − 3x + __ 7x 3 −__1 1 f9(x) = 3x2 − __x 2 + __ 7 2 kx + 5 (b) ______ x2 5 kx f(x) = ___2 + ___2 = kx−1 + 5x−2 x x f9(x) = −kx−2 − 10x−3

2

1 Given that y =

(3 marks)

−3

(x + 3) _______ x

, x ≠ 0, find

(3 marks)

(3 marks)

dy ___

dx (4 marks)

Multiply out the brackets, then write the −1 function in the form y = ax + b + cx before differentiating.

_

6 ___

1 __

√x → x2

x2

→ 6x −2

Constant terms differentiate to zero, and x terms differentiate to a constant.

f(x) = 7 → f9(x) = 0

This rule works for any value of n, including fractions and negative numbers.

dy 1 Given that y = 3x6 − 8 + __3 , x ≠ 0, find ___ x dx

Write every term in a polynomial in the form ax n before differentiating.

f(x) = 3x + 1 → f9(x) = 3

1 __ −3 Start by rewriting x 3 as x . Remember that you are multiplying by −3 when you differentiate this term, so the new term is negative.

With respect to x just means tha t x is the variable. You should treat any oth er letters in the function as constants. It’s OK to leave powers as negativ e numbers or fractions in your final answers. The answer to part (a) could be in either of these forms: 3 −_21 1 3__ _1 3x 2 − _ +_ 3x2 − ___ 2x +7 7 2√ x

__

2 + 5√x 2 (a) Write ________ in the form 2x p + 5x q, x where p and q are constants. (2 marks) __

2

(b) Given that y = 3x + 1 − dy find ___ dx

2 + 5√x ________ x

,

(4 marks)

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Differentiation 2 y

You can use the derivative or gradient function to find the rate of change of a function, or the gradient of a curve.

P (2, 28)

This curve has equation y = x 3 + 5x 2. Its gradient function dy has equation ___ = 3x 2 + 10x. You can find the gradient at dx any point on the graph by substituting the x-coordinate at that point into the gradient function.

–5

0

x

Gradient at P = 32 At the point P: x=2 dy ___ = 3(2)2 + 10(2) = 12 + 20 = 32 dx

__

f(x) = (10 + 2√x )2, x > 0 __ (a) Show that f(x) = 100 + k√x + 4x, where k is a constant to be found. _ (2 marks) _ _ 2 2 f(x) = 10 + 20√x_ + 20√x + ( 2√x ) = 100 + 40√x + 4x k = 40 (2 marks)

(b) Find f 9(x). 1 __ f(x) = 100 + 40x2 + 4x 1 −__ f 9(x) = 20x 2 + 4

(1 mark)

(c) Evaluate f 9(25).

f 9(25) = 20( 25 2 ) + 4 1 = 20(__ 5) + 4 =4+4=8 1 −__

Second-order derivatives You can differentiate twice to find the second-order derivative. You write the second-order derivative as d2y ____ or f 0(x). dx 2

Evaluating f9(x) f 9(x) tells you the rate of change of the function for a given value of x. You can calculate f 9(x) for a given value of x by substituting that value of x into the derivative.

1 __1 1 _____ = ______1 = ___ = 5 √ 2 25 25 For a reminder about using the index laws to simplify powers have a look at page 1. −__21

25

y = 5x 3

Differentiate dy = 15x 2 dx Differentiate d2y = 30x dx2

1 f(x) = 3x 3 + 5x, x > 0 (a) Differentiate to find f9(x). (2 marks) (b) Given that f9(x) = 41, find the value of x. (3 marks) 2 y d 2 Given that y = 2x 2 + 4x−2, find ____2 dx (4 marks)

37

y = x 3 + 5x 2

__

Given that y = 8√x − 3x 2 + 5x, x > 0, d2y ____ (4 marks) find 2 dx 1 __

y = 8x2 − 3x 2 + 5x 1 dy −__ ___ = 4x 2 − 6x + 5 dx 3 d2y −_2_ ____ = −2 x −6 dx 2

3 The curve C has equation y = x(x − 1)(x + 3) dy (2 marks) (a) Find ___ dx (b) Sketch C, showing each point where C meets the x-axis. (3 marks) (c) Find the gradient of C at each point where the curve meets the x-axis. (2 marks)

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Tangents and normals You can use differentiation to work out the equations of tangents and normals. The curve drawn in black has equation y = x 2 − 2x + 4 You can differentiate this to work out the gradient function: dy ___ = 2x − 2 dx There is more about finding gradient functions on page 36.

y

y = x 2 – 2x + 4

8

The tangent to the curve at P is a straight line that just touches the curve at P .

7 6

At the point P(2, 4) the gradient of the curve is 2, so the gradient of the tangent is also 2.

5

The tangent passes through (2, 4) and it has equation y = 2x

3

4

The normal to the curve at P is a straight line that is perpendicular to the tangent.

P

2

The normal is perpendicular to the tangent, so it has 1 gradient −__ 2.

1

The normal also passes through (2, 4) and it has equation 1 y = −__ 2x + 5 There is more about parallel and perpendicular lines on page 18.

0

1

2

3

4

5

6

7

8 x

Golden rules If a curve has gradient m at point P:

1 2

The tangent at P also has gradient m 1 The normal at P has gradient −__ m

Follow these steps: 1. Differentiate to find the gradient function. dy 2. Substitute x = 2 into ___. The gradient dx at point P is −1. 3. Use y − y1 = m(x − x1) to find the equation of a straight line with gradient −1 that passes through (2, 3).

The curve C has equation x>0 y = x3 – 3x2 – x + 9, The point P with coordinates (2, 3) lies on C. Find the equation of the tangent to C at P, giving your answer in the form y = mx + c, where m and c are constants. (5 marks) dy ___ = 3x 2 − 6x − 1 dx = 3(2)2 − 6(2) − 1 = 12 − 12 − 1 = −1 y − y1 = m(x − x1) y − 3 = (−1)(x − 2) y − 3 = −x + 2 y = −x + 5

There is more on finding equations of straight lines on page 17.

The curve C has equation y = f(x), x > 0, where __ dy 8 ___ = √x + ___2 − 5 dx x Given that the point P (4, 11) lies on C, find the equation of the normal to C at point P, giving your answer in the form ax + by + c = 0 (4 marks)

dy ___ You have been given dx , so start by substituting the x-coordinate of P to find the gradient of the tangent at P. The normal will be perpendicular to this.

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Stationary points 1 You can use calculus to find the stationary points of a graph or function in your exam. You need to be confident with differentiation – have a look at pages 36 and 37 for a reminder.

Using differentiation The stationary points of a graph or function are the points where dy the derivative, ___ or f9(x), is equal dx to zero.

y P

This graph has stationary points at P and Q. The slope of the curve is 0 at both points.

Q 0

x y = f(x )

Find the coordinates of the stationary point on the curve with equation y = 3x2 + 12x + 5 (4 marks) dy ___ = 6x + 12 dx dy When ___ = 0, 6x + 12 = 0 dx 6x = –12 x = –2 2 So y = 3 × (–2) + 12 × (–2) + 5 = –7 Stationary point is (–2, –7).

The diagram shows part of the curve with 12 equation y = 3x + ___2 2 15 x y

O

12 y 5 3x 1 x2 2 15

x

Use calculus to show that y is increasing for x > 2 (4 marks) y = 3x + 12x –2 – 15 dy 24 ___ = 3 – 24x –3 = 3 – ___ dx x3 24 If x > 2 then x 3 > 8 and ___ 0 x3 dy So if x > 2, ___ > 0 therefore y is increasing. dx You need to show that the derivative is nonnegative for all values of x in the range given.

39

To find the coordinates of the stationary point using calculus: dy ___ 1. Differentiate to find dx . dy ___ = 0. 2. Set dx 3. Solve the equation to find the value or values of x. 4. Find the corresponding value of y for each value of x.

Increasing and decreasing functions You can use the derivative to decide if a function is increasing or decreasing in a given interval: If f 9(x) > 0 for a < x < b then f(x) is increasing in the interval a < x < b. If f 9(x) < 0 for a < x < b then f(x) is decreasing in the interval a < x < b. The sign of f 9(x) (+ or –) must be the same (or O) in the whole interval, otherwise the function is neither decreasing nor increasing.

1 Find the coordinates of the stationary point on the curve C with equation y = x2 – 8x + 3 (4 marks) 2 Use calculus to find the x-coordinates of the stationary points on the curve with (4 marks) equation y = x3 – 5x2 + 8x + 1 There are two stationary points. Differentiate, then solve a quadratic equation by factorising.

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Stationary points 2 There are different types of stationary points on graphs. You need to be able to decide on the d2y nature of a particular point. You can do this by finding the value of the second derivative, ____ or dx2 f 99(x ) at that point.

Maximum or minimum?

1

y

dy d2y If ___ = 0 and ____ < 0 then the stationary point is a maximum. dx dx 2

P

dy d2y At P, ___ = 3 × (−2)2 − 12 = 0 and ____ = 6 × (−2) = −12 < 0 so P is a maximum. dx dx 2

2

If

dy ___ dx

= 0 and

d y ____ 2

dx 2

2

–2 0

> 0 then the stationary point is a minimum.

dy d2y At Q, ___ = 3 × (2)2 − 12 = 0 and ____ = 6 × (2) = 12 > 0 so Q is a minimum. dx dx 2 d y For a reminder about finding ____ have a look at page 37. dx 2 2

__

The curve C has equation y = 12√x − 2x, x > 0 (a) Use calculus to find the coordinates of the turning point of C. (4 marks) 1 __

y = x 3 – 12 x

y = 12x2 − 2x 1 dy − __ 6_ ___ = 6x 2 − 2 = ___ −2 dx √x dy 6_ − 2= 0 When ___ = 0, ___ _ dx √x 6 = 2√x x=9 __ So y = 12√9 − 2 × 9 = 18 Turning point is (9, 18). d 2y (2 marks) (b) Find ___2 dx 2 3 _ _ d y − ____ = −3x 2 dx2 (c) State the nature of the turning point. (1 mark) 3 d2y −_2_ At turning point, x = 9, so ____ 2 = −3 × 9 dx 1 = −__ 9 d2y ____ < 0 so the turning point is a maximum. dx2

x

Q

dy ___ dx

d y ____ 2

= 3x 2 − 12

dx 2

= 6x

Sketching gradient functions You can use the features of y = f(x) y to sketch y = f9(x). Positive gradient, so y = f9(x) is above the x-axis Stationary point, so y = f9(x) cuts the x-axis Negative gradient so y = f9(x) is below the x-axis

0

x y = f(x )

y

Vertical asymptotes in the same places on both graphs

0

x y = f9(x )

A turning point is a stationary point which is a local maximum or a local minimum. d2y ____ of value the out work (c), part For dx 2 when x = 9. Write down whether it is greater or less than 0 and state whether the turning point is a maximum or a minimum.

The diagram shows a sketch of the curve with equation y = 5x2 − 3x − x3 The curve has stationary points at A and B. (a) Use calculus to find the coordinates of A and B. (6 marks) 2 y d (b) Find the value of ___2 at B, and hence verify that B is a maximum. (2 marks) dx

y O

B

A

x

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Modelling with calculus You can use calculus to solve real-life problems involving maxima and minima.

y

3x

x

A cardboard box is made in the shape of an open-topped cuboid, with volume 18 000 cm3. The base of the cuboid has width x cm and length 3x cm. The height of the cuboid is y cm. (a) Show that the area, A cm2, of cardboard needed to make the box is given by 48 000 (4 marks) A = 3x2 + ______ x Volume = 3x2y = 18 000 6000 y = ______ x2 2 A = 3x + 2 × xy + 2 × 3xy = 3x 2 + 8xy 6000 = 3x2 + 8x ______ x2 48 000 = 3x2 + ________ x (b) Use calculus to find the value of x for which A is stationary. (4 marks) 48 000 dA ___ = 6x − ________ dx x2 48 000 dA ___ When = 0, 6x − ________ =0 dx x2 3 6x = 48 000 x3 = 8000 x = 20

(

)

(c) Show that A is a minimum at this point and find its value. (4 marks) 2 96 000 d A ____ = 6 + ________ dx2 x3 96 000 d2A = 6 + ________ = 18 > 0 When x = 20, ____ dx2 203 So A is a minimum. 48 000 48 000 A = 3x2 + ________ = 3 × (20)2 + ________ x 20 = 3600 This is the least area of cardboard needed to make the box.

41

! (a) Use the information given about the volume of the cuboid to write y in terms of x. Then write A in terms of x and y, and substitute your first expression to get A in terms of x only.  (b) Because A is a function of x only, you can differentiate with respect to x to find dA ___ . Find the stationary point of A by dx setting this equal to 0 and solving to find x. d2A (c) You need to find ____2 to determine the dx nature of the stationary point. You will need to use problem-solving skills throughout your exam – be prepared!

1 An oil well produces x barrels of oil each day. It models its profit, £P each x2 day, using the formula P = 80x − ___ 50 dP ___ (a) Find (2 marks) dx (b) Hence show that P has a stationary point at x = 2000 and use calculus to determine the nature of that stationary point. (4 marks) 2 The diagram shows a container in the shape of an open-topped x cylinder, with height x m and radius r m. The cylinder has a r capacity of 100 m3. (a) Show that the area of sheet metal, A m2, needed to make the tank is given 200 by A = p r2 + ____ (4 marks) r (b) Use calculus to find the value of r for which A is stationary. (4 marks) (c) Prove that this value of r gives a minimum value of A. (2 marks) (d) Hence calculate the minimum area of sheet metal needed to make the tank. (2 marks) When you differentiate, remember that p is a constant.

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Integration Integration is the opposite of differentiation. You can use this rule to integrate terms which are written in the form ax n. You increase the power by 1 ... This is the symbol for integration.

xn + 1

The constant of integration When you differentiate, any constant terms disappear. So lots of functions have the same derivative.

... then divide by the new power. This rule doesn’t work if the original power is −1

y = x2 + 5

You have to add the constant of integration.

y = x 2 − 19

∫ x ndx = _____ + c, n ≠ −1 n+1 You are integrating with respect to x.

To integrate a function, write each term in the form ax n, then integrate one term at a time.

y = x2

Find ∫(12x + 6x − 15x ) dx, giving each term in its simplest form. (5 marks) 3

∫( 12x 3 + 6x − 15x_23_ ) dx =

12x 4 _____ 4

+

6x 2 ____ 2

_5_

15x3 − _____ + _5_

(3)

c

2

+ 1 = _35_. Dividing by _5_ is the sam e as 3 dividing by 5 then multiplying by 3. 3

Golden rules

1 2 3

Write every term in a polynomial in the form ax n before integrating. Remember to include the constant of integration.

1 Given that y = ___3 − 3x 5, x ≠ 0, find ∫y dx x (3 marks) 6 3x x−2 − ____ + c ∫(x−3 − 3x 5) dx = ____ −2 6 1 −2 1 6 = −__ − __ 2x 2x + c

Simplify any coefficients if possible.

1 Find ∫(1 − 3x 3) dx 2 Find ∫(3x + 1)2 dx

Integrate term-by-term and don’t forget to add the constant of integration. For each term: • increase the power by 1 • divide by the new power. _2_

_5_

= 3x + 3x − 9x3 + c 4

dy = 2x Integrate y = x 2 + c dx

When you integrate you don’t know the constant. You write ‘+ c ’ at the end to show this. This is called indefinite integration.

_2

3

Differentiate

Be careful with negative powers. For the first term, you have to increase the power of −3 by 1 to get −2, then divide by the new power, −2.

(3 marks) (4 marks)

__

3 Given that y = 6x 2 + 5x√x , x > 0, (3 marks) find ∫y dx.

Expand the brackets first.

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Finding the constant If you know one point on the original curve, or one value of f(x), then you can calculate the constant of integration.

Using substitution

y = x2 + 4 y = x2 y

All three of these curves have the same gradient function: dy ___ = 2x dx But only one passes through the point (2, −1). You can find its equation by integrating, then substituting x = 2 and y = −1 to find the value of c. y = x2 + c −1 = 22 + c c = −5 y = x2 − 5

dy ___

1__ = 7 + ___ ,x>0 dx √x Given that y = 26 at x = 4, find y in terms of x. (6 marks) dy ___ dx

y

1 −__ 2

=7+x

1 __

x2 = 7x + ___ + 1 __

(2)

c

1 __

= 7x + 2x2 + c 26 = = c= y=

1 __

7(4) + 2(4)2 + c 28 + 4 + c −6 1 __ 7x + 2x2 − 6

y = x2 – 5

0

(2, –1)

x

Given that f(−1) = 9 and (5 marks) f 9(x) = 6x2 − 10x − 3, find f(x). 3 2 6x 10x f(x) = ____ − _____ − 3x + c 3 2 = 2x3 − 5x2 − 3x + c f(−1) = 2(−1)3 − 5(−1)2 − 3(−1) + c = −2 − 5 + 3 + c = −4 + c 9 = −4 + c c = 13 f(x) = 2x 3 − 5x 2 − 3x + 13

f(−1) = 9 is the same as saying tha t the curve with equation y = f(x) passes through the point (−1, 9).

Three key steps

1

Integrate, and remember to include the constant of integration.

2

Substitute the values of x and y you know and solve an equation to find c.

1 The curve C has equation y = f(x), x > 0, and the point (4, 17) lies on C. __ 2 + 3√x ________ , Given that f 9(x) = 3 − x2 find f(x). (5 marks)

43

3

Write out the function including the constant of integration you’ve found.

dy (x2 + 5)2 ,x≠0 2 ___ = ________ dx x2 dy (a) Show that ___ = x2 + 10 + 25x−2 dx

(2 marks) (b) Given that y = −13 at x = 1, find y in terms of x. (6 marks)

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Definite integration In your exam, you might have to find an integral with limits. This is called definite integration. You should make sure you are confident with indefinite integration before revising this – check page 42 for a reminder.

Evaluating a definite integral A definite integral has a numerical answer. Integrate (6x + 1) in the normal way and write the integral in square brackets. You can ignore the constant of integration.

upper limit of the integral

∫2 (6x + 1)dx 5

lower limit of the integral

= = = =

[3x

2

(3 × 5 + 5) − (3 × 22 + 2) 80 − 14 Evaluate the … and subtract the integral at the value of the integral at upper limit… the lower limit. 66

Use calculus to find the value of

∫ (2x + 6 x ) dx ∫ ( 2x + 6x )dx = [ x 9

√

1 9

__

1 __ 2

(5 marks) 2

+ 4x 2 ] 1

_3_ 9

= ( 92 + 4 × 92 ) − ( 12 + 4 × 12 ) = 189 − 5 = 184

1

_3_

Write the limits next to the square brackets.

+ x ]52 2

_3_

If you are using calculus to do definite integration you will usually need to give an exact answer. If your answer isn’t a whole number or a fraction, write it in simplified form. ______ surd __ __ __ ___ √ √ √ 3 4 = 3 4 √ 2 = √ 3 × 4 2 12 = 2 Have a look at page 3 for a reminder about surds.

Some calculators have a key like the one shown here which can work out numerical integration. You cannot just use this key and write down your answer. You need to use calculus and show your working, or you will get a maximum of 1 mark. You could use this key to check your answer, but it’s safer to stay away from it entirely!

Find the exact value of

∫

( x−21 )dx

12 1

__

= [ 2x2 ] 1

1 12 __

∫ ( ___1x )dx (4 marks)

___

12

1

√

__

_

= (2√12 ) − (2√1 ) __

= 4√3 − 2

6 Start by writing __ as 6x−2. x2 1 Use calculus to find the exact value of 3 6 3x2 − 7 + __2 dx (5 marks) x 1 2 2__ 6 , find f(x) dx, giving 2 Given that f(x) = __3 − ___ x √x __1 √ your answer in the form a − b 2 , where a and b are constants. (5 marks)

∫(

)

∫

Be careful when you are subtracting the value of the integral at x = 1. Use brackets to make sure you don’t make a mistake with negative numbers.

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Area under a curve y

You can use definite integration to find the area between a curve and the x-axis. The area between the curve y = f(x), the x-axis and the lines x = a and x = b is given by

A=

A

b

∫a f(x) dx

a

0

b

x y = f(x )

Look at the previous page for a reminder about definite integration.

The diagram shows part of the curve with equation y = (x + 3)(1 − x)

The diagram shows part of the curve with equation y = x2(x − 6) y

y

R O 23

O

1

Use calculus to find the exact area of the shaded region, R. (5 marks)

y = 3 − 2x − x 2

∫

1

[

Find the area of the shaded region, R. (6 marks) 3 2 y = x − 6x 6 6 x4 (x 3 − 6x 2) dx = ___ − 2x 3 4 0 0

]

x3 1 (3 − 2x − x2) dx = 3x − x2 − __ 3 −3 –3 1 = (3 − 1 − __) − (−9 − 9 + 9) 3 _5_ = − (−9) = 10 _23_ 3 Area of R = 10 _23_

45

]

= (324 − 432) − (0 − 0) = −108 Area of R = 108

Negative areas When you use a definite integral to find an area below the x-axis, the answer will be negative. If you are asked to find an area, make sure you give your final answer as a positive number.

Be careful with this question – you can’t 4 just find y dx because an area below the

∫

y

Use calculus to find the total O area of the shaded regions, between x = 1 and x = 4 and bounded by C, the x-axis, and the line x = 1

[

∫

The curve crosses the x-axis at −3 and 1. So the limits for your definite integral will be −3 and 1. Always put the right-hand boundary as the upper limit and the left-hand boundary as the lower limit. You can give your exact answer as a fraction, mixed number or decimal.

The diagram shows part of the curve C with equation y = x(x − 2)(x − 4)

x

R 6

x

C

1 2

4x

1

x-axis will produce a negative integral. You need to work out two separate definite integrals to find these two areas, then add the areas together: A1 =

(9 marks)

A1 1 2 A2 4

∫

2

1

y dx

∫

4

A2 = − y dx 2

Total area = A 1 + A2

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More areas You can use areas of triangles and trapeziums, together with definite integration, to find areas enclosed by curves and straight lines. Here are three examples: y

1

y

2

y = f(x ) Q

A R (a, 0) x

0

P

Q

A

P

y

3

y = f(x )

P (a, 0)

A 0 R (b, 0)

x

0

Q (b, 0) x

a b

∫ a f(x )dx A = OPQR

–

a

∫o f(x )d x

A=

b

∫ a f(x )dx

The diagram shows part of the curve C with 8 equation y = __2 + x − 2, x > 0 x y

C

R Q 1

x

4

The points P and Q lie on the curve and have x-coordinates 1 and 4 respectively. The region R is bounded by the curve and the line segment PQ. Find the exact area of R. (8 marks)

A=

PQ R

–

OPQ

If you have to find the area between a curve and a line like this, plan your answer before you start. Work out how you can use triangles, trapeziums and rectangles to work out the area. y

y

y

Everything in blue is part of the answer.

P

O

–

–

= 0

x

x

0

x

0

P Before you can work out the need you ium, trapez the area of Q 7 to know the y-coordinates of 2.5 P and Q. It’s sometimes easier 3 to work in decimals rather than ns. fractio __1 Use the formula A = 2(a + b)h to work out the area of the trapezium. Then work out 4

∫ y dx and subtract it from the area of the 1

trapezium, to find the shaded area.

When x = 1, y = 8 + 1 − 2 = 7 So coordinates of P are (1, 7). 1 1 __ When x = 4, y = __ 2 + 4 − 2 = 22

So coordinates of Q are (4, 2.5). 1 __

Area of trapezium = 2 (7 + 2.5)(4 − 1) = 14.25 4

∫ (8x 1

−2

4 1 2 + x − 2) dx = [ −8x−1 + __ 2 x − 2x ] 1

= (−2 + 8 − 8) − (−8 + 1 = −2 − (−9 __ 2)

1 __ 2

− 2)

= 7.5 So area of R = 14.25 − 7.5 = 6.75

The straight line with equation y = x cuts the curve with equation y = x(5 − x) at the points O and A.

y y5x R O

A x

y 5 x(5 2 x)

(a) Find the coordinates of A.

(2 marks)

(b) Use calculus to find the exact area of the shaded region R. (7 marks)

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Exponential functions You need to be able to sketch the graph of y = a x. You can only sketch this graph when a is a positive number. y = 3x x y

y=a

Passes through (0, 1).

y = 2x

y = 0 is an asymptote. 1

y=

() 1 2

If a > 1 graph curves upwards.

x

If 0 < a < 1 graph curves downwards.

x

0

Sketch the graph of (a) y = (_2 ) + 3 1 x

(3 marks)

(3 marks)

(b) y = 2x – 3 y

y

4

y=

( 21 )

x

+3 1 8

3 x

0

There is more about transformations of graphs on pages 13 and 14.

The letter ‘e’ represents a constant number. It is approximately equal to 2.718 28… This is the graph of y = ex

Golden rule

(3, 1)

0

This is a translation of the graph y = (__1 )x 2 by the vector (0 . The asymptote at y = 0 3) is translated to y = 3.

y = ex

y=

2x –3

x

x This is a translation of the graph y = 2 3 by the vector (0). You should label any points where the curve cuts the axes.

__1 When x = 0, y = 2−3 = 8

y y = ex 1

0

x

Differentiate with respect to x x _ (ii) e3x (iii) 5 e 2 (3 marks) (i) ex 1 __ y = ex y = e 3x y = 5e 2x dy dy dy 5 __1 ___ ___ ___ = ex = 3 e 3x = __ e 2x dx dx dx 2

dy

kx If y = ekx then ___ dx = k e

1 Match each of these equations to one of the graphs on the right. (1 mark) (a) y = 0.2 x 2 2 y y y A B C x (1 mark) (b) y = 2(5 ) x (1 mark) (c) y = 2x 2 1 O x O O _1 −x (1 mark) (d) y = ( 3 ) 2 Differentiate with respect to x x _ (b) 2 e5x (1 mark) (c) e 3 (1 mark) (a) e22x (1 mark)

47

D x

y

O

x

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Logarithms Logarithms (or logs) are a way of writing facts about powers. These two statements mean the same thing: You say ‘log to the base a of b equals x ’.

ax = b

loga b = x

a is the base of the logarithm. For example: log3 9 = 2

32 = 9

Remembering the order The key to being confident in log questions is remembering the basic definition. Start at the base, and work in a circle.

… to get b

loga b = x Raise a to the power x …

Laws of logarithms Learn these four key laws for manipulating expressions involving logs. These laws all work for logarithms with the same base.

1

loga x + loga y = loga (xy)

3

1 loga __ = –loga x x

log4 8 + log4 2 = log4 16 = 2 (since 42 = 16)

()

1 1 __ log8 (__ 2 ) = –log8 2 = – 3

Find (a) the positive value of x such that (2 marks) logx 49 = 2 2 x = 49 x=7 (b) the value of y such that log5 y = 22 (2 marks) –2 5 =y 1 y = ___ 25

Write down the corresponding power fact. Remember: ax = b loga b = x

2

x loga x – loga y = loga _y_

4

loga (x n) = n loga x

( )

1 __ 1 2 log9 18 – log9 6 = log9 3 = __ 2 (since 9 = 3)

log5 (253) = 3 log5 25 = 3 × 2 = 6

Express 3 loga 2 + loga 10 as a single logarithm to base a. (3 marks) 3 3 loga 2 + loga 10 = loga (2 ) + loga 10 = loga (23 × 10) = loga 80 3 Use law 4 to write 3 loga 2 as loga (2 ), then use law 1 to combine the two logarithms.

Special cases Learn these two special cases to save time:

1 Find (a) the value of y such that log3 y = –1 (2 marks) (b) the value of p such that logp 8 = 3 (2 marks) (2 marks) (c) the value of log4 8

1

loga a = 1

2

loga 1 = 0

2 Express as a single logarithm to base a (2 marks) (a) 2 loga 5 (2 marks) (b) loga 2 + loga 9 (3 marks) (c) 3 loga 4 – loga 8 3 Show that 2 log8 6 – log8 9 = _3 2

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Equations with logs If you see an equation involving logarithms in your exam, you will probably need to rearrange it using the laws of logarithms, which are covered on page 48.

Two steps to solving log equations

Undefined logs

Follow these two steps to solve most log equations in your exam: Group the log terms on one side, then use the laws of logs on page 48 to write them as a single logarithm. Rewrite loga f(x) = k as f(x) = a k and solve the equation to find x.

The value loga b is only defined for b > 0. You can’t calculate loga 0 or the log of any negative number. If an equation contains loga x or loga kx then ignore any solutions where x < 0.

1 2

Solve the equation 2 log5 x – log5 3x = 2 (4 marks) log5 x2 – log5 3x = x2 ___ log5 ( ) = 3x x2 ___ = 3x x2 = x 2 – 75x = x(x – 75) = 0 so

2

If there are solutions to ignore in an exam question, you might be given a range of possible values for x.

Follow the two steps given above: 1. Rearrange the left-hand side into a single logarithm. 2. Write the corresponding power fact: 2 log5 f(x) = 2 ⇒ f(x) = 5

2 52 = 25 75x 0 x = 0 or x = 75

You need to solve two simultaneo us equations. You can ignore the neg ative square root because p and q are positiv

e.

Given that 0 , x , 2 and log3 (2 2 x) 2 2 log3 x = 1, find the value of x. log3 (2 – x) – log3 (x ) = 1 2–x log3 _____ =1 x2 2–x _____ = 31 x2 2 – x = 3x2 3x2 + x – 2 = 0 (3x – 2)(x + 1) = 0

(6 marks)

2

(

x = _23_

)

x = –1

Find the exact values of p and q.

Ignore this solution because 0 < x < 2.

1 Solve log2 (x + 1) – log2 x = log2 5 (3 marks) 2 Solve the equation log6 (x – 1) + log6 x = 1 (4 marks)

49

p and q are positive constants, with p = 5q 1 log5 p + log5 q = 2 2 From 2 : log5 (pq ) = 2 Substituting 1 : log5 (5q2) = 5q2 = q2 = q= p= Substituting into 1 :

(6 marks)

2 52 = 25 5 __ √5 __ 5√5

3 Solve log3 (x – 1) = 21 4 Find the values of x such that 2 log4 x – log4 (x – 3) = 2

(2 marks) (5 marks)

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Exponential equations You can find unknown powers in equations using the log functions on your calculator. Make sure you write down any logarithms you are working out. You can use this key to work out logs to any base.

(a) Solve the equation 4x = 13, giving your answer to 3 significant figures. (3 marks) x = log4 13 = 1.85 (3 s.f.) (b) Find, to 3 significant figures, the value (3 marks) of y for which 5y = 4 y log (5 ) = log 4 y log 5 = log 4 log 4 y = _____ = 0.861 (3 s.f.) log 5 Use the fact that 52x = (5x)2 to write a quadratic equation in 5x. For a reminder on the laws of indices have a look at page 1. It might help to write the equation as Y 2 – 3Y + 2 = 0, with Y = 5x. Factorising gives you two values for 5x. Each of these gives you a value for x. Remember that loga 1 = 0 for any base, so log5 1 = 0.

x

x–3

Solve the equation 2 = 5 , giving your answer to 3 significant figures. (4 marks) x x  – 3 log 2 = log 5 x log 2 = (x – 3) log 5 3 log 5 = x log 5 – x log 2 3 log 5 = x(log 5 – log 2) 3 log 5 x = ____________ = 5.27 (3 s.f.) log 5 − log 2

1 Find, to 3 significant figures (a) the value of b for which 2b = 15 (3 marks) (b) the value of x for which 6x = 0.4 (3 marks) 2 (a) Solve the equation 32x + 3x = 6, giving your answer to 2 decimal places. (6 marks) (b) Explain why there is only one solution to (1 mark) the equation 32x + 3x = 6

This key means log10. You can take logs of both sides of an equation and solve it using this key.

Make sure you write down the logarithm you need to find even if you are working it out on your calculator in one go. Part (a) key. shows a method using the logs of taking Part (b) shows a method by both sides and using the laws of logs. This works for any base, so you can use the key on your calculator.

Solve the equation 52x – 3(5x) + 2 = 0 giving your answers to 2 decimal places where appropriate. (6 marks) x 2 x (5 ) – 3(5 ) + 2 = 0 (5x – 2)(5x – 1) = 0 5x = 2 x = log5 2 = 0.43 (2 d.p.)

5x = 1 x = log5 1 = 0

! This exponential equation has different bases on each side. You can still take logs of both sides but you must take logs to the same base on each side. You will need to use problem-solving skills throughout your exam – be prepared!

3 Solve 3x – 1 = 22x + 1, giving your answer correct to 3 significant figures. (4 marks) x2 x–1 g(x) = 2 , x ∈ R 4 f(x) = 6 , x ∈ R Show that the curves y = f(x) and y = g(x) do not intersect. (5 marks) 2 Set 6 x = 2x – 1 and use the discriminant of the resulting quadratic equation in x.

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Natural logarithms A logarithm to the base e is sometimes called a natural logarithm. The graph of y = ln x is the graph of y = ex reflected in the line y = x. It has an asymptote at x = 0 and crosses the x-axis at (1, 0).

y

y=x

Golden rule

1

ln x works just like any other logarithm: e

ln x

Look for the e

y = ex

=

ln(ex)

=x

and ln functions on your calculator to find

0

y = ln x 1

x

values of ex and ln x.

Find the exact solutions to the equation (4 marks) ex + 3e−x = 4 x 2 x (e ) + 3 = 4e x 2 (e ) − 4ex + 3 = 0 (ex − 3)(ex − 1) = 0 ex = 3 x = ln 3

ex = 1 x = ln 1 = 0

You can write this as a quadratic equation using the substitution u = ex: _ u + _3 u =4 u2 – 4u + 3 = 0 If a question asks for exact solutio ns then you should leave your answers as logs, or powers of e, and simplify them as much as possible.

! The laws of logarithms work exactly the same with ln as they do with log10 and loga You can’t combine the 2x with the e3x + 1 easily, so just take logs of both sides. You can then use the laws of logs to simplify the left-hand side. Group the x terms together, then factorise to get x on its own. You will need to use problem-solving skills throughout your exam – be prepared!

1 Given that f(x) = ln x, x > 0, sketch, on separate axes, the graphs of (a) y = f(x – 2) (2 marks) (b) y = –f(x) (2 marks) (c) y = f(3x) (2 marks) 2 The point P with y-coordinate 6 lies on the curve with equation y = 3 e2x − 1. Find, in terms of ln 2, the x-coordinate of P. (2 marks) 3 Solve (a) ln (x + 1) − ln x = ln 5 (2 marks) (5 marks) (b) e4x + 3 e2x = 10 (c) ln (6x + 7) = 2 ln x, x > 0 (4 marks)

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Solve 2x e3x + 1 = 10, giving your answer in the a + ln b form _______ where a, b, c and d are integers c + ln d to be found. (5 marks) x 3x + 1 ln(2 e ) = ln 10 x 3x + 1 ) = ln 10 ln 2 + ln (e x ln 2 + 3x + 1 = ln 10 x (ln 2 + 3) = ln 10 − 1 −1 + ln 10 x = __________ 3 + ln 2 a = −1, b = 10, c = 3, d = 2 4 Find the exact solution to the equation (5 marks) 3x e2x − 5 = 7 5 The function f is defined by 3x 2 − 7x + 2 , x ≠ ±2 f : x ↦ ___________ x2 − 4 3x − 1 (a) Show that f(x) = ______ (3 marks) x+2 (b) Hence, or otherwise, solve the equation ln (3x2 − 7x + 2) = 1 + ln (x2 − 4), x > 2 giving your answer in terms of e. (4 marks) Start with 6 = 3 e2x−1. Divide both sides by 3 then take the natural logarithm of both sides.

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Exponential modelling

You can use exponential functions to model lots of real-life situations.

1

Growth models A typical growth model can be described as N = C + N0 ekt N

2

Decay models A typical decay model can be described as N = C + N0 e−kt N C 1 N0

C +N0 C

C

0

t

This model describes the temperature P °C of the water in a kettle t minutes after it has boiled: P = 25 + A e−kt, t > 0 where A and k are positive constants. (a) Given that the initial temperature of the water was 100 °C, find the value of A. (2 marks) When t = 0, P = 25 + A e0 100 = 25 + A A = 75

After 10 minutes, the water in the kettle has cooled down to 40 °C. 1 (3 marks) (b) Show that k = __ 10 ln 5

40 = 25 + 75 e −10k 15 = 75 e −10k 1 __ = e −10k 5 1 −10k = ln(__ = ln (5 −1) = −ln 5 5)

1 ln 5 So k = __ 10

(c) Find the temperature of the water after 18 minutes, in °C to 1 decimal place. (2 marks) P = 25 + 75 e −(10ln5)×18 = 29.1 °C (1 d.p.) 1 __

1 The number of cells, N, in a bacterial culture at a time t hours after midday is modelled as N = 100 e0.8t, t > 0 (a) Write down the number of cells in the culture at midday. (1 mark) (b) Show that the rate of change of N at time t is directly proportional to the size of the population. (3 marks)

0

t

plete You will sometimes be given a com it. In exponential model and asked to use the of two this example, you are not given m: constants in the model. To find the n in the • Substitute the information give model. the of question into the equation t Make sure you substitute the righ e. plac variable in the right nown • Solve an equation to find the unk constants.

Rates of change You might have to find the rate of change of a quantity in an exponential model. You can do this by differentiating with respect to time. In the model in the Worked example on the left, the rate of change of temperature with time would be given by dP ___ = −kA e −kt dt The units would be °C/min. The fact that the rate of change is negative tells you that the temperature is decreasing.

2 The mass, M grams, of a sample of radon after t hours is modelled using the equation M = 250 e−kt, t > 0 where k is a positive constant. (a) What was the initial mass of the sample? (1 mark) After 90 hours the sample has lost half its mass. (b) Find the value of k to 3 significant figures. (4 marks)

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Modelling with logs You can use logarithms to determine the constants in some exponential and polynomial models. There are two different cases that you need to know about:

1

The polynomial model y = ax n

2

In this model a and n are constants. If x and y satisfy this model then the graph of log y against log x will be a straight line: log y = n log x + log a gradient = n

The exponential model y = kbx

In this model k and b are constants. If x and y satisfy this model then the graph of log y against x will be a straight line: log y = (log b)x + log k

y-intercept = log a

gradient = log b

y-intercept = log k

A scientist models the mass, m grams, of a fluorine sample and the log10 m time elapsed, t hours, using the equation m = pqt, where p and q are constants. (0, 2.7) (10, 1.1) She observes the actual mass over a period of 10 hours, and plots the graph shown on the right, of log m against t. (a) Find an equation for the line. (2 marks) t O 1.1 − 2.7 _________ For part (b), the safest approach = − 0.16 Gradient = is to 10 rearrange the equation of the line into log m = –0.16t + 2.7 the form m = pqt, then compare values. (b) Determine the values of p and q in the Check it! model to 3 significant figures. (4 marks) 501 × 0.69210 = 12.6 –0.16t + 2.7 m = 10 log 12.6 = 1.1 ✓ = 102.7 × (10–0.16)t t = 501 × 0.692 So p = 501 and q = 0.692 (3 s.f.) (c) Interpret these values in the context of the model. p = 501 is the initial mass of the sample in grams. q = 0.692 is the proportional change in the sample each hour.

(2 marks)

(d) Use the model to predict the mass of the sample after 3 days. 3 days = 72 hours, so m = 501 × 0.69272 = 1.54 × 10−9 g The mass of the sample will be about 1.54 × 10−9 grams.

(1 mark)

(e) Give one reason why this prediction may not be accurate. (1 mark) The model is based on 10 hours of data, so it may not be accurate over a longer period.

A computer algorithm is used to allocate medical students to hospitals. When there are N students, the runtime of the algorithm, x milliseconds, is expected to follow the rule x = aN b, where a and b are constants. (a) Show that this relationship can be written in the form log x = k log N + c, giving k and c in terms of a and b. (2 marks)

53

The algorithm is run a number of times and the following values of x and N are found: N x

1000 1500 2000 2500 3000 3500 4000 460 980 1660 2510 3520 4680 5990

(b) Plot a graph of log x against log N, and comment on the accuracy of the expected model. (3 marks) (c) Find the values of a and b, giving your answers to 2 significant figures. (4 marks)

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You are the examiner! In your exam you might be asked to identify errors in working. You should also be confident checking your own work. Each of these students has made a key mistake in their working. Can you spot them all? __

1

5− 2√3 __ Simplify _______ giving your answer in the √3 − 1 __

form p + q √3 , where p and q are rational numbers. (4 marks) __

__

√3 + 1 5− 2√3 _______ _________ __ × __ √3 − 1 √3 + 1

__

__

5 − 2√3 (√3 + 1) __ __ __ = ______________________ 2 √ √ √ ( 3) − 3 + 3 − 1

2

The equation x2 + 3px + p = 0, where p is a non-zero constant, has equal roots. Find the value of p. (4 marks)

b2 − 4ac = 0 3p 2 − 4p = 0 p (3p − 4) = 0 _ p = 0 or p = _4 3

__

=

5 − 6 − 2√3 ______________ 1 __

2

__

= − 2 − √3

3

Find ∫(12x5 − 8x 3 + 3) dx, giving each term in its simplest form. (4 marks)

∫(12x 5 − 8x 3 + 3) dx

5

3x 8x 4 12x 6 = _____ − ____ + ___ 6 4 1 6 4 = 2x − 2x + 3x

Given that 0 < x < 4 and log5 (4 − x) − 2 log5 x = 1, find the value of x. 4−x log5 ______ = 1 2x 4 −x ______ =5 2x 4 − x = 10x 11x = 4 4 x = __ 11

(

)

4

Find the first 3 terms, in ascending powers of x, of the binomial expansion of (3 − x)6 and simplify each term. (4 marks) 6 (3 − x)6 = 36 + × 35 × −x 1 6 + × 34 × −x 2 … 2 = 729 − 1458x − 1215x 2 …

( ) ( )

Checking your work (6 marks)

If you have time left in your exam you should check back through your working: Check you have answered every question part. Make sure your answers are given in the correct form. Double-check numerical calculations such as binomial coefficients. Cross out any incorrect working with a single neat line and underline the correct answer.

Find the mistake in each student answer on this page, and write out the correct working for each question. Turn over for the answers.

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You are still the examiner! Before looking at this page, turn back to page 54 and try to spot the key mistake in each student’s working. Use this page to check your answers. The corrections are shown in red and these answers are now 100% correct. __

1

5− 2√3 __ Simplify _______ giving your answer in the √3 − 1 __

2

The equation x2 + 3px + p = 0, where p is a non-zero constant, has equal roots. Find the value of p. (4 marks)

form p + q √3 , where p and q are rational numbers. (4 marks) __

b 2 − 4ac = 0 (3p )2 − 4p = 0 p (3p − 4) = 0 _ p = 0 or p = _4 3

__

√3 + 1 5− 2√3 ________ _________ __ × __ √3 − 1 √3 + 1

__

__

(5 − 2√3 )(√3 __ + 1) __ __ = ______________________ 2 (√3 ) − √3 + √3 − 1 __

=

5 − 6 − 2√3 ______________

2 __ = − 2 − √3 1 __

9p2 − 4p = 0 p(9p − 4) = 0 4_ p = 0 or p = _9

__

=

3√3 − 1 _________

2 3 __ = − 2 + _2_√3 1 __

Use brackets when you are substituting into any formula. And be especially careful when dealing with squares, fractions or negative numbers. Look at page 7 for more on the discriminant.

Be really careful with your algebra. If you have to multiply a fraction by an expression, use brackets to term. __ __ make sure__you__multiply every __ __ √ 3 √ (5 − 2√3 )( 3 + 1) = 5 __ − 2√3 √3 − 2√3 + 5 = 3√3 − 1 Surds are covered on page 3.

Find ∫(12x5 − 8x 3 + 3) dx, giving each term in its simplest form. (4 marks)

∫(12x5 − 8x3 + 3) dx

8x 4 3x 12x 6 = _____ − ____ + ___ + c 6 4 1 = 2x 6 − 2x 4 + 3x + c

If you’re doing indefinite integration you’ll lose a mark if you forget the constant of integration. For a reminder about integration, look at page 42.

5

Given that 0 < x < 4 and log5 (4 − x) − 2 log5 x = 1, find the value of x. (6 marks) 4−x log5 ______ = 1 log5 (4 − x) − log5 x 2 = 1 2x 4−x 4−x ______ log5 ______ =1 =5 x2 2x 4−x ______ =5 4 − x = 10x x2 11x = 4 4 − x = 5x2 4 5x2 + x − 4 = 0 x = __ 11 (5x − 4)(x + 1) = 0 x = _45_ because x cannot equal −1 as log5 (−1) is not defined.

(

55

)

(

4

Find the first 3 terms, in ascending powers of x, of the binomial expansion of (3 − x)6 and simplify each term. (4 marks) 6 (3 − x)6 = 36 + × 35 × (−x) 1 6 × 34 × (−x)2 … + 2 = 729 − 1458x − 1215x 2 …

( ) ( )

−

3

The binomial expansion in the formulae booklet is for (a + b)n. If b is negative you need to use brackets when you substitute. The terms will alternate between + and −. Have a look at page 24 for a reminder.

)

Make sure you are confident with the laws of logs as they’re not in the formulae booklet. Here you have to write 2 log5 x as log5 x2 first. Revise this topic on page 48.

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Proof by contradiction

You can use the following steps to prove something by contradiction:

Assume…

...the negation of the original statement.

For example, to prove that there is no largest odd number, assume that there exists a largest odd number, n .

Use…

…logical steps to show that this assumption leads to a contradiction. If n is an odd number then n + 2 is also an odd number. You also know that n + 2 > n.

Negation The first step in a proof by contradiction is to assume the negation of a statement. The negation of a statement is a statement that asserts its falsehood. Statement

Negation

All mice are white

There exists a mouse that is not white

There are infinitely many prime numbers

There are finitely many prime numbers

There is no smallest rational number

There exists some smallest rational number, n

√2

√2

__

is an irrational number

__

Conclude…

…that the assumption was incorrect and the original statement is true. This contradicts the assumption that n is the largest odd number. You can conclude that there is no largest odd number.

Rational or irrational? You might have to use contradiction to prove statements involving rational numbers.

A rational number is a number that can a be written in the form __, where a and b b are integers. 12 _2_ __ Examples are 4, 3 , − 7 and 0.

An irrational number is a number that a cannot be expressed in the form __, b where a and b are integers. You can by contradiction to show __ __ use proof that √ 2 and √ 3 are irrational numbers.

is a rational number

!

Prove by contradiction that if p + q is an irrational number, then at least one of p and q is an irrational number. (3 marks) Assumption: p + q is an irrational number and both p and q are rational numbers. c a Write p = __ and q = __, where a, b, c and d are d b all integers. ad + bc c a Then p + q = __ + __ = ________ d b bd ad + bc Since a, b, c and d are all integers, _________ bd must be rational, and hence p + q is rational. Therefore at least one of p and q must be irrational.

1 Prove that if p is a non-zero rational number and q is an irrational number, then pq is an irrational number. (3 marks)

When completing a proof by contradiction, you should always state your assumption. Use the word ‘assumption’ or ‘assume’ to show that you understand the steps of the proof. If you have assumed that a number is rational, then a good starting point is to write it in the a form __ where a and b are integers. In some b cases you might also need to assume that a and b have no common factors. You will need to use problem-solving skills throughout your exam – be prepared!

You can use the symbol to indicate a contradiction. You could also write out ‘this contradicts the assumption that p + q is an irrational number’.

2 Prove that there is no possible value θ for which 2, sin θ and tan θ are three consecutive terms in a geometric sequence. (3 marks)

geometric sequence. See page 69 for a definition of a

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Algebraic fractions

You will usually be able to simplify algebraic fractions in your exam using these steps:

Factorise…

Cancel…

…expressions in the numerators or denominators into linear factors.

…common factors in the numerator and denominator of each fraction.

If you factorise and cancel first then your fractions will be easier to add or subtract.

Add or subtract… …any fractions by finding a common denominator, to make a single fraction.

You might need to do steps 1 and 2 again once you have added or subtracted your fractions.

Common denominators To add or subtract algebraic fractions with different denominators you need to find a common denominator. Here are two examples:

1 2

3(2 3(2 x + 1) x + 1) + 5x 5x 5 _______ _________ _________ _______________ x + 2x + 1 = x(2x + 1) + x(2x + 1) = x(2x + 1)

_3_

2 _____

x+1

The common denominator is the product of the two denominators.

2(x − 2) 2(x − 2) − 2x 2x 2x − _____________ = _____________ − ______________ = ______________ (x + 1)(x − 2) (x + 1)(x − 2) (x + 1)(x − 2) (x + 1)(x − 2) The denominators already share a factor of (x + 1), so you only have to change the first fraction.

Once you are confident you might be able to skip the step shown in green above.

After this step, you can simplify the fractions further by expanding the brackets in the numerator and collecting like terms.

2(3x + 2) ______ 2 Express ________ − 2 3x + 1 9x − 4 as a single fraction in its simplest form.

(4 marks)

2(3x + 2) __________

2(3x + 2) 2 2 − _______ = _________________ − _______ 3x + 1 (3x + 2)(3x − 2) 3x + 1 9x − 4 2 2 = _______ − _______ 3x − 2 3x + 1 2

2(3x + 1) − 2(3x − 2) = ______________________ (3x − 2)(3x + 1) 6 = ________________ (3x − 2)(3x + 1)

3x 2 − 8x − 3 1 Simplify fully ___________ x2 − 9

(3 marks)

squares. Watch out for the difference of two 2 Use a2 − b = (a + b)(a − b)

57

You will probably have to do multiple steps so get used to writing out your working neatly. You can show which factors you are cancelling by drawing a neat line through them. After you have added or subtracted your fractions you should expand the brackets in the numerator and simplify again: 2(3x + 1) − 2(3x − 2) = 6x + 2 − 6x + 4 =6 You can leave the final denominator factorised like this, or multiply it out.

x+5 1 − ______ 2 Express ___________ 2x 2 + 7x − 4 2x − 1 as a single fraction in its simplest form. (4 marks)

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Partial fractions

Many algebraic fractions can be written as the sum of simpler fractions. This technique is called writing a fraction in partial fractions. In your A-level exam you might have to use partial fractions with the binomial expansion or when integrating. You can revise these topics on pages 75 and 105.

7 − 2x f(x) = ______________ (2x − 1 ) (x + 1 ) Express f(x) in partial fractions. (3 marks) 7 − 2x A B _____________ = ______ + _____ (2x − 1)(x + 1 ) 2x − 1 x + 1 7 − 2x = A(x + 1) + B(2x − 1) 1 1 1 __ 7 − 2( __ Let x = __ 2: 2 ) = A( 2 + 1) A=4 Let x = −1: 7 − 2(−1) = B(2(−1) − 1) B = −3 4 3 ______ _____ − f(x) = 2x − 1 x + 1

Golden rule Find as many missing values as possible by substituting values for x to make some of the factors equal to zero. The more factors you can find this way, the easier it will be to equate coefficients later.

The denominators on the right-hand side are factors of the original denominator. If all the factors are different, then each one appears as a denominator once.

Cover up and calculate If the expression has no repeated factors, you can use this quick method to find numerators. Choose a factor, and work out the value of x which makes that factor equal to zero. Then cover it up with your finger, and evaluate what’s left of the fraction with that value of x.

f(x) =

x (2x − 1)(x + 1)

7−2 __________________

7 − 2(−1) 9 ___________ = ____ = −3 2(−1) − 1 −3

Covering up (x + 1) in the Worked example above and evaluating what’s left with x = −1 gives you B.

! You need to do a bit more work if there is a repeated factor, like (x − 3)2, or if the fraction is improper. You should always work out any values you can by substituting first. Here you can work out one value by substituting x = 3. To work out the other values you need to equate coefficients. You could multiply out both sides first: 2x2 − 1 = Ax2 + (−6A + B)x + (9A − 3B + C) You will need to use problem-solving skills throughout your exam – be prepared!

C 2x − 1 B _______ _______ + _______2 , x ≠ 3 2 =A+ 2

(x − 3) (x − 3) (x − 3) Find the values of A, B and C. (4 marks) 2 2 2x − 1 = A(x − 3) + B(x − 3) + C Let x = 3: 2(3)2 − 1 = C C = 17 2 Equate x  terms: A =2 Equate constant terms: −1 = 9A − 3B + C −1 = 18 − 3B + 17 −36 = −3B B = 12

need one fraction with denominator There is a repeated factor, so you 2 ominator (x + 2) . (x + 2) and another fraction with den C B 6x 2 − 1 8 x2 2 ______________ = A + ______ + _____ 1 g(x) = ______________2 (2x − 3)(x + 1) 2x − 3 x + 1 (3x − 2 ) (x + 2 ) Find the values of A, B and C. (4 marks) Express g(x) in partial fractions. (4 marks)

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Algebraic division

You might need to find missing coefficients when a cubic or quartic expression is divided by a quadratic expression. You can use long division, but make sure you set your work out neatly. 3x 4 − 6x 3 + x − 2 Here is the working for __________________ : x2 − 1

The x coefficient is 0, so write + 0x

You need to multiply (x 2 – 1) by 3x 2 to get the term 3x 4, so the first term in your answer is 3x 2

3x 2 – 6x + 3 x + 0x – 1 3x 4 – 6x 3 + 0x 2 + x 3x 4 + 0x 3 – 3x 2 –6x 3 + 3x 2 + x –6x 3 + 0x 2 – 6x Always line up terms with the same power of x. 3x 2 – 5x 3x 2 + 0x 2

If you are dividing by a quadratic, the remainder will be a linear term.

–2 –2 –2 –3

–5x + 1

The x 2 coefficient is 0, so write + 0x2 3x 2 × (x 2 + 0x – 1) = 3x 4 + 0x 3 – 3x 2

Be careful with negative terms when subtracting: –2 – (–3) = 1

Quotient

1 − 5x 3x 4 − 6x 3 + x − 2 So __________________ = (3x 2 − 6x + 3) + _______ x2 − 1 x2 − 1

Divisor

Given that dx + e 3x 4 − 2x 3 − 5x 2 − 4 _________________ ≡ ax 2 + bx + c + ______ , x≠±2 x2 − 4 x2 − 4 find the values of the constants a, b, c, d and e. (4 marks) 4 3 2 2 2 3x − 2x − 5x − 4 ≡ (ax + bx + c)(x − 4) + dx + e ≡ ax 4 + bx 3 + cx 2 − 4ax 2 − 4bx − 4c + dx + e ≡ ax 4 + bx 3 + (c − 4a)x 2 + (d − 4b)x + (e − 4c) x 4 terms → a = 3 x 3 terms → b = −2 x 2 terms → c − 4a = −5 c − 12 = −5 c=7

x terms → d − 4b = 0 d+8=0 d = −8 Constant terms → e − 4c = −4 e − 28 = −4 e = 24

Given that dx + e 2x 4 + 4x 2 − x + 2 _______________ ≡ ax 2 + bx + c + ______ , x ≠ ±1 x2 − 1 x2 − 1 find the values of the constants a, b, c, d and e. (4 marks)

59

You can also compare coefficients to find the missing coefficients. Follow these steps: 1. Multiply both sides by the divisor. 2. Expand the brackets carefully then collect like terms. 3. Compare coefficients on both sides, starting with the highest power of x.

As long as you write down what each constant is equal to, you don’t need to write out the whole expression at the end.

e Whichever method you use, make sur you either: • write out the expression in full with the constants in place, or • write a = … , b = … , etc.

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Domain and range

A function maps numbers in its domain onto numbers in its range. Here is an example: f is the name of the function. You can use any letter, but f and g are the most common.

_________

f(x) 5 √ x 2 2 ,

This is the domain of the function. The function is only defined for these input values. The range of this function is f(x) > 0. This tells you all the possible output values for the function.

x>2

This tells you what the function does to x. _______ f(18) = √___ 18 − 2 = √ 16 = 4

x is the input. You say ‘f of______ x’. √x 2 2 You can also write f : x →______ and say ‘f maps x onto √ x 2 2 ’.

Composite functions If you apply two functions one after the other, you can write a single function which has the same effect as the two combined functions. This is called a composite function. g

f

x

The function gf has the same effect as applying function f then applying function g.

f(x ) gf

The functions f and g are defined by f : x ↦ 1 + 4x 3, x ∈ ℝ 1 g : x ↦ 2 − _x_, x ∈ ℝ, 0 , x < 4 (a) Show that the composite function gf is 1 + 8x 3 (4 marks) gf : x ↦ _______3 1 + 4x 1 gf(x) = 2 − ________3 1 + 4x 2(1 + 4x 3) − 1 = ______________ 1 + 4x 3 1 + 8x 3 = ________3 1 + 4x (b) Solve gf(x) = 0 1________ + 8x 3 =0 so 1 + 4x 3

gf(x )

The order is important. The function being applied first goes closest to the x.

(2 marks) 1 + 8x 3 = 0 8x 3 = −1 1 x 3 = − __ 8

x

1 = − __ 2

The functions f and g are defined by f : x ↦ 3x + 2, x ∈ ℝ x g : x ↦ _____, x ∈ ℝ,, x ≠ −2 x+2

need to state Don’t worry about the domain. If you in the question. a domain you will be told to do so

gf(x) means you do f first, then g. To find an expression for gf(x) you need to substitute the whole expression for f(x) into the expression for g(x): _1_ g:x ↦ 2 − x Substitute the whole expression for f(x) here.

Real numbers Most of the functions you encounter will be defined on some subset of the real numbers. Here are some domains of functions: x∈ℝ –6 –5 – 4 –3 –2 –1

0

1

2

3

4

5

6

0

1

2

3

4

5

6

0

1

2

3

4

5

6

x ∈ ℝ, x > 0 –6 –5 – 4 –3 –2 –1

x ∈ ℝ, 0 < x ø 4 –6 –5 – 4 –3 –2 –1

You can write this as (0, 4]. The square bracket shows that 4 is included.

(a) Find gf(−2).

(2 marks)

(b) Show that the composite function fg is 5x + 4 (4 marks) fg : x ↦ ______ x+2 (c) Solve the equation f(x) = [f(x)]2 (4 marks)

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Graphs and range

You can use graphs of the form y = f(x) or y = g(x) to represent functions. The y-coordinates tell you the output values of the function (the range).

The function f has domain −5 < x < 7. A sketch of the graph of y = f(x) is shown below. y

as the domain. Use the same type of inequality

−8

f(x)

4

Use f(x) or y as the variable. e. Do not use x when writing the rang

4 2 O

5

values of The range is the possible output inates ord the function. This is all the y-co on the graph of the function:

5

7

(a) Write down the range of f. (1 mark) −8 f(x) 4 (b) Find ff(5). (2 marks) ff(5) = f(0) = 2 ff(5) means f[f(5)] When x = 5, y = 0, so f(5) = 0 When x = 0, y = 2, so f(0) = 2

x

8

Work out the value of the function near the edges of the domain. If you have a graphing calculator you could plot the gra ph to get an idea of its shape.

Find the range of the functions defined by: x+1 (2 marks) (a) f(x) = _____, x > 5 x−3 6 x+1 When x = 5, _____ = __ = 3 2 x−3 x+1 As x → ∞, _____ → 1 but is always > 1. x−3 Range is 1 < f(x) < 3 (b) g(x) = x2 − 6x + 5, x ∈ ℝ, 0 < x < 4 g(x) = (x − 3)2 − 4 (2 marks) So g(x) has a minimum at (3, −4). Maximum value of g(x) occurs at x = 0. g(0) = 02 − 6 × 0 + 5 = 5 Range is −4 g(x) 5

This is a quadratic function. You can find the minimum value by completing the square. Look at the edges of the domain to find the maximum. y 5

4

x

0

–4

y

1 The function f has domain −5 < x < 8 and is linear from (−5, −6) to (0, 4) and from (0, 4) to (8, 0).

4 22 O

(a) Write down the range of f. (b) Find ff(−1).

61

8

x

(1 mark) (2 marks)

2 The function g is defined by 3x − 7 2 − _____, x ∈ ℝ, x > 3 g : x ↦ __________ x2 − 5x + 6 x − 3 1 (a) Show that g(x) = _____, x > 3 (3 marks) x−2 (b) Find the range of g. (2 marks)

You need to use the fact that the two sections are linear to work out the values of the function between the given points.

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Inverse functions

For a function f, the inverse of f is the function that undoes f. You write the inverse as f21. If you apply f then f21, you will end up back where you started.

f

x

f(x)

If you apply f, then f−1 you have applied the composite function f−1f. The output of f−1f is the same as the input. You can write: f−1f(x) = ff−1(x) = x

f –1

Finding the inverse To find the inverse of a function given in the form f(x) 5 … you need to: Write the function in the form y5… Rearrange to make x the subject Swap any y’s for x’s and rewrite as f21(x) 5 …

You aren’t asked to state the dom ain here, so you don’t need to include it in you r answer.

(3 marks)

(a) Find g (x).

4 y = _____ x+1 y(x + 1) = 4 xy + y = 4 xy = 4 − y 4−y x = _____ y

Find f−1(x). y=

y (4 − x) = 4y − xy = 4y − 1 = 4y − 1 = x= 4x − 1 f−1(x) = _______ 2+x

(3 marks)

x+1 2 _______ 4−x 2x + 1 2x + 1 2x + xy x(2 + y) 4 y−1 _______ 2+y

You can often use this rule to find the domain of an inverse function.

! so

4−x g−1(x) = _____ x

(b) Find the domain of g−1. When x = 3,

2x + 1 _____ , x ∈ ℝ, x ≠ 4 4−x

The range of a function is the domain of its inverse, and vice versa.

4 ____ , x ∈ ℝ, x > 3 x+1 −1

f :x ↦

Golden rule

The function g is defined by: g:x ↦

The function f is defined by:

4 _____

(2 marks)

=1 x+1 4 As x → ∞, _____ → 0 but is always > 0. x+1 So range of g is 0 < g(x) < 1 So domain of g−1 is 0 < x < 1

Find the range of g to work out the domain of g−1. If you have a graphing calculator you can 4 sketch the graph of y = _____ to see what it x+1 looks like. y

5

O

5

10 x

You are interested in the y-values for x > 3.

The function h is defined by: x+5 h : x ↦ _____ x , x ∈ ℝ, x > 1 (a) Find h−1(x). (b) Write down the range of h−1. (c) Find the domain of h−1.

You will need to use problem-solving skills throughout your exam – be prepared!

(3 marks) (1 mark) (2 marks)

x+5 Write y = _____ x then rearrange to make x the subject. x appears twice on the right-hand side so you will need to factorise to get x on its own.

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Inverse graphs

The graph of y = f−1(x) is the graph of y = f(x) reflected in the line y = x.

You need to reflect the curve in the line y = x. y

The diagram shows part of the curve with equation y = f(x). The curve intersects the coordinate axes at (−5, 0) and (0, 2). y

y=x

y = f(x )

2 –5 O

2 25

y = f –1(x)

–5

x

O

Sketch the curve with equation y = f−1(x). (2 marks) y

x

2

You can reflect a point in the line y = x by swapping the x- and y-coordinates: (−5, 0) → (0, −5) and (0, 2) → (2, 0)

Existence of the inverse O

2

x

–5

The inverse of a function only exists if the function maps each point in its domain to a unique point in its range. This is sometimes called a one-to-one function. If f is a oneto-one function then a horizontal line drawn anywhere on the graph of y = f(x) will never cross the graph more than once. y

The function h is defined by h : x ↦ x 2 + 1, x ∈ ℝ Explain why the function h does not have an inverse. (1 mark) h is not a one-to-one function. Both of these answers are correct: • h is not a one-to-one function • h is a many-to-one function. You can also use the word ‘mapping’ instead of ‘function’ in your answer.

y = g(x )

x

O

g is NOT a one-to-one function because a horizontal line crosses the graph of y = g(x) more than once.

y

The function f has domain −8 < x < 4 and is linear from (−8, −5) to (5, 0) and from (5, 0) to (6, 4). A sketch of the graph of y = f(x) is shown.

O

Sketch the graph of y = f −1(x). Show the coordinates of the points corresponding to A and B. (3 marks) B(28, 25)

63

A(6, 4)

5

x

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Modulus

The modulus of a number is its positive numerical value. You write the modulus of x as |x|. You can use the graph of y = f(x) to sketch the graphs of y = |f(x)| and y = f(|x|).

y = f(x)

y = |f(x)|

y = f(|x|)

y

y

y

14

14

x

O

x

O

(4, –2) Any points below the x -axis are reflected in the x -axis. Every point on the curve must have a non-negative y -coordinate.

The diagram shows a sketch of the curve with equation y = f(x). y

B(5, 5)

(– 4, –2)

O

(4, –2)

x

Replace the curve to the left of the y -axis with a reflection of the curve to the right of the y -axis.

On separate diagrams sketch the following graphs, showing the coordinates of the points corresponding to A and B. (a) y = |f(x)| (3 marks) (b) y = f(|x|) (3 marks) y

y (5, 5)

x

O

14

(4, 2)

(–5, 5)

(5, 5)

(–3, 2)

A(23, 22)

0

x

x

0

Sketching y = |ax + b | You can sketch the modulus of a linear function by sketching the graph, then reflecting any points that are below the x-axis. y

y

y = 21 x + 1 1

1 –2 0

y =|21 x + 1|

x

–2 0

x

The graph of y = |ax + b| is always a V-shape.

The function f is defined by f : x ↦ 3|x| − 5, x ∈ ℝ State the range of f. (2 marks) f(x)

−5

|x| is always greater than or equal to 0.

y

The diagram shows a sketch of the curve with equation y = f(x). On separate diagrams sketch the following graphs. (a) y = |f(x)| (3 marks) (b) y = f(|x|) (3 marks) In each case, show the coordinates corresponding to the turning points A and B.

B(4, 0) O

x

A(21, 29)

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Modulus transformations

You revised these transformations of the graph of y = f(x) on pages 13 and 14: • y = f(x) + a Translation (0 a) −a • y = f(x + a) Translation ( 0 ) • y = a f(x) Vertical stretch, scale factor a

• y = −f(x)

1 Horizontal stretch, scale factor _a_ Reflection in the x-axis

• y = f(−x)

Reflection in the y-axis.

• y = f(ax)

Golden rule Carry out transformations in this order: Anything ‘inside’ the function brackets Multiples or modulus of the whole function Addition or subtraction outside the function brackets.

You need to be able to combine these transformations and use the modulus function to sketch more complicated transformations.

The diagram shows part of the graph of y = f(x), x ∈ ℝ. The graph consists of two line segments that meet at the point P(4, −3). y

1 f(|x|) + 4 y = __ 2

For part (a) you need to carry out a translation −4 l stretch with scale ( 0) followed by a vertica factor 2. y

y

y = 2f(x + 4)

y = f(x + 4)

4

x

O 23

P

Sketch, on separate diagrams, the graphs of (a) y = 2f(x + 4) (3 marks) (b) y = |f(−x)| (3 marks) On each diagram, show the coordinates of the point corresponding to P. (a)

y

(0, – 6)

x

0

x

0 (0, – 6)

For part (b) you need to carry out a reflection in the y-axis followed by a modulus. y

y = f(–x )

y

(b)

y

y = |f(–x )|

0

x

(– 4, 3)

(– 4, 3)

0

x

0 (0, – 3)

0

x

x

(– 4, –3) Have a look at page 64 for a reminder about sketching the modulus of a function.

y

The diagram shows a sketch of y = f(x). The graph has turning points at P and Q. (a) Write down the coordinates of the point to which Q is transformed on the curve with equation (i) y = 2f(2x) (ii) y = |f(x + 4)| (4 marks) (b) Sketch, on separate diagrams, the graphs of (i) y = f(−x) + 3 (ii) y = −|f(x)| (6 marks) Indicate on each diagram the coordinates of any turning points.

65

P(4, 8)

O

x Q(10, 23)

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Modulus equations

Solving an equation involving a modulus function is a bit like solving two equations. You need to consider the situations when the argument (the part inside the modulus) is positive and negative separately. You can use a graph to check that your answers make sense.

! The function f is defined by f : x ↦ |3x − 6|, x ∈ ℝ (a) Sketch the graph with equation y = f(x), showing the coordinates of the points where the graph cuts or meets the axes. (2 marks) y

To solve |3x − 6| = x you need to solve two equations: • Positive argument: 3x − 6 = x • Negative argument: −(3x − 6) = x Use your graph to check that the answers definitely exist: y y 5 | 3x 2 6 |

6

0

2

(b) Solve f(x) = x 3x − 6 = x 3x = x + 6 2x = 6 x=3

y5x

x

6

(3 marks) −(3x − 6) = x −3x + 6 = x 6 = 4x _ x = _3 2

Here is a foolproof way of solving equations involving a modulus: 1. Rearrange the equation so the modulus is on one side. 2. Solve the equation with a positive argument. 3. Solve the equation with a negativ e argument. 4. Use a graph or plug the answer s back into the original equation to check that they exist.

O

If y = x crosses y = |3x − 6| twice then there are two solutions. You will need to use problem-solving skills throughout your exam – be prepared!

Solve 4 − |x + 2| = _12_x 1 |x + 2| = 4 − __ 2x 1 x + 2 = 4 − __ 2x _3_ 2x = 2 _ x = _4 3

2 Solve 2x + 1 = 5 − |x − 1|

(5 marks)

(5 marks) 1 −(x + 2) = 4 − __ 2x 1 −x − 2 = 4 − __ 2x 1 −6 = __ 2x x = −12

10 1 _4_ _ __ _2_ __ Check: 4 − _4 3 + 2 = 4 − 3 = 3 = 2(3) ✓ 4 − |−12 + 2| = 4 − |−10| 1 = 4 − 10 = −6 = __ 2 (−12) ✓

|

1 The function f is defined by f : x ↦ |2x + 4|, x ∈ ℝ (a) Sketch the graph with equation y = f(x), showing the coordinates of the points where the graph cuts or meets the axes. (2 marks) (b) Explain why the equation f(x) = x has no solutions. (1 mark) (c) Solve f(x) = −x (3 marks)

x

2

|

Be careful. This equation has only one solution. Find separate solutions for the positive and negative arguments, then plug them both back into the equation to check which one is valid.

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Arithmetic sequences An arithmetic sequence is a sequence where the difference between consecutive terms is constant. You usually use a to represent the first term and d to represent the common difference. Here are two examples:

1 2

+3

+3

+3

+3

11, 14, 17, 20, 23 … a = 11, d = 3, n th term = 11 + 3(n − 1) − 21

− 21

− 21

− 21

1, 21 , 0, − 21 , −1 … a = 1, d = − 21 , n th term = 1 − 21 (n − 1)

The first term of an arithmetic sequence is a and the common difference is d. The 13th term of the sequence is 8 and the 16th 1 term of the sequence is 12_2 (a) Write down two equations for a and d. (2 marks) a + (13 − 1)d = 8 so a + 12d = 8 1 1 1 a + (16 − 1)d = 12 __ so a + 15d = 12 __ 2 2 2 (b) Find the values of a and d. (2 marks) 1 __ 2 − 1 : 3d = 4 2 1 __ d=1 2 1 1 Substitute d = 1 __ into 1 : a + 12 1 __ = 8 2 2 a + 18 = 8 a = −10

( )

1 The first term of an arithmetic sequence is a and the common difference is d. The 9th term of the sequence is 3 and the 11th term of the sequence is −4 (a) Write down two equations for a and d. (2 marks) (b) Find the values of a and d. (2 marks) 2 An arithmetic sequence has first term p2 + 1 and common difference p, where p > 0 The 7th term of the sequence is 24 Work out the value__ of p. Give your answer in the form a + b√2 , where a and b are integers. (5 marks)

67

Finding the nth term If an arithmetic sequence has first term a and common difference d, then the n th term, or general term is

a + (n − 1)d

You could also work out the common difference by writing down a few terms of the sequence: 1

1 1 +1 2 +1 2 +1 2

…,

8,

?,

?,

12 21 , …

en the 13th There are three jumps betwe __ 1 __ 1 term and the 16th term. 12 2 − 8 = 4 2 __1 __ 1 So each jump is 4 2 ÷ 3 = 1 2 Solve the two equations simultaneously to find a and d. Number each equation to keep track of your working. You can work in decimals or mixed numbers. You could give your answer as a = −10, d = 1.5

3 Beth is saving money for a deposit on a house. In the first month she saves £300. Each month she increases the amount she saves by £20 (a) Show that in the fifth month Beth saves £380 (1 mark) (b) Find an expression in terms of n for the amount Beth saves in the nth month. (2 marks) Use the information given to writ ea quadratic equation involving p. You need the answer in surd form so solve your equation by completing the square . There is more about this on pag e 4.

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Arithmetic series In a series, the terms are always added together. A series is arithmetic if its terms have a common difference, like 8 + 10 + 12 + 14 + … . Here is another example: −5

a + ( n − 1)d = 20 70 − 5( n − 1) = 20 n = 11 −5

Sum to n terms

70 + 65 + 60 + … + 20 You could use the formula on the right to work out the sum of this series. Use a = 70, d = −5 and n = 11. The sum is: 1 2(11)[2(70)

1

+ (11 − 1)(−5)] = 2 (11)(90) = 495

An arithmetic series has first term a and common difference d. Prove that the sum of the first n terms of the series, S, is _1 n[2a + (n − 1)d ] (4 marks) 2 1 S = a + (a + d )+ ... + [a + (n − 1)d ] 2 S = [a + (n − 1)d ] + … + (a + d) + a 1 + 2: 2S = [2a + (n − 1)d ] + ... + [2a + (n − 1)d ] 2S = n[2a + (n − 1)d ] 1 S = __ 2 n[2a + (n − 1)d ]

Substitute the values of a and d into the formula to get an expression for the sum in terms of n. Set this expression equal to 175 and simplify to get a quadratic equation in n.

Solve your quadratic equation by factorising. n is the number of terms in the series, so it must be a positive integer. This means you can ignore the negative solution.

If an arithmetic series has first term a and common difference d, then the sum of the 1 first n terms is __ 2 n[2a + (n − 1)d ]. This formula is given in the booklet. But you also need to know its proof and this is not in the formulae booklet. Look at the first Worked example below to see the proof in action.

To prove this you need to write the sum forwards and backwards. You can then add pairs of terms to get an expression for 2S: + [a + (n − 1)d] = 2a + (n − 1)d a (a + d) + [a + (n − 2)d] = 2a + (n − 1)d (a + 2d) + [a + (n − 3)d] = 2a + (n − 1)d ... and so on. Every pair of terms adds up to 2a + (n − 1)d, and there are n pairs, so 2S = n[2a + (n − 1)d]

An arithmetic series has first term 13 and common difference 4 The sum of the first n terms of the sequence is 175 (3 marks) (a) Show that n satisfies 2n2 + 11n − 175 = 0 1 a = 13, d = 4, S = 175, Sn = __n[2a − (n − 1)d ] 2 1 __ n[26 + 4(n − 1)] = 175 2 13n + 2n2 − 2n = 175 2n2 + 11n − 175 = 0

(b) Hence find the value of n. (2n + 25)(n − 7) = 0 25 n = −___ or n = 7 2

1 An arithmetic series has first term −3 and common difference d. The sum of the first 20 terms of the series is 320 Find the value of d. (4 marks) _1 2 Prove that the sum of the first n natural numbers is 2n(n + 1) (3 marks)

(3 marks)

You can’t use the formula because it says ‘prove’. Try to apply the technique in the first Worked example.

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Geometric sequences To get from one term to the next in a geometric sequence you multiply by the same number each time. You usually use a to represent the first term and r to represent the common ratio. So the sequence is of the form a, ar, ar 2, …, ar n – 1, … Here are two examples:

1

×2

3,

×2

6,

12,

×2

24,

48 …

a = 3, r = 2, n th term = 3 × 2n − 1

×(− 2 ) ×(− 2 ) ×(− 2 ) ×(− 2 ) 1

2

×2

80, −40,

1

1

20, −10,

1

5…

a = 80, r = − 21 , n th term = 80 × (−21 )n − 1

The fifth term of a geometric sequence is 324 and the eighth term is 12 1 (a) Show that the common ratio is _3 (2 marks) ar 4 = 324 1 2 ar 7 = 12 1 ___ 3 2 ÷ 1: r = 27 1 __ r= 3 (b) Find the first term of the sequence. (2 marks) 1 4 a __ = 324 3 1 ___ a = 324 81 a = 26244

( )

In a geometric sequence the ratio between consecutive terms is constant. This gives you this useful relationship: u4 u3 u2 __ __ __ =… = = u3 u2 u1 Substituting u1 = 9, u2 = k and u3 = (k + 4) into this relationship gives you a quadratic equation, which you can solve to find two possible values of k.

1 A geometric sequence has first term a = 420 5 and common ratio r = _6 Find the 20th term of the sequence. Give your answer to 3 significant figures. (2 marks)

k _____ In part (c), the common ratio will be k – 6

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General term If a geometric sequence has first term a and common ratio r, then the n th term is

un = ar n − 1 You need to learn this formula – it’s not given in the exam.

To get from the fifth term to the eighth term you multiply by r three times, so in 3 total you have multiplied by r : ×r 3 ×r

×r

×r

… , 324, ?, ?, 12, … Use the formula for the nth term to write two equations. Divide one equation by the other to eliminate a, then take the cube root of both sides to work out r. You can substitute your value of r into one of your equations to find the value of a.

The first three terms of a geometric sequence are 9, k and (k + 4) respectively. Find the two possible values of k. (5 marks) k+4 = ______ 9 k k2 = 9(k + 4) k2 – 9k – 36 = 0 (k – 12)(k + 3) = 0 k = 12 or k = –3 k __

2 The first three terms of a geometric sequence are (k – 6), k and (2k + 5) respectively, where k is a positive constant. (4 marks) (a) Show that k2 – 7k – 30 = 0 (b) Hence show that k = 10 (2 marks) (c) Write down the common ratio for this sequence. (1 mark)

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Geometric series In a series, the terms are added together. The terms in a geometric series have a common ratio. You write a for the first term and r for the common ratio. Here are two examples.

1

×3

×3

×3

Sum to n terms

2 + 6 + 18 + 54 + 162 + … a = 2 and r = 3 The sum of the first 5 terms is S 5 = 242

×(−2) ×(−2) ×(−2) 1

2

×3

1

1

30 − 15 + 7.5 − 3.75 + …

a = 30 and r = −21 . When r is negative, the terms alternate between + and −. You can still use the formula to work out Sn :

[ ( __ ) ] ___________

30 1 –

S4 =

1–

–

1 2

( ) –

1 __

4

If a geometric series has first term a and common ratio r, then the sum of the first n terms is a(1 – r n) a(r n – 1) Sn = ________ or Sn = ________ 1–r r–1 You can use either of these versions. The first one appears in the formulae booklet. You also need to know its proof and this is not in the formulae booklet. Look at the Worked example below to see the proof in action.

= 18.75

2

! To prove this you need to write out Sn and then multiply every term by r. If you subtract rSn from Sn most of the terms cancel: a + ar + ar2 + … + ar n – 2 – ar n – 1 – ar – ar2 – … – ar n – 2 – ar n – 1 – ar n You are left with Sn – rSn = a – ar n. You can factorise the left-hand side and divide by (1 – r) to get Sn on its own. You will need to use problem-solving skills throughout your exam – be prepared!

10

Find ∑ 50(2 ) k

k=1

(3 marks)

10

∑ 50(2k ) = 100 + 200 + 400 + … + 51 200

k =1

So a = 100 and r = 2 100(1 – 210) S10 = ___________ = 102 300 1–2

A geometric series has first term a and common ratio r. Prove that the sum of the first n terms of a(1 – r n) (4 marks) the series, Sn, is ________ 1–r 1 Sn = a + ar + ar 2 + ... + ar n – 1 2 rSn = ar + ar 2 + … + ar n – 1 + ar n 1 – 2 : Sn – rSn = a – ar n Sn(1 – r) = a(1 – r n) a(1 – r n) Sn = ________ 1–r

For a reminder about sigma notation have a look at page 72. It’s a good idea to write out a few terms to check that you have the correct values for a and r. You can enter the calculation in one go on your calculator using the keys. and You can’t use the formula. Try to apply the technique in the first Worked example.

1 A geometric series has first term a = 7 and 3 common difference r = _2 . Find the sum of the first 20 terms of the series, giving your answer to the nearest whole number. (2 marks)

2 In the geometric series 1 + 2 + 4 + 8 + … each term is twice the previous term. Prove that the sum of the first n terms of this (4 marks) series is 2n – 1

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Infinite series

1 1 1 This diagram shows the geometric series 1 + __ + __ + … with first term 1 and common ratio __ 2 4 2 1 1 1 + + + 81 + … 2 4

0

1

2

The sum gets closer and closer to 2, but never reaches it. You say that the sum of infinity of this series is 2. You write the sum to infinity as S∞.

Convergent series

Sum to infinity For a geometric series with first term a and common difference r, the sum to infinity is a S∞ = ____ where –1 < r < 1 1–r 1 In the example above, a = 1 and r = __, 2 1 so S∞ = _______1 = 2 1– 2

The sum to infinity of a geometric series only exists if the series is convergent. This only happens if the common ratio, r, is between –1 and 1. In the formulae booklet this condition is written as |r| < 1.

Be careful – the inequalities are strict. Geometric series with r > 1, r = 1, r < –1 or r = –1 are not convergent, so S∞ doesn’t exist. 1 In the example above, r = __ so the series is convergent. 2

Find the sum to infinity of the geometric series 6 12 _3_ 1 ___ 1 ___ 1 … (3 marks) 5 15 45 6 3 3 2 a = __, r = ___ ÷ __ = __ 5 15 5 3 _3_ 5 9 S∞ = _____ = __ 5 2 1 – __ 3 Use the formula for S∞ to write an equation and solve it to find r. Remember that r can be positive as well as negative. This is the convergent geometric series 432 144 ____ _____ 80 – 48 + 5 – 25 + … 80 __ = 50 ______ which has sum to infinity _3_ 1– – 5

( )

You need to find the values of a and r before you can use the formula for S∞. a is the first term and r is the common ratio, so u2 6 ÷ _3_. You can check your ___ __ = r= 5 15 u1 value of r using u2 and u3: u3 6 = _2_ ✓ 12 ÷ ___ ___ __ = 3 15 45 u2

The first term of a geometric series is 80 and the sum to infinity is 50. Find the common ratio, r. (3 marks) 80 ____ = 50 1–r 80 = 50(1 – r) 50r = –30 3 r = – __ 5

1 Find the sum to infinity of the geometric series 15 + 12 + 9.6 + … (3 marks) 2 The first term of a geometric series is 2 and the common ratio is k. The sum to infinity of the series is 3k + 4, where k is a constant. Find the value of k. (5 marks)

71

Use the formula for S∞ to write a quadratic equation involving k. Remember that in order for the sum to infinity to exist, k must be between –1 and 1.

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Recurrence relations A recurrence relation tells you how to find one term in a sequence from the previous term. You can use the recurrence relation and the first term to generate the sequence. You can use any letter to represent a term in a sequence.

This number tells you the term number. The first term is 5.

x1 = 5 xn + 1 = 2xn + 3, n > 1

The term-to-term rule is The term numbers in a series ‘multiply by 2 then add 3’. are all positive integers.

So for this sequence: x2 = 2x1 + 3 = 2(5) + 3 = 13 x3 = 2x2 + 3 = 2(13) + 3 = 29 ... and so on.

Increasing, decreasing or periodic? You need to be able to identify the following properties of some sequences: A sequence is increasing if xn + 1 > xn for all n ∈ ℕ.

2, 4, 6, 8… is an increasing sequence

A sequence is decreasing if xn + 1 < xn for all n ∈ ℕ. __1 __1 __1 __1 2, 3, 4, 5…

is a decreasing sequence

A sequence is periodic if its terms repeat in a cycle. For a periodic sequence there exists an integer k such that xn + k > xn for all n ∈ ℕ. The value of k is called the period or order of the sequence. −1, 0, 1, 0, −1, 0, 1, 0… is a periodic sequence with order 4

A sequence a1, a2, a3, … is defined by a1 = k, an + 1 = 3an + 1, n > 1, where k is a constant. (a) Find the set of values of k for which the sequence is increasing. (3 marks) an + 1 > an 3an + 1 > an 1 an > −__ 2 1 So the sequence is increasing for all k > −__ 2 3

(b) Given that ∑ ar = 44, show that k = 3 r =1 (4 marks) a1 = k and a2 = 3k + 1 a3 = 3a2 + 1 = 3(3k + 1) + 1 = 9k + 4 3

∑ ar = a1 + a2 + a3

r =1

= k + (3k + 1) + (9k + 4) = 13k + 5

13k + 5 = 44 13k = 39 so k = 3

If the sequence is increasing then an + 1 > an Use the recurrence relation to rearrange this __1 into the form an > − 2 . This must be true for all values of n, so it must be true for a1 = k

Sigma notation

The Greek letter sigma ( ∑ ) is used to show the sum of a set of terms in a sequence. The numbers above and below the ∑ tell you which terms to add together:

4

∑

r =1

You are adding the first four terms of the sequence.

ur = u1 + u2 + u3 + u4 Substitute values of r from 1 up to 4, and add the terms. 3

ar in terms of k. Write an expression for r∑ =1

1 A sequence a1, a2, a3, … is defined by a1 = 5, an + 1 = 2an − 4, n > 1 (a) Find the values of a2 and a3, and state whether the sequence is increasing, decreasing or neither. (3 marks) 5

(b) Calculate the value of ∑ ar

(3 marks)

r =1

Write a quadratic equation and solve it by factorising. Remember that p is positive.

2 A sequence x1, x2, x3, … is defined by x1 = 2, xn + 1 = pxn − 1, n > 1 where p is a positive constant. (a) Write down an expression for x2 in terms of p. (1 mark) (2 marks) (b) Show that x3 = 2p2 − p − 1 (c) Given that x3 = 9, find the value of p. (3 marks)

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Modelling with series You can model a variety of real-life situations using sequences or series.

Nisha is buying a car. She pays for it monthly over 12 months. The garage offers two repayment plans. Both plans form an arithmetic sequence. Plan A: First month payment £X Payments decrease by £2Y each month Plan B: First month payment £(X − 1100) Payments decrease by £Y each month (a) Show that the total amount paid under Plan A for the 12-month period is £(12X − 132Y) (2 marks) a = X, d = −2Y, n = 12 1 Sn = __ n[2a + (n − 1)d ] 2 1 = __(12)[2X − 2Y (12 − 1)] 2 = 6(2X − 22Y ) = 12X − 132Y

(b) For the 12-month period, the total paid is the same for both plans. Find the value of Y. (4 marks) 1 __ For Plan B, Sn = n[2a + (n − 1)d ] 2 1 = __(12)[2(X − 1100) − Y (12 − 1)] 2 = 12X − 66Y − 13200 12X − 66Y − 13200 = 12X − 132Y 66Y = 13200 Y = 200 (c) Under Plan A, Nisha’s final payment would be £300. Work out the value of X. (3 marks) X + (12 − 1)(−400) = 300 X − 4400 = 300 X = 4700

Hanna is planning a training regime. On the first day she does 25 sit-ups. Each day she increases the number of sit-ups by 6. (a) Find an expression for the number of sit-ups Hanna does on the nth day of her regime. (2 marks)

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The payments decrease, so the values of d in your arithmetic sequences will be negative. For Plan A: a = X, d = −2Y and n = 12 For Plan B: a = (X − 1100), d = −Y and n = 12 Write down the formula for the sum of the first n terms of an arithmetic series. This formula is given in the formulae booklet. Then substitute for a, d and n and simplify your expression.

Write your expression for the total paid under Plan B, make it equal to 12X − 132Y and solve an equation to find Y.

! The monthly payments decrease by 2Y each month, so d = −400. Make sure you use the formula for the nth term, not for the sum. Write an equation and solve it to find the value of X. You will need to use problemsolving skills throughout your exam – be prepared!

(b) Show that the total number of sit-ups she had done after n days is 3n2 + 22n (3 marks) (c) On the nth day of her regime, Hanna does 79 sit-ups. How many sit-ups has she done in total in the n days of her regime? (4 marks)

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Series and logs You can use logarithms to answer questions involving geometric sequences and series. For a reminder about working with exponentials and logs, look at page 50.

Modelling with geometric sequences A car costs £16 000 new and depreciates in value by 20% each year. Its value each year 4 produces a geometric sequence with first term a = 16 000, and common ratio r = 0.8, or __ 5 Year 1

× 0.8

£16000

Year 2

× 0.8

£12800

Year 3

£10240

…

Year n

_________ log 0.8 n > 7.212…

Don’t try to write log0.8 0.25 < n – 1 and use the calculator – it’s safer to take logs of both sides.

key on your

log 0.8 is negative so when you divide both sides by log 0.8 you have to reverse the inequality.

n must be a whole number, so the answer is n = 8

! A geometric series has first term 5 and 3 common ratio _2 . Find the smallest value of n for which the sum of the first n terms of the series exceeds 12 000 (4 marks) a________ (1 – r n ) a = 5, r = 1.5, Sn = 1–r 5(1 – 1.5n ) __________ > 12 000 1 – 1.5 5(1 – 1.5n ) < –6000 1 – 1.5n < –1200 1.5n > 1201 log (1.5n ) > log 1201 n log 1.5 > log 1201 log 1201 n > ________ = 17.488 log 1.5 n = 18

Be really careful with inequalities. • If you multiply or divide by a negative number, change the direction of the inequality sign. • If a < 1 then log a is negative. • Check that your answer makes sense. You’re usually looking for the next integer greater than the value you calculate. • Double-check whether you should be using the formula for the nth term, arn – 1, or the a(1 2 r n) formula for the sum to n terms, ________ 12r You will need to use problem-solving skills throughout your exam – be prepared!

1 A biologist models the number of trout after n years as 200 × (1.4)n – 1. Her model predicts that after k years, the number of trout in the lake will exceed 600. log 3 (2 marks) (a) Show that k > ______ + 1 log 1.4 (b) Find the smallest possible value of k. (2 marks)

2 A geometric series has first term 4 and 9 common ratio __ 10 . Find the smallest value of n for which the sum of the first n terms of the series exceeds 30. (4 marks)

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Binomial expansion 2

You need to be able to use the binomial theorem to find a series expansion of expressions in the form (a + bx)n, where n is any real number. You need to use this version of the binomial series, which is given in the formulae booklet: The expansion is

1) r r +___ −___ − 1).. n(n ___ ___ ___.(n n(n − 1)_ 2 ___ x + ... (⎪x⎪< 1, n ∈ ℝ) n = 1 + nx + ______ x + ... + r × ... × 2 × x) 1 (1 + 1×2

only valid for values of x between −1 and 1.

If you are given an expression in the form (a + bx)n you will need to rearrange it by taking out a factor n bx bx ___ _a_ of an like this: (a + bx)n = a n (1 + ___ a ) . In this case the expansion is valid for a < 1 or |x| < b

| |

1 ______ f(x) = _______ √ 4 + 5x (a) Find the binomial expansion of f(x), in ascending powers of x, as far as the term in x3, and state the range of values of x for which it is valid. (6 marks) _1

−2

f(x) = (4 + 5x )

5x 1 = __ 1 + ___) 2( 4

=4

1 −__ 2

_1

−2

(1 + 4 ) 5x ___

1 −__ 2

1 1 −__ −_3_ −__ −_3_ −_5_ 3 2 ( 2 ) ___ 2 ( 2 )( 2 ) ___ 5x 5x 2 5x ________ _____________ 1 1 ___ __ __ = 2 [1 + (− 2 )( ) + ( + ( 1 × 2 × 3 )( 4 ) + ...] 1 × 2 )( 4 ) 4 75 2 625 3 1 _ = __ 1 − _5 x + ___ x − ____ x + …] 2[ 8 128 1024 75 2 5 625 3 1 = __ − __ x + ___ x − ____ x +… 2 16 256 2048

4 Valid for |x| < _5_

3+x _____ (4 marks) (b) Hence find the coefficient of x in the series expansion of _______ √ 4 + 5x 3+x 75 5 625 _________ 1 _______ = (3 + x) __ − __x + ____x2 − _____x3 + … (2 ) 16 256 2048 √ 4 + 5x

5 1 x term in expansion = (3)(−__ x + (x)(__ 16 ) 2)

7 15 1 = (−__ + __ x = −__ x 16 2) 16

7 So coefficient of x is −__ 16

Start by writing the expression in the form (a + bx)n, then take out a factor of a n. You can now use the binomial series from the formulae booklet, with 1 ing n = − __ 2 , replac 5x x with ___ 4 A common mistake is not to square or 5x cube all of ___ , so 4 use brackets when you write out the series. You can work out the coefficients in one go on your calculator using brackets and the key.

! In your exam, you might be asked to find a binomial series in lots of ways: • ‘Find the binomial expansion of …’ • ‘Use the binomial theorem to expand …’ • ‘Find the series expansion of …’ • ‘Expand f(x) in ascending powers of x …’ You will usually be told the highest power of x you need to find. You will need to use problem-solving skills throughout your exam – be prepared!

75

3

______

(a) Expand √ 1 − 6x , in ascending powers of x up to and including the x3 term, simplifying each term. State the range of values of x for which the expansion is valid. (4 marks) (b) Use your expansion, with a suitable value ____ 3 of x, to obtain an approximation to √ 0.94 . Give your answer to 6 decimal places. (2 marks)

You must use your answer to part (a). To find a suitable value of x, solve 1 – 6x = 0.94

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Radians, arcs and sectors If you have to solve length or area problems involving circles and triangles in your A-level exam, any angles will usually be measured in radians.

Writing radians Make sure you know whether you are working in radians or degrees. These two angles are measured in radians.

There are π radians in a semicircle, or 2π radians in a full turn.

π rad

π radians = 180° so 1 radian = 180° π

180 × π

Radians

If the angle is written in terms of π, then it is definitely in radians.

Degrees ×

π 3

0.7

This means 0.7 radians. This angle could also be written as 0.7 rad.

π 180

Radians in length and area formulae Radians are really useful for working out arc lengths and sector areas. Here are two formulae that only work if the angle θ is measured in radians:

1

Arc length , = rθ

This is the Greek letter θ. It is sometimes used to represent an unknown angle.

The diagram shows ABC, a sector of a circle with centre A and radius 9 cm. B

2

r

θ

ℓ A r

_1

Sector area A = 2r 2θ

Neither of these formulae is in the booklet, so make sure you learn them.

The angle, θ, is given in radians, so you can use the formulae for arc length and sector area given above.

9 cm 1.3 rad A

C

Given that the size of /BAC is 1.3 radians, find (a) the length of the arc BC (2 marks) ℓ = rθ = 9 × 1.3 = 11.7 cm (b) the area of the sector ABC. (2 marks) 1 2 1 __ __ 2 × 9 × 1.3 = 52.65 cm2 A= r θ= 2 2

Remember to write out the formula first. That way, you will get some credit even if you make a mistake when you’re substituting your values.

Notation: ITC Stylus for Pearson ED, Bold, 9/11pt ITC Stylus for Pearson ED XBd f(x) 5 x3 12345,. 2 x2 P(A ∩ ∩ A) ∈ℝ a bc x y θ α β δ η µ π σ ϕ x∈ =+–×÷> 0 and 0 < α < __ 2 (4 marks) ______ ___

R = √5 2 + 3 2 = √34 3 α = arctan (_5_) = 0.5404… ___ 5 cos x − 3 sin x = √34 cos (x + 0.5404…)

(b) Hence, or otherwise, solve the equation 5 cos x − 3 sin x = 4 for 0 < x < 2p, giving your answers to 2 decimal places. (5 marks) ___

√ 34 cos (x + 0.5404...)

=4 4 ___ cos (x + 0.5404...) = _____ √ 34 x + 0.5404... = 0.8148... x = 0.8148... − 0.5404... = 0.2744... or x + 0.5404... = 2π − 0.8148... = 5.4683... x = 5.4683... − 0.5404... = 4.9279... x = 0.27, 4.93 (2 d.p.)

Be careful. In part (b) of this questio n, the principal value from your calculator will not give you an answer in the require d range.

1 (a) Express 3 sin 2θ + 2 cos 2θ in the form R sin (2θ + α), where R > 0 and p (4 marks) 0 < α < __ 2 (b) Hence, or otherwise, solve the equation 3 sin 2θ + 2 cos 2θ + 1 = 0 for 0 < θ < p, giving your answers to 2 decimal places. (5 marks)

and

b α = arctan (_a_)

above, you If you don’t want to learn the rules page 82: can use the addition formulae from R sin x sin α 5 cos x − 3 sin x = R cos x cos α − R cos α ➀ Equate coefficients of cos x: 5 = ➁ R sin α Equate coefficients of sin x: 3 = 2 2 2 2 2 5 +3 ➀2 + ➁2: R (cos α + sin α) = ________2 R = √ 52 + 3 _3_ R sin ___α_ ___ ➁ ÷ ➀: R cos α = tan α = 5 _3_ α = arctan ( 5 )

! Pay attention to these three Rs when solving trig equations in your exam: • Rounding − Don’t round any values until the end of your calculation. Learn how to use the ‘STORE’ or ‘MEMORY’ functions on your calculator so you can use unrounded values, or write down values to at least 4 decimal places. • Radians − Look at the range to decide whether you should be working in degrees or radians. Make sure your calculator is in the correct mode. • Range − Check that all your solutions are within the specified range, and check that you have found every possible solution within that range. You will need to use problem-solving skills throughout your exam – be prepared!

2 The function f is __ defined by y = f : x ↦ √ 3 cos x + sin x (a) Given that f(x) = R cos (x − α), where R > 0 and 0 < α < 90°, find the value of R and the value of α. (4 marks) (b) Hence sketch the graph of y = f(x) for 0 < x < 360°, showing clearly the coordinates of any maxima or minima, and any points where the graph meets the coordinate axes. (5 marks)

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Trig modelling Tides, circular motion, pendulums and springs are just a few examples of real-life situations that can be modelled using trigonometric functions.

(a) Express 2 sin θ − 1.5 cos θ in the form p R sin (θ − α), where R > 0 and 0 < α < __. 2 Give the value of α to 4 decimal places. (3 marks) ________ R = √2 2 + 1. 5 2 = 2.5 1.5 α = arctan ( ___ 2 ) = 0.64350… 2 sin θ − 1.5 cos θ = 2.5 sin (θ − 0.6435…)

(b) (i) Find the maximum value of 2 sin θ − 1.5 cos θ (ii) Find the value of θ, for 0 < θ < p, at which this maximum occurs. (3 marks) (i) 2.5 (ii) sin (θ − 0.6435…) = 1 π θ − 0.6435… = __ 2 θ = 2.2143 (4 d.p.)

Tom models the height of sea water, H metres, on a particular day by the equation 4pt 4pt H = 6 + 2 sin (____) − 1.5 cos (____), 0 < t < 12, 25 25 where t is the number of hours after midday. (c) Calculate the maximum value of H predicted by this model and the value of t, to 2 decimal places, when this maximum occurs. (3 marks)

4πt H = 6 + 2.5 sin (____ − 0.6435…) 25 Hmax = 8.5 4πt Occurs when ____ = 2.2143… 25 so t = 4.41 (2 d.p.)

(d) Calculate, to the nearest minute, the times when the height of sea water is predicted, by this model, to be 7 metres. (6 marks) 4πt 6 + 2.5 sin (____ − 0.6435…) = 7 25 4πt sin (____ − 0.6435…) = 0.4 25 4πt ____ − 0.6435… = 0.4115... or π − 0.4115... 25

t = 2.0988… = 2.06 pm

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t = 6.7115… = 6.43 pm

Maximum and minimum

A function in the form R cos (θ ± α ) or R sin (θ ± α) has a maximum value of R and a minimum value of −R. y R

0

y = R sin (θ – a)

π a 2+

3π α 2 +

θ

–R

The maximum value of R sin (θ − α) occurs when sin (θ − α) = 1. π 5π This occurs when θ − α = __, ___, … 2 2

There will often be a connection bet ween different parts of the same exam question. Look for similarities betwee n the complicated equation in part (c) and the expression in part (a). you find Part (d) says ‘times’ so make sure of the all possible solutions for the range utes. model. Remember to convert to min : 60 Multiply the decimal part by 0.0988… × 60 = 5.9…

The displacement, d cm, of a pendulum at time t seconds is modelled by the equation d = 4 cos 1.2t + 2 sin 1.2t, t > 0 (a) Given that d = R cos (1.2t − α), where p R > 0 and 0 < α < __, find the value of R 2 and the value of α. (3 marks) (b) Write down the maximum displacement of the pendulum. (3 marks) (c) Find the times in the interval 0 < t < 5 when the displacement of the pendulum is 0. (2 marks)

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Parametric equations 1

You can define a curve by giving the x-coordinate and y-coordinate as separate functions of another variable, called a parameter. y t=0

π t =–2

π t =2

x

0 t=π

Golden rules Each value of t generates a point on the curve. The value of t is neither the x- nor the y-coordinate of the curve.

The curve with parametric equatio ns y = 3 cos t + 2, x = 3 sin t + 1, −π < t < π is a circle, centre (1, 2) with radius 3.

Converting to cartesian Cartesian equations only involve x and y. To convert to cartesian form you need to eliminate the parameter, t. If the equations contain trig functions use an identity like sin2 t + cos2 t ≡ 1 Otherwise, write t in terms of either x or y then substitute into the other equation.

The curve crosses the y-axis whe n x = 0. Substitute x = 0 into the x-equat ion to find the value of t at this point. Then sub stitute this value of t into the y-equation to find the value of y at this point. ms of x Find an expression for cos t in ter of y. You ms and an expression for sin t2 in ter 2 eliminate can then use sin t + cos t ≡ 1 to ation t. Finally, rearrange the cartesian equ n. stio into the form asked for in the que

The curve C has parametric equations x = ln t, y = t2 − 3 Find a cartesian equation of C. (3 marks)

y

P 21 O

3

x

The curve C shown has parametric equations x = 1 + 2 cos t, y = sin t, 0 < t < p (a) Find the exact coordinates of the point P where the curve crosses the y-axis. (3 marks) When x = 0, 1 + 2 cos t = 0 1 cos t = − __ 2 2π t = ___ 3__ √3 2π 2π When t = ___, y = sin ___ = ___ 2 3 3 __ √ 3 P is (0, ___) 2

(b) Find a cartesian equation of C in the form (3 marks) y2 = f(x). x−1 sin t = y cos t = _____ 2 2 2 cos t + sin t = 1 x_____ −1 2 2 ( 2 ) +y = 1 x−1 2 y2 = 1 − ( _____) 2

t = ex y = (ex)2 − 3 = e2x − 3

Start by using the addition formula for sin (A + B) to write y in terms of sin t and cos t.

p p p x = sin t, y = sin (t + __), − __ < t < __ 3 2 2 Show that a cartesian equation of the curve is __ ______ √ 3 y = _12_ x + ___ √ 1 − x2 , −1 < x < 1 (3 marks) 2

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Differentiating sin x and cos x You need to be able to differentiate sin and cos from first principles. For a reminder about differentiation from first principles, have a look at page 35.

! Prove, from first principles, that the derivative of cos x is −sin x. You may assume that as cos h − 1 sin h h → 0, ____ → 1 and ________ → 0. (5 marks) h h Let f(x) = cos x f(x + h) − f(x) _____________ f9(x) =hlim →0 h cos(x + h) − cos(x) ___________________ =hlim →0 h (cos x cos h − sin x sin h) − cos x _______________________________ =hlim →0 h cos h − 1 sin h _________ ____ =hlim ) cos x −( h ) sin x) →0(( h cos h − 1 sin h Since _________ → 0 and ____ → 1 the limit h h tends to 0 × cos x − 1 × sin x = −sin x So f9(x) = −sin x as required.

1. Write out the rule for differentiation from first principles – this is given in the formulae booklet. 2. Substitute f(x) = cos x. 3. Use the addition formula on cos(x + h). 4. Rearrange the expression in the limit to sin h cos h − 1 get the expressions ________ and ____ on h h their own – this will allow you to use the conditions given in the question. 5. State how you are using the conditions given in the question. 6. Write down what you have proved. You will need to use problem-solving skills throughout your exam – be prepared!

Small angle approximations The proof shown above makes use of small angle approximations for sin and cos. When x is small and measured in radians: 1 2 sin x ≈ x cos x ≈ 1 − __ tan x ≈ x 2x You can use these approximations to deduce the limits given in the proof above. When h is small: sin h h ____ ≈ __ = 1 and h h 1 2 1 − __ cos h − 1 2h − 1 1 _________ ____________ ≈ = −__ 2h h h This tends to 0 as h → 0

1 (a) Given that f(x) = sin x, show that cos h − 1 sin h f9(x) = lim((________)sin x + (____) cos x) h→0 h h (3 marks) (b) Hence prove that f9(x) = cos x (2 marks) 2 Given that θ is small and measured in 1 − cos 3θ 9 radians, show that _________ ≈ __ (4 marks) tan θ sin θ 2

87

Show that, for small values of θ measured in 1 − 4 cos θ + sin θ (4 marks) radians, ________________ ≈ −1 3 + tan 2θ 1 − 4 cos θ + sin θ __________________ 3 + tan 2θ

≈

1 2 1 − 4(1 − __ 2θ ) + θ ___________________

3 + 2θ 2 2θ + θ−3 = _____________ 3 + 2θ (2θ + 3)(θ − 1) _______________ = 2θ + 3 =θ−1 θ − 1 → −1 as θ → 0

that small If θ is small then you can assume are also small. 3θ multiples of θ such as 2θ and __ 3 f(x) = cos x √3 _π_ ___ Prove, from first principles, that f9( ) = − 3 2 (5 marks)

In question 3, follow the same ste ps that are shown in the worked example above, but replace x with _π_. 3

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The chain rule

The chain rule lets you differentiate a function of a function. The chain rule is not in the formulae booklet, so make sure you know how to use it confidently. Follow this step-by-step method: Substitute the ‘inside’ function with u . _____

y = √x 3 + 1 u = x3 + 1 __ 1 __ y = √u = u 2

Treat u as a single variable and differentiate.

Multiply the result by the derivative of u .

dy ___

du ___

du

1 − __1 = __ u 2 2

= 3x 2 dx dy du dy 3 x2 ___ ______ = ___ × ___ = _________ 3 dx du dx 2 √x + 1

1 __ 1 3 = __ ( x + 1 )− 2 2 1 _________ _____

2 √x 3 + 1

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

=

This is how the chain rule is sometimes stated.

2 f(x) = ______ , x > 0 3x + 1 Differentiate f(x) and find f 9(1). (3 marks) −1 y = f(x) = 2(3x + 1) = 2u−1 with u = 3x + 1 dy du dy ___ = ___ × ___ = −2u −2 × 3 dx du dx = −6(3x + 1)−2 −6 −6 _ f9(x) = _________2 f9(1) = ___________2 = −_3 8 (3x + 1 ) (3 × 1 + 1 )

Splodge You can use the splodge method to apply the chain rule quickly. The ‘inside’ function dy is splodge. You work out ___________ , then d(splodge) multiply by the derivative of splodge! y = ( 5x – 1 )9 dy = 5 × 9( 5x – 1 )8 dx Derivative of splodge

this You don’t have to write down all of your exam. working to apply the chain rule in aight str p If you’re confident you can jum from the first line of working to: −2 f 9(x) = −2(3x + 1) × 3

dy d(splodge)

Functions of y C is a curve with equation 2y3 + 10y + 2 = x dy (3 marks) (a) Find ___ in terms of y. dx d y dx 1 ___ = 6y 2+ 10 so ___ = ________ dy dx 6y 2 + 10

You can use this version of the chain rule to differentiate equations where x is given in dy 1 terms of y: ___ = _____ dx dx ___ ( dy )

(b) Find the gradient of the curve at the point (−8, −2). (2 marks) d y 1 1 = ___ y = −2, so ___ = ____________ 34 dx 6(−2 )2 + 10

is given in terms of y so you need to dx substitute the y-coordinate to fi nd the gradient.

dy 1 Given that y = , find ___ 2 dx √ x − 3x + 1 (3 marks)

2 f(x) = (√ x + 6)6. Differentiate f(x) and find f 9(8), writing your answer as a power of 2. (4 marks)

1 ___________ __________

dy ___

3

__

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Differentiating standard functions You need to learn these four derivatives for your A-level exam. They’re not given in the formulae booklet and you need to be really confident using them.

Trigonometric functions d ___ ( dx

Exponential functions d ___ (

sin x ) = cos x

dx

d ___ ( dx

ex ) = ex

d ___ (

cos x ) = −sin x

dx

These rules only work for angles measured in radians.

1 ln x ) = _x_

ex is the only function that is the same when differentiated.

Round in circles Differentiate with respect to x (a) sin (x2 + 1) d ___ [sin (x 2 + 1)] = 2x cos (x2 + 1) dx (b) sec x d d ___ ___ (sec x) = (cos x ) −1 dx dx = −(cos x ) −2 × (−sin x ) sin x = ______ = sec x tan x cos2 x

(2 marks)

If you keep differentiating sin x you will end up back where you started. d dx

(3 marks)

sin x d dx

– cos x

cos x

d dx – sin x

d dx

these. You need the chain rule for both of

The function f is defined by f : x ↦ ln (4 − x 2 ), 0 < x , 2 Find the exact value of the gradient of the curve y = f(x) at the point where it crosses the x-axis. (5 marks)

2x 1 f9(x) = ______2 × (−2x ) = ______ 2 4−x x −4 When f(x) = 0, ln(4 − x 2) = 0 4 − x 2 = 1 __ __ __ x = √ 3 or x = −√ 3 __ __ 2 √3 f9(√3 ) = _________ = −2 √3 __ 2 (√ 3 ) − 4

The value of a boat, £V, after t years is modelled as _1_ V = 12 000 e −5t, t > 0 dV (a) Find ___ (2 marks) dt 1 __ dV 1 ___ = 12 000 e − 5t × (− __ 5) dt 1 __ = −2400 e − 5t dV (b) Find the exact value of V when ___ = −800 dt (3 marks) 1 __ −800 = −2400 e − 5t 1 __ 1 e − 5t = __ 3 1 1 __ − __ 5 t = ln ( 3 ) = −ln 3 t = 5 ln 3

Use

1 Differentiate with respect to x (a) sin 2x − cos 4x (b) x2 − e4x − 3 (c) ln (3x2 + 1)

89

(2 marks) (3 marks) (2 marks)

dy ___ dx

1 2 = _____, and sin2 2y + cos 2y = 1 x d ___ ( dy )

2 Given that x = sin 2y, show that dy ________ 1 ___ = ______ dx 2 √ 1 − x 2 3 Use the chain rule to show that d ( 2 ) ___ sin x = sin 2x dx

(4 marks)

(3 marks)

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The product rule The product rule lets you differentiate two functions multiplied together. It’s not given in the formulae booklet so you need to learn it. Here’s an example of a function that can be differentiated using the product rule:

y = ex (x2 − 2) v

Differentiate with respect to x (a) y = x4 ln 3x

(3 marks)

u=x v = ln 3x d du v 1 ___ ___ = 4x3 = _x_ dx dx dy dv du 1 ___ = u ___ + v ___ = x 4( _x_) + (ln 3x) (4x3) dx dx dx = x 3(1 + 4 ln 3x) 4

(b) y = ex(1 + cos 2x)

(3 marks)

u= v = 1 + cos 2x d d u v ___ ___ = ex = −2 sin 2x dx dx dy du dv ___ = u ___ + v ___ = e x (−2 sin 2x) + (1 + cos 2x) e x dx dx dx = e x (1 − 2 sin 2x + cos 2x ) ex

Formulae booklet Unless you are specifically asked to show them, you can quote these results from the formulae booklet:

f(x) tan kx sec kx cot kx cosec kx

If y = uv, where u and v are functions of x, then

dy dv du = u ____ + v ____ dx dx dx

____

In function notation, if h(x) = f(x) g(x), then h9(x) = f(x) g9(x) + g(x) f 9(x)

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ ⎫ ⎬ ⎭

u

Golden rule

f 9(x) k sec2 kx k sec kx tan kx −k cosec2 kx −k cosec kx cot kx

For any constant, k 1 × k = 1_ d (ln kx ) = __ ___ x kx dx It’s easy to get in a mess with the product rule. du d Start by writing out u, v, ___ and ___v. d x d x Then use brackets when you substit ute to make sure you don’t make a mistak e.

(a) Given that h(x) = e2x cosec x, find h9(x). (4 marks) h(x) = f(x) g(x) f(x) = e2x g(x) = cosec x 2x f 9(x) = 2e g9(x) = −cosec x cot x h9(x) = f(x) g9(x) + g(x) f9(x) = e 2x (−cosec x cot x ) + (cosec x)(2e 2x ) = e 2x cosec x (2 − cot x )

p (b) Solve, in the interval 0 < x < __, the 2 equation h9(x) = 0. (4 marks) e 2x cosec x (2 − cot x) = 0

For part (b) you need to look at the factors of h9(x). e2x and cosec x can never equal 0, so the factor (2 − cot x ) must equ al 0.

1 Differentiate with respect to x __ (a) √ x sin 5x (b) ln xx

Write ln x x as x ln x

(3 marks) (3 marks)

e 2x ≠ 0

cosec x ≠ 0

2 − cot x = 0 1 tan x = __ 2 x = 0.464 (3 s.f.)

2 A curve has equation y = e x(3x2 − 4x − 1) dy (4 marks) (a) Find ___ dx (b) Find the x-coordinates of the turning points on the curve. (4 marks)

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The quotient rule

If one function is divided by another you can differentiate them using the quotient rule. One version is given in the formulae booklet:

f(x)_ ___ g(x)

function

) − f(x) g9(x) (g(x))2

derivative

But the version on the right is often easier to use, so you should learn it for your exam.

sin 5x Differentiate _____ with respect to x. (3 marks) x2 u = sin 5x v = x2 dv du ___ ___ = 5 cos 5x = 2x dx dx du dv v ___ − u ___ dy (x 2)(5 cos 5x) − (sin 5x )(2x) dx dx ___ __________ ___________________________ = = dx v2 (x 2) 2 5x cos 5x − 2 sin 5x = __________________ x3

rule using This solution shows the quotient answer to e function notation. You will hav in the questions about curves or graphs tions func ut abo ns form y = … and questio a to ide d in the form f : x ↦ … so it’s a goo be comfortable with both versions.

Be really careful with signs when using the quotient rule. The minus sign on top means you have to subtract all of f(x) g9( x).

91

u If y = _v_ , where u and v are functions of x, then

du − u ____ dv v ____ d y dx dx ____ = ______________ 2 dx v

g(x___________ f 9(x)___ ___

1 Differentiate with respect to x cos__x (a) _____ √x 2x _____ (b) _______ √ 3x + 1 x2 + 1 (c) _________ ln (x 2 + 1)

Golden rule

(3 marks) (3 marks) (4 marks)

The minus sign on top of the fraction means the order is important.

by writing When using the quotient rule, start dv ___ du ___ ts when out u, v, dx and dx . Then use bracke don’t make you substitute to make sure you a mistake. Make sure you are really confiden t differentiating expressions like sin 5x using the chain rule.

The function h is defined by ex + 2 , x ∈ ℝ, x ≠ ln 3 h : x ↦ ______ ex − 3 −5 e x (3 marks) Show that h9(x) = ________ (e x − 3) 2 f(x) h(x) = ____ g(x) f(x) = ex + 2 g(x) = ex − 3 g9(x) = ex f 9(x) = ex f9(x) g(x) − f(x) g9(x) h9(x) = _________________ [g(x)] 2 x x e (e − 3) − (e x + 2) e x = _______________________ (e x − 3) 2 −5e x = _________ (e x − 3) 2

2 Use the quotient rule to show that d ___ (4 marks) (cot x) = −cosec 2 x dx 9 3 3 − ______________ 3 f(x) = _____ − ______ 2 2 x − 1 x + 2 (x + 2)(x − 1) 3x (a) Show that f(x) = ______ (4 marks) 2 x +2 (b) Hence, or otherwise, find f 9(x) in its simplest form. (3 marks)

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Differentiation and graphs

dy In your exam you might have to use ___ to find the gradient of a curve at a given point. You can use dx this information to find turning points, and to find the equations of tangents and normals to the curve.

A curve C has equation 5 3 y = ________2 , x ≠ __ 3 (5 − 3x) The point P on C has x-coordinate 2. Find an equation of the normal to C at P in the form ax + by + c = 0, where a, b and c are integers. (7 marks) y = 3(5 − 3x ) −2 dy ___ = −6(5 − 3x ) −3 × (−3) dx = 18 (5 − 3x ) −3 18 = _________3 (5 − 3x ) d y 18 At P, ___ = ___________3 = −18 dx (5 − 3(2)) 3 ____________ =3 At P, y = (5 − 3(2)) 2 so P is the point (2, 3) 1 −1 Gradient of normal at P = _____ = ___ −18 18 1 y − 3 = ___ (x − 2) 18 18y − 54 = x − 2 x − 18y + 52 = 0

Tangents and normals If the point P (x0, y0) lies on the curve with equation y = f(x): The tangent at P has gradient f9(x0 ) An equation for the tangent could be y − y0 = f9(x0)(x − x0)

−1 The normal at P has gradient _____ f9( x 0 )

An equation for the normal could be −1 y − y0 = _____ (x − x0 ) f9(x 0 )

! It’s easier to use the chain rule than the quotient rule to differentiate a function like this. Rewrite it with a negative power. Make sure you read the question carefully. You need to find the normal and not the tangent, and you will lose a mark if you don’t give your final equation in the correct form. You will need to use problem-solving skills throughout your exam – be prepared!

The curve y = f(x) has turning poi nts when = 0. To find the coordinates of P , you need dx to solve the equation e2x (1 + tan 2 x) = 0 dy ___

y

The diagram shows a sketch of the curve with p p equation y = e2x tan x, − __ < x < __ 2 2 The curve has a turning point at P. dy (a) Show that ___ = e2x (1 + tan x)2 dx (b) Find the exact coordinates of P.

(3 marks) (3 marks)

π 22

O P

π x 2

(c) Find an equation of the tangent to the curve at the point where x = 1. Give your answer in the form y = ax + b where a and b are constants given to 3 significant figures. (3 marks)

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Parametric differentiation You can use this version of the chain rule to find the gradient of a curve defined by parametric equations.

Golden rule dy dy dx ____ = ____ ÷ ___ dx dt dt

This formula gives you the derivative in terms of t. You need to know the value of t for a particular point on the curve to find the gradient of the curve at that point.

A curve C has parametric equations p x = sin2 t, y = 2 tan t, 0 < t < __ 2 dy ___ (a) Find in terms of t. (4 marks) dx d y dx ___ = 2 sin t cos t, ___ = 2 sec 2 t dt dt 2 d y 2 sec t 1 ___ = ___________ = __________ dx 2 sin t cos t sin t cos 3 t p The tangent to C at the point where t = __ 3 cuts the x-axis at the point P. (b) Find the x-coordinate of P. (6 marks) _π_ At the point where t = : __ 3 √3 2 π_ 3_ ___ 2 _ x = sin ( ) = ( ) = _4 2 3 __ π y = 2 tan(__) = 2 √ 3 3 d y 16 1 1 ___ __ __ = ______________ = _________ = ___ π _π_ _ _ dx 3 √ √ 3 3 1 ___ __ sin( ) cos ( ) 3 3 ( 2 )( 8 ) Equation of tangent: __ 16 3 __ x − __ y − 2 √ 3 = ___ ( 4) √3 __ 16 3 __ x − __ At P, y = 0, so −2 √ 3 = ___ ( 4) √3 _ _3_ −_3 8 = x−4

given in If you have to differentiate a curve asked to parametric form you will usually be dy ___ and dx ___ dy ___ to find dx in terms of t. Find dt and dt dy ___ dx ___ . ÷ out k wor n The n. dow m the write dt dt wer, but You don’t have to simplify your ans r in the it might speed up your working late question.

Here is a sketch of the curve C and the __ tangent at the point where t = π : 3 y

5 4

π t =3 ( 43 , 2 3)

3 2 1

O

C P ( 83 , 0) 0.5

1

x

_ x = _3 8

93

The diagram shows a sketch of the curve C with parametric equations p 0 < t < 2p x = 2 cos (t − __), y = 3 sin 2t, 3 dy (a) Find ___ in terms of t. (4 marks) dx (b) Find an equation of the normal to the curve at the point where t = 0. (4 marks) dy (c) Find the coordinates of all the points on C where ___ = 0. dx (5 marks) d y Find all the values of t where ___ = 0, then find dx the x- and y-coordinates at each of these points.

y C O

x

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Implicit differentiation An implicit equation is one that cannot easily be written in the form y = f(x) or x = g(y). You can differentiate an implicit equation to find dy an expression for ___ in terms of x and y. Follow dx these steps: • Differentiate every term on both sides of the equation with respect to x. dy • Collect terms involving ___ on one side and the dx remaining terms on the other side. dy • Factorise to get ___ on its own. dx

Golden rules These rules will help you with implicit differentiation in your exam: dy d ___ [f(y)] = f 9(y) ___ dx dx d ___ dx

dy [e 2y] = 2 e 2y ___ dx

dy [g(x)y] = g9(x)y + g(x) ___ dx dx

d ___ d ___ dx

dy [x 3 y] = 3x 2 y + x 3 ___ dx

! Make sure you differentiate the constant term on the right-hand side: d ___ [1] = 0 dx You can use golden rule 1 above to differentiate 3 cos 2y with respect to x. Remember that you have x and y in your dy expression for ___ so you probably won’t be dx able to simplify it very much. You will need to use problem-solving skills throughout your exam – be prepared!

A curve C is described by the equation 3x2 + 4y2 − 2x + 6xy − 5 = 0 Find an equation of the tangent to C at the point (1, −2), giving your answer in the form ax + by + c = 0, where a, b and c are integers. (7 marks)

dy dy 6x + 8y ___ − 2 + 6y + 6x ___ = 0 dx dx d y d y 8y ___ + 6x ___ = 2 − 6y − 6x dx dx d y ___ (8y + 6x) = 2 − 6y − 6x dx 2 − 6y − 6x dy ___ = ____________ dx 8y + 6x 1__________________ − 3y − 3 x = 4y + 3x dy 1 − 3(−2) − 3(1) 4 At the point (1, −2), ___ = __________________ = − _5_ dx 4(−2) + 3(1)

A set of curves satisfy the equation 6 sin 2x + 3 cos 2y = 1 dy (3 marks) Find ___ in terms of x and y. dx dy 12 cos 2x − 6 sin 2y ___ = 0 dx dy 2 cos 2x ___ = ________ dx sin 2y

can Be careful with the 6xy term. You 2 rule n use the product rule (golde ted above). Once you have differentia the every term, you needdy to rearrange ___ on one equation so all the dx terms are dy ___ side. You can then factorise to get dx er on its own, then divide by the oth ) −2 (1, at factor. To find the gradient you need to substitute x = 1 and ___ dy . for n sio res exp r you y = −2 into dx

1 The point P with coordinates (3, −1) lies on the curve with equation x3 + y2 + 3x2y = 1 dy 9 (5 marks) Show that at P, ___ = − __ 25 dx 2 The curve C is described by the equation 2x2 − y2 = y e3x (a) Show that the point (0, −1) lies on C. (1 mark) (b) Find an equation of the tangent to C at the point (0, −1). (7 marks)

4

Equation of tangent: y − (−2) = −_5_ (x − 1) 4x + 5y + 6 = 0

d ___

dx (

dy y e 3x) = 3y e 3x + e 3x ___ dx

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Differentiating ax

d You can use ___(e x) = e x to differentiate dx expressions involving ex. You need to be more careful when differentiating powers of constants other than e.

d (a x )= a x ln a ____ dx

Two methods You can use the laws of logs to differentiate a x. y = a x = ex ln a dy ___ = ln a ex ln a = a x ln a dx

You can differentiate a x implicitly.

You can learn this rule but you need to know how to derive it as well. Follow the steps in the box on the right to use ex to differentiate a x.

dy Given that y = 3x, show that ___ = 3 x ln 3 dx (2 marks) x ln 3 y=e d y ___ = ln 3 ex ln 3 dx = 3x ln 3

You might have to differentiate a x as part of an implicit differentiation. As long as you’re not asked to show it, you can use the rule: d x ___ (a ) = a x ln a dx

uting Be really careful when you’re substit e you sur ke with implicit differentiation. Ma correct substitute the x- and y-values in the dy ___ places in your expression for dx . iation on page 94. There’s more on implicit different

y = ax ln y = ln a x = x ln a dy _1_ ___ y dx = ln a dy ___ = y ln a = a x ln a dx

ting Make sure you’re confident conver x and ex ln a: between a x x ln a a x = e ln a = e

A curve has equation 2x + y2 = 2xy dy Find the exact value of ___ at the point on C dx with coordinates (3, 2). (7 marks) dy dy 2x ln 2 + 2y ___ = 2y + 2x ___ dx dx d y d y 2y ___ − 2x ___ = 2y − 2x ln 2 dx dx y d ___ (2y − 2x) = 2y − 2x ln 2 dx 2y − 2 x ln 2 dy ___ = __________ dx 2y − 2x dy 2(2) − 2 3 ln 2 At the point (3, 2), ___ = _____________ dx 2(2) − 2(3) = −2 + 4 ln 2

2 A curve has equation xy + (_21_) = 2 Find an equation of the normal to the curve at the point (0, −1). (6 marks) y

d 1 Find ___ (4 x sin x) dx

Use the product rule. Look at page 90 for a reminder.

95

(3 marks)

–1__ ___ The gradient of the normal will be ___ dy ( dx )

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Points of inflexion

You can use second derivatives to determine where a curve is concave or convex, and to find points of inflexion. A point of inflexion occurs when the curve changes from being concave to convex, or vice versa. dy = 3x2 – 18 x + 24 dx d2y and 2 = 6x – 18 dx d2y So ø 0 for all dx 2 values of x ø 3. This means the curve is concave on this interval.

y 30

Golden rules

y = x 3 – 9x 2 + 24x point of inflexion

The curve y = f(x) is convex on the interval [a, b] if and only if f0(x) > 0 for all a < x < b.

d2y dx 2 ù 0 for all values of x ù 3. This means the curve is convex on this interval. The point where it changes from being concave to convex is the point of inflexion.

20 10

–4 –3 –2 –1 O

The curve y = f(x) is concave on the interval [a, b] if and only if f0(x) < 0 for all a < x < b.

1

2

3

4

5

6

7

x

Use the product rule twice.

The curve C has equation y = x ex (a) Show that C is convex on the interval [0, 1]. (5 marks) dy ___

= x ex + ex = (x + 1)ex

dx 2 d y ____

= (x + 1)ex + ex = (x + 2)ex

dx ex > 0 for all x ∈ R and x + 2 > 0 for all d2y x ∈ [0, 1], so ____ > 0 for all x ∈ [0, 1] dx2 so C is convex on that interval. 2

(b) Find the exact coordinates of the point of inflexion on C. (3 marks) d y ____

d2y ____ = 0 ⇒ x = −2. When x < −2, −2, ____ > 0 so x = −2 is a dx2 point of inflexion. At x = −2, y = −2e−2, so the coordinates are (−2, −2e−2). 2

1 The curve C has equation y = x + x − x + 2 dy d2y (a) Find ___ and ____2 (4 marks) dx dx (b) Find the coordinates of the stationary points on C and determine their nature. (5 marks) (c) Find the coordinates of the point of inflexion on C. (3 marks) 3

2

Watch out! There are two things to be careful of when finding points of inflexion: A stationary point on a graph can be a local maximum, a local minimum, or a point of inflexion. But, in general, a point of inflexion does not have to be a stationary point. Although f0(x) must equal 0 at a point of inflexion, having f0(x) = 0 doesn’t necessarily guarantee a point of inflexion. You also need the sign of f0(x) to change on either side of that point. For example, when x = 0, the curve y = x4 has d2y ____ = 0. But x = 0 is a local minimum on this curve, dx2 d2y is never negative. not a point of inflexion, because ____ dx2

2 The curve C has equation y = (x2 − 2)ex Show that this curve has two stationary points and find their coordinates. (8 marks) 3 The curve C has equation y = x4 + 3x3 − 6x2 + 2 (a) Show that C has two points of inflexion, and find the values of x at these points. (7 marks) (b) State, with a reason, whether C is concave or convex on the interval [1, 2]. (2 marks)

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Rates of change You can model lots of physical or financial situations by describing how a variable changes with time. Equations involving rates of change are called differential equations. You can revise how to form differential equations on this page, and how to solve them on page 109.

If you need to use the formula for the volume of a cone in your exam it will be given to you with the question. You can use similar triangles to find the rela tionship between r and h. Draw a sketch to help you.

16 cm

r 24 cm h

A container is made in the shape of a hollow inverted right circular cone. The height of the container is 24 cm and the radius is 16 cm, as shown in the diagram. Water is flowing into the container. When the height of water is h cm, the surface of the water has radius r cm and the volume of water is V cm3. 4p h 3 (2 marks) (a) Show that V = _____ 27 [The volume V of a right circular cone with vertical height h and base radius r is given by the formula V = _13_ pr 2h.] _r_

h

=

16 ___ 24

so r = 1 __

r

dV Use your answer to part (a) to find ___. dh The rate the water is flowing into the dV container is ___ = 8 cm3 s–1. The rate of dt dh change of h is ___ . dt

For part (a), remember that θ is a constant.

24 cm

h

Water flows into the container at a rate of 8 cm3 s−1. (b) Find, in terms of p, the rate of change of h when h = 12. (5 marks) 4p h 2 dV 12p h 2 ___ = _______ = ______ dh 27 9 4p h 2 dV dV dh ___ = ___ ÷ ___ = 8 ÷ ______ dt dt dh 9 9 = 8 × ______2 4p h 18 ______ = p h2 When h = 12: 18 dh 1 ___ = _________2 = _____ dt 8 p p (12)

97

The chain rule allows you to multiply and divide derivatives in the same way as fractions. dV dV dh dh dV dV ___ = ___ × ___ so ___ = ___ ÷ ___ dt dt dt dh dt dh

16 cm

2h ___

3 2 2h V = 3 πr 2h = 3 π (___) h 3 3 4π h = ______ 27 1 __

Chain rule

θ

h 6 cm

40 cm θ

The diagram shows a section of gutter in the shape of a prism. The cross-section of the gutter is a symmetrical trapezium. Water is flowing into the gutter. When the depth of the water is h cm, the volume of the water is V cm3. dV (a) Show that ___ = 240 + 80h cot θ (3 marks) dh Water flows into the gutter at a constant rate of 40 cm3 s−1. (b) Given that when h = 2.5 the rate of change of h is 0.1 cm s−1, find the value of θ correct to 1 decimal place. (5 marks)

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Iteration

An iterative process is one where an answer is fed back in as a new starting value. You can use an iterative formula to find numerical answers to a given degree of accuracy.

Change of sign 3

You can use a change of sign (from positive to negative, or vice versa) to show that a particular interval contains a root of an equation.

2

f(x) = x + 2x − 3x − 11 (a) Show_______ that f(x) = 0 can be rearranged as 3x + 11 x = _______ , x ≠ −2 x+2

√

(2 marks)

y = f( x ) x -axis

x + 2x − 3x − 11 = 0 3

2

x 2(x + 2) − 3x − 11 = 0

1.645

_______ x 2(x + 2) = 3x + 11 3 x + 11 3x + 11 _______ ______ 2 so x = x = x+2 x+2

√

The equation f(x) = 0 has one_______ positive root, α. 3x n+ 11 The iterative formula xn+1 = _______ is used to xn + 2 find an approximation to α. (b) Taking x1 = 0, find, to 3 decimal places, the values (3 marks) of x2, x3 and x4.

√

________

√ 3x + 11 = √ ________ = 2.03732494... = 2.037 (3 d.p.) x +2

3(0) + 11 x 2 = ________ = 2.34520788... = 2.345 (3 d.p.) (0) + 2

You can iterate quickly using the Ans func  tion on your calculator: 3Ans + 11 Ans + 2

2

√

2 ________

3x 3 + 11 x 4 = _______ = 2.05874811... = 2.059 (3 d.p.) x3 + 2

2.03732494

(c) Show that α = 2.057 correct to 3 decimal places. (3 marks) f(2.0565) = (2.0565)3 + 2(2.0565)2 − 3(2.0565) − 11 = −0.01378… Negative f(2.0575) = (2.0575)3 + 2(2.0575)2 − 3(2.0575) − 11 = 0.0041401… Positive Change of sign so 2.0565 < α < 2.0575, so α = 2.057 correct to 3 d.p.

f(x) = ln (x + 1) − 2x + 2, x > 0 The equation f(x) = 0 has one root, α. (a) Show that 1 < α < 2. (2 marks)

1.655

f(1.645) is positive and f(1.655) is negative so f(x) = 0 has a root, α, between 1.645 and 1.655. All values in this interval round to 1.65, so α = 1.65 to 2 decimal places.

________

x3

α

You can visualise this iteration usin g a cobweb diagram. This shows the xn values converging on the root. y =x

(x3, x4) (x3, x3)

y =

3x + 11 x+2 (x2, x3)

(b) Use the iterative formula xn+1 = _12_ ln (xn + 1 ) + 1, x0 = 1.5 to calculate values of x1, x2 and x3, giving your answers to 5 decimal places. (3 marks) (c) Show that α = 1.4475 correct to 4 decimal places. (3 marks)

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The Newton–Raphson method You can use the Newton–Raphson method to find a numerical solution to an equation of the form f(x) = 0. The method works by using tangents to get closer and closer to a root. Each improved approximation, xn + 1, is the point where the tangent to the curve y = f(x) at x = xn crosses the x-axis.

y

y = f(x)

Tangent to y = f(x) at x = x0

x1 x2

O

x0

Tangent to y = f(x) at x = x1

f(x) = x4 − ex The equation f(x) = 0 has a root, α, in the interval [1, 2]. Taking x0 = 1.5 as your starting value, apply the Newton-Raphson process once to find a second approximation to α. (5 marks)

x

Root of f(x) = 0

Golden rule The formula for the Newton–Raphson method is given in the formulae booklet:

f(xn ) xn+1 = xn − _____ f9(x n)

f9(x) = 4x3 − ex With x0 = 1.5: f(x0) 1.54 − e1.5 x1 = x0 − _____ = 1.5 − _______________ f9(x0) 4 × 1.53 − e1.5 = 1.436 (3 d.p.)

Start by differentiating f(x). Make sure you check that your final answer is in the correct interval.

Failure cases The choice of starting value is important in the Newton–Raphson method. Here, three very similar starting values are used to try and find an approximation to α: y y y y = f(x) y = f(x) y = f(x)

O

x0

x

α

The process will converge on a different root of f(x) = 0.

y

The diagram shows part of the curve with equation y = f(x) where__ f(x) = x3 − 2√ x3 O

x0

O

α

x0 is a stationary point of the curve, so f9(x0) = 0. The tangent is horizontal and never intersects the x-axis.

y 5 x3 2 2 x3

x

(a) Show that f(x) = 0 has a root, α, in the interval [1, 2]. (2 marks)

99

x

O

x0 α

x

The gradient of the tangent is small so the point of intersection is far from α. The process will converge slowly.

(b) Find f9(x). (2 marks) (c) Explain why x0 = 1 would not be a suitable first approximation when applying the Newton–Raphson procedure to find α. (1 mark) (d) Taking x0 = 2 as a first approximation, apply the Newton–Raphson procedure twice to f(x) to obtain an approximate value of α. (5 marks)

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Integrating standard functions Make sure you are really familiar with these four integrals. They’re not given in the formulae booklet and you will use them a lot in your exam.

Trigonometric functions

∫ cos x dx = sin x + c

∫ e x dx = e x + c

∫ sin x dx = −cos x + c

∫

Remember that indefinite integrations always include a constant of integration, c.

+ b)

Integrating f9(ax

This is a useful integration rule to learn:

∫

Exponential functions

f9(ax + b) dx = _1_ f(ax + b) + c

a

This is an application of the chain rule ‘in reverse’. There are more examples of this on the next page.

( x ) dx

1

__

= ln|x| + c

1 You use a modulus sign when you integrate _x_, because ln x only exists for positive values of x.

(a) Find ∫ cos 2x dx

(2 marks)

∫cos 2x dx = __21 sin 2x + c

∫

6 (b) Find ______ dx 3x + 1

(2 marks)

∫ 3x6+ 1 dx = 2 ln|3x + 1| + c ______

1 dx 6 dx = 6 ______ ∫ ∫ ______ +1 x 3 3x + 1

This is a definite integration. You don’t need to use a constant of integra tion.

Show that 1 _____

__

4 √2 − 2 ∫ √2 − x dx = _______ 3 1 _____

0

1

∫ √2 − x dx = ∫ (2 − x) dx

(3 marks)

1 __ 2

0

[

0

2_ = −_3 (2 − x ) 2

]0

_3_ 1

_3_ 2_ _( ) _32_ (2) 2) = (−_2 1 ) − (−_3 3 __ __ 4 √2 − 2 _2_ _2_( √ ) _________ = −3 + 3 2 2 = 3

1 (a) Find ∫ 4 sin 2x dx x _ (b) Find ∫ e 6 dx 2 Find the exact value of 3 Find

∫ 1 − 4x dx 2 ______

∫

(2 marks) (2 marks)

_p_ 2

0

2 sin _1 θ dθ 2

(3 marks)

You can use the rule for integrating f(ax + b) to integrate any of these functions: f(x) = ___1___ f(x) = (ax + b)n ax + b f(x) = sin(ax + b) f(x) = cos(ax + b) Remember to multiply by _1_ and leav e a ax + b inside the function. You can always check your answers by differentia ting.

2___ dx = 2 ∫ ___1___ dx ∫ ___ 1 − 4x 1 − 4x

Use the rule for integrating function s of the form fˇ(ax + b), and be carefu l with the signs.

(2 marks)

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Reverse chain rule You can use the chain rule in reverse to integrate some expressions. Here are two useful examples. If you can spot these integrations you can save time in your exam.

∫f9(x)[f(x) ] dx = n +1 1 [f(x)] n

_______

n+1

For example:

∫2x (x

2

For example:

n = 10 1 2 11 + 1) dx = __ 11 (x + 1) + c

f9(x) = 2x

∫ f(x)

f9(x) ____ dx = ln|f(x)| + c

+c

f9(x) = 6x2

∫ 2x6x− 1 dx = ln|2x 2

10

_______

3

3

− 1| + c

f(x) = 2x3 − 1

f(x) = x2 + 1

Adjusting constants

∫ 5x

x (a) Find _______ dx 2

(2 marks)

+1

∫ 5x x+ 1 dx = ∫ 5x10+x 1 dx 1 _______ __ 10 2

_______ 2

If you can guess the form of the integral, but can’t work out what constant to multiply it by, then you can: Differentiate your guess for the integral.

1 = __ ln|5x2 + 1| + c 10 _______

(b) Find ∫ x2 √ 1 − 4 x3 dx

∫x

2√

_______ 3

∫

1 __

1 − 4x dx = x 2(1 − 4x 3) 2 dx _3_ 3 2

Try y = (1 − 4x ) 1 1 __ __ dy ___ = −12x 2 × _32_ (1 − 4 x 3) 2 = −18x 2 (1 − 4x 3) 2 dx _______ _3_

∫

1 ( So x 2 √ 1 − 4x 3 dx = − __ 1 − 4x 3) 2 + c 18

! Watch out for common standard derivatives, especially when trigonometric functions are involved. You might need to use results from the formulae booklet to spot the form of the integral.

f(x) sec kx

f9(x) k sec kx tan kx

∫ x + 2x − 3 2 cos 3x Find ∫ _______ dx sin 3x

Adjust the constant if necessary.

constant For part (b) it’s hard to see what , write a ead to take outside the integral. Inst guess for the integral as ‘y = …’ __ 1 dy 2 ( 1 − 4x 3 ) 2 . ___ x with it e par com Now find dx and nd, so It is −18 times the original integra divide y by −18.

Find ∫ 2 tan x sec5 x dx

∫2 tan x sec

∫

(3 marks)

x dx = 2 tan x sec x (sec x )4 dx Try y = sec5 x dy ___ = sec x tan x × 5 sec4 x dx = 5 tan x sec5 x 5

You will need to use problem-solving skills throughout your exam – be prepared!

∫

2_ So 2 tan x sec5 x dx = _5 sec5 x + c

x+1 1 Find __________ dx 2

(3 marks)

2

(3 marks) x Try y = ln (2e − 1)

101

Compare it with the original expression.

(3 marks)

3 The function f is defined by ex , x>0 f : x ↦ _______ x 2e − 1 Find the area enclosed by the curve with equation y = f(x), the line x = 1 and the coordinate axes. Give your answer correct to 3 decimal places. (4 marks)

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Identities in integration You can use trigonometric identities to simplify integrations. If you’re not sure which identity to use, have a look at the standard integrals in the formulae booklet. See if you can write the integral in terms of one of these functions.

of sec2 You can write the integral in terms using the identity 2 sec2 θ ≡ 1 + tan θ

2 tities involving sec Have a look at page 80 for iden 2. and cosec

x Find tan2(__) dx 2

∫

(2 marks)

∫tan (__2x ) dx = ∫(sec (__2x ) − 1) dx 2

2

x = 2 tan (__) − x + c 2

Use this result from the formulae booklet 1 with k = __ : 2 f(x)

∫ f(x) dx

sec2 kx

_1_ tan

k

kx

sin2 x and cos2 x You can integrate sin2 x and cos2 x using the double angle formulae for cos:

∫

2 + 3p Show that __p sin x dx = ______ π

∫ _ sin π 4

2

p 4

∫

2

π

8

(5 marks)

1 1 __ x dx = _π ( __ 2 − 2 cos 2x) dx 4

cos 2A ≡ 1 − 2 sin2 A

π 1 1 __ = [ __ 2 x − 4 sin 2x ] _π_ 4

π π 1 π 1 = (__ − __ sin 2π) − (__ − __ sin __) 2 4 2 8 4 3π 1 = ____ + __ 4 8 2 + 3π = ________ 8

1 Find the exact value of

∫

p __ 12

0

sin 3x cos 3x dx (5 marks)

2 (a) By writing sin 7x as sin (4x + 3x) and by writing sin x as sin (4x − 3x), show that sin 7x + sin x ≡ 2 sin 4x cos 3x (4 marks) (b) Hence, or otherwise, find ∫ sin 4x cos 3x dx

cos 2A ≡ 2 cos2 A − 1

Have a look at page 83 for a reminder about these identities.

Use identity 2 from the box above to write sin2x in terms of cos 2x. The questio n says ‘show that’, so make sure you cle arly show the integrated function before sub stituting your limits and evaluating the inte gral.

Use the identity sin 2A ≡ 2 sin A cos A with A = 3x Use the addition formulae on pag e 82 and add together the two expressions. Part (b) says ‘Hence’ so you can save a lot of time by using your answer to part (a) to simplify the integral.

(2 marks)

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Integration by substitution

You can use a substitution to turn a complicated integral into a simpler one. If you have to use a substitution in your exam it will usually be given with the question. e3x Follow these steps to find _________ dx using the substitution u = e3x : (e3x + 1)2

∫

du Find ___ and write an expression dx for dx in the form f(x) du

u = e 3x

Swap dx for f(x) du in the integral and simplify if possible.

∫ (e e+ 1)

du ___

This is an expression for dx in the form f(x) du.

= 3e 3x dx 1 dx is equivalent to ( _____ du 3e 3x ) so

3x

_________ 3x

2

e3x 1 1 _____ dx = _________ du = __________ du 3x (e + 1)2 ( 3e3x ) 3(e3x + 1)2

∫

∫

∫

∫

1 1 −2 du = ________2 du = __ 3 (u + 1) 3(u + 1)

Substitute every x to get an integral involving only u and du.

Integrate with respect to u.

1 ( ) −1 + c = − __ 3 u+1

Use your substitution in reverse to get an answer in terms of x.

1 ( 3x ) = − __ 3 e +1

−1

The new integral should be easier to find.

+c

Transforming limits Use the substitution x = sin θ to find the exact _1 1 dx (7 marks) value of 2 ________ 2 _3 0 (1 − x ) 2

∫

dx x = sin θ so ___ = cos θ, so dx = cos θ dθ dθ 1 1 __ _π_ When x = __ 2 , θ = arcsin 2 = 6 When x = 0, θ = arcsin 0 = 0 _π __________ 1 __ 1 1 2 ________ 6 d x = cos θ dθ _3_ 2 _23_ 2 0 (1 − x ) 0 (1 − sin θ) 2 _π 1 = 6 _________3_ cos θ dθ 2 0 ( cos θ) 2

∫

∫ ∫ = sec θ dθ ∫ _π 6

0

2

_π_

__

= [tan θ] = 3 0 6

√3 ___

1 Use the substitution u = 3 + sin x to show that sin 2x 6 _________ ________ 2 dx = 2 ln (3 + sin x) + 3 + sin x + c (3 + sin x)

∫

where c is a constant.

103

(5 marks)

When you use substitution for a definite integral you need to use the substitution to transform your limits. Transform limits from values of x to values of θ.

∫

1 __

x=2

x =0

1 ________ _3_

(1 − x2) 2

dx →

∫

_π

θ= 6

θ=0

sec2 θ dθ

You can now use your values of θ with the integrated expression to evaluate the integral.

You might need to use trigonomet ric identities in an integration by sub stitution. You can use sin2 θ + cos2 θ ≡ 1 to write the whole integral in terms of cos θ. 1 __ ______ cos θ ≡ ___1___ cos θ ≡ ___1___ (cos2 θ) _23_ cos3 θ cos2 θ

2 Use the substitution u2 = 2x + 1 to find the exact value of 4 _______ 4x ______ dx (7 marks) 0 √ 2x + 1

∫

ntiation You will need to use implicit differe and u to find the relationship between d reminder. dx. Have a look at page 94 for a

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Integration by parts You can integrate some functions written as a product of two functions using integration by parts. This rule is given in the formulae booklet:

du

dv u ___ is the function you want to integrate. You have to dx work out which part of the function to set as u and dv which part to set as ___ . dx

If there’s no ln x, set x or x2 as u

(a) Use integration by parts to find ∫ x sin 3x dx (3 marks) u=x du ___ =1 dx x sin 3x dx =

∫

!

u = ln x du 1 ___ = _x_ dx

∫

3 1

dv ___

=1 dx v=x

ln x dx = [x ln x] − 3 1

∫

3 1

1 dx

= [x ln x − x ] 13 = (3 ln 3 − 3) − (1 ln 1 − 1) = 3 ln 3 − 2

= sin 3x dx 1 v = − __ 3 cos 3x

1 1 __ (x)(− __ 3 cos 3x ) − (− 3 cos 3x )(1) dx 1 1 __ = − __ 3 x cos 3x + 3 cos 3x dx

= − 3 x cos 3x +

1 __ 9

∫

∫

sin 3x + c

(b) Using your answer to part (a), find ∫ x2 cos 3x dx (3 marks)

u = x2 du ___ = 2x dx x 2 cos 3x dx

∫

dv ___

= cos 3x dx 1 v = __ 3 sin 3x

1 1 __ = (x 2)( __ 3 sin 3x ) − ( 3 sin 3x) (2x) dx

= =

You will need to use problem-solving skills throughout your exam – be prepared!

1

dv ___

1 __

You can integrate ln x by writing it as ∫(lnx) (1) dx du 1_ . You always set u = ln x, so ___ = x dx This means that the second part of the formula becomes: 1 ∫(x)(_x_ ) dx = ∫1 dx

3

dx

dv Set ex, sin x or cos x as ___ dx

Always write down which part of the function you are setting as u and dv ___ which part you are setting as dx . du ___ Then calculate dx and v and write them down before substituting. You don’t need to include a constant of integration when you work out v but you should include one at the end of your final integral.

∫ lnx dx

dv ___

Follow these rules for choosing which parts of dv the function to set as u and ___ in your exam: dx always set ln x as u

dv dx = uv − v ___ dx ∫ dx ∫ u ___ dx

Find the exact value of

Choosing u and

=

(4 marks)

1 2 __ 3 x sin 3x 1 2 __ 3 x sin 3x 1 2 __ 3 x sin 3x

−

∫

∫

_2_ 3 x sin 3x dx 1 1 _2_ __ __ 3 − 3 x cos 3x + 9 sin 3x + 2 _2_ __ 9 x cos 3x − 27 sin 3x + c

− ( +

1 Use integration by parts to find 1 __ ln x dx x2 2 (a) Find ∫ x ex dx (b) Hence show that

∫

∫ x e dx = e − 2 1

2 x

0

)

c

(4 marks) (3 marks) (4 marks)

dv ___ x You always set e as dx

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Integrating partial fractions You can integrate some functions by writing them as partial fractions. You can revise this technique on page 58.

4x 3 + 5 = 1 + _5__ + ___________ −8 + ________________ 11 __________________

Once it is written in partial fractions, you can integrate the expression term-by-term.

x (2x − 1) 2

x

This integrates to 5 ln |x|

2x − 1

(2x − 1) 2

This integrates to −4 ln |2x − 1|

Write this as 11(2x − 1)−2 to integrate it. The result is __ (2x − 1)−1 − 11 2

∫

4x3 + 5 __ (2x − 1) −1 + c So _________2 dx = x + 5 ln|x | − 4 ln|2x − 1| − 11 2 x(2x − 1)

5x + 3 (a) Express ______________ in partial fractions. (2x − 3 ) (x + 2 ) (3 marks) 5x + 3 _______________

A B = _______ + _____ (2x − 3)(x + 2) 2x − 3 x + 2 5x + 3 = A(x + 2) + B (2x − 3) 3 3 3 Let x = _2_: 5(_2_) + 3 = A(_2_ + 2) A=3 Let x = −2: 5(−2) + 3 = B (2(−2) − 3) B=1 5x + 3 3 1 So _______________ = _______ + _____ (2x − 3)(x + 2) 2x − 3 x + 2

(b) Hence find the exact value of

!

5x + 3 dx, ∫ _____________ (2x − 3)(x + 2) 6

2

giving your answer as a single logarithm. (5 marks)

∫

_______ + _____) dx = 2 ( 2x − 3 x+2 6

3

1

The question says ‘hence’ so you need to use your partial fractions from part (a) to work out the integration. If you are doing definite integration, make sure you write out the integral before doing any substitutions. You can get method in marks even if you make a mistake your working.

6 _3_ [ 2 ln|2x − 3| + ln|x + 2|] 2 3 3 = (_2_ ln 9 + ln 8) − (_2_ ln 1 + ln 4)

= ln 27 + ln 8 − ln 1 − ln 4 = ln 54

Use the laws of logs to simplify your answer: __ 3 _3_ _3_ (√ ) 2 2 ln 9 = ln 9 = ln 9 = ln 27 Remember that ln a + ln b = ln ab a and ln a − ln b = ln __ so: b 27 × 8 ln 27 + ln 8 − ln 1 − ln 4 = ln (_______) 1×4 = ln 54 You will need to use problem-solving skills throughout your exam – be prepared!

C 18 x 2 + 10 B f(x) = _________ = A + ______ + ______ 2 3x + 1 3x −1 9x − 1 (a) Find the values of the constants A, B and C. (4 marks) (3 marks) (b) Hence find ∫ f(x) dx

∫

2

(c) Find f(x) dx, giving your answer in 1

the form 2 + ln k where k is a constant to be found. (3 marks)

105

9x 2 – 1 is a difference of two squ ares. Factorise it using (a2 – b2) = (a + b) (a – b).

r Use the laws of logs to simplify you that k expression for part (c). Remember doesn’t have to be an integer.

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Area between two curves You can find the area between two curves using integration. In some cases you can use a formula to find the area between two curves. If f(x) > g(x) for all values of x in the range a < x < b, then b

A=

∫

y

y = f(x) A

y = g(x )

(f(x) − g(x)) dx

a But watch out. If the curves intersect then you need to find the point of intersection and do two separate calculations.

O

a

∫

0

−4

y 5 12x 2 x2 R2

–4

Everything in blue is part of the answer.

(x 3 − (12x − x 2)) dx

!

1 4 1 3 0 __ 2 = [__ 4 x − 6x + 3 x ] −4

1 1 __ 4 2 3 = 0 − (__ 4 (−4) − 6(−4) + 3 (−4) )

= Area R2 =

160 ___ 3

∫

3

0

(12x − x 2 − x 3) dx

1 3 1 4 3 __ = [6x 2 − __ 3 x − 4 x ]0

99 1 1 __ ___ 3 4 = 6(3)2 − __ 3 (3) − 4 (3) − 0 = 4

937 160 99 ___ ____ Total area = ___ 3 + 4 = 12

x

3

O R1

x 3 = 12x − x 2 3 2 x + x − 12x = 0 x(x + 4)(x − 3) = 0 Curves intersect when x = −4, x = 0, x = 3 Area R1 =

y 5 x3

y

The diagram shows the curves with equations y = x3 and y = 12x − x2. Find the total area of the shaded region bounded by the two curves. (6 marks)

x

b

The curves intersect at the origin. To use the formula given above, you need to consider the two regions separately. Label the regions R1 and R2, and work out the points of

intersection, then use A = ∫ (f(x) − g(x)) dx a twice. b

You will need to use problem-solving skills throughout your exam – be prepared!

y

The diagram shows the curves with equations y = sin x __ 3p and y = √ 3 cos x in the range 0 < x < ___. The curves 2 intersect at the points P and Q. (a) Find the coordinates of P and Q. (3 marks) (b) Find the area of the finite region R bounded by the curves and the y-axis. (5 marks) (c) Find the area of the finite region S bounded by the curves between P and Q. (7 marks)

R O

P x

S Q

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Areas and parametric curves y

You can use integration to find the area under a parametric curve. It’s almost always easier to integrate with respect to the parameter, t. For a reminder about parametric equations have a look at page 86.

R=

∫

x=b x=a

y dx =

∫

t=μ t=λ

t=λ t=µ R

dx y ___ dt dt

0

a

b

x

! y C R O

25

x

4

The curve C with parametric equations y = t(3 − t2) x = 3t2 − 5 cuts the x-axis at (−5, 0) and (4, 0). The region R is enclosed by the curve and the x-axis. Use integration to find the area of R __ in the form k√ 3 where k is a constant. (5 marks) When x = −5, t = 0 __ When x = 4, t = √ 3 dx ___ = 6t so dx = 6t dt dt __

R=

∫

x=4

x=−5

∫ =∫

y dx =

t=√ 3

t=0

t(3 − t 2)(6t) dt

__

t=√ 3 t=0

= [6 t

3

(18 t 2 − 6 t 4) dt __

3 6 − _5_ t 5] 0

__ 3

√

6

__ 5

= 6 (√ 3 ) − _5_ (√ 3 )

When you evaluate a definite integral with respect to the parameter t, you must transform the limits into values of t. The first steps of your working should be writing down the values of t at the limits, and dx finding ___ so you can write down the dt relationship between dx and dt. You will need to use problem-solving skills throughout your exam – be prepared!

When you’ve been practising lots of complicated integrations, don’t get caught out by an easy one. The expression in t is just a polynomial, so multiply out and inte grate. or You can write your limits as ‘x = …’ ch. Make ‘t = …’ so you know which is whi r (dx or sure your limits match your operato l. dt) before evaluating the integra result For part (b) you will need to use this from the formulae booklet:

__ 36 √ = __ 3 5

The diagram shows the curve with parametric equations p p x = tan t y = 2 cos t − 1 − __ < t < __ 2 2 (a) Find the coordinates of the points A and B where the curve cuts the coordinate axes. (3 marks) The region R is bounded by the curve, and the x- and y-axes. (b) Use integration to find the area of R. Give your answer correct to 3 decimal places. (4 marks)

107

∫ f(x) dx f(x) _1 sec kx k ln |sec kx + tan kx|

y A

C R

O

B

x

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The trapezium rule There are some definite integrals that are difficult or even impossible to evaluate using calculus. You can use a numerical method called the trapezium rule to find an approximation. The formulae booklet contains this formula for using the trapezium rule with n strips strips.. y0, y1, y2, … and so on are values of y which you will calculate in a table of values. There are n + 1 values for y in total.

Numerical Methods

b b−a _1 h{(y + y ) + 2(y + y + … + yn − 1)}, where h = _____ ≈ 2 dx y 1 rule: n ium n 0 The trapez 2

∫

a

(a) Given that y = sec x, complete the table with the values of y corresponding to p p p x = ___, __ and __. (2 marks) 4 16 8 p 3p ___ _p_ _p_ ___ x 0 4 16 8 16 y 1 1.01959 1.08239 1.20269 1.41421 (b) Use the trapezium rule, with all the values for y in the completed table, to obtain an

∫ secx dx. Show all the steps _p_

estimate for

4

3p ___ so make The value for 16 is given to 5 d.p. same sure you give your answers to the degree of accuracy.

Increasing accuracy The trapezium rule becomes more accurate as you increase the number of strips used. The tops of the trapezia are closer to the curve so the estimate is more accurate.

y

y

0

of your working and give your answer to 4 decimal places. (3 marks)

π π n = 4, a = 0, b = __, h = ___ 4 16 _π_ 4 π 1 sec x dx ≈ __ × ___ [(1 + 1 . 41421) + 2 0 16 2(1.01959 + 1.08239 + 1.20269)]

∫

= 0.8859 (4 d.p.)

∫ sec x dx is ln (1 + √2 ). _p_

The exact value of

__

4

0

(c) Calculate the % error in using the estimate you obtained in part (b). (2 marks) __

0.8859 − ln (1 + √2 ) ____________________ __

ln (1 + √2 )

0

a

b

x

0

a

b

x

If you are asked to work out the error in an estimate use: • Absolute error = Estimate − Exact value • Percentage error = Estimate − Exact ______ value ______ ______ ___ × 100% Exact value

× 100% = 0.51% (2 d.p.)

The diagram shows a sketch of the curve with equation y = 3x2 ln x, x > 0.

Remember to change the width of each strip (h) for part (b)(ii).

∫

y

O

1

x

2

(b) Given that I = 3x2 ln x dx, use the trapezium 1 rule (i) with values of y at x = 1, x = 1.5 and x = 2 to find an estimate of I (ii) with all the values from the table to find another estimate of I. (5 marks)

(a) Complete the table of values for values of y corresponding to x = 1.25, 1.5 and 1.75 (c) Explain why increasing the number of values (2 marks) improves the accuracy of your estimate. (1 mark) x 1 1.25 1.5 1.75 2 y 0 8.31777

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Solving differential equations You revised how to form differential equations on page 97. You can use your integration skills to solve differential equations: Separate the variables onto different sides.

Integrate both sides. Only include one constant of integration.

Substitute the boundary conditions to find the value of c.

Rearrange the solution into the form needed.

Separating variables You can’t integrate expressions containing a mixture of different variables. Before you integrate both sides of a differential equation you need to separate the variables. dy 1 ____ If ___ = f(x) g(y) then dy = f(x) dx dx g(y)

Water is being heated in a kettle. At time t seconds, the temperature of the water is θ °C. The rate of increase of the temperature of the water at any time t is modelled by the differential equation dθ ___ θ < 100 = λ(120 − θ), dt where λ is a positive constant. Given that θ = 20 when t = 0, (a) solve this differential equation to show that θ = 120 − 100e −λt (8 marks)

∫

For example: dy 1 ___ = 2x 2 (1 − 5y) 3 ⇒ ________3 dy = 2x 2 dx dx (1 − 5y)

∫

________

When the temperature of the water reaches 100 °C, the kettle switches off. (b) Given that λ = 0.01, find the time, to the nearest second, when the kettle switches off. (3 marks) 100 = −0.01t e = −0.01t = t=

120 − 100e−0.01t 0.2 ln 0.2 = −1.6094… 161 (nearest second)

1 (a) Find a general solution to the differential dx (5 marks) equation (sec 2t) ___ = x dt (b) Find a particular solution to this equation p (2 marks) given that x = 2 when t = __ 4

109

∫

can keep Remember that λ is a constant. You n you it on either side of the equation whe e it on separate variables. It’s easier to leav + c. λt get the ‘dt ’ side then integrate to

∫ 1201 − θ dθ = ∫ λ dt

−ln (120 − θ) = λt + c When t = 0, θ = 20: −ln (120 − 20) = λ(0) + c c = −ln 100 So −ln (120 − θ) = λt − ln 100 ln (120 − θ) = −λt + ln 100 120 − θ = e −λt + ln 100 = 100e −λt θ = 120 − 100e −λt

∫

Once you have integrated you hav e found a general solution to the differentia l equation. You could rearrange it at this point:

−ln (120 − θ) = λt + c 120 − θ = e −λt − c = A e −λt θ = 120 − A e −λt A is a constant equal to e−c. You cou ld substitute the boundary conditions at this stage to find the particular solutio n asked for in the question.

Separate the variables to get 1_ dx = cos 2t dt x

∫

∫

2 (a) Find ∫ (2y + 1 )−3 dy

(2 marks)

(b) Given that y = 0.5 at x = −8, solve the dy (2y + 1) 3 , differential equation ___ = ________ dx x2 giving your answer in the form y = f(x). (6 marks)

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You are the examiner! In your exam you might be asked to identify errors in working. You should also be confident checking your own work. Each of these students has made a key mistake in their working. Can you spot them all?

6 3 1 Express ______ − ______ + _______ as 2x + 3 2x − 3 4x 2 − 9 a single fraction in its simplest form. (4 marks)

3 _______ 2x + 3

6 1 − _______ + ________________ 2x − 3 (2x + 3 )(2x − 3 ) = =

3(2x − 3) − 2x + 3 + 6 _______________________ (2x + 3)(2x − 3)

4x ________________

dy dy 3x 2 + 4y ___ − 4y − 4x ___ = 1 dx dx dy ___ 2 (4y − 4x) = 4y − 3x + 1 dx 4y − 3 x 2 + 1 dy ___ = ____________ dx 4y − 4x

∫

du u = x − 1 so ___ = 1, so du ≡ dx dx 4 4 4 1 1 __ __ u +__ 1 x ______ _____ _____ dx = du = (u 2 + u −2) du 2 √x − 1 2 √u 2 1 4 _3_ __ = [_23_u 2 + 2u 2]

∫

∫

2

=(

α = arctan(_3_) = 63.43° (2 d.p.) 6

__

+ 4) − ( 3

4√ 2 _____

__ = _23_( 14 − 5√2 )

Given that x2 − 1 , y = ______ e3x

dy x > 1, find ___ . dx

(3 marks)

dy ___

(x 2 − 1)(3e3x ) − (e3x )(2x ) = ________________________ dx (e3x )2 3x 2 − 2x − 3 = ______________ e3x

Use the substitution u = x − 1 to find 4 x _____ dx (5 marks) the exact value of ______ 2 √x − 1

16 __ 3

______

R = √3 2 + 6 2 = 6.71 (2 d.p.)

(2x + 3)(2x − 3)

A curve C has equation x3 + 2y2 − 4xy = 1 dy Find ___ in terms of x and y. (4 marks) dx

∫

Given that 3 cos θ − 6 sin θ = R cos (θ + α), where R > 0 and 0 < α < 90°, find the exact value of R and the value of α correct to 2 decimal places. (3 marks)

+ 2√ 2 ) __

Checking your work If you have time left in your exam you should check back through your working: Check you have answered every question part. Double−check any trigonometric identities, making sure you have + and − in the correct places. Cross out any incorrect working with a single neat line and underline the correct answer. Make sure your working is neat and legible and any sketch graphs are clearly labelled.

Find the mistake in each student answer on this page, and write out the correct working for each question. Turn over for the answers.

110

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You are still the examiner! Before looking at this page, turn back to page 110 and try to spot the key mistake in each student’s working. Use this page to check your answers. The corrections are shown in red and these answers are now 100% correct.

6 3 1 Express ______ − ______ + _______ as 2x + 3 2x − 3 4x 2 − 9 a single fraction in its simplest form. (4 marks)

2x + 3

6 1 − _______ + ________________ 2x − 3 (2x + 3 )(2x − 3 ) =

R = √3 2 + 6 2 = 6.71 (2 d.p.) = 3√5

3(2x − 3) − (2x + 3) + 6 _______________________

α = arctan(_3_) = 63.43° (2 d.p.) 6

(2x + 3)(2x − 3) 4x = ________________ (2x + 3)(2x − 3) 2(2x − 3) = ________________ (2x + 3)(2x − 3) 2 _______ = 2x + 3

asks for Read the question carefully. If it wer as an exact value then leave your ans a surd. Have a look at page 84.

with Always use brackets when working algebraic fractions.

and the Always write out the product rule also quotient rule before substituting. It dv . u d ___ and ___ dx helps to write down your u, v, dx

Revise this topic on page 57.

3

e 91. Revise the quotient rule on pag

2

A curve C has equation x + 2y − 4xy = 1 dy Find ___ in terms of x and y. (4 marks) dx

Given that x2 − 1 , y = ______ e3x

dy dy 3x 2 + 4y ___ − 4y − 4x ___ = 1 = 0 dx dx dy ___ 2 (4y − 4x) = 4y − 3x + 1 dx 4y − 3 x 2 + 1 dy ___ = ____________ dx 4y − 4x

u = x2 − 1

du ___

dy x > 1, find ___ . dx

(3 marks)

v = e3x

dv ___

= 2x = 3e3x dx du dv v ___ − u ___ d y (x 2 − 1)(3e3x ) − (e3x )(2x ) dx dx ___ = ___________ = ________________________ 2 dx v (e3x )2 3x 2 − 2x − 3 ______________ −3x 2 + 2x + 3 = ______________ = 3x e e3x ⎤ ⎪ ⎪ ⎪ ⎪ ⎪ ⎦

dx

m including You have to differentiate every ter e. the constant on the right-hand sid n see page 94. For more on implicit differentiatio

⎤ ⎪ ⎪ ⎦

3 _______

Given that 3 cos θ − 6 sin θ = R cos (θ + α), where R > 0 and 0 < α < 90°, find the exact value of R and the value of α correct to 2 decimal places. (3 marks) ______ __

Use the substitution u = x − 1 to find the 4 x _____ dx (5 marks) exact value of ______ 2 √x − 1

∫

du u = x − 1 so ___ = 1, so du ≡ dx dx 4 43 43 1 1 __ __ u +__ 1 x _____ ______ _____ dx = (u 2 + u − 2 ) du du = 2 √x − 1 2 1 √u 21 1 43 _3_ __ = [_23_u 2 + 2u 2]

∫

=

∫

__

(2√ 3

+

2√ 3 )

__ = 4√ 3 − _83_

111

__

∫

21

__

+ 4) − ( 3 __ _2_ ( = 3 14 − 5√ 2 )

− ( + 2) = ( _2_ 3

16 __ 3

4√ 2 _____

+ 2√ 2 ) __

For definite integration using substitution, you need to change the limits to values of u before calculating.

ered on Integration by substitution is cov page 103.

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Statistics

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Sampling In statistics, a population is a group of people you are interested in. A sample is a smaller group chosen from a larger population. In order to select a sample, you need to name or number the population to create a list called a sampling frame. Here are four types of sample you need to know about:

Random samples

1 2

Simple random sample – pick names out of a hat or generate random numbers. Systematic sample – number the population, and select sampling units at regular intervals. The first sampling unit is selected randomly.

A chicken farmer wants to test the shell thickness of a batch of 100 eggs. Give one reason why a census might not be suitable for this test. (1 mark) The farmer might need to break the eggs to test them, and he would have none left to sell.

If the testing process destroys or consumes the sampling unit, then a census may not be appropriate.

Non-random samples

3 4

Sample vs census A census observes or measures every member of a population. Carrying out a census can be time-consuming and expensive, and can produce a large amount of data to clean and process. A sample is quicker and cheaper than a census, but if the sample is not representative of the population then it can introduce bias.

Stratified sampling If a population is split into groups, you can select a stratified sample by ensuring the number of elements selected from each group is proportional to the size of the group. To work out how many members to select from each group, find the sampling fraction and multiply this by the size of each group. Sample size Sampling fraction 5 ______________ Population size

Quota sample – select a predetermined number of sampling units with certain defined characteristics. Opportunity sample – select most conveniently available sampling units that meet required criteria.

Boys

Girls

For a sample of size 6 from this pop ulation of size 24 you need to select 6 6 8 × ___ = 2 girls and 16 × ___ = 4 boys. 24 24

This table shows the number of employees in an engineering company. The CEO of the company wants to select a random sample of 50 employees, stratified by gender and office location. (a) How many female employees from Bristol should she select for her sample? (2 marks) (b) Describe a suitable sampling frame. (1 mark)

Male Female Total

Bristol 119 152 271

London 91 74 165

Total 210 226 436

A stratified sample is an example of a random sample – all items have an equal chance of selection.

112

Statistics

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Mean

_

The mean is a measure of central tendency. The mean is sometimes written as x . You might have to calculate an estimate of the mean of grouped data given in a frequency table.

Formulae for the mean

Grouped data

Learn these two formulae for the mean – they’re not given in the formulae booklet.

1 2

For n discrete data values, the mean is: _ ∑x The sum of the data values x = ____ n The number of data values For data given in a frequency table: The sum of (frequency × data value) _ ∑fx or (frequency × midpoint) x = _____ ∑f The total frequency

You can use formula 2 on the left with a grouped frequency distribution. The value of x for each group is the midpoint of that group. Height, h (cm) 0 0. Find (a) the time at which P comes instantaneously to rest. (6 marks)

∫

v = (4 − 0.2t) dt = 4t – 0.1t2 + c When t = 0, v = 17.6 17.6 = 4(0) – 0.1(0)2 + c ⇒ c = 17.6 So v = 4t – 0.1t2 + 17.6 Set v = 0 and solve: –0.1t2 + 4t + 17.6 = 0 t = –4 or t = 44 P comes to rest after 44 seconds. (b) the distance travelled by the particle in the first two seconds of its motion. (4 marks) 2 s = (4t − 0.1 t2 + 17.6) dt

∫

0

1 = [2 t 2 − ___ t 3 + 17.6t] 30 0 1 = (2 (2) 2 − ___ (2) 3 + 17.6(2)) − 0 30 = 42.9 m (3 s.f.) 2

A target in a fairground shooting range moves along a straight line. At time t seconds, its distance from a fixed point O is x m and its velocity, v m s–1, _ is given by v = 0.1√ t − 0.2t, 0 < t < 6 When t = 0, x = 4. (a) Find the acceleration of the target when t = 4. (4 marks)

An indefinite integral always produces a function with a constant of integration. You need to find this constant in order to use the function to solve numerical problems. When you integrate to find velocity or displacement, you might need to use a given value to find the constant of integration. If this is the value when t = 0 it is called an initial condition. There is more about finding functions by integrating on page 43.

! Follow these steps for part (a): 1. Integrate the expression for the acceleration of P to find an expression for the velocity. 2. Use the initial conditions to find the value of the constant of integration. 3. Set the velocity equal to 0 and solve. The model is only valid for t > 0 so you can reject the negative root of the equation. You will need to use problem-solving skills throughout your exam – be prepared!

You can use definite integration to find the distance travelled between two given times. This is the same as the area under the velocity–time graph between t = 0 and t = 2. v m s–1

O

xm

P

(b) Find an expression for x, giving your answer in the form x = f(t). (4 marks) (c) Find the value of x when t = 5. (1 mark)

160

ics

Mechan

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Deriving suvat formulae You can use integration to derive the suvat formulae for motion with constant acceleration.

A particle moves in a straight line with constant acceleration, A m s–2. At time t seconds the velocity of the particle is v m s–1 and its displacement relative to a fixed origin is s m. Given that the initial velocity of the particle is U m s–1 and its initial displacement is 0 m, prove that (a) v = U + At (3 marks) v = A dt = At + c When t = 0, v = U ⇒ c = U So v = At + U = U + At as required.

∫

(b) s = Ut + _2 At2 1

∫

(3 marks)

∫

s = v dt = (U + At) dt 1 2 = Ut + __ 2 At + c

When t = 0, s = 0 ⇒ c = 0

In this example capital letters are used to represent letters that can be treated as constants when you integrate.

Given that v = u + at

1

s = ut + _2 at2 1

2

prove that v2 = u2 + 2as. You may not make use of any formulae other than those given above. (3 marks) Squaring both sides of 1 : v 2 = (u + at)2 = u2 + 2uat + a2t2 1 2 = u2 + 2a(ut + __ 2 at )

Substituting 2 into the above: v 2 = u 2 + 2as as required.

1 2 So s = Ut + __ 2 At as required.

Integrate, then use the given initial conditions to find the constants of integration. Remember to write your final answers in the form asked for in the question.

Have a look at pages 150 and 151 for a reminder about the five suvat formulae.

Using a velocity–time graph

s

You can also derive the suvat formulae from a velocity–time graph. The diagram shows a velocity–time graph for a particle accelerating constantly from velocity U m s−1 to velocity V m s−1 in time T seconds. The acceleration, a m s−2, of the particle is equal to the gradient of the graph: V−U a = _____, so V = U + aT T

V

1 A particle moves in a straight line along the x-axis. At time t seconds, its distance from the origin is given by s = 10t – kt2, t > 0 where k is a constant. (a) Show that the particle is moving with constant acceleration. (4 marks)

161

You can use the two suvat formulae in the example on the left to derive the other suvat formulae.

U O

T

t

(b) Given that the particle is instantaneously at rest when t = 4, find the value of k. (3 marks) 2 Use the velocity–time graph above to show that the distance travelled, s m, by the particle in time T seconds is given by U+V (4 marks) s = (______) T 2

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Mechani

cs

You are the examiner! In your exam you might be asked to identify errors in working. You should also be confident checking your own work. Each of these students has made key mistakes in their working. Can you spot them all?

1

A stone is thrown vertically upwards with speed 16 m s−1 from a point h metres above the ground. The stone hits the ground 4 s later. Find (a) the value of h (3 marks) (b) the speed of the stone as it hits the ground. (3 marks) s = h, u = 16, v = ?, a = 9.8, t = 4 1 2 (a) s = ut + __ 2 at 1 2 = 16 × 4 + __ 2 × 9.8 × 4 = 142.4 h = 140 (2 s.f.) (b) v = u + at = 16 + 9.8 × 4 = 55 (2 s.f.)

3

A

P

B

Two particles A and B of masses 7 kg and 11 kg respectively are connected by a light inextensible string which passes over a smooth pulley. Particle A rests on a rough horizontal table and particle B hangs vertically. The system is released from rest. (a) By writing two separate equations of motion for A and for B, find the acceleration of the system. (5 marks) (b) State how you have used the fact that the pulley is smooth in your calculations. (1 mark) (a) 7 + 11 = 18 kg F = ma 11g = 18a 11g a = ____ = 5.99 m s−2 (3 s.f.) 18 (b) There is no friction in the pulley.

2

In this question i and j represent the unit vectors due east and north respectively. A particle is acted on by two forces, (2i − 5j) N and (7i + j) N. Find the magnitude and bearing of the resultant force acting on the particle. (4 marks) Resultant force = (2i − 5j) + (7i + j) = (9i − 4j) N 4j θ

9i

4 tan θ = __ 9 θ = 23.96…° Bearing of force = 90° − 23.96…° = 066.0° (1 d.p.)

4

The acceleration, a m s–1 of a particle at time t seconds is given by a = 6t2 + 1, t > 0. Given that the particle has an initial velocity of 5 m s−1, find an expression for the velocity, v m s−1, at time t seconds. (3 marks)

∫

v = (6t2 + 1) dt = 2t3 + t

Checking your work If you have any time left at the end of your exam, you should check back through your working: Check you have answered every part and given all the information asked for. Check your accuracy. Give answers to 2 significant figures if you use g = 9.8 in your calculation, and 3 s.f. everywhere else. Draw large, well-labelled diagrams and make sure you show all the forces acting on an object. If you’ve written working for a question somewhere else or on extra paper, make sure you say where it is under the question.

Find the mistakes in each student’s answer on this page, and write out the correct working for each question. Turn over for the answers.

162

ics

Mechan

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You are still the examiner! Before looking at this page, turn back to page 162 and try to spot the key mistakes in each student’s working. Use this page to check your answers. The corrections are shown in red and these answers are now 100% correct.

1

A stone is thrown vertically upwards with speed 16 m s−1 from a point h metres above the ground. The stone hits the ground 4 s later. Find (a) the value of h (3 marks) (b) the speed of the stone as it hits the ground. (3 marks) Taking upwards as the positive direction s = −h, u = 16, v = ?, a = −9.8, t = 4 1 2 __1 2 (a) s = ut + __ 2 at = 16 × 4 − 2 × 9.8 × 4 __1 2 = 16 × 4 + 2 × 9.8 × 4 = −14.4 = 142.4 h = 14 (2 s.f.) h = 140 (2 s.f.) (b) v = u + at = 16 − 9.8 × 4 = 16 + 9.8 × 4 = −23.2 = 55.2 m s−1 Speed = 23 m s−1 (2 s.f.)

Direction is important in suvat questions. If your positive direction is up, then acceleration is negative. Revise motion under gravity on page 152.

3

2

In this question i and j represent the unit vectors due east and north respectively. Particle A is acted on by two forces, (2i − 5j) N and (7i + j) N. Find the magnitude and bearing of the resultant force acting on the particle. (4 marks) Resultant force = (2i − 5j) + (7i + j) = (9i − 4j) N N 9i

4j θ

θ

9i

R

________ 2 √ 2

–4j

|R| = 9 + 4 = 9.85 N (3 s.f.) 4 tan θ = __ 9 θ = 23.96…° Bearing of force = 90° − ¹ 23.96…° = 066.0° (1 d.p.) = 114.0° (1 d.p.) Make sure you provide all the information asked for in the question. Be careful with negative coefficients of i and j in forces given as vectors. There is more about forces as vectors on page 154.

Two particles A and B of masses R 7 kg and 11 kg respectively are T P A connected by a light inextensible string which passes over a smooth 7g T pulley. Particle A rests on a rough horizontal table and particle B B hangs vertically. The system is 11g released from rest. (a) By writing two separate equations of motion for A and for B, find the acceleration of the system. (5 marks) (b) State how you have used the fact that the pulley is smooth in your calculations. (1 mark) (a) 7 + 11 = 18 kg For A: T = 7a 1 F = ma For B: 11g – T = 11a 2 11g = 18a 1 + 2 : 11g = 18a 11g ____ a= = 5.99 m s−2 (3 s.f.) 6.0 m s−2 (2 s.f.) 18 (b) There is no friction in the pulley. The tension in the string is the same on both sides of the pulley.

4

The acceleration, a m s−1 of a particle at time t seconds is given by a = 6t2 + 1, t > 0. Given that the particle has an initial velocity of 5 m s−1, find an expression for the velocity, v m s−1, at time t seconds. (3 marks)

∫

v = (6t2 + 1)dt = 2t3 + t + c When t = 0, v = 5 so c = 5 v = 2t3 + t + 5

When you integrate you should include a constant of integration then use the initial conditions to determine its value. For more see page 160.

separate equations of motion. You have used g so give If the question specifies a method you should use that method, so write and 157. To revise modelling assumptions see page 148. your answer to 2 s.f. For more on connected particles see pages 156

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Moments 1 A moment is a measure of the turning effect of a force on a body. You can use moments to answer questions about rods in equilibrium.

Taking moments If a force of magnitude F N acts at a perpendicular distance x m from a point P, the moment of F about P is Fx:

moment = force × distance The units of a moment are newton-metres (N m). The turning effect of a moment can be either anticlockwise or clockwise.

This force produces an anticlockwise moment about A of 100 × 7 = 700 N m 100 N 7m

0.5 m

A

B 40 N This force produces a clockwise moment about B of 40 × 0.5 = 20 N m

Equilibrium

4m 28 N

If a rod is horizontal and in equilibrium, then

sum of clockwise = sum of anticlockwise moments moments ∑ =∑ ↻

↺

1.2 m

B

1m

mg

6g

A uniform rod AB has length 2 m and mass 6 kg. A particle of mass m kg is attached to the rod at A. The rod is supported at C, where CB = 1.2 m, and the system is in equilibrium with AB horizontal. Find the value of m. (4 marks) moment about C = 6g × 0.2 moment about C = mg × 0.8 moment = moment 1.2g = 0.8mg m = 1.5kg ↺

↻

If you take moments about C you can ignore the weight of the beam.

30 N

6N

This rod is uniform. That means that its full weight acts at the midpoint of the rod. You can revise non-uniform rods on page 166.

↻

2m R

C

2m

↺

The rod on the right has weight 30 N and is held in place by a vertical force of 28 N. A particle of weight 6 N is on the end of the rod. ∑ moments about P = 30 × 2 + 6 × 4 = 84 N m ∑ moments about P = 28 × 3 = 84 N m

A

3m

P

You can take moments about any point. Whichever point you choose, you can ignore any forces acting at that point. In this question there is a normal reactio n R acting at C. The magnitude of this reactio n will be (mg + 6g) N, the total weight of the whole system. The easiest way to answer this question is to take moments abo ut C. This means you can ignore the normal reaction R.

C

A

B

xm 7m

A seesaw consists of a beam AB of length 7 m supported by a smooth pivot at its centre C. Emma has mass 40 kg and sits on the end B. Paul has mass 50 kg and sits at a distance x m from C. The beam is modelled as a uniform rod. Find the value of x for which the seesaw rests in horizontal equilibrium. (3 marks)

164

↺ ↻

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Moments 2 If a rod is hanging in equilibrium, then the resultant force acting on it must be zero. You sometimes need to resolve vertically to find unknown forces in moments questions.

Tensions and reactions There are two ways a rod can be supported in a moments question:

2T

T

9m A

R1

C

dm

A steel girder AB has weight 210 N. It is held in equilibrium in a horizontal position by two vertical cables. One cable is attached to the end A. The other cable is attached to the point C on the girder, where AC = 9 m. The girder is modelled as a uniform rod, and the cables as light inextensible strings. Given that the tension in the cable at C is twice the tension in the cable at A (a) find the tension in the cable at A (2 marks) R(↑): T + 2T − 210 = 0 T = 70 N (b) show that AB = 12 m. (4 marks) Taking moments about A: 210 × d = 2T × 9 210d = 1260 d=6 AB = 2 × d = 12 m A small load of weight W newtons is attached to the girder at B. The load is modelled as a particle. The girder remains in equilibrium in a horizontal position. The tension in the cable at C is now three times the tension in the cable at A. (c) Find the value of W. (7 marks) ↺

↻

3U

A

210 N

W

R(↑): U + 3U − 210 − W = 0 1 4U − W = 210 Taking moments about B: U × 12 + 3U × 3 = 210 × 6 21U = 1260 U = 60 Substituting into 1 : 4 × 60 − W = 210 W = 30 N ↺

↻

165

W This rod is supported by ropes or cables, which have a tension.

The calculations work in exactly the same way in both situations.

Start by labelling the diagram. The tension at C is twice the tension at A so label them T and 2T. (a) Resolve vertically for the whole system to work out the value of T. (b) If you take moments about A you can ignore the tension in this cable. The rod is uniform so its weight acts at its midpoint. Find the distance from A to this point, then multiply by 2 to find the length of AB. (c) This is a new situation so draw a new diagram showing the forces on the rod. The tensions will be different – choose a different letter from T to make sure you don’t make a mistake.

A

C

B 5m

B

C

6m

W This rod is resting on supports, which produce a normal reaction on the rod.

T2

2m 3m

9m

T1

B

210 N

U

R2

A uniform rod AB has mass 6 kg and length 5 m. The rod rests in horizontal equilibrium on two smooth supports. One support is at the end B. The other is at a point C on the beam, where AC = 2 m. (a) Find the reaction on the beam at C. (3 marks) A particle of mass 18 kg is attached to the rod at a point D. The beam remains in equilibrium. The reactions on the beam at C and B are now equal. (b) Find the distance AD. (7 marks)

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Centres of mass The centre of mass of an object is the point at which you can assume the weight of the object acts. In a uniform rod, the centre of mass is the midpoint of the rod. If a rod is non-uniform, the weight can act at a different point.

2X

A

Golden rule

0.8 m

If a rod is non-uniform, do not assume the weight acts at its midpoint.

↺

↻

You will need to use problem-solving skills throughout your exam – be prepared!

4m A plank AD weighs D 20 kg and has A B C length 4 m. 1.6 m 2m The plank rests horizontally on two smooth supports at B and C, where BC = 2 m and CD = 1.6 m. A particle of mass m kg is attached to the plank at D. The plank is initally modelled as a uniform rod. With this model, the plank is on the point of tilting about C.

0.6 m

4.5g

3m

A non-uniform rod AB has length 3 m and mass 4.5 kg. The rod rests in equilibrium in a horizontal position, on two smooth supports at P and Q, where AP = 0.8 m and QB = 0.6 m. The centre of mass of the rod is at G. Given that the magnitude of the reaction of the support at P on the rod is twice the magnitude of the reaction of the support at Q on the rod, find (a) the magnitude of the reaction of the support at Q on the rod (3 marks) R(↑): 2X + X − 4.5g = 0 X = 1.5g = 15 N (2 s.f.) (b) the distance AG. (4 marks) Taking moments about A: 4.5g × AG = 2X × 0.8 + X × 2.4 = 4X 4.5g × AG = 6g 4 AG = __ m 3

On the point of tilting If a rod is on the point of tilting, one of the reactions on the rod will be equal to zero. This rod is being slid to the right. Initially it rests on two supports A and B. Just before it tips over to the right, it will be resting only on B. The reaction at A will be zero.

B

↻

Resolve vertically for the whole system to find X. The rod is non-uniform so its centre of mass G is not necessarily the midpoint of the rod. You need to assume that the weight of the rod acts at G, and use moments to find the length AG. You could take moments at any point to solve this problem. For example: Taking moments about P: 4.5g × PG = X × 1.6 = 2.4g 8 PG = __ 15 m 4 AG = 0.8 + PG = _3 m

Q

G

↺

!

P

X

R A

Supported equally by A and B

B O

A

R

2R

On the point of tilting about B

B

(a) The rod is on the point of tilting about C so the reaction at B will be zero. (b) Assume that the weight of the plank acts at the centre of mass.

(a) Find the value of m. (6 marks) The plank is now modelled as a non-uniform rod. With the new model, the reaction at C is 75 N greater than the reaction at B. (b) Find the distance of the centre of mass of the plank from D. (6 marks)

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Resolving forces You can answer some questions by resolving forces into perpendicular components.

Using sin and cos To resolve a force into components, imagine sketching a right-angled triangle, with the force as its hypotenuse. For example, a force of magnitude X acting at an angle θ to the horizontal can be resolved horizontally and vertically: R(↑): X sin θ R(→): X cos θ You can use the notation R(→) and R(↑) to show which direction you are resolving in. The arrow points in the positive direction.

This force has the same effect as a horizontal force of magnitude X cos θ and a vertical force of magnitude X sin θ.

X sin θ

X θ

X cos θ

X cos θ is adjacent to the angle, and X sin θ is opposite the angle.

Just cos A box of mass 5 kg is lying on a smooth plane which is inclined at an angle of 30° to the horizontal. The box is released from rest and accelerates at a m s−2 down the slope. Modelling the box as a particle, find (a) the value of a (3 marks) R

40°

30° 5g

5g cos 30°

R(↙): 5g sin 30° = 5a 2.5g = 5a a = 0.5g = 4.9 m s−2 (b) the magnitude of the normal reaction of the slope on the box. (2 marks) R(↖): R − 5g __ cos 30° = 0 5√3 R = _____ g = 42 N (2 s.f.) 2

300 cos 50° N

300 N

Only the component of the force which acts down the slope contributes to the acceleration of the box. You need to resolve the weight of the box (5g N) into components parallel and perpendicular to the slope. Use R(↙) to show that the positive direction is down the slope. The normal reaction always acts perpendicular to the plane. It is equal in magnitude to the component of the weight that is perpendicular to the plane, and acts in the opposite direction.

A truck is pulling a car using a light, inextensible tow-rope. The tow-rope is at an angle of 40° to the road. Together, the truck and car accelerate constantly at a rate of 2.5 m s−2. The car has mass 800 kg and experiences a constant resistance to motion of magnitude 500 N. Find (a) the tension in the tow-rope (4 marks) (b) the normal reaction of the road on the car. (2 marks)

167

300 cos 40° N

50°

a

5g sin 30°

30°

If the direction you are resolving in makes an angle of θ with the line of action of the force F, then the component in that direction is F cos θ. You can use this to resolve without using sin.

The car is being accelerated by the horizontal component of the tension in the tow-rope.

40°

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Friction Friction is a force which acts in the opposite direction to the motion of a particle on a rough surface.

Coefficient of friction

R = 49 N

The coefficient of friction, µ, between an object and a surface is a measure of roughness. For a moving object the force of friction has magnitude

F = R

F = µR

20 N

5 kg

 = 0.1

where R is the magnitude of the normal reaction between the object and the surface.

5 g = 49 N This box is accelerating along a rough surface. The coefficient of friction between the box and the surface is 0.1. The normal reaction is 49 N so F = µR = 0.1 × 49 = 4.9 N So the resultant force on the box is 20 − 4.9 = 15.1 N.

If µ = 0 the surface is smooth – there is no friction. Any surface with µ > 0 is called a rough surface.

! 100 N

R

30° F = R

Don’t confuse constant speed with constant acceleration. Constant acceleration requires a resultant force. If an object is moving with constant speed, the resultant force on it is zero.

Everything in blue is part of the answer.

mg

A small box is pushed along a floor. The floor is modelled as a rough horizontal plane and the box is modelled as a particle. The coefficient of friction between the box and the floor is _12. The box is pushed by a force of magnitude 100 N which acts at an angle of 30° with the floor. Given that the box moves with constant speed, find the mass of the box. (7 marks)

You will need to use problem-solving skills throughout your exam – be prepared!

You need to resolve the 100 N force horizontally and vertically:

R(→): 100 cos 30° − F = 0 __ F = 50√3 N __ 1 F = µR = __ R, so R = 100√3 N 2 R(↓): mg +__100 sin 30° − R = 0 100√3 − 100 sin 30° m = ____________________ g = 12.571… = 13 kg (2 s.f.)

A block of mass 20 kg is being pulled along rough horizontal ground at a constant speed using a rope. The tension in the rope is P N and it makes an angle of 15° with the ground. PN The coefficient of friction between the ground and 15° the block is 0.5.

30°

100 sin 30° N

100 cos 30° N

100 N

Use the horizontal component to work out the size of the frictional force, and then use this to work out the normal reaction, R. Then use the vertical component to work out the mass of the box, m. Remember that there are two forces acting vertically downwards on the block – its weight (mg) and the vertical component of the 100 N force. Resolve vertically and horizontally to find two simultaneous equations in P and the normal reaction, R.

(a) Find the value of P. (8 marks) The rope is cut. (b) Explain whether the normal reaction between the ground and the block will increase or decrease. (2 marks)

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Sloping planes Questions involving friction and sloping (or inclined) planes are really common. For a reminder about resolving forces parallel and perpendicular to a sloping plane have a look at page 167. An object moving freely on a rough plane inclined at an angle θ to the horizontal has three forces acting on it: • weight = mg • normal reaction, R = mg cos θ • friction, F = µR. If the component of the weight that acts down the slope (mg sin θ) is greater than the friction, the object will accelerate.

A small brick of mass 0.5 kg is placed on a rough plane which is inclined to the horizontal at an angle θ, where tan θ = _43, and released from rest. The coefficient of friction between the brick and the plane is _13 Find the acceleration of the brick. (9 marks) R

0.5g sin θ θ

F =

1 3R

θ 0.5g

A box of mass 3 kg is pulled up a rough plane by a force of magnitude 25 N acting parallel to the line of greatest slope of the plane. The plane is inclined at an angle of 40° to the horizontal. The coefficient of friction between the box and the plane is 0.2

F = µR

mg θ

Trig ratios If you are given one trigonometric ratio you can work out the values of the other two by sketching a triangle. Here are two examples:

1

4 tan θ = _3_

2

_3_ 5

32 + 42 = 5 θ

sin α =

5 α

4

4 sin θ = _5_

cos θ = _53_

3

3

52 − 32 = 4 0.5g cos θ

R(↖): R − 0.5g cos θ = 0 R = 0.5g cos θ 3 Using cos θ = __ : R = 0.3g 5 R(↙): 0.5g sin θ − F = 0.5a 1 0.5g sin θ − __R = 0.5a 3 Substituting the value of R into this equation: 1 0.5g sin θ − __ × 0.3 g = 0.5a 3 4 Using sin θ = __ : 0.4g − 0.1g = 0.5a 5 0.3g = 0.5a 3g ___ = 5.9 m s−2 (2 s.f.) a= 5

169

R = mg cos θ

4 cos α = _5_

tan α = _34_

Ratios involving 3-4-5 triangles like these are common. They are designed to make the question easier to answer, so don’t work out 4 3 tan−1 ( _3_ ) or sin−1 ( _5_ ) to get the angle as a decimal.

Friction depends on the normal reaction so it’s usually a good idea to resolve perpendicular to the plane first. This gives you the magnitude of the normal reaction, which you can use to find the frictional force. Subtract the friction from the component of the weight that acts down the plane to find the resultant force down the plane acting on the brick. There is more about resultant forces and F = ma on page 153.

By modelling the box as a particle, find (a) the normal reaction of the plane on the box (3 marks) (b) the acceleration of the box. (5 marks)

25 N

40°

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Projectiles You can model a projectile as a particle moving freely under gravity. This means that you ignore air resistance, and any rotational forces acting on the particle. You resolve the initial velocity (or velocity of projection) into vertical and horizontal components.

u sin θ m s–1

u m s–1

θ

At its highest point, the vertical component of the velocity will be zero. 0 m s–2

u cos θ m s–1

θ is the angle of elevation (or angle of projection).

g m s–2

At any point, the horizontal component of acceleration is zero, and the vertical component of acceleration is g = 9.8 m s–2

! A particle is projected from a point O on a horizontal plane with speed 15 m s–1, at an angle of θ above the horizontal, where 4 tan θ = _3 Find (a) the time taken for the particle to hit the plane (4 marks) R(↑): uy = 15 sin θ = 12 m s−1 s = 0, u = 12, v = ?, a = −9.8, t = ? 1 2 s = ut + __ 2 at 0 = 12t − 4.9t2 t = 0 or 2.448… The particle hits the plane after 2.4 s (2 s.f.) (b) the horizontal distance travelled by the particle. (2 marks) R(→): ux = 15 cos θ = 9 m s−1 s = vt = 9 × 2.448… = 22 m (2 s.f.)

1 A stone is projected at an angle of elevation of 40° from the top of a building, 28 m above level ground. The stone strikes the ground 6 seconds later. By modelling the stone as a particle moving freely under gravity, find (a) the speed of projection (5 marks) (b) the distance from the base of the building to the point where the stone strikes the ground. (2 marks)

In projectiles questions you usually start by analysing the vertical motion of the particle. The vertical motion has constant acceleration so you can use the suvat formulae. Revise these on pages 150 and 151. The horizontal motion has constant speed (zero acceleration) so you can use s = vt (or distance = speed × time). You will need to use problem-solving skills throughout your exam – be prepared!

When resolving the initial velocity into components you can use ux to represent the horizontal component and uy to represent the vertical component. In part (a), you could also use symmetry to deduce that the vertical component of the velocity will be −12 when the particle hits the plane: v = u + at −12 = 12 − 9.8t t = 2.4 (2 s.f.)

2 A particle is projected horizontally from a height of 10 m above a level plane, at a speed of 18 m s−1. Find the speed of the particle at the point where it hits the plane. (6 marks)

Watch out! Speed is the magnitude of the velocity. You need to consider both the vertical and horizontal components.

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Projectile formulae You might be asked to derive general formulae associated with projectile motion in your exam.

! A particle is projected from a point O on level ground at a speed of 7 m s−1 and an angle of α above the horizontal. When the particle has moved a horizontal distance x m, its height above the point of projection is y m. 1 2 2 (a) Show that y = x tan α − __ 10 x (1 + tan α). (7 marks) R(↑): uy = 7sin α R(→): ux = 7cos α 1 2 Using s = ut + __ 2 at for the vertical motion: 1 y = 7sin αt − 4.9t2 Using s = vt for the horizontal motion: x = 7cos αt 2 x _______ Rearranging 2 : t = 7cos α Substituting into 1 : 2 x x y = 7sin α(_______) − 4.9(_______) 7cos α 7cos α 1 2 __ 2 = x tan α − 10 x sec α 1 2 2 = x tan α − __ 10 x (1 + tan α)

(b) Given that when x = 3, y = 1.5, find two possible values for α. (5 marks) 2 1.5 = 3 tan α − 0.9(1 + tan α) Let u = tan α: 1.5 = 3u − 0.9(1 + u2) 3u2 − 10u + 8 = 0 (3u − 4)(u − 2) = 0 u = 2 or _43_, so tan α = 2 or _43_ So α = 63.4° or 53.1° (3 s.f.)

A particle is projected from a point O on level ground at a speed of U m s−1 and an angle of α above the horizontal. Show that (a) the time of flight of the particle is given 2U sin α (4 marks) by ________ g (b) the horizontal range of the particle U 2 sin 2α (4 marks) is given by ________ g

171

You can write expressions for y and x in terms of t and α. These are parametric equations for the path of the projectile, with parameter t. You can convert them into cartesian equations by eliminating t. Revise parametric equations on page 86. You will need to use problem-solving skills throughout your exam – be prepared!

Use trigonometric identities to get your equation in the required form. sin θ 1 tan θ ≡ _____ sec θ ≡ _____ cos θ cos θ sec2 θ ≡ 1 + tan2 θ Substitute x = 3 and y = 1.5 into the equation from part (a). This is a quadratic equation in tan α. There are two possible values of tan α, which correspond to the two possible paths which pass through the point (3, 1.5). y

tan α = 2 (3, 1.5) tan α =

O

4 3

x

Parabolic path The example on the left is one particular case of the general formula for the path of a projectile. If a particle is projected from the origin with initial velocity U and angle of elevation α, the equation of its trajectory will be: 1 + tan2 α y = x tan α − gx 2(__________ 2U 2 ) This is a quadratic curve (or parabola). __1 (a) Use s = ut + 2 at2, leaving your answer in terms of g. (b) Use your answer from part (a) and the identity 2sin θ cos θ ≡ sin 2θ

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Static particles A particle is in equilibrium if the resultant of all the forces acting on it is zero. So the resultant in any direction will also be zero.

Finding unknown forces You can resolve forces horizontally and vertically to find an unknown force. The particle on the right is in equilibrium with three forces acting on it: 80 N, X N and Y N. R(↑): 80 sin 30° − X sin 45° = 0 80 sin 30° X = __________ sin __ 45° = 40√2 N __ R(→): 80 cos 30° + 40√2 cos 45° − Y = 0 Y = 109 N (3 s.f.)

XN YN

30°

A particle of mass 5 kg is held in equilibrium by two light inextensible strings. One string is horizontal. The other string is inclined at an angle of 30° to the horizontal. The tension in the horizontal string is Y newtons and the tension in the other string is X newtons. Find (a) the value of X (3 marks) (b) the value of Y. (3 marks)

YN

80 N

30°

80 cos 30°

45°

X cos 45°

XN

X sin 45°

There are three forces acting on the particle: the tensions in the two strings (which will be identical) and the weight of the particle. Label all three forces on the diagram. Because the particle is in equilibrium, the resultant of the forces in the vertical direction will be zero.

A particle of mass 3 kg A 60° 60° B is attached at C to two identical light inextensible strings AC and BC, each TN TN inclined at 60° to the horizontal. The particle C hangs in equilibrium. Find the tension in each string. (3 marks) 3g R(↑): 2T sin 60° − 3g = 0 3g T = _________ 2 sin 60° __ = g√3 = 17 N (2 s.f.)

1

80 sin 30°

Triangles of forces If you are confident using vectors to represent forces, 30° T N you could draw a triangle of forces for this problem. You 3g can use trigonometry to find the magnitude of T. 30° T N There is more about writing forces as vectors on page 154.

2

O 50° P

15 N

A particle P is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point O. A horizontal force of magnitude 15 N is applied to P. The particle P is in equilibrium, with the string taut and OP making an angle of 50° with the downward vertical. Find (a) the tension in the string (3 marks) (b) the weight of P. (4 marks)

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Limiting equilibrium If a particle on a rough surface is in limiting equilibrium (or on the point of moving), then • it is static (so the resultant of all the forces acting on it is zero)

R 1.2 N

Have a look at page 172 for more on dealing with static particles.

F = R

• the frictional force is F = µR.

40°

Look at page 168 for a reminder about friction.

Add the weight of the ring, the normal reaction and the frictional force to the diagram. The ring is in limiting equilibrium so you know that • the resultant force in any direction will be 0 • F = µR . Resolve vertically to find the normal reaction, then resolve horizontally to find the value of µ.

Remember to use unrounded values in later calculations. In this question, you could lose marks by using R = 1.7 N in part (b). Use the button on your calculator to enter the exact value, or use at least 4 decimal places. You have used g = 9.8 m s−2 in this calculation, so you need to round any final answers to 2 significant figures.

0.25g

A small ring of mass 0.25 kg is threaded on a fixed rough horizontal rod. The ring is pulled upwards by a light string which makes an angle of 40° with the horizontal, as shown. The string and the rod are in the same vertical plane. The tension in the string is 1.2 N and the coefficient of friction between the ring and the rod is μ. Given that the ring is in limiting equilibrium, find (a) the normal reaction between the ring and the rod (4 marks) R(↑): R + 1.2 sin 40° − 0.25g = 0 R = 0.25g − 1.2 sin 40° = 1.6786… = 1.7 N (2 s.f.) (b) the value of μ. (6 marks) R(→): 1.2 cos 40° − F = 0 µR = 1.2 cos 40° 1.2 cos 40° µ = ___________ 1.6786… = 0.5476… = 0.55 (2 s.f.)

F < μR If an object on a rough plane is not moving, the value of F is only as large as it needs to be to resist motion. You sometimes write FMAX = µR.

This block is resting on a sloping plane. The component of its weight acting down the slope is 3g sin 30° = 14.7 N. This is less than µR = 2.1g cos 30° = 18 N, so the block does not move. It is in equilibrium, with F = 14.7 N.

R

F ø R g

3k

3g sin 30°

3g

30°

 = 0.7 3g cos 30°

Check whether the particle is on the point of moving up or down the slope. Here it is on the point of moving up the slope, so the frictional force will act down the slope.

A particle P of mass 5 kg is held at rest on a rough plane. The plane is inclined to the horizontal at an angle α where cos α = _45 The coefficient of friction between the particle and the plane is 0.5 The particle is held in place by a horizontal force of magnitude X N. The particle is in equilibrium and on the point of moving up the plane. Find the value of X. (7 marks)

173

XN α

P

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Static rigid bodies You might need to combine your knowledge of moments, forces and friction to solve problems involving rigid bodies in equilibrium. One of the most common questions of this type involves a ladder leaning against a wall.

! A uniform ladder of mass 15 kg and length 6 m rests against a smooth vertical wall, with its lower end resting on rough horizontal ground. The coefficient of friction between the ground and the ladder is 0.4, and the ladder is inclined at an angle α to the horizontal, where tan α = 2. A builder loads bricks through a window into a bucket attached to the top of the ladder. Find the maximum mass of bricks that can be loaded before the ladder starts to slip. (8 marks) Let the maximum mass of bricks be m kg. S

3m

R3m

mg

15g

It is essential to draw a diagram for problems such as this. Make sure you include: • the normal reactions at the wall and the ground These are different so use different letters.

• the frictional force for any rough surface Frictional forces act so as to oppose the direction of motion.

• the weight of the ladder. If the ladder is uniform this will act at its midpoint.

Then you need to resolve horizontally and vertically, and choose a point to take moments about. In this example, you need to find m, so you can’t take moments about the top of the ladder. Take moments about the bottom of the ladder, and remember that: Moment = force × perpendicular distance You will need to use problem-solving skills throughout your exam – be prepared!

α F

R(→): F = S R(↑): R = 15g + mg Taking moments about the bottom of the ladder: 15g × 3 cos α + mg × 6 cos α = S × 6 sin α 2__ 1__ 1__ 45g ___ + 6mg ___ = 6S ___ (√5 ) (√5 ) (√5 ) 1 45g + 6mg = 12S When the ladder is on the point of slipping (in limiting equilibrium), F = µR. Since S = F = µR = 0.4R, and R = 15g + mg, equation 1 becomes: 45g + 6mg = 12 × 0.4(15g + mg) 45 + 6m = 72 + 4.8m m = 22.5 kg

When the value of the coefficient of friction is a minimum, the ladder will be in limiting equilibrium.

A uniform ladder AC of mass m and length 3k rests on rough horizontal ground, leaning against a smooth horizontal wall. The angle between the ladder and the ground is 3 α, where tan α = _4 . A builder of mass 5m stands on the ladder at a point B, a distance k from the bottom. C 3k k A

B α

(a) Given that the builder and the ladder are in equilibrium, find the least possible value of the coefficient of friction between the ladder and the ground. (8 marks) (b) State how you have used the fact that the ladder is uniform in your calculations. (1 mark)

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Connected particles 2 When two particles are connected via a pulley, one particle might be on a plane. If the plane is rough, you’ll have to consider friction as well. Have a look at page 168 for a reminder. 5 __ You have been given tan α = 12. You can sketch a right-angled triangle to work out exact values of sin α and cos α:

R T

P

T

Q

F 2.6g

α

mg

Two particles P and Q have masses 2.6 kg and m kg respectively. The particles are attached to the ends of a light inextensible string that passes over a smooth pulley which is fixed at the top of a rough plane. The plane is inclined to the 5 horizontal at an angle α, where tan α = __ 12 The coefficient of friction between P and the plane is 0.5. The particle P is held at rest on the inclined plane and the particle Q hangs freely below the pulley with the string taut. The system is released from rest and Q accelerates vertically downwards at 1.8 m s−2. Find (a) the magnitude of the normal reaction of the inclined plane on P (2 marks) R F α

T α 2.6g cos α 2.6g

R(↖): R − 2.6g cos α = 0 12 12 Using cos α = ___ : R = 2.6g × ___ 13 13 = 24 N (2 s.f.)

(b) the value of m. (8 marks) R(↗): Using F = ma for P to find T: T − F − 2.6 g sin α = 2.6 × 1.8 5 1 Using sin α = ___ : T − F = 14.48 13 F = µR = 0.5 × 24 = 12 Substituting into 1 : T − 12 = 14.48 T = 26.48 N R(↓): Using F = ma on Q: mg − T = 1.8m 9.8m − 26.48 = 1.8m 8m = 26.48 m = 3.31 = 3.3 kg (2 s.f.)

175

122 + 52 = 13 α

5

12

5 __ sin α = 13

__ 12 cos α = 13

You might also be able to find these fractions on your calculator: 5 sin tan-1 12

5 13

Resolve parallel to the plane to write an equation of motion for P, and resolve vertically to write an equation of motion for Q.

Two particles P and Q of masses 4 kg and Q (2 kg) 2 kg respectively are P (4 kg) attached to the ends θ of a light inextensible string, which runs over a smooth pulley. Q is held at rest on a rough plane inclined at an 3 angle u to the horizontal, where tan u = _4 . The coefficient of friction between Q and the 1 plane is _4 . The particles are released from rest and Q accelerates up the plane. (a) Find the acceleration of Q. (10 marks) After 0.6 s, particle P hits the floor and remains there. Particle Q continues up the slope, reaching its highest point after a further T s. In this motion Q does not reach the pulley. (b) Find the value of T. (6 marks)

You need to use the suvat formulae for part (b). u will be the speed of Q after 0.6 s, and v will be 0.

Have a look at pages 150 and 151 for a reminder.

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Vectors in kinematics Velocity and displacement are both vector quantities. This means they have magnitude and direction. You can use unit vectors i and j to describe the position or velocity of an object.

Position and velocity Make sure you don’t confuse position vectors and velocity vectors:

1

Position (or displacement) vectors tell you where an object is relative to a fixed origin. Distance is the magnitude of a position vector. i + b j) m, If an object has position vector r = (a________ its distance from the origin is |r| = √a2 + b2 m

P moves with velocity vector (7i + 3j) m s–1 Its speed is 72 + 32 = 58 m s–1 (7i + 3j) m s–1

P

2

Velocity vectors tell you what direction an object is moving in and how fast. Speed is the magnitude of a velocity vector. If an object has velocity vector v = (p i + q j ) m s−1, ________ its speed is |v| = √p2 + q2 m s−1

j

(–2i + 5j) m P has position vector (–2i + 5j) m i O Its distance from O is (–2)2 + 52 = 29 m

i is the horizontal unit vector and j is the vertical unit vector.

Velocity and time If an object starts at a point with position vector r0 m and moves for time t seconds with velocity vector v then its new position vector r will be given by

Q t=O v = (3i – 2j) m s−1 t=1

r = r0 + vt

In the diagram, Q starts with initial position vector r0 = (2i + 9j ) m and moves with velocity vector v = (3i − 2j ) m s–1. After 3 seconds: r = r0 + vt = (2i + 9j ) + 3(3i − 2j ) = (2 + 3 × 3)i + (9 − 3 × 2)j = 11i + 3j

t=2 r0 = (2i + 9j) m

A particle P is moving with constant velocity (−5i + 4j) m s−1. At time t = 9 seconds, P is at the point with position vector (−12i − 2j) metres. Find the distance of P from the origin at time 5 s. (5 marks) r = r0 + ~vt ~ ~ (−12i~ − 2j ) = r0 + 4(−5i~ + 4j ) ~ ~ ~ r0 = (−12i~ − 2j ) − 4(−5i~ + 4j ) ~ ~ ~ = 8i~ − 18j ____________ ~

|r0| = √82 + (−18)2 = 19.7 m (3 s.f.) ~

A ship S moves with constant velocity (−10i − 5j) km h−1. It passes a buoy with position vector (12i + 25j) km at midday. Find (a) the speed of the ship (2 marks) (b) its position vector at 3 pm. (3 marks)

O

t=3 Q r = (11i + 3j) m j i

You know the position of P at t = 9 s and you want to know its position 4 seconds earlier. Let r0 be the position of P at 5 s, and use r = r0 + vt with −1 t = 4, v = (−5i + 4j) m s and r = (−12i − 2j) m. The distance from the origin is the vector. tude of the magni______ _ |r0| = √a 2 + b2

Velocity and position vectors have units. The units of velocity here are km h−1 so measure time in hours.

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Vectors and bearings If you have to solve a vector problem involving bearings, i will be the unit vector due east and j will be the unit vector due north.

Finding a bearing Two ships P and Q are moving with constant velocities. Ship P moves with velocity (2i − 3j) km h−1 and ship Q moves with velocity (3i + 4j) km h−1. (a) Find, to the nearest degree, the bearing on which Q is moving. (2 marks)

You can use trigonometry to find the bearing an object is moving on if you know its velocity vector. Remember that bearings are measured clockwise from north and you should always give bearings to the nearest degree. N

N θ θ

4j

3 tan θ = __ so θ = 36.86…° 4 Bearing is 037° (nearest degree)

2i

A

3i

At 2 pm, ship P is at the point with position vector (i + j) km and ship Q is at the point with position vector (−2j) km. At time t hours after 2 pm, the position vector of P is p km and the position vector of Q is q km. (b) Write down expressions, in terms of t, for (i) p (ii) q___ › (5 marks) (iii) PQ. (i) p = ( i + j) + t (2i − 3j) (ii) q = (−2j) + t (3i + 4j) (iii) q − p = [(−2j) + t (3i + 4j)] − [( i + j) + t (2i − 3j)] = (−i − 3j) + t ( i + 7j) (c) Find the time when (i) Q is due north of P (ii) Q is north-west of P. (4 marks) ___›

PQ = (−i − 3j) + t ( i + 7j) = (−1 + t)i + (−3 + 7t)j (i) i component = 0 −1 + t = 0 t = 1, so time is 3 pm (ii) i component = −j component −1 + t = −(−3 + 7t) 8t = 4 t = 0.5, so time is 2.30 pm

In part (b), find expressions for the position vectors of A and B, rA and rB in terms of t. To show that A and B will collide you need to find a single value of t that makes the i components equal and the j components equal.

177

θ v

A is moving with velocity vector (2i − j ) m s−1 1 tan θ = __ 2 so θ = 26.56…° Bearing = 90° + 26.56…° = 117° (nearest degree) –j

For part (b), use r = r0 + vt to write expressions for the position vectors of P and Q in terms of t. ___›

For part (c), write the vector PQ in terms of its i component and its j component. ___› of P the When Q is due north is› i component of PQ will be 0. When Q___ PQ of nent compo i north-west of P the will be minus the j component. Q

Q

PQ = k j P

kj

PQ = −k i + k j

−k i

P

A particle A moves with constant velocity (−i + 4j) m s−1. At time t = 0, A is at the point with position vector (3i − 8j) m. (a) Find the direction in which A is moving, giving your answer as a bearing. (2 marks) At time t = 0, a second particle B is at the point with position vector (−21i + 16j) m, and is moving with velocity (5i − 2j) m s−1. (b) Show that A and B will collide at a point P and find the position vector of P. (5 marks)

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Variable acceleration 3 You might need to use A-level calculus techniques to model the motion of a particle moving in one dimension with acceleration which varies with time. You can revise the relationships between displacement, velocity and acceleration for a particle moving in one dimension on page 160.

! A particle P is moving in a straight line. At time t seconds, t > 0, the velocity of P, v m s−1, is given by 1 + 2 sin pt, 0 < t < 10π v= cos 2pt, 10π < t < 20π where p is a constant. The initial acceleration of P is 0.2 m s−2. (a) Find the value of p. (3 marks) For the motion from 0 < t < 10π dv a = ___ = 2p cos pt dt When t = 0, a = 0.2, so 0.2 = 2p cos 0 p = 0.1 (b) Find the displacement of the particle from its starting position after (i) 10π seconds (ii) 1 minute. (8 marks) For the motion from 0 < t < 10π:

{

∫

(i) s = v dt =

∫

10π 0

This is a piecewise function. You use the top part of the function for 0 < t < 10π and the bottom part of the function for 10π < t < 20π. v

3

v = 1 + 2 sin 0.1t

2

v = cos 0.2t

1 0

10π

–1

20π t

In part (a) you only need to consider the motion of the particle at t = 0, so you can just use the top part of the function. You will need to use problem-solving skills throughout your exam – be prepared!

To find acceleration, differentiate the expression for velocity with respect to time. Remember that p is a constant.

(1 + 2 sin 0.1t) dt

= [t − 20 cos 0.1t] 0

10π

= (10π − 20 cos π) − (0 − 20 cos 0) = 10π + 20 − 0 + 20 So displacement after 10π seconds = 10π + 40 = 71.4159… m (ii) For the motion from 10π < t < 20π:

∫

s = v dt =

∫

60 10π

(cos 0.2t) dt

= [5 sin 0.2t]

60 10π

= 5(sin 12 − sin 2π) = −2.6828… m So displacement after 60 seconds is 71.4159… − 2.6828… = 68.7 m (3 s.f.)

In part (a) you will need to use the product rule for differentiation, and in part (b) you will need to use integration by parts. You can revise these on pages 90 and 104 respectively.

The diagram shows the velocity–time graph for a particle moving on the positive x-axis. v

O

t

The velocity of the particle, v m s−1, at time t seconds is given by t ,t>0 v = ___ e3t (a) Find the maximum velocity of the particle. (5 marks) (b) Find the total distance travelled by the particle in the first 2 seconds of its motion, giving your answer correct to 3 significant figures. (4 marks)

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Calculus with vectors You can use the relationships between displacement, velocity and acceleration to solve problems in two dimensions. Your vectors should be given in terms of the unit vectors i and j, and you are always integrating or differentiating with respect to time.

Displacement = f(t )i + g(t )j Differentiate

Velocity = f9(t )i + g9(t )j Differentiate

A particle P of mass 0.6 kg is moving in a horizontal plane under the action of a single force F N. The velocity of the particle v m s−1 at time t seconds is given by _ v = 80 √ t i + t3j, t > 0 Find (a) the acceleration of P when t = 4 (3 marks) dv 40 __ i + 3t2j a = ___ = ___ dt √t When t = 4, 40 __ i + 3(42)j = 20i + 48j a = ___ √4 (b) the magnitude of F when t = 4. (3 marks) F = m a = 0.6(20i + 48j) = 12i + 28.8j

Integrate

Integrate

Velocity = f0(t )i + g0(t )j

Golden rules Differentiate or integrate each component of a vector separately. Read questions carefully – if you are asked to find speed or distance you will need to find the magnitude of the velocity or displacement vector. If you are integrating a vector quantity, your constant of integration should be a vector.

_____________

|F| = √ 122 + 28.82 = 31.2 N

You need to include a constant of integration, which will be a vector. Use the fact that r = 12i − 3j when t = 3 to find the constant c. It’s a good idea to write your final answer giving the i component and j component as separate functions of t.

A particle is moving in a horizontal plane with velocity (3i + 10tj) m s−1. Given that the particle passes through the point with position vector (12i − 3j) m at time t = 3 seconds, find an expression for the displacement of the particle, r m, at time t seconds. (3 marks)

∫

∫

r = v dt = (3i + 10tj) dt

A particle Q is moving in a horizontal plane. The acceleration of the particle, a m s−2 at time t seconds is given by a = (4t − 1)i − j, t > 0 When t = 0, the particle is at the origin and moving with velocity −3i m s−1. (a) Find the velocity of the particle when t = 2. (4 marks) (b) Find the time at which the particle is moving parallel to the unit vector j. (3 marks) (c) Find the distance of the particle from the origin when t = 2. (6 marks)

179

= 3ti + 5t2j + c When t = 3, r = 12i − 3j, so 9i + 45j + c = 12i − 3j c = 3i − 48j So r = (3t + 3)i + (5t2 − 48)j

Watch out! You need to integrate all of (4 – 2t) and use the initial condition to find the i component of the velocity. The particle will be moving parallel to the vector j when this component is equal to zero.

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You are the examiner! In your exam you might be asked to identify errors in working. You should also be confident checking your own work. Each of these students has made key mistakes in their working. Can you spot them all?

1

A uniform ladder of length 10 m and weight 180 N rests on rough horizontal ground, leaning against a smooth vertical wall. The ladder makes an angle of 60° with the ground, and the coefficient of friction 1 between the ladder and the ground is _4 A builder of weight 700 N climbs the ladder. Find the maximum distance from the ground the builder can climb before the ladder starts to slip. (6 marks) S

xm 5m

700 N 180 N

R

60° F

R(↑): R = 180 + 700 = 880 N R(→): F = S 1 Limiting equilibrium, so F = µR = __ 4 × 880 = 220 N Taking moments about base of ladder: 180 × 5 + 700x = 10S 900 + 700x = 2200 x = 1.9 m (2 s.f.)

3

A particle P moves with constant acceleration (2i − 5j) m s−2. At time t = 0, P has speed u m s−1. At time t = 3 s, P has velocity (−6i + j) m s−1. Find the value of u. (5 marks) v = u + at (−6i + j) = u + 3(2i − 5j) u = (−6i + j) − 3(2i − 5j) = (−12i + 16j) m s−1

2

PN A small box of mass 15 kg rests on a rough horizontal plane. The coefficient of RN friction between the box FN and the plane is 0.2 15 kg 50° A force of magnitude 15g N P newtons is applied to the box at 50° to the horizontal. The box is on the point of sliding along the plane. Find the value of P, giving your answer to 2 significant figures. (9 marks) R(→): P cos 50° − F = 0 F = µR = 0.2R so 1 P cos 50° − 0.2R = 0 R(↑): R − 15g = 0 R = 15g Substitute into 1 : P cos 50° − 3g = 0 3g P = ________ = 46 (2 s.f.) cos 50°

Checking your work If you have time left in your exam you should check back through your working: Check you have answered every question part and given all the information asked for. Draw large, clearly labelled diagrams showing any forces acting on an object. If you have chosen a positive direction in a kinematics question, make sure your quantities all have the correct sign. Make sure that your answers make sense. Scalar quantities such as distance and speed must be positive.

Find the mistakes in each student’s answer on this page, and write out the correct working for each question. Turn over for the answers.

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You are still the examiner! Before looking at this page, turn back to page 180 and try to spot the key mistakes in each student’s working. Use this page to check your answers. The corrections are shown in red and these answers are now 100% correct.

1

A uniform ladder of length 10 m and weight 180 N rests on rough horizontal ground, leaning against a smooth vertical wall. The ladder makes an angle of 60° with the ground, and the coefficient of friction 1 between the ladder and the ground is _4 A builder of weight 700 N climbs the ladder. Find the maximum distance from the ground the builder can climb before the ladder starts to slip. (6 marks)

2

S

xm 5m

700 N 180 N

R

60° F

R(↑): R = 180 + 700 = 880 N R(→): F = S 1 Limiting equilibrium, so F = µR = __ 4 × 880 = 220 N Taking moments about base of ladder: 180 × 5cos 60° + 700x cos 60° = 10S sin 60° 900 + 700x = 2200 x = 1.9 (2 s.f.) 450 + 350x = 1905.255… x = 4.2 m (2 s.f.)

To find a moment about a point you multiply the magnitude of the force by the perpendicular distance between the point and the line of action. Revise moments on pages 164 and 174.

Check whether you are asked for the velocity or the speed. Speed is the magnitude of the velocity vector. Acceleration and speed vectors are covered on page 176.

181

PN A small box of mass 15 kg rests on a rough horizontal plane. The coefficient of RN friction between the box FN and the plane is 0.2 15 kg 50° A force of magnitude 15g N P newtons is applied to the box at 50° to the horizontal. The box is on the point of sliding along the plane. Find the value of P, giving your answer to 2 significant figures. (9 marks) R(→): P cos 50° − F = 0 F = µR = 0.2R so 1 P cos 50° − 0.2R = 0 R(↑): R − 15g = 0 P sin 50° + R − 15g = 0 R = 15g P sin 50° + R = 15g 2 Substitute into 1 : P cos 50° − 3g= 0 3g P = ________ = 46 (2 s.f.) cos 50° 5× 1 5P cos 50° − R = 0

P sin 50° + R = 5P cos 50° + P sin 50° = P(5 cos 50° + sin 50°) = P=

+ 2

15g 15g 15g 37 N (2 s.f.)

Don’t forget to include all the forces when you’re resolving. Learn more about this on page 167.

3

A particle P moves with constant acceleration (2i − 5j) m s−2. At time t = 0, P has speed u m s−1. At time t = 3 s, P has velocity (−6i + j) m s−1. Find the value of u. (5 marks) v = u + at (−6i + j) = u + 3(2i − 5j) u = (−6i + j) − 3(2i − 5j) = (−12i + 16j) m s−1 _____________

u = √ (−12)2 + 162 = 20 m s−1

WORKED SOLUTIONS

Worked solutions PURE MATHEMATICS

(b) 8x6 + 7x3 − 1 = 0 8(x3)2 + 7x3 − 1 = 0

1. Index laws

(8x3 − 1)(x3 + 1) = 0

_1

x3 = −1 or x3 = _​​ 8 ​​ so x = −1 or x = _​​ 2 ​​  1

_3

1 x(4​x​−​ 2 ​ ​)3 = x1(43​x−​ ​ 2 ​ ​) _3 _1 = 64​x−​ ​ 2 ​  + 1​= 64​x−​ ​ 2 ​ ​ 3

2 (9y ​ )​​ = ​9​ ​​y​

__

  3y15 = 27y15 = (​√9 ​) __ _1 √ 5 + 2​ x ​  __ 5 2​x​ ​2 ​ ​ 3 ​ ________  = ​  2  ​ + ___ ​  2 ​  2 ​   x x x _1 = 5x−2 + 2​x​ ​2 ​  − 2​

2. Expanding and factorising 1 (x + 2)(x + 1)2 = (x + 2)(x2 + 2x + 1) = x3 + 2x2 + x + 2x2 + 4x + 2 = x3 + 4x2 + 5x + 2 b = 4, c = 5, d = 2 2 3x3 − 2x2 − x = x(3x2 − 2x − 1) = x(3x + 1)(x − 1) 3 25x2 − 16 = (5x + 4)(5x − 4)

__

__

__

__

__ __

1 (x + ​√3 ​)(x   −√ ​ 3 ​)   = x2 + x​√ ​ 3 ​​  − x​√ ​ 3 ​​  − √ ​​ 3 ​​ √3 ​​​​​  = x2 − 3 ___

___

__

__

√98 ​ = √ ​ 49 ​ × √ ​ 2 ​ = 7​√2 ​  2 ​

__

__

__   8(3 − √ ​ 5 ​) 24 − 8​√5 ​  8 __ _______________ __ __  ​ = ________ 3 (a) ______ ​     ​   = 6 − 2​√5 ​  =    ​     ​   ​  4 3+√ ​ 5 ​  (3 + √ ​ 5 ​)  (3 − √ ​ 5 ​)   __

__

__

__

__ ​ 5 ​)   (4 + √ ​ 5 ​)  (2 + √ 4+√ ​ __ 5 ​  _______________ 13 + 6​√5 ​  __ __  ​ = ________ (b) ​ ______   =    ​  ​   ​  = −13 − 6​√5 ​   ​  −1 2−√ ​ 5 ​  (2 − √ ​ 5 ​)  (2 + √ ​ 5 ​)  

4. Quadratic equations 1

2

2(x − 3) + 3x = 14

2(x2 − 6x + 9) + 3x = 14 2x2 − 12x + 18 + 3x = 14 2x2 − 9x + 4 = 0 (2x − 1)(x − 4) = 0 x = ​ _2 ​   or  x = 4 1

2 (a) x2 − 10x + 7 = (x − 5)2 − 52 + 7 = (x − 5)2 − 18



__

p = 1, q = −18

__

__

__

​​ x ​​ = 2 so x = 25 or x = 4​​ ​​ x ​​ = 5 or √ __ 2 4​​√x ​​ + x = 3 __ __ (​​√x ​​)  2 + 4​​√x ​​ − 3 = 0 __ (​​√x ​​ + 2)2 − 4 − 3 = 0 __ __ √ ​​ x ​​ + 2 = ​± ​√7 ​​  __ __ √ ​​ x ​​ = −2 ​±  ​√7 ​​  __ __ __ __ so √ ​​  x ​​ = −2 + √ ​​ 7 ​​  or √ ​​  x ​​ = −2 − √ ​​  7 ​​  __ x = (−2 + ​​√ 7 ​​ )2 __ = 11 − 4​​√ 7 ​​  a = 11, b = 4 f(x) = 0 3 4 2 x − 4x − 5 = 0 (x2 − 5)(x2 + 1) = 0 x2 = 5 or x2 = −1 __ x [ R, x , 0 so the only root is x = −​​√5 ​​  √

_3

3. Surds

__

(​​√x ​ ​ − 5)(​​√x ​​ − 2) = 0

​ 2 ​ ​  = 5x−2 + 2​x− ​ 



2

(​​√x ​​)  − 7​​√x ​​ + 10 = 0

​

__

x + 10 = 7​​√x ​​ 

(c)

_3 10 × ​ _3 ​  ​ 2 ​  2

_ 10 ​ 2 ​ 

1

__

6. Sketching quadratics 1 (a)

y

y 5 (x 2 2)2

(b)

y

y 5 (x 2 2)2 1 k

41k 4 O

(2, k)

x

2

O

x

2 (a) x2 + 6x + 15 = (x + 3)2 − 32 + 15 = (x + 3)2 + 6 a = 3, b = 6 y (b) 15 (23, 6) O

x

(c) The graph does not intersect the x-axis so the equation has no real solutions.

(b) x2 − 10x + 7 = 0 (x − 5)2 − 18 = 0

7. The discriminant

(x − 5)2 = 18 ___ __ x − 5 = ±​√18 ​ = ±3​√2 ​  __ x = 5 ± 3​√2 ​ 

1 (−2)2 − 4 × 3 × (−5) = 64 2 b2 − 4ac = 0 22 − 4p × (−3) = 0 4 + 12p = 0 1 p = −​ _3 ​ 

a = 5, b = 3

5. Functions and roots 1 (a)

x4 − 3x2 − 4 = 0

(x2)2 − 3x2 − 4 = 0 (x2 + 1)(x2 − 4) = 0 x2 = −1 or x2 = 4 so x = 2 or x = −2

3 (a) (k + 5)2 − 4 × 1 × 2k = k2 + 10k + 25 − 8k = k2 + 2k + 25 2 (b) k + 2k + 25 = (k + 1)2 − 12 + 25 = (k + 1)2 + 24 p = 1, q = 24

182

WORKED SOLUTIONS (c) (k + 1)2 > 0 for all k, so discriminant > 0 for all k, so f(x) = 0 has distinct real roots.

2 (a)

b2 − 4ac > 0

(k − 3)2 − 4 × 1 × (−4k) > 0 k2 − 6k + 9 + 16k > 0

8. Modelling with quadratics

k2 + 10k + 9 > 0

(a) 42 m (b) 0 = 42 + 0.7d − 0.14d 2

(b) (k + 9)(k + 1) > 0 y

__________________

2 − 0.7 ± ​√0.7     − 4(−0.14 ) (42) ​ ______________________ d =     ​​      ​​ 2(−0.14) d = −15 or d = 20 d > 0 so the horizontal distance is 20 m.

y 5 (k 1 9)(k 1 1)

21 O

29

(c) h = 42 + 0.7d − 0.14d 2

k

= 42 − 0.14(d 2 − 5d) = 42 − 0.14((d − 2.5)2 − 2.52)

k < −9 or k > −1

= 42.875 − 0.14(d − 2.5)2 0.14(d − 2.5)2 > 0 so the maximum height is 42.9 m (3 s.f.)

(a), (d)

9. Simultaneous equations 1 x + y = 5

x2 + 2y2 = 22

11. Inequalities on graphs y

1

6

2

y 5 6 2 x2

y5

From 1 : x = 5 − y Substitute into 2 :

(5 − y)2 + 2y2 = 22 2

25 − 10y + y + 2y = 22

O

22

3y2 − 10y + 3 = 0

11

1

2 6

2

1 2

x

6

6 − x2 = _​​ 2 ​​ x + 1 1

(3y − 1)(y − 3) = 0 1 1 14 ​  3 ​  y = ​ _3 ​  ⇒ x = 5 − _​ 3 ​ = __ y=3⇒x=5−3=2 14 _1 Solutions: x = ​ __ 3 ​,  y = ​  3 ​  and

(b)

x = 2, y = 3

(c) (i) {x  :  x < − 2.5} ∪ {x   :   x > 2}

2 (a) y = x + 6

2x2 + x − 10 = 0 (2x + 5)(x − 2) = 0 x = −  ​​ _2 ​​  or x = 2 5

1

(ii) −2.5 < x < 2

xy − 2x2 = 7 2 Substituting 1 into 2 : x(x + 6) − 2x2 = 7 2

2

x + 6x − 2x = 7

12. Cubic and quartic graphs 1 (a) x3 − 9x = x(x2 − 9) = x(x + 3)(x − 3)

−x2 + 6x − 7 = 0

(b)

x2 − 6x + 7 = 0

y y 5 x(x 1 3)(x 2 3)

(b) (x − 3)2 − 32 + 7 = 0 (x − 3)2 = 2 __ x = 3 ± ​√2 ​  __ __ x=3+√ ​ 2 ​  ⇒ y = 9 + √ ​ 2 ​  __ __ x=3−√ ​ 2 ​  ⇒ y = 9 − √ ​ 2 ​  __ __ Solutions: x = 3 + √ ​ 2 ​,   y=9+√ ​ 2 ​  and __ __ x = 3 − √ ​ 2 ​,   y=9−√ ​ 2 ​ 

23 O

2

y y 5 (2x 2 1)(x 2 3)2

10. Inequalities 1

x

3

x(x − 5) < 14 x2 − 5x − 14 < 0

O

(x − 7)(x + 2) < 0

1 2

3

x

29

y y 5 (x 2 7)(x 1 2)

3 22 O

7

y y 5 x(5 2 x)(2x 1 1)(x 1 4)

x

−2 < x < 7 24

183

2 12 O

5

x

WORKED SOLUTIONS 13. Transformations 1

16. Points of intersection

(a)

(a)

y

(4, 6)

O

y 5 x2(x 2 3)

O

x

8

y

3

y 5 2f(x)

x

8

y 5 x(8 2 x)

(b)

y

(b) x2(x − 3) = x(8 − x) x(x − 3) = 8 − x   x ≠ 0

(24, 3)

x2 − 2x − 8 = 0

x

O

28

(x + 2)(x − 4) = 0 x = −2 ⇒ y = −2 × (8 − (−2)) = −20 or x = 4 ⇒ y = 4 × (8 − 4) = 16

y 5 f(2x)

(c)

y

(0, 0), (−2, −20), (4, 16)

(42k, 3)

2k O

17. Equations of lines

x

82k

1 y − y1 = m(x − x1) 1 y − (−5) = −​ _3 ​ (x − 6)

y 5 f(x 1 k)

3y + 15 = −x + 6 x + 3y + 9 = 0 y2 − y1 ________ 11 − 2 3  ​ = ​     ​  = _​   ​  2 m = ​ ______ x2 − x1  8 − (−4) 4 y − y1 = m(x − x1) 3 y − 11 = ​ _4 ​ (x − 8)

14. Transformations 2 (a)

y y 5 f(x 1 1)

y = _​ 4 ​ x + 5 3

23

3 (a) 3(1) + 4(5) − k = 0

x

O

k = 23 (b) 3y + 4x − 23 = 0 3y = −4x + 23

x 5 22

​ 3 ​  y = −​ _3 ​ x + __ 4

(b) When x = 0, f(x + 1) = f(1) (1 + 2)2 _9 1 f(1) = _____ ​​     ​​  = ​​  2 ​​  = 4​​ _2 ​​  1+1

Gradient = −​  3 ​ 

18. Parallel and perpendicular

(−3, 0) and ​​(0, 4​  2  ​)​ _1

1 (a) 10 − 3x = 10 − 3(4) = 10 − 12 = −2 1 1 (b) m = − ​ ___   ​ = __ ​   ​  −3 3 y − y1 = m(x − x1) 1 y − (−2) = ​ _3 ​ (x − 4)

​15. Reciprocal graphs 1

y

3y + 6 = x − 4

4 y 5 2x O

2 (a)

23

_4

x − 3y − 10 = 0 1 2 4x − 5(0) − 1 = 0 ⇒ x = ​ _4 ​  1 A is the point (​ _4 ​ , 0) 4 1 4x − 5y − 1 = 0 ⇒ y = ​ _5 ​ x − _​ 5 ​  _4 Gradient of L1 = ​ 5 ​  5 Gradient of L2 = −​ _4 ​ 

x

(b)

y y5

O

y − y1 = m(x − x1)

y

3 x2

y − 0 = −​ _4 ​  (x − _​ 4 ​ ) 5

3

x

O

y5

3 (x 1 1)2 x

1

​ 16  ​  y = −​ _4 ​  x + __ 5

5

19. Lengths and areas (a) x − 2(0) + 6 = 0 ⇒ x = −6 P is (−6, 0)

x 5 21

(c) y = 0 and x = −1

(0) − 2y + 6 = 0 ⇒ y = 3 Q is (0, 3)

______

___

_____

__

PQ = √ ​ 62 + 32 ​  = ​√45 ​ = ​√9 × 5 ​ = 3​√5 ​ 

184

WORKED SOLUTIONS So has equation y + 3 = _​​ 3 ​​(x − 4) or 3y − 4x + 25 = 0

(b) x − 2y + 6 = 0 ⇒ y = _​ 2 ​ x + 3

4

1

Gradient of L1 =

_1 ​ 2 ​ 

Substitute y = 5 into this equation: 3(5) − 4x + 25 = 0 so x = 10 Centre of circle is (10, 5). _______ ____ Radius of circle is √ ​​ 102 + 52 ​  =√ ​ 125 ​​  So equation of circle is (x − 10)2 + (y − 5)2 = 125

Gradient of L2 = −2 y − 3 = −2(x − 0) y = −2x + 3 (c) 0 = −2x + 3 ⇒ x = ​ _2 ​  3

R is (​ _2 ​ , 0) 3

23. The factor theorem

PR = 6 + _​ 2 ​ = __ ​  2 ​  3

15

1 (a) f(2) = 23 − 7(2)2 − 14(2) + 48

Area = ​ _2 ​ × __ ​  2 ​ × 3 = __ ​ 4 ​  1

15

45

20. Equation of a circle __________________ 2 2

___

__

(a) AB = √ ​ (2    − (−6)) + (4 − 0)  ​= √ ​ 80 ​ = 4​√5 ​  −6 + 2 _____ 0+4 (b) ​ ______ ​   ​   ​    ​= (−2, 2)  ​,  2 2 __ (c) Centre (−2, 2), radius 2​√5 ​ 

(

)



= 8 − 28 − 28 + 48



=0

So (x − 2) is a factor. (b) 2

1

__

 2 (x − (−2))2 + (y − 2)2 = (2​√5 ​) (x + 2)2 + (y − 2)2 = 20

21. Circle properties

−7

−14

48

2

−10

−48

−5

−24

0

f(x) = (x − 2)(x2 − 5x − 24)

= (x − 2)(x − 8)(x + 3)

2 (a)

f(−4) = 0

(a) (x − 5) + (y + 2) = 100

2(−4)3 − 3(−4)2 − 65(−4) − a = 0

(b) (−1 − 5)2 + (6 + 2)2 = (−6)2 + 82 = 36 + 64 = 100



2

2

So (−1, 6) lies on C. (c) Radius between (−1, 6) and (5, −2) has gradient (−2) − 6 ___ −8 4  ​ = ​   ​ = − ​ _3 ​  ​ ________  5 − (−1) 6 3 Tangent is perpendicular to radius, so has gradient _​ 4 ​ . y − y1 = m(x − x1)

y − 6 = ​ _4 ​ (x + 1) 3

4y − 24 = 3x + 3 3x − 4y + 27 = 0

22. Solving circle problems

185

1

1 x2 + (x + 7 + 2)2 = 45 x2 + x2 + 18x + 81 = 45 x2 + 9x + 18 = 0 (x + 6)(x + 3) = 0 So x = −6 or x = −3 Points of intersection are (−6, 1) and (−3, 4). 2 2x − y + 2 = 0 so y = 2x + 2 Equation of circle: (x − k)2 + y2 = 4 So points of intersection satisfy: (x − k)2 + (2x + 2)2 = 4 x2 − 2kx + k2 + 4x2 + 8x + 4 = 4 5x2 + (8 − 2k)x + k2 = 0 Two distinct solutions so b2 − 4ac > 0: (8 − 2k)2 − 20k2 > 0 64 − 32k + 4k2 − 20k2 > 0 64 − 32k − 16k2 > 0 k2 + 2k − 4 < 0 __ Solutions to k2 − 2k − 4 = 0 are k = −1 ± √ ​​ 5 ​​  So values of k that satisfy inequality are: __ __ −1 − √ ​​ 5 ​​  < k < −1 + √ ​​ 5 ​​  3 Perpendicular bisector of (0, 0) and (0, 10) is y = 5 Perpendicular bisector of (0, 0) and (8, −6) passes through 1 4 (4, −3) and has gradient ​​ ____    ​​ = _​​  3 ​​. − 3 __ (​ ​  4  )​ ​

−128 − 48 + 260 − a = 0



a = 84

(b) −4

2

−3

−65

−84

2

−8

44

84

−11

−21

0

f(x) = (x + 4)(2x2 − 11x − 21)

= (x + 4)(2x + 3)(x − 7)

24. Binomial expansion 1

( ) ( )

( )

( ) ( )

( )

9 9 1 (a) (1 + 3x)9 = 19 + ​  ​     ​   ​× 18 × 3x + ​      ​   ​   ​× 17 × (3x)2 1 2 9  + ​  ​     ​   ​× 16 × (3x)3 + … 3 = 1 + 27x + 324x2 + 2268x3 + … 4 4 (b) (2 + 5x)4 = 24 + ​  ​     ​   ​× 23 × 5x + ​      ​   ​   ​× 22 × (5x)2 1 2 4  + ​  ​     ​   ​× 2 × (5x)3 + … 3 = 16 + 160x + 600x2 + 1000x3 + … 12 12 ​   ​   ​× 310 × (−x)2 ​   ​   ​× 311 × (−x) + ​        (c) (3 − x)12 = 312 + ​        1 2 12       9 3    + ​  ​   ​   ​× 3 × (−x) + … 3 = 531 441 − 2 125 764x + 3 897 234x2   − 4 330 260x3 + … 5 5 ​   ​   ​× 23 × (kx)2 + … 2 (a) (2 + kx)5 = 25 + ​  ​     ​   ​× 24 × kx + ​      1 2 = 32 + 80kx + 80k2x2 + …

( ) ( )

( )

( )

( )

(b) 80k = 48 k = ​ _5 ​  3

​  5 ​  (c) 80k2 = 80 × (​​  ​ _5 ​  )​​ ​ = ___ 3 2

144

25. Solving binomial problems

( )

30 1 (a) x3 term = ​        ​   ​   ​× 127 × (2x)3 = 32 480x3 3 p = 32 480 30 ​   ​   ​× 126 × (2x)4 = 438 480x4 (b) x4 term = ​        4 q 438 480 __ 27 q = 438 480 and __ ​   ​  = _______ ​   ​  = ​  2 ​  p 32 480

( )

WORKED SOLUTIONS

(

( ) ( )

)

( ) ( ) ( )

( )

x 12 x x 2 12 12 2 (a) ​​  1 + __ ​   ​   ​​ ​ = 112 + ​        ​   ​   ​× 111 × ​  __ ​   ​   ​+ ​        ​   ​   ​× 110 × ​​  __ ​   ​   ​​ ​ 1 2 4 4 4 x 3 12   + ​  ​       ​   ​× 19 × ​​  __ ​   ​   ​​ ​+ … 3 4 33 55 = 1 + 3x + __ ​ 8 ​ x2 + __ ​ 16 ​ x3 + … x 12 ​   ​   ​​ ​= (1.025)12 (b) x = 0.1 so ​​  1 + __ 4 33 55 (1.025)12 ≈ 1 + 3 × (0.1) + __ ​ 8 ​  × (0.1)2 + __ ​ 16 ​ × (0.1)3 + …

(



29. Trigonometric graphs (i) (a)

y 1

)

y 5 sin (2x)

O

= 1.3447 (4 d.p.)

π 2

3π 2

π

2π

x

21 π

(b) (0, 0), (​ _2  ​, 0), (π, 0), (​ __ ​,  0), (2π, 0) 2  

26. Proof 1 (a) No statement of proof at the end. (b) f(−2) = 0 so, by the factor theorem, (x + 2) is a factor of f(x). 2 (a) 2(1)2 + 11 = 13; prime 2(2)2 + 11 = 19; prime 2(3)2 + 11 = 29; prime 2(4)2 + 11 = 43; prime 2(5)2 + 11 = 61; prime 2(6)2 + 11 = 83; prime 2(7)2 + 11 = 109; prime 2(8)2 + 11 = 139; prime 2(9)2 + 11 = 173; prime 2(10)2 + 11 = 211; prime (b) e.g. n = 17: n2 + n + 17 = (17)2 + 17 + 17 = 323 = 17 × 19 3 Equivalently, need to prove that (4 − x)2 − 7 + 2x > 0 for all real values of x. (4 − x)2 − 7 + 2x = 16 − 8x + x2 − 7 + 2x = x2 − 6x + 9 = (x − 3)2 2 (x − 3) > 0 for all real values of x So (4 − x)2 − 7 + 2x > 0 for all real values of x So (4 − x)2 > 7 − 2x for all real values of x, as required.

27. Cosine rule (a) a2 = b2 + c2 − 2bc cos A PR2 = 262 + 152 − 2 × 26 × 15 × cos 0.8 = 357.5687… PR = 18.9 cm (3 s.f.) QR2 + QP2 − PR2 (b) cos ∠PQR = ​ ________________        ​ 2 × QR × QP 202 + 132 − 18.9092        ​= 0.4065 … = _______________ ​  2 × 20 × 13 −1 ∠PQR = cos  (0.4065…) = 1.15 rad (3 s.f.)

28. Sine rule

7π π π ___ ∠CAB = π − __ ​   ​  − __   ​   ​  = ​   ​  3 5 15 AC BC _____ _____   ​    ​  ​  = ​  sin B sin A AC _____ 13 ​ ____    ​  = ​  7π π  ​   sin ​ _3  ​ sin ​ __ ​ 15   π

13 sin ​ _3  ​ AC = _______ ​   ​  = 11.3 cm (3 s.f.) 7π sin ​ __ ​ 15  

3π

(ii) (a)

(

y 5 tan x 1

y

π 2

)

3 2 1 O 21

π 2

x

3π 2

22 23 x5π

(b)

π (​ _2  ​,

0),

3π (​ __ ​,  0); 2  

x 5 2π

asymptotes at x = 0, x = π and x = 2π

30. Trigonometric equations 1 1 (a) y 1

O

21

90°

180°

270°

360°

x

y 5 sin x

(b) sin−1 (−0.3) = −17.5° (1 d.p.) −17.5 + 360 = 342.5 180 − (−17.5) = 197.5 x = 342.5°, 197.5° (1 d.p.) 2 (a)

3 cos θ = 1 1 cos θ = _​ 3 ​ 1 cos−1  (​ _3 ​) = 1.23 rad (2 d.p.) θ = 1.23 rad, −1.23 rad (2 d.p.) (b) tan θ + 2 = 0 tan  θ = −2 −1 tan  (−2) = −1.11 rad (2 d.p.) −1.11 + π = 2.03 θ = −1.11 rad, 2.03 rad (2 d.p.)

31. Trigonometric identities 1 1 3 cos2 x − 9 = 11 sin x 3(1 − sin2 x) − 9 = 11 sin x 3 − 3 sin2 x − 9 = 11 sin x 3 sin2 x + 11 sin x + 6 = 0 (3 sin x + 2)(sin x + 3) = 0 2 sin x cannot equal −3, so sin x = −​ _3 ​ 2 sin−1  (−​ _3 ​) = −0.7297… rad −0.7297… + 2π = 5.5534… π − (−0.7297…) = 3.8713… x = 5.553 rad, 3.871 rad (3 d.p.)

186

WORKED SOLUTIONS 5 sin x = 2 tan x 2 sin x 5 sin x = ​ ______ ​  cos x 5 sin x cos x = 2 sin x 5  sin  x cos x − 2 sin x = 0 sin  x(5 cos x − 2) = 0

(b) sin x = 0 ⇒ x = 0°, 180°

5 cos x − 2 = 0 ⇒ cos x = _​ 5 ​ 2

cos−1 ​( _​ 5 ​  )​= 66.42…° 2

360 − 66.42… = 293.57… x = 0°, 180°, 66.4°, 293.6° (1 d.p.)

32. Trigonometric equations 2 1 (a) sin (x − 40°) = −​ _2 ​, 0 < x < 360° 1

−40° < x − 40° < 320° −1

_1

sin   (−​ 2 ​) = −30° 180 − (−30) = 210 x − 40 = −30, 210 x = 10°,__ 250° ​√3 ​  (b) cos (2x) = ___ ​   ​,  0 < x < 360° 2 0 < 2x 0. dy ___ 2 ​    ​ = 4x − 8x−3 dx 2

dy ____

  ​= 4 + 24x−4 dx2 3 (a) y = x3 + 2x2 − 3x dy ​ ___  ​ = 3x2 + 4x − 3 dx ​ 

WORKED SOLUTIONS (b)

41. Modelling with calculus

y y 5 x(x 2 1)(x 1 3)

23

O

x

1

dy (c) At (−3, 0): ___ ​    ​ = 3(−3)2 + 4(−3) − 3 = 27 − 12 − 3 = 12 dx dy ​    ​ = 3(0)2 + 4(0) − 3 = −3 At (0, 0): ___ dx dy ​    ​ = 3(1)2 + 4(1) − 3 = 4 At (1, 0): ___ dx

38. Tangents and normals

y − y1 = m(x − x­1) 2 y − 11 = ​ _5 ​ (x − 4) 5y − 55 = 2x − 8 2x − 5y + 47 = 0

39. Stationary points 1 1 y = x2 − 8x + 3 dy ​ ___  ​ = 2x − 8 dx dy When ___ ​    ​ = 0, 2x − 8 = 0 dx x = 4

√

y = 42 − 8(4) + 3 = 16 − 32 + 3 = −13 Turning point is at (4, −13). 2 y = x3 − 5x2 + 8x + 1 dy ​ ___  ​ = 3x2 − 10x + 8 dx dy When ___ ​    ​ = 0, 3x2 − 10x + 8 = 0 dx (3x − 4)(x − 2) = 0

42. Integration 3x4 3 1 x − ​ ___ ​ + c = x − ​ _4 ​ x4 + c 4 9x3 6x2 2 ∫(9x2 + 6x + 1) dx = ___ ​   ​ + ​ ___ ​ + x + c 3 2 3 = 3x + 3x2 + x + c _3 ​ 2 ​  2 3 y = 6x + 5​x​​ 5 ​ _ ​  _3 6x3 5​x​2​ ∫(6x2 + 5​x​ ​2 ​ ​) dx = ___   ​ + c ​   ​ + ​ ___ 3 (​  _​ 52 ​  )​

Stationary points at x = _​ 3 ​  and x = 2. 4

40. Stationary points 2 (a) y = 5x2 − 3x − x3 dy ___ ​    ​ = 10x − 3 − 3x2 dx dy When ___ ​    ​ = 0, 10x − 3 − 3x2 = 0 dx 3x2 − 10x + 3 = 0 2

_1

_1

3

_5

= 2x3 + 2​x​ ​2 ​ ​ + c



43. Finding the constant __

2 3​√x ​  1 f 9(x) = 3 − ​ __2  ​ − ____ ​  2 ​   x x _3 = 3 − 2x−2 − 3​x−​ ​ 2 ​ ​

(3x − 1)(x − 3) = 0 _1

2 (a) πr2x = 100 100 x = ____ ​  2 ​  πr A = πr2 + 2πrx 100 ​  2 ​  ​ = πr2 + 2πr​  ____ πr 200 ____ 2 = πr + ​  r ​    (b) A = πr2 + 200r−1 dA 200 ___ ​   ​ = 2πr − ____ ​  2 ​  dr r dA 200 ___ When ​   ​ = 0, 2πr − ____ ​  2 ​  = 0 dr r 200  ​  2πr = ​ ____ r2 3 2πr = 200 100 r3 = ​ ____     ​ π____ 3 100 ​  π    ​ ​ = 3.17 (3 s.f.) r = ​  ____ 2 A d 400  ​ = 2π + ____ ​  3 ​  (c) ​ ____ dr2 r d2A 400 When r = 3.17, ____ ​  2 ​ = 2π + _____ ​    ​ = 18.84… > 0 dr 3.173 So A is a minimum. 200 200 ​  r ​   = π(3.17)2 + ____ ​    ​ = 94.7 m2 (3 s.f.) (d) A = πr2 + ____ 3.17

( )

__ dy 8 5 1 When x = 4, ___ ​    ​ = √ ​  2  ​­ − 5 = 2 + _​ 2 ​  − 5 = −​ _2 ​  ​ 4 ​  + __ dx 4 5 Gradient of tangent = − ​ _2 ​  _2 Gradient of normal = ​ 5 ​ 

_1

x2 1 (a) P = 80x − ___ ​    ​  50 dP x ​ ___ ​ = 80 − ​ ___  ​  dx 25 dP x (b) When ___ ​   ​ = 0, 80 − ​ ___  ​  = 0 dx 25 x 80 = ​ ___  ​  25 x = 2000 d2P 1 ____ ​  2 ​ = − ​ ___  ​ < 0 dx 25 Hence P is a maximum at x = 2000.

−​ _ ​  1

13 __

_1

3​x​ 2​ 2x−1 _____ f(x) = 3x − ____ ​   ​ − ​  _1    ​ + c −1 ​( −​   ​  )​

13 __

A: x = ​ 3 ​ , y = 5(​ 3 ​ ) − 3(​ 3 ​ ) − (​ 3 ​ ) = −​ 27 ​ , coordinates (​ 3 ​ , −​ 27 ​ ) B: x = 3, y = 5(3)2 − 3(3) − 33 = 9, coordinates (3, 9) d2y (b) ​ ____2  ​ = 10 − 6x dx d2y At B, x = 3, so ____ ​  2  ​ = 10 − 6 × 3 = −8 dx d2y ____ ​  2  ​ < 0 so B is a maximum. dx



2

6 ​  __   ​ + c = 3x + ​ x  ​ + ___ ​√x ​  2 __

f(4) = 17

6 2 ​  __   ​ + c = 17 So 3(4) + __ ​   ​  + ___ 4 √ ​ 4 ​ 

12 + _​ 2 ​ + 3 + c = 17 1

c = ​ _2 ​  6 2 3 ​  __   ​ + _​ 2 ​  f(x) = 3x + __ ​ x  ​ + ___ √ ​ x ​  3

188

WORKED SOLUTIONS 47. Exponential functions

dy (x2 + 5)2  ​     2 (a) ​ ___  ​ = ________ ​  dx x2

1 (a) C

x4 + 10x2 + 25     ​   = ​ _____________ x2 = x2 + 10 + 25x−2 x3 25x−1 (b) y = __ ​   ​ + 10x + _____ ​   + c  ​  3 −1 25 1 ​  x ​ + c = ​ _3 ​ x3 + 10x − ___

2 (a) –2 e

(c) D 1 __1 (c) __ ​​   ​ ​ e​​  ​  3 ​ x​​ 3

5x



(b) 10 e

1 (a) y = 3−1 = _​ 3 ​  1

(b) p3 = 8 ⇒ p = 2

(c) log4 8 = _​ 2 ​  3

2 (a) loga (52) = loga 25 (b) loga (2 × 9) = loga 18

​ _3 ​  + 10 − 25 + c = −13 1

_1

_5

c = 2 − ​ 3 ​  = ​ 3 ​  25 5 1 ​  x ​ + _​ 3 ​  y = ​ _3 ​ x3 + 10x − ___

44. Definite integration

6 6 ∫1 ​( 3x2 − 7 + __ ​  2  ​  )​  dx = [​​  x3 − 7x − __ ​ x  ​  ]​1 ​​  x 3

3

( )

43 ​   ​   ​= loga 8 (c) loga (43) − loga 8 = loga ​  __ 8 3 ​ 2 ​log​  8​​  6 – ​log​  8​​ ​9​  ​​​  = ​log​  8​​  (​6​​  2​)  – ​log​  8​​  9 62 = ​log​  8​​ ​  ​ __ ​   ​ = ​log​  8​​  4 9 2 = ​ _3 ​​

( )

__ 2

since ​​8​​  ​  3 ​ ​​ = ​​​(​√  8 ​ )​​​  ​​ = 4 3

_2



= (27 − 21 − 2) − (1 − 7 − 6)

49. Equations with logs



= 4 − (−12)

1 log2 (x + 1) − log2 x = log2 5



= 16 __ 2 2 _1 −3 ​  2 ​ − 4​√x ​   ​ ​​  2 ∫1 (​  6x−3 − 2​x−​ ​ 2 ​ ​  )​  dx = ​​  ___ x 1 __ −3 √2 ​)  − (−3 − 4) = (​ __  ​   − 4​ 4

[

]

__

= __ ​  4 ​  − 4​√2 ​  25



(

)

x+1 log2 ​  _____ ​   ​    ​ = log2 5 x x+1 ​ _____  = 5 x ​   x + 1 = 5x x = ​ _4 ​  1

2 log6 (x − 1) + log6 x = 1

45. Area under a curve

log6 [(x − 1)x] = 1

y = x3 − 6x2 + 8x

x(x − 1) = 6

A1 above the x-axis:

∫1 (x3 − 6x2 + 8x) dx = ​​[ 2

1  _​ 4 ​ x4

− 2x3 +

]

2 4x2  ​​1​​ 

_1



= (4 − 16 + 16) − (​ 4 ​ − 2 + 4)



= 1.75

x2 − x − 6 = 0 (x + 2)(x − 3) = 0 x = 3 because log6 (−2) is not defined. 3 log3 (x − 1) = −1 x − 1 = _​ 3 ​  1

A2 below the x-axis:

∫24 (x3 − 6x2 + 8x) dx = [​​  _​ 14 ​ x4 − 2x3 + 4x2 ]4​​2​​ 

= (64 − 128 + 64) − (4 − 16 + 16)

So A2 = 4

= −4

x = _​ 3 ​  4

4 2 log4 x − log4 (x − 3) = 2 log4 (x2) − log4 (x − 3) = 2

(

)

46. More areas

x2 log4 ​  _____ ​     ​  ​ = 2 x−3 x2 ​ _____    ​   = 16 x−3 x2 = 16x − 48

(a) 5x − x2 = x

x2 − 16x + 48 = 0

Total area = 1.75 + 4 = 5.75

5 − x = 1, x ≠ 0

(x − 12)(x − 4) = 0

x=4

x = 12 or x = 4

Coordinates of A are (4, 4). (b) Area of R = area under curve − area of triangle Area under curve:

∫04(5x − x2) dx = [​​  _​ 52 ​ x2 − _​ 13 ​ x3 ]4​​0​​ 



= (40 − __ ​ 3 ​ ) − (0) 64

_2

= 18 ​ 3 ​ 

Area of triangle = ​ _2 ​  × 4 × 4 = 8 1

_2

_2

Area of R = 18 ​ 3 ​  − 8 = 10 ​ 3 ​ 

189

(d) A

48. Logarithms

At x = 1, y = −13: 25 _1 3 ​    ​ + c = −13 ​ 3 ​ (1) + 10(1) − ___ (1)

1

(b) B –2x

50. Exponential equations 1 (a) 2b = 15 b = log2 15 = 3.91 (3 s.f.) (b) 6x = 0.4 x = log6 0.4 = −0.511 (3 s.f.) 2 (a) 32x + 3x = 6 x 2 (3 ) + 3x − 6 = 0 (3x − 2)(3x + 3) = 0 3x = 2 ⇒ x = log3 2 = 0.63 (2 d.p.)

WORKED SOLUTIONS (b) ln (3x2 − 7x + 2) − ln (x2 − 4) = 1 3x2 − 7x + 2  ​ ln​​( ​ __________      ​​ = 1 x2 − 4 ) 2 3x − 7x + 2 ​​ __________  ​​   =e    x2 − 4 3x − 1 ​​ ____ ​​   =e x+2 3x − 1 = ex + 2e 3x − ex = 1 + 2e x (3 − e) = 1 + 2e 1 + 2e x = ​​ _____ ​​  3−e

(b) The other factor gives 3x = −3, but 3x is positive for all values of x so this solution does not exist. 3 3x – 1 = 22x + 1 (x – 1)log 3 = (2x + 1) log 2 x log 3 – 2x log 2 = log 2 + log 3 x (log 3 – 2 log 2) = log 2 + log 3 log  2  +  log   3 x = _____________ ​​    ​​    log  3 − 2 log  2



= –6.23 (3 s.f.) 4 If curves intersect then x-coordinate satisfies 2 ​​6​​  ​x​​  ​​  = ​2​​  x−1​ ​x​​  2​ log  6 = (x − 1 )  log 2 (log   6) ​x​​  2​ − (log  2) x + log  2  =  0​ Discriminant = (–log 2)2 – 4(log 6)(log 2) = –0.8463… So no real solutions, so curves do not intersect.

52. Exponential modelling 1 (a) 100 (b) ​ N = 100 ​e​​  0.8t​ dN ___ ​   ​ = 0.8 × 100 ​e​​ 0.8t​ dt = 0.8N​

51. Natural logarithms (b)

(c)

y

y

y

y 5 f(3x)

y 5 f(x 2 2) O

2 3

x

x

O 1 y 5 2f(x)

Asymptote x = 2

2 (a) 250 grams (b) 125 = 250 e−90k e−90k = 0.5 −90k = ln 0.5 k = 0.00770 (3 s.f.)

Asymptote x = 0

2 6 = 3 e2x − 1 2 = e2x − 1 ln 2 = 2x − 1 1 1 x = ​​ _2 ​​  + ​​ _2 ​​  ln  2 x + 1 3 (a) ​ln ​(_____ ​  x ​     ​​ = ln 5 ) x + 1 ​ ​ ____     5 x ​​ = x + 1 = 5x 1 x = ​​ _4 ​​  (b) (e2x)2 + 3 e2x − 10 = 0 (e2x + 5)(e2x − 2) = 0 e2x = −5 or e2x = 2 2x = ln 2 1 x = ​​ _2 ​  ln  2​

1 3

Asymptote x = 0

x

53. Modelling with logs (a) x = aNb log x = log (aNb) = log (Nb) + log a = b log N + log a So k = b and c = log a (b) log x 4 3.8 3.6 3.4 3.2 3 2.8 2.6 2.4 2.2 2

2.6

2.8

3

3.2

3.4

3.6

3.8

4 log N

Graph of log x against log N is approximately linear so the model is accurate. (c) Equation of graph is approximately log x = 1.85 log N – 2.9, so k = 1.85 and c = –2.9 So a = 10–2.9 = 0.0013 (2 s.f.) and b = 1.85

56. Proof by contradiction 1 Assumption: p is a non-zero rational number, q is an irrational number and pq is a rational number. c a ​​    ​,​ where a, b, c, d are integers, a​ ≠ 0​. Write p = ​​ __ ​ ​ and pq = __ b d pq c __ a cb Then q = ___ ​​  p ​ = __ ​    ​ ÷ ​   ​  = ___ ​    ​​which is rational. d b ad Hence pq must be irrational.

‖

(c) ln (6x + 7) = ln x2 6x + 7 = x2 2 x − 6x − 7 = 0 (x − 7)(x + 1) = 0 x = 7 or x = −1 4 ln (3x e2x − 5) = ln 7 x ln 3 + ln (e2x − 5) = ln 7 x ln 3 + 2x − 5 = ln 7 x(2 + ln 3) = 5 + ln 7 5 + ln 7 x = _______ ​​   ​​  2 + ln 3 3x2 − 7x + 2  ​​     5 (a) f(x) = __________ ​​  x2 − 4 (3x − 1)(x − 2)    = ___________ ​​     ​​ (x + 2)(x − 2) 3x − 1 _____  ​​  = ​​  x+2

O

‖

1 (a)

2 Assumption: 2, sin θ and tan θ are consecutive terms of a geometric sequence. tan θ sin θ  ​​   ​​   and r = _____ ​​    So r = ​​ ____ 2 sin θ sin θ _____ tan θ ​​ ____  ​​   ​​   = ​​    2 sin θ 1 1 ​ _2 ​  sin  θ = _____ ​​ ​​       cos θ

190

WORKED SOLUTIONS

‖

‖

1 1 1 ​​ > 1 for all values of θ. But _​  2 ​  sin θ < _​  2 ​ and _____ ​​       cos θ

Hence 2, sin θ and tan θ are not consecutive terms of a geometric sequence.

57. Algebraic fractions (3x + 1)(x − 3) 3​x​​  2​ − 8x − 3 _____________  ​   = ​    1 ​ ___________       ​ (x + 3)(x − 3) ​x​​  2​ − 9 3x + 1 _____ = ​​   ​​​  x + 3 x + 5 x + 5 1 1 2 ​ ___________       ​  − ​ ______    ​         ​  − ​ ______    ​  = ​ _____________ 2​x​​  2​ + 7x − 4 2x − 1 (2x − 1)(x + 4) 2x − 1 x + 5 x + 4 = ​ _____________       ​  − ​ _____________       ​ (2x − 1)(x + 4) (2x − 1)(x + 4) x + 5 − (x + 4) _____________ 1 _____________    ​  = ​      ​​ = ​    (2x − 1)(x + 4) (2x − 1)(x + 4)

58. Partial fractions C 8​x​​  2​ A B ​​  1 ​ ​ ______________       ​​ = _______    ​   + ​ ______    ​   + ​ _______    ​​  (3x − 2)​(x  +  2)​​  2​ 3x − 2 x + 2 ​(x  +  2)​​  2​ 8x2 = A(x + 2)2 + B(3x − 2)(x + 2) + C(3x − 2) Let x = −2: 8(−2)2 = C(3(−2) − 2) 32 = −8C C = −4 Let x = ​​ _3 ​​ : 8​​(_​  3 ​ )​​ = A​(​ _​  3 ​  + 2)​​ 2

2 2

2

2

__ ​​ __ 9 ​​   = ​​  9 ​​ A 32



64

A = ​​ _2 ​​  1

Equate x2 terms: 8 = A + 3B 1 8 = ​​ _2 ​​  + 3B

B = ​ _2 ​  2 8x 5 1 4 ​​ ______________ ​​        ​​ = ________    ​​     ​​     ​​  + _______ ​​  − _______ ​​  (x + 2)2(3x − 2) 2(3x − 2) 2(x + 2) (x + 2)2 2 ​6​x​​  2​ − 1 = A(2x − 3)(x + 1) + B(x + 1) + C(2x − 3)​ Let x = −1: 6(−1)2 − 1 = C(2(−1) − 3) 5 = −5C C = −1 3 3 2 3 Let x = ​​ _2 ​​ : 6​​(_​  2 ​ )​​ − 1 = B​​(_​  2 ​  + 1)​​ 5

_ ​​ __ 2 ​​   = ​​  2 ​​ B B=5 Equate x2 terms: 6 = 2A A=3 25

5

(c) 3x + 2 = (3x + 2)2 3x + 2 = 9x2 + 12x + 4 0 = 9x2 + 9x + 2 0 = (3x + 1)(3x + 2) 1 2 x = − ​​ _3 ​​  or x = − ​​ _3 ​​ 

61. Graphs and range 1 (a) −6 < f(x) < 4 (b) ff(−1) = f(2) = 3 3x − 7 2 2 (a) g(x) = ​ __________ − ​ _____   ​      ​  ​x​​  2​ − 5x + 6 x − 3 2(x − 2) 3x − 7 = ​ ____________       ​  − ​ ____________      ​ (x − 2)(x − 3) (x − 2)(x − 3) 3x − 7 − 2(x − 2)        ​ = ​ _______________ (x − 2)(x − 3) x − 3 = ​ ____________       ​ (x − 2)(x − 3) 1 = ​​ _____    ​​  x−2 1 (b) When x = 3, g(x) = _____ ​​     ​​  =  1 3 − 2 As x → ∞, g(x) → 0 but is always > 0. Range is 0 < g(x) < 1

62. Inverse functions (a)

x + 5 y = ​​ _____   ​  x ​

xy = x + 5 xy − x = 5 x(y − 1) = 5 5 x = _____ ​​     ​​  y − 1 5    ​​  h−1(x) = ​​ _____ x − 1 (b) Range of h−1 = domain of h, so h−1(x) > 1 (c) When x = 1, h(x) = 6 As x → ∞, h(x) → 1 but is always > 1. Range of h is 1 < h(x) < 6 Domain of h−1 is 1 < x < 6

63. Inverse graphs y

59. Algebraic division 2​ x​​  4​ + 4​x​​  2​  −  x + 2 ≡ (​x​​  2​ − 1)(a​x​​  2​  +  bx + c) + dx + e ≡ a​x​​  4​  +  b​x​​  3​  +  c​x​​  2​  −  a​x​​  2​  −  bx − c + dx + e ≡ a​x​​  4​  +  b​x​​  3​ + (c − a)​x​​  2​ + (d − b)x + (e − c) x4 terms → a = 2 x3 terms → b = 0 x2 terms → c − a = 4 so c = 6 x terms → d − b = −1 so d = −1 Constant terms → e − c = 2 so e = 8

60. Domain and range

191

(a) f(−2) = 3(−2) + 2 = −6 + 2 = −4 −4 −4    ​   = ​ ___  ​  =  2​ gf(−2) = g(−4) = ______ ​​  −4 + 2 −2 x ​     ​  ​  +  2​ (b) fg(x) = ​3​(_____ x + 2 ) 2(x + 2) 3x = ​​ _____     ​   + ​ _______  ​​  x + 2 x + 2 5x + 4 = ​​ ______ ​​  x + 2

A9 (4, 6)

5 x

O

B9 (25, 28)

64. Modulus (a)

y A9 (21, 9)

O

(b)

B (4, 0)

x

B9 (24, 0) O

65. Modulus transformations (a) (i) (5, −6) (ii) (6, 3)

y

B (4, 0) x

WORKED SOLUTIONS 68. Arithmetic series

y

(b) (i) P9 (24, 11)

1 ​ _2 ​(20)[2(−3) + (20 − 1)d ] = 320 10(−6 + 19d) = 320 −6 + 19d = 32 19d = 38 d=2 1

O x

Q9 (210, 0) y (ii) O

2 S = 1 + 2 + ... + (n − 1) + n 1 S = n + (n − 1) + … + 2 + 1 2 1 + 2 : 2S = (n + 1) + (n + 1) + … + (n + 1)

x Q9 (10, 23)

2S = n(n + 1) 1  S = _​ 2 ​n(n + 1)

P9 (4, 28)

69. Geometric sequences

66. Modulus equations

1 un = 420 × (  ​​ _​ 6 ​  )​​ ​= 13.1 (3 s.f.) k 2k + 5 2 (a) ​ _____    ​       = ______ ​   ​ k−6 k 2 k = (2k + 5)(k − 6) 5 19

y

1 (a)

4

k2 = 2k2 − 7k − 30 k2 − 7k − 30 = 0

x

22 O

(b) (k + 3)(k − 10) = 0 (b) The graph of y = x does not intersect the graph of y = f(x) so the equation f(x) = x has no solutions. (c)  2x + 4  = −x Positive argument: 2x + 4 = −x 4 x = − ​​ _3 ​​  Negative argument: −(2x + 4) = −x x = −4 The graph of y = −x would intersect the graph of y = f(x) twice so both solutions exist. 2  x − 1  = 4 − 2x Positive argument: x − 1 = 4 − 2x 5 x = ​​ _3 ​​

Check: 2x + 1 = ​2​(_​  3 ​)​  +  1​ = ​​ __ 3  ​​ 5

13

_ 5 −  x − 1  = ​5 − ​| ​ _3 ​  − 1|​  = ​ __ 3  ​​ so x = ​​  3 ​​is a solution. Negative argument: −(x − 1) = 4 − 2x x=3 Check: 2x + 1 = 7 5 −  x − 1  = 3 so x = 3 is not a solution. 5



13

67. Arithmetic sequences 1 (a)

a + 8d = 3

1

a + 10d = −4

2

(b) 2 − 1 : 2d = −7 d = −​ _2 ​ 7

_7

⇒ a + 8(−​  2 ​) = 3

a − 28 = 3 a = 31 2

p2 + 1 + 6p = 24 p2 + 6p − 23 = 0 (p + 3)2 − 32 = 0 ___ p + 3 = ±​√32 ​  __ p = −3 + 4​√2 ​ because p > 0.

3 (a) 300 + 4 × 20 = 380 (b) 300 + (n − 1) × 20 = 20n + 280

5

k is positive, so k = 10 10 5 k ​   ​ = _​ 2 ​ (c) r = _____ ​     ​ = ___ k−6 4

70. Geometric series

(  (  ) )

3 20 7​ 1 − ​​ _​  2 ​  ​​ ​  ​ ___________ 1 S20 = ​     ​  = 46 540 (nearest whole number) 3 1 − _​ 2 ​

2

Sn = 1 + 2 + 4 + 8 + … + 2n − 1 n

2Sn = 2 + 4 + 8 + 16 + … + 2 2 − 1 :

1 2

Sn = 2n − 1

71. Infinite series _ 1 a = 15, r = ​​ __ 15 ​​ = ​​  5 ​​ 15 S∞ = _____ ​   _4  ​ = 75 1 − ​ 5 ​ 12

4

2 a = 2, r = k

2 ​     ​  S∞ = _____ = 3k + 4 1−k 2 = (3k + 4)(1 − k) 2 = 4 − k − 3k2 3k2 + k − 2 = 0 (3k − 2)(k + 1) = 0 S∞ exists, so −1 < k < 1, so k cannot equal −1. k = _​ 3 ​ 2

72. Recurrence relations 1 (a) a2 = 2(5) − 4 = 6 a3 = 2(6) − 4 = 8 Increasing (b) a4 = 2(8) − 4 = 12 a5 = 2(12) − 4 = 20 5

 

​∑     ​ ar​ r=1

= 5 + 6 + 8 + 12 + 20 = 51

192

WORKED SOLUTIONS 2 (a) x2 = 2p − 1

(b) x = 0.01

= 0.979587 (6 s.f.)

(a) l = rθ 27 = 12θ θ = __ ​ 12 ​= 2.25 rad 27

(b) A = _​ 2 ​r2θ 1

73. Modelling with series

= _​ 2 ​× 122 × 2.25 1

(a) 25 + (n − 1) × 6 = 6n + 19

= 162 cm2

(b) _​ 2 ​n[2a + (n − 1)d ] = _​ 2 ​n[2(25) + 6(n − 1)] = 25n + 3n(n − 1) = 3n2 + 22n 1

77. Areas of triangles (a) A = _​ 2 ​r2θ 1

(c) 6n + 19 = 79 6n = 60 n = 10 S10 = 3(10)2 + 22(10) = 300 + 220 = 520

= _​ 2 ​× 2.082 × 1.05 1



= 2.27 m2 (2 d.p.) (b)

∠BCA = ∠CAD = 1.05 rad

AC = AD = 2.08 (radii of same circle) sin ∠ABC _______ sin 1.05        ​   = ​   ​ ​ __________ 2.08 1.95

74. Series and logs

2.08 sin 1.05    ​  = 0.9252… sin ∠ABC = ​ ___________ 1.95 ∠ABC = 1.1816… rad

1 (a) 200 × (1.4)k − 1 > 600 (1.4)k − 1 > 3 log (1.4)k − 1 > log 3 (k − 1) log 1.4 > log 3 log 3 k − 1 > ______ ​     ​  log 1.4 log 3 +1 k > ​ ______  ​  log 1.4 (b) k > 4.265… k=5

40

76. Radians, arcs and sectors

(c) 2p2 − p − 1 = 9 2p2 − p − 10 = 0 (2p − 5)(p + 2) = 0 5 p = _​ 2 ​  or  p = −2

1

____

​​  3 ​​ (0.01)3 ​​√  0 . 94 ​  ≈​1 − 2(0.01) − 4(0.01)2 − __

3

(b) x3 = px2 − 1 = p(2p − 1) − 1 = 2p2 − p − 1

∠BAC = π − 1.05 − 1.1816… = 0.91 rad (2 d.p.)

(c) Area of triangle = _​ 2 ​ab sin θ 1



= _​ 2 ​× 1.95 × 2.08 × sin 0.91



= 1.6009…

1

Total area = 1.6009… + 2.27 = 3.87 m2 (2 d.p.)

78. Sec, cosec and cot

(  (  ) ) n

9 ​  10  ​  ​​ ​  ​ 4​ 1 − ​​ __ a(1 − rn) ___________ ​     ​   =    ​   ​  2 Sn = ________ 9 1−r 1 − __ ​ 10  ​

1

y

(  (  ) ) n

9 ​ 10  ​  ​​ ​  ​ 4​ 1 − ​​ __ ___________  ​   > 30    ​  9 1 − __ ​ 10  ​

( 

)

​​ __ ​ 10  ​  )​​ ​  ​ > 3 4​1 − (  9

n

O

(  ) n 9 1 ​​( __ ​ 10  ​  )​​ ​ < ​ _4 ​ n 9 1 ​ 10  ​  )​​ ​ < log ​ _4 ​ log ​​( __



n

9 3 1 − ​​ __ ​ 10  ​  ​​ ​ > ​ _4 ​

2

_ n log ​ __ 10  ​ < log ​  4 ​ 9

1

log ​ _4 ​    ​    (with a change of direction because n > ​ _____ 9 9 log ​ __ 10  ​ log __ ​ 10  ​is negative) 1

n > 13.157

π 2

π

3π 2π 2

x

y

2 2360 2180 O

180

360 θ

n = 14

75. Binomial expansion 2

_ _ (​ ​  3 ​)​​(− ​  3 ​)​ _1 _1 ______ ​  3 ​ ​​ (a) ​​​(1 + (−6x)) ​ ​​ =  1  + ​​(​  3 ​)​​(−6x) + ​​(​   ​ ​​​(−6x)​​  2​  + 1 × 2 ) 1

2

​​(​ 

    ​​​(−6x)​​  3​  + …  ​ 1 × 2 × 3 )

( )(

)(

)

5 2 − ​ _3 ​ ​​ − ​ _3 ​ ​ __________



1 ​ _​  3 ​ ​​

= 1 − 2x − 4x2 − __ ​​  3  ​​  x3 + … 40

Expansion is valid for ​|−6x |  < 1​or ​|x |  < _​ 6 ​​

193

1

79. Trigonometric equations 3 1 cot2 x − 3 cot x + 2 = 0 (cot  x − 2)(cot x − 1) = 0  cot x = 2 cot x = 1 1 tan  x = _​​  2 ​​  tan  x = 1 x = 26.6°, 206.6° x = 45.0°, 225.0°

WORKED SOLUTIONS __

2 0 < x < π so 0 < 2x < 2π 2 sec 2x = __ ​  __  ​  √ ​  3 ​ __ √ ​  3 ​  cos 2x = ___ ​​   ​​  2 11π π ​   ​     2x = ​ __ ​ ,  ____ 6 6 π 11π   ​   ​  x = ​ ___  ​ , ____ 12 12

(b) ​​√ 2 ​​   cos (x + 45°) = 0.5 1    ​​  cos (x + 45°) = ___ ​​  __ 2​√ 2 ​  0 < x < 360° so 45° < x + 45° < 405° x + 45° = 69.3°, 290.7° x = 24.3°, 245.7° (1 d.p.)

83. Double angle formulae

80. Trigonometric identities 2 1 cosec2 x − sec2 x ≡ (1 + cot2 x) − (1 + tan2 x) ≡ cot2 x − tan2 x 2 2 3(sec  θ − 1) + 7 sec θ = 3 3 sec2 θ + 7 sec θ − 6 = 0 (3 sec θ − 2)(sec θ + 3) = 0 2 sec θ = ​​ _3 ​​   7 sec  θ = −3  3 1 cos  θ = −​​ _3 ​​  θ = 109.5°, 250.5° (1 d.p)

81. Arcsin, arccos and arctan π 1 (a) ​​ __ ​​    3

π (b) ​​ __ ​​    3

π (c) ​​ __ ​​   4

π 1 1 2 (a) g​​(_​  2 ​)​​ = arcsin ​​ _2 ​​ − __ ​​   ​​   4 π π __ __

= ​​   ​​  − ​​   ​​   6 4 π = − ​​ ___  ​​  12 π (b) arcsin x − ​​ __ ​​   = 0 4 π arcsin  x = ​​ __ ​​   4 1__ x = ​​ __   ​​  √ ​  2 ​  π (c) y = arcsin x − ​​ __ ​​   4 π y + ​​ __ ​​    = arcsin  x 4 π x = sin​​(y + ​ __ ​ )  ​​ 4 3π π π g−1(x) = sin​​(x + ​ __ ​  )​​, − ​​ ___ ​​   < x < __ ​​   ​​   4 4 4 y (d)

1 cos 2A ≡ cos (A + A) ≡ cos A cos A − sin A sin A ≡ cos2 A − sin2 A ≡ cos2 A − (1 − cos2 A) ≡ 2 cos2 A − 1 2 (a) cos 3x = cos (2x + x) = cos 2x cos x − sin 2x sin x = (2 cos2 x − 1) cos x − (2 sin x cos x) sin x = 2 cos3 x − cos x − 2 sin2 x cos x = 2 cos3 x − cos x − 2(1 − cos2 x) cos x = 2 cos3 x − cos x − 2 cos x + 2 cos3 x = 4 cos3 x − 3 cos x (b) 2(4 cos3 x − 3 cos x) − 1 = 0 1 4 cos3 x − 3 cos x = _​​  2 ​​ 

cos 3x = ​​ _2 ​​  0 < x < 90° so 0 < 3x < 270° 3x = 60°   x = 20° 1

84. a cos O- ± b sin O______

___

1 (a) R = √​​  ​3​​  2​  + ​2​​  2​ ​   = ​√ 13 ​​ 

α = arctan​​(_​  3 ​)​​ = 0.5880… 2

___

3 sin 2θ + 2 cos 2θ = √​​  13 ​​  sin  (2θ + 0.5880…) (b) ___ 3 sin 2θ + 2 cos 2θ = −1 √ 13 ​  sin (2θ + 0.5880…) = −1 ​ 1 ___    ​​  sin (2θ + 0.5880…) = ​− ​ ___ √ ​  13 ​  0 < θ < ​π​so 0.5880… < 2θ + 0.5880… < 2​π​+ 0.5880… 2θ + 0.5880… = −0.2810…, 3.4226…, 6.0021… θ = 1.42, 2.71 (2 d.p.) ________ 2 (a) R = √ ​​  ​​√ 3 ​​​   ​  + ​1​​  2​ ​   = ​√ 4 ​   =  2​ 1__ __ α = arctan ​​ ​    ​  ​​ = 30° (√ ​  3 ​  ) (b) y = 2 cos (x − 30°) __ 2

2 3π 4

2π 4

O

π 4

x

__

y 2

82. Addition formulae 1 sin 15° = sin (45° − 30°) = sin 45° cos 30° − cos 45° sin 30° __ √ ​  3 ​  __ 1 1__ __ 1__ __ __ = ​​    ​  × ​   ​  − ​    ​  × ​   ​​  √ √ ​ __  2 ​  2 ​  2 ​  2 ​√ 3 ​  − 1 __ ​​    = ______ ​​  2​√ 2 ​  __ __ √ ​  6 ​  − ​√ 2 ​    = ______ ​​   ​​  4 1 4 __   __ ​​      ​   cosec  15° = ______ ​​  = ​ ______ sin 15° √​  6 ​  − ​√ 2 ​  2 (a) cos (x + 45°) = cos x cos 45° − sin x sin 45° 1__ 1 = cos x × __ ​​  __  ​​ − sin  x × ​​ __   ​​  √ √ ​  2 ​  ​  2 ​  cos x − sin  x __ ​​  = __________ ​​    √ ​  2 ​ 

(30°, 2)

3

O

120°

22

300°

360°

x

(210°, 22)



85. Trig modelling ______

___

__

(a) R = √​​  ​4​​  2​  + ​2​​  2​ ​   = ​√ 20 ​  = 2​√ 5 ​​ 

α = arctan ​​(_​  4 ​)​​ = 0.4636… 2

__

(b) 2​​√ 5 ​​   cm

194

WORKED SOLUTIONS __

(c) 2​​√ 5 ​​   cos  (1.2t − 0.4636) = 0 cos (1.2t − 0.4636) = 0 0 < t < 5 so −0.4636 < 1.2t − 0.4636 < 5.5364 π 3π 1.2t − 0.4636 = __ ​​   ​​ ,  ___ ​​   ​​    2 2 t = 1.70, 4.31 (2 d.p.)

86. Parametric equations π π y = sin t  cos ​​ __ ​​  + cos  t  sin ​​ __ ​​   3__ 3 √   3  ​   ​ 1 ​​   ​​  cos  t = ​​ _2 ​​  sin  t + __ 2__ √  3 ​  ________ ​ 1 ​​   ​​  √​  1  − ​sin​​  2​t ​​  = _​​  2 ​​  sin  t + __ 2 __ ​ √ ​  3 ​  _____2 _1 __ = ​​  2 ​  x + ​   ​ ​ √ 1 − ​x​​  ​ ​​  2

89. Differentiating standard functions 1 (a) 2 cos 2x + 4 sin 4x (b) 2x − 4 e4x − 3 6x (c) ​​ _______    ​​  3​x​​  2​ + 1 dx 2 ​​ ___ ​  = 2 cos 2y dy _________ = 2​√ 1  − ​sin​​  2​  2y ​  _____

= 2​√ 1 − ​x​​  2​ ​​​  dy 1 1 _____ ​​ ___  ​ = ​ _____    ​     ​​   = ​ ________ dx dx √ 1 − ​x​​  2​ ​  2​ ___ ​ ​   ​  ​ ( dy ) d d 3 ___ ​​    ​( ​​sin​​  2​x)​  = ​ ___  ​( ​​ sin x)​​​  2​ dx dx = 2 sin x × cos x = sin 2​x

87. Differentiating sin x and cos x f(x + h) − f(x) _____________ 1 (a) f ′(x) = ​lim​​ ​ ​      ​   h→0 h + h) − sin  sin  ( x (x) ________________ = ​lim​​ ​ ​      ​   h→0 h (sin x cos h + cos x sin h) − sin x = ​lim​​ ​ ​  __________________________         ​ h→0 h cos h − 1 sin h = ​lim​​ ​​ (​(________ ​   ​     ​ ____ ​   ​     ​cos x)​ )​sin x + ( h→0 h h ) cos h − 1 sin h (b) Since ________ ​   ​    → 0 and ​ ____  ​   → 1 the limit tends to h h 0 × sin x + 1 × cos x = cos x So f ′(x) = cos x as required.

1 − (​ 1 − ​ _2  ​ ​(3θ)​​  2​)​ 1 − cos 3θ ______________  ​ ≈    ​    2 ​​ _________ ​​  tan θ sin θ ​θ​​  2​ 9 ​θ​​  2​ = ____ ​  2 ​  2 ​θ​​  ​ 1

= _​  2 ​ 9

π

π

f( ​  _3  ​ + h) − f( ​ __3   ​) π __ _____________   ​ = ​lim​​ ​ ​      ​   3 f ′​(​   ​ ) 3 h→0 h π π __ cos  ( ​  3   ​ + h) − cos ( ​ __3   ​)     ​   = ​lim​​ ​  ^ ​ _________________ h→0 h π π π __ __ (cos ​  3   ​  cos  h − sin ​ 3   ​  sin  h) − cos ( ​ __3   ​) ____________________________     ​ = ​lim​​ ​  ​      h→0 h __ √ ​  3 ​  cos h − 1 __ sin h ___ 1 ________ ____  ​      ​  ​  ​   − ​ ​   ​      ​  ​ = ​​lim​​ ​​ ​(​(​  ) 2 ( h ) 2 ​ )​ h→0 h sin h cos h − 1  ​    → 0 and ​ ____  ​   →   1 the limit tends to Since ________ ​  h __ h __ √ √ ​  3 ​  ___ ​  3 ​  1 0 × ​ __ ​  − 1 × ___ ​   ​=   −​   ​  2 2 __ 2 ​√ 3 ​  π So f ′​(__ ​   ​  )​ = − ​ ___ ​ as required. 3 2

90. The product rule __

1 (a) u = ​​√ x ​​   v = sin 5x dv du ____ 1 ___ ___ ​    ​​  ​     ​ = 5 cos 5x​ ​​    ​​ = ​​  __ dx 2​√ x ​  dx d dv du ​  ___    ​(uv) = u​  ___   ​ + v​  ___  ​ dx dx dx __ sin 5x √ ​​  __ ​​  = 5​  x  ​ cos  5x + ______ 2​√ x  ​ 10x cos 5x__+ sin 5x _________________ =    ​​   ​​  2​√ x  ​ (b) u = x v = ln x dv __ du 1 ___ ​___ ​     ​ = ​ x  ​​ ​​    ​​ = 1 dx dx dv du d ​  ___    ​(uv) = u​  ___   ​ + v​  ___  ​ dx dx dx 1 ​  x  ​)​ +  ln x =  x​(__ = 1 + ln x 2 (a) u = ex v = 3x2 − 4x − 1 du dv ​​  ___  ​​= ex ​ ​___   ​​= 6x − 4 dx dx dy dv du ​​ ___  ​ = u​  ___   ​ + v​  ___  ​​ dx dx dx = ex(6x − 4) + ex(3x2 − 4x − 1) = ex(3x2 + 2x − 5) dy (b) When ___ ​​    ​​= 0, ex(3x2 + 2x − 5) = 0 dx ex ≠ 0 so 3x2 + 2x − 5 = 0 5

91. The quotient rule

88. The chain rule

​√ x ​(  −sin x) − (cos x)​(_​  2 ​ ​ x​​  − ​  2 ​ )​ ​ _______________ −2x sin x − cos x 1 (a) ​​ _______________________         ​​  __ 2 ​​ = ​​    _3 (​​ √​  x ​)  ​​​  ​ 2​x​​ ​  2 ​​ __

_1

1 y = ​​​(​x​​  2​ − 3x + 1)​​ ​− ​ 2 ​​​ dy _3 1 ​​ ___  ​​ =  − ​​ _2 (​​​​​ ​x​​  2​ − 3x + 1)​​ ​− ​ 2 ​​​  ×  (2x − 3) dx 3 − 2x ___________ = ​ ______________       ​​ 2 ( √ 2​  ​​   ​x​​  ​ − 3x + 1)​​​  3​ ​

______

6



2 f(x) = ​​​(​x​​ ​  3 ​​ + 6)​​​  ​​ _1

5

1 f′(x) = ​6​​(​x​​ ​  3 ​​ + 6)​​​  ​  × ​ _3 ​​x​​ − ​ 3 ​​​ _1

__

_2

6​(​√  x ​ + 6)​ = ​​  _________  ​​    3 __ 2 3​(​√  x ​) ​ 3

6​(​√  8 ​ + 6)​ _____ 6​​(8)​​​  5​ ______ 6​​(​2​​  3​)​​​  5​ _____ 2​(​2​​  15​)​ 14 f′(8) = _________ ​​   ​  = ​   ​  = ​  2 ​    ​​    = ​   = ​2​​  ​ 3 __ 2 2 2 3​​(2)​​​  ​ 3​​(2)​​​  ​ ​2​​  ​ 3​(​√  8 ​ )​ 3

195

__

5

5

(3x + 5)(x − 1) = 0

x = ​− ​ _3 ​​ or x = 1



1

_1

3 (​​ 2)​ − ​(2x)(​​ _ ​  3x + 1 ​)  ​  2 )​ (​​​ 3x + 1)​​ ​− ​ 2 ​​ ________ ​(√ 3x + 2 __________________________       ​​  (b) ​​     ______ 2 ​  = ​  _3 (​​ √ ) ​  3x + 1 ​  ​​​  ​ ​​(3x + 1)​​ ​​  2 ​​ 2x ​ ​​  2​ + 1))​​(2x)​ − ​(​x​​  2​ + 1)​​(______ ​  2    ​  ​ ​(ln (x ​ ​​  2​ + 1) − 1)​ 2x​(ln (x ​x​​  ​ + 1 ) ________________ ______________________________  ​  = ​   ​​          (c) ​​     (​​ ln (x​ ​​  2​ + 1))​​​  2​ (​​ ln (x​ ​​  2​ + 1))​​​  2​ _1

WORKED SOLUTIONS d d cos x 2 ​​ ___  ​ (cot x) = ​ ___  ​​ (_____  ​ ​ ​  dx dx sin x ) (sin x)(−sin x) − (cos x)(cos x) _________________________  ​    = ​     (sin x2)

−​(sin​​  2​x  + ​cos​​  2​x) _______________  ​ = ​      ​sin​​  2​x −1   ​  = ​ _____ ​sin​​  2​x =  − ​cosec​​  2​​x



2

3(x − 1) 3(​x​​  ​ + 2) 9  ​​   ​  − ​ _____________      ​  − ​ _____________      3 (a) f(x) = _____________ ​​     (​x​​  2​ + 2)(x − 1) (​x​​  2​ + 2)(x − 1) (​x​​  2​ + 2)(x − 1) 3(​x​​  2​ + 2) − 3(x − 1) − 9 ____________________  ​​    =    ​​  (​x​​  2​ + 2)(x − 1) 3x(x − 1)  ​​      = ​​ _____________ (​x​​  2​ + 2)(x − 1) 3x = ______ ​​  2    ​​  ​x​​  ​ + 2 ​(​x​​  2​ + 2)​​(3)​ − ​(3x)​​(2x)​ ___________________  ​​    (b) f′(x) = ​​    (​​ ​x​​  2​ + 2)​​​  2​

= ​​ 

2

6 − 3​x​​  ​ ________

 ​​    (​​ ​x​​  2​ + 2)​​​  2​

dy (c) ​​  ___  ​  = 0​ means 6 cos 2t = 0 dx cos 2t = 0, 0 < t < 2​π​ 5π ___ 7π π 3π ___    ​, ​      ​​    t = ​​ __ ​ ,  ​ ___ ​, ​   4 4 4 4 __ __ √ ​  6 ​  + ​√ 2 ​  π π π ​   ​   − ​ __ ​  )​​ = ______ ​​     ​​  t = ​​ __ ​​ :  x = 2 cos ​​(__ 4 4 3 2 π y = 3 sin 2​​(__ ​   ​ )  ​​ = 3 4 __ __ √ ​  6 ​  − ​√ 2 ​  3π __ 3π π ___   x = 2 cos ​​(___ ​   ​    − ​   ​  )​​ = ______ ​​     ​​  t = ​​   ​​:   3 4 4 2 3π ​   ​    ​​ = −3 y = 3 sin 2​​(___ 4) __ __ − ​√ 6 ​  − ​√ 2 ​  5π __ 5π π ___ ___ ________   x = 2 cos ​​(​   ​    − ​   ​  )​​ = ​​     ​​  t = ​​   ​​:   3 4 4 2 5π ​   ​    ​​ = 3 y = 3 sin 2​​(___ 4) __ __ − ​√ 6 ​  + ​√ 2 ​  7π __ 7π π   x = 2 cos ​​(___ ​   ​    − ​   ​  )​​ = ________ ​​     ​​  t = ​​ ___ ​​:   3 4 4 2 7π ​   ​    ​​ = −3 y = 3 sin 2​​(___ 4)

Coordinates are ​​(​ 

92. Differentiation and graphs dy (a) ​​ ___  ​ = ​(​e​​  2x​)​​(​sec​​  2​x)​ + ​(tan x)​​(​2  e​​  2x​)​ dx = ​e​​  2x​​(​sec​​  2​x + 2 tan x)​ = ​e​​  2x​​(​tan​​  2​x + 2 tan x + 1)​ = ​e​​  2x(​​​ 1 + tan x)​​​  2​​​ (b) ​​e​​  2x(​​​ 1 + tan x)​​​  2​ =  0​ 1 + tan x = 0 tan  x = −1 π x = −​ ​ __ ​​   4 π __ π π When x = − ​​ __ ​​ ,  y = ​​e​​ 2​(− ​ 4  ​)  ​​  tan ​(− ​ __ ​  )​​ 4 4 π __ −  = ​− ​e ​​ ​  2  ​​​  π __ π Coordinates of P are ​​(− ​ __ ​ ,   − ​e ​​ − ​ 2  ​ ​)​​ 4 (c) When x = 1: y = e2 tan 1 = 11.5077… dy ​​ ___  ​  = ​e​​  2(​​​ 1 + tan 1)​​​  2​​ = 48.3268… dx Equation of tangent: y − 11.5077… = 48.3268…(x − 1) y = 48.3x − 36.8

93. Parametric differentiation dx π dy (a) ​​  ___ ​  = −2 sin ​(t − ​ __ ​  )​​, ​​  ___ ​  = 6 cos 2t​ 3 dt dt dy _____________ 6 cos 2t ___   ​​    ​ = ​    π  ​​ dx −2  sin ​ t − ​ __  ​   ​ ( 3) π (b) When t = 0, x = ​2 cos ​(− ​ __ ​  )​​ = 1, y = 3 sin 0 = 0 3 __ dy ____________ 6 cos 0 6 ___ __________ ​​    ​ = ​     __  ​   = 2​√ 3 ​​  = ​  π  ​   dx −2  sin ​ − ​ __ √ ​  3 ​  ( 3 ​  )​ −2​(− ​ __ ​ )​ 2 −1 ___ Gradient of normal = ​​  __  ​​  2​√ 3 ​  Equation of normal: −1__ y − 0 = ​​ ___   ​​( x − 1) 2​√ 3 ​  __ x + 2​​√ 3 ​​y   − 1 = 0

__

__

__

__

 3)​​, ( ​​ ​   −3)​​,  ​,   ​,  2 2__ __ __ − ​√ 6 ​  − ​√ 2 ​  − ​√ 6 ​  + ​√ 2 ​  ​   3)​​, ( ​​ ________ ​   −3)​​  ​,   ​,  ​​(________ 2 2 √ ​  6 ​  + ​√ 2 ​  ______

√ ​  6 ​  − ​√ 2 ​  ______

__

94. Implicit differentiation dy dy 1 3x2 + 2y ​___ ​    ​​+ 6xy + 3x2 ​___ ​    ​​= 0 dx dx dy ___ ​​    ​​(2y + 3x2) = −3x2 − 6xy dx dy − 3​x​​  2​ − 6xy  ​​    ​​ ___  ​ = ​ ___________ dx 2y + 3​x​​  2​ dy _________________ −3​(3)​​  2​ − 6(3)(−1) 9 At P, ​​ ___  ​ = ​     ​  =  − ​ ___  ​​     25 dx 2(−1)  +  3​(3)​​  2​ 2 (a) LHS: 2(0)2 − (−1)2 = −1 RHS: −1 e3(0) = −1 So (0, −1) lies on C. dy dy (b) 4x − 2y ​​ ___  ​​ = 3y e3x + e3x ​___ ​    ​​ dx dx dy ___ 3x 3x 4x − 3y e = ​​    ​​  (2y + e ) dx dy 4x − 3y​  e​​  3x​ ​​ ___  ​ = ​  _________  ​​    dx 2y  + ​e​​  3x​ dy 4(0)  −  3(−1) ​e​​  3(0)​ _______________ At (0, −1), ​​ ___  ​ = ​        ​ = −3​ dx 2(−1)  + ​e​​  3(0)​ Equation of tangent: y − (−1) = −3(x − 0) 3x + y + 1 = 0



95. Differentiating ax d 1 ​​ ___  ​  (​4​​  x​ sin x) = ​4​​  x​  cos  x  + ​4​​  x​ ln 4 sin x dx = ​4​​  x​(cos  x + ln 4 sin x​) dy 1 y 1 dy 2 y + x ​___ ​    ​​ + (​​ _​  2 ​ )​ ​  ln ​​ _2 ​​ ​​ ___  ​​= 0 dx dx dy y 1 1 ​​ ___  ​​ ​​(x + (​ _​  2 )​ ​ ln ​ _2 )​ ​​ = −y dx dy −y    ​​  ​ ___  ​ = ​​ __________ dx x + ​(_​  1 ​)y​ ln ​ _1 ​ 2

2

196

WORKED SOLUTIONS



97. Rates of change

dy −(−1) −1     1  ​​  = ​ _____   ​  At (0, −1), ​ ___  ​  = ​​ ___________ dx 0 + ​(_​  1 ​ )−1 ​ ln ​ _ ​  2 ln 2 2

(a)

2

x θ

Gradient of normal = 2 ln 2 = ln 4 Equation of normal: y − (−1) = ln 4(x − 0) (ln 4)x − y − 1 = 0

h

h ​tan  θ = ​ __ x  ​​ so x = h cot ​θ​

V = _​​  2 ​ h[6 + (6 + 2h cot θ)]​× 40 1

96. Points of inflexion

V = 240h + 40h2 cot θ dV ​​ ___ ​ ​= 240 + 80h cot θ dh

dy 1 (a) ​ ___  ​ = 3x2 + 2x − 1 dx ​d​​  2​y ​ ____2  ​ = 6x + 2 d​x​​  ​

dh dV dV (b) ​​ ___ ​  = ​ ___ ​  ÷ ​ ___ ​ ​ dt dt dh

dy (b) ​ ___  ​ = 0 ⇒ 3x2 + 2x − 1 = 0 dx (3x − 1)(x + 1) = 0 x = _​  3 ​ or x = −1 1

​ 27 ​ When x = _​ 3 ​, y = (​ _3 ​)3 + (​ _3 ​)2 − _​ 3 ​+ 2 = __ 1

1

1

1

49

When x = −1, y = (−1)3 + (−1)2 − (−1) + 2 = 3 ​  27 ​ ) and (−1, 3). So coordinates of stationary points are (​ _3 ​ , __ 2 ​ d ​​  y ​ 1 1 When x = _​ 3 ​, ____ ​  2  ​ = 6(​ _3 ​) + 2 = 4 > 0 so stationary point d​x​​  ​ is a minimum. d2y When x = −1, ​​ ____2  ​​= 6(−1) + 2 = −4 < 0 so stationary dx point is a maximum. 1 49

d 2y (c) ​​ ____2  ​​ = 0 ⇒ 6x + 2 = 0 dx 1 x = −​ _3 ​

​ 27 ​ When x = −​ _3 ​, y = (−​ _3 ​)3 + (−​ _3 ​)2 − (−​ _3 ​) + 2 = __ 1

1

40      0.1 = ​​ ________________  ​​ 240 + 80(2.5) cot θ 24 + 20 cot θ = 40 cot  θ = 0.8 tan  θ = 1.25 θ = 51.3° (1 d.p.)

1

1

65

1 65 ​  27 ​). So point of inflexion is at (−​ _3 ​, __

dy 2 ___ ​    ​ = (x2 − 2)ex + 2x ex dx = (x2 + 2x − 2)ex dy ___ ​    ​ = 0 ⇒ (x2 + 2x − 2)ex = 0 dx ex is never equal to zero so x2 + 2x − 2 = 0 x = 0.732… or x = −2.732… When x = 0.732…, y = −3.044… When x = −2.732…, y = 0.355… So coordinates of stationary points are (0.73, −3.04) and (−2.73, 0.36) (2 d.p.). dy 3 (a) ​​ ___  ​​= 4x3 + 9x2 − 12x dx d2y ​​ ____2  ​​ = 12x2 + 18x − 12 dx d2y ____ ​​  2  ​​= 0 ​⇒​ 12x2 + 18x − 12 = 0 dx 2x2 + 3x − 2 = 0 (2x − 1)(x + 2) = 0 x = _​​  2 ​​ or x = −2 1

(b) f′′(x) > 0 for all values of x in [1, 2] so C is convex on this interval.

98. Iteration (a) f(1) = ln (1 + 1) − 2(1) + 2 = 0.693… Positive f(2) = ln (2 + 1) − 2(2) + 2 = −0.901… Negative Change of sign so 1 < α < 2 (b) x1 = 1.45815 x2 = 1.44970 x3 = 1.44798 (c) f(1.44745) = ln (1.44745 + 1) − 2(1.44745) + 2 = 0.000146… Positive f(1.44755) = ln (1.44755 + 1) − 2(1.44755) + 2 = −0.0000124… Negative Change of sign, so 1.44745 < α < 1.44755 So α = 1.4475 (4 d.p.)

99. The Newton–Raphson method ___

(a) f(1) = 13 − 2​​√ ___ ​1​​  3​ ​​ = −1 < 0   2.343… > 0 f(2) = 23 − 2​​√ ​2​​  3​ ​​=

Change of sign so f(x) has a root in [1, 2] _3

(b) f(x) = x3 − 2​​x​​  ​  2 ​​​ f ′(x) = 3x2 − 3​​x​​  ​  2 ​​​ _1

__

(c) f ′(1) = 3(1)2 − 3​​√ 1 ​​  = 0

So the tangent would be horizontal and would never intersect the x-axis. f(​x​  0​​)  ​​  (d) x1 = ​​x​  0​​ − _____ ​    f′(​x​  0​​) ___ ​2​​  3​− 2 ​√ ​2​​  3__​ ​     ​​  = ​2 − ​ __________ 3​(​2​​  2​)​− 3 ​√ 2 ​  = 1.6979… f(​x​  1​​) x2 = ​​x​  1​​ − _____  ​​  ​    f′(​x​  1​​) __________ 1.6979..​.​​  3​− 2 ​√ 1.6979..​ .​​  3​ ​  _______________________ _________ = ​1.6979... −     ​      ​​ 3​(1.6979..​.​​  2​)​− 3 ​√ 1.6979... ​  = 1.5987 (4 d.p.)

197

WORKED SOLUTIONS 100. Integrating standard functions 1 (a) −2 cos 2x + c π __

​  2 ​    2 ​ ​ ​   ​​​2  0

∫ 

104. Integration by parts

x _

π __

dv 1 1 u = ln x ​​ ___  ​ ​ = ___ ​​    ​​  dx ​x​​  2​ du 1 1 ​​ ___  ​​ = __ ​​    ​​ v = ​− ​ __ x  ​​ dx x 1 1 1 __ 1 ___ __ ​  2  ​   ln x  dx = (ln x)​(− ​ __ x  ​)​− ​(− ​  x  ​)​​(​  x  ​)​dx ​x​​  ​ ln x 1 = − ​ ____  + ​(___ ​  2  ​ )​dx x ​ ​x​​  ​ ln x __ 1 = ​− ​ ____  ​ − ​​  x  ​​ + c x ​ dv 2 (a) u = x ​​ ___  ​ ​ = ex dx du ___ ​​    ​​ = 1 v = ex dx ​ ​ ∫x ​e​​  x​  dx = ​​​ x ex − ∫​e​​  x​ ​​​ dx ​ ​ x x = x e − e + c = ex(x − 1) + c dv 2 (b) u = x ​​ ___  ​ ​= ex dx du ___ ​​    ​​ = 2x v = ex dx

(b) ​6 ​e​​ ​  6 ​​​ + c

sin ​​ _2 ​​ θ dθ = [​​ −4  cos ​ _2 ​ θ] π = (−4  cos ​​ __ ​​ )  − (−4  cos  0) 4 __ = −2​​√ 2 ​​  + 4 2 1 ​​     ​​   dx = − _​  2 ​ ln |1 − 4x| + c 3 ______ 1 − 4x 1

​   ​  2 ​​  0​  ​

1

∫

∫

101. Reverse chain rule

∫ ∫

2x + 2 1 1 dx = ​ _2 ​  ln   ​x​​  2​ + 2x − 3   + c 1 ​​ _2 ​​ __________ ​  2   ​   ​x​​  ​ + 2x − 3 2 3  cos 3x 2 ​       ​  dx  = ​​ _3 ​​  ln   sin 3x   + c 2 ​​ _3 ​​ _______ sin 3x 1 1 2​e​​  x​ 1 ​​​​  x    ​   dx 3 ​​∫ ​   ​​​f(x) dx = ​​ _2 ​​​ ∫ ​   _______ 0 0 2​e​​  ​ − 1 = ​​ _2 ​​ ​​​​ [ln   2​e​​  x​ − 1  ]​​  0​​​  1

1



= ​ _2 ​  ln  (2e  −  1)  −  ​​ _2 ​​  ln  (1)



= 0.745 (3 d.p.)

1

1

1



π __ ​  12  ​ 

​  12  ​ ​  12  ​ ​ ​ ​ 1 1 1 ​​∫ ​   s​​ in  3x  cos 3x​​ ​dx = ​∫ ​   _​​​  2 ​  sin  6x​​ ​dx = ​​[− ​ __ ​ ​​  ​  ​​ 12  ​  cos  6x​] ​ ​ ​0 0 0 π __

__ __ = ​​(− ​ __ 12  ​  cos ​  2 ​ )  ​ − ​(− ​  12  ​  cos  0)​​ 1



π

1

∫sin 4x cos 3x dx = ∫​(_​  2 ​  sin  7x + ​ _2 ​  sin  x)​dx (b) 1

1

_ = − ​​ __ 14  ​​  cos  7x − ​​  2 ​​  cos  x + c 1

1

103. Integration by substitution

du 1 1 u = 3 + sin x so ___ ​​    ​ = cos x​ and dx = (​​ _____ ​  cos  x ​) ​​  du dx sin 2x sin 2x _____ 1 __________  ​  dx =  _____ ​     ​  2 ​  ​        ​   du ​(3 + sin x)​​  2​ ​u​​  ​ cos x 2 sin x cos x _____ 1  ​  ​      du =  z​ __________ cos  x ​  ​u​​  2​ 2(u − 3)  ​   du ​    =  _______ ​u​​  2​ 2 6 ​  u ​  − ​ ___2  ​ )​​  du =  ​(__ ​u​​  ​

∫

∫ ∫ ∫ ∫​

= 2 ln    u   + 6​u​​  −1​  +  c

6    ​  = 2 ln (3 + sin x) + ​ ________  + c 3 + sin x du 2 u2 = 2x + 1 so 2u ​​ ___  ​​= 2 dx So dx = u du When x = 4, u = 3 When x = 0, u = 1 3 2(​u​​  2​ − 1) 4 4x ​​∫ ​   _______ dx​ = ​​∫ ​   ________ ​​​  ______    ​   ​​​      u ​u du​ 0 √ 1 ​  2x + 1 ​  3 ​ = ​​∫ ​   (​​​​​​ 2​u​​  2​ − 2)​  du​ 1 ​ 3 2 = [​​​ _​  3 ​​u​​  3​ − 2u]​​  1​​​ 

= ​​(_​  3 ​ (3)3 − 2(3))​​ − (​​ _​  3 ​ (1)3 − 2(1))​​ 4 = 12 − ​​(− ​ _3 ​ )​​ 2

2

1

1

0

0

= [​​​x2 ex − 2 ex(x − 1)​]​  0​​​  1

= [​ex(x2 − 2x + 2)​]​  0​​​  =e−2 1



1

= ​​ __ 12  ​​  2 (a) sin 7x = sin (4x + 3x) = sin 4x cos 3x + cos 4x sin 3x sin  x = sin (4x − 3x) = sin 4x cos 3x − cos 4x sin 3x  1 + 2 sin 7x + sin x = 2 sin 4x cos 3x



∫

​​∫ ​   ​​​x​​  2​​  e​​  x​  dx​ = [​ ​x2 ex​​]​  0​​​  − 2​​∫ ​   ​​x​  e​​  x​  dx​

102. Identities in integration π __

∫

1 2

105. Integrating partial fractions (a) 18x2 + 10 = A(3x + 1)(3x − 1) + B(3x − 1) + C(3x + 1) 1 1 2 1 Let x = _​​  3 ​​ : 18​​(_​  3 ​ )​​ + 10 = C​​(3​(_​  3 ​ )​+ 1)​​ 12 = 2C C = 6 1 1 2 1 Let x = −​​ _3 ​​ : 18​​(−​ _3 ​ )​​ + 10 = B​​(3​(−​ _3 ​ )​− 1)​​ 12 = −2B B = −6 Equate x2 terms: 18 = 9A A = 2 6 6 (b) (​ 2 − ​ ______    ​      ​ ​  dx = 2x − 2 ln    3x + 1  + 2 ln    3x − 1  + c + ​ ______ 3x + 1 3x − 1 ) 3x − 1 2 ​  = 2x + ln ​(______  ​  )​  + c 3x + 1 2 2 2 3x − 1 (c) ​​∫ ​   ​​​f(x) dx = ​​[2x + ln ​​(______ ​   ​ ​​​  ​ ​​  ​ ​ 3x + 1 ) ] 1 1 25 1 = ​(4 + ln ​(___ ​   ​ ))​ ​ − ​(2 + ln ​(__ ​   ​ ))​ ​ 4 49

∫



100 ​   ​ )​​ = 2 + ln ​(____ 49

106. Area between two curves __

(a) sin x = ​​√ 3 ​​ cos x   __

tan x = √ ​​  3 ​​  π 4π ​​   ​​    x = __ ​​   ​​ ,  ___ 3 3

__

__

​  3 ​  π √ 4π ​√ 3 ​  Coordinates are P​​(__ ​   ​ , ___ ​   ​  ​​ and Q​​(___ ​   ​,  −​ ___ ​ )​​ 3 2 ) 3 2 __ π __ (b) Area of R = ∫ ​​ 0 ​  ​ 3 ​ ​  ​(​√ 3 ​  cos  x − sin  x)​​​dx __

π __

​  3 ​ 

​= ​​[√ ​  3 ​   sin  x + cos x]​​  0​ ​ __ __ π π = ​(​√ 3 ​   sin ​ __ ​  + cos ​ __ ​ )​− (​ ​√ 3 ​  sin  0 + cos  0)​ 3 3 = 2 − 1 = 1​

​​  3 ​​  = 13​​ _3 ​​  = __ 1

40

198

WORKED SOLUTIONS __

109. Solving differential equations 1 1 (a) ∫__ ​ x  ​ dx = ∫cos 2t dt

​  3 ​  (​ sin x − √ (c) Area of S = ​​∫  ​  3 ​  cos  x)​dx​​ π __ 4π ​ __  ​ 

​  3  ​ 

__

4π __ ​  3  ​  π ​ __3  ​ 

​  3 ​  sin  x]​​  ​  ​ ​= [​​ − cos x − √ __ 4π √__ ___ 4π π π = ​(−  cos ​ ___ ​   − ​  3 ​  sin ​   ​    ​− ​ −  cos ​ __ ​  − √ ​  3 ​  sin ​ __ ​  )​ 3 3) ( 3 3 = 2 − (−2 )  = 4​

1

​ln  x = ​ _2 ​  sin  2t + ln 2 −   ​ _2 ​  ​ 1

(a) When x = 0, tan t = 0 So t = 0, and y = 2 cos (0) − 1 = 1 A is the point (0, 1) When y = 0, 2 cos t − 1 = 0 __ π π So t = ± ​​ __ ​​ ,  and x = tan  ± ​​ __ ​​  = ± ​​√ 3 ​​  3 3 __ From the diagram, B is the point (​​√ 3 ​​,  0). dx (b) ​​ ___ ​   = ​sec​​  2__​t​so dx = sec2 t dt π dt t=​ __3   ​ x=​√  ​   3 ​ ​ ​​   ​   (​​​ 2 cos t − 1)(​​ ​sec​​  2​t)​  dt​ ​​​ R=∫ ​​   ​   ​​y​ ​dx​ = ∫ ​ ​ x=0 t=0 π t=​ __3   ​ ​ 2 ) ( = ​​∫  ​   ​​​ 2 sec t  − ​sec​​  ​t ​  dt​ ​​​ ​ t=0 π __ ​  3  ​  = [​​​ 2  ln      sec x + tan x   −  tan t]​​  0​ ​​ __ __ = (2 ln (2 + √​​  3 ​​)  − √​​  3 ​​)  − (2 ln (1) − 0)) __ __ = 2 ln (2 + √​​  3 ​​)  − √​​  3 ​​ = 0.902 (3 d.p.)

y

​− ​ _4 ​​ (2(0.5) + 1)−2 = ​− ​ __ −8  ​​ + c 1



1

1.25

1.5

1.75

2

0

1.04599

2.73689

5.14147

8.31777

_1

(b) (i) I ≈ ​​  2 ​​  × 0.5[(0 + 8.31777) + 2(2.73689)] = 3.4479 (4 d.p.)

(ii) I ≈ _​​  2 ​​  × 0.25[(0 + 8.31777) + 2(1.04599 + 2.73689 + 5.14147)] = 3.2708 (4 d.p.) (c) Because the tops of the trapezia are closer to the curve 1

199

1

​ 1 2 (a) ∫​(2y  +  1)​​  −3​ ​​​ dy = − ​ _4 ​ (2y + 1)−2 + c ​ 1 ∫​(2y  +  1)​​  −3​  dy = ∫___ (b) ​  2  ​  dx ​x​​  ​ 1 1 ​− ​ _4 ​​ (2y + 1)−2 = ​− ​ __ x  ​​ + c When y = 0.5, x = −8:

108. The trapezium rule x

1

c = ln 2 − _​​ 2 ​​ 

107. Areas and parametric curves

(a)

​​ln x = ​ _2 ​  sin  2t + c​ ​ π 1 ​   ​  )​  +  c (b) ln 2 = ​ _2 ​  sin  2​(__ 4 1 = ​ _2 ​  +  c

1

1 _1 ​− ​ __ 16  ​   = ​  8 ​  +  c​ 3 c = ​− ​ __ 16  ​​ 

3 1 __ 1 ​− ​ _4 ​​ (2y + 1)−2 = ​− ​ __ x  ​​ ​− ​  16  ​​  1 4 3 ​​ ________ = __ ​​    ​ + ​ __ ​​     ​​   ​(2y  +  1)​​  2​ x 4 16 + 3x 1 = _______ ​​     ​​      ​ ​  ​​ ________ 4x ​(2y  +  1)​​  2​ _______ 4x ​     ​ ​  ​  2y + 1 = ​±​ ​​  _______ 16 + 3x _______ x 1 ​     ​ ​  ​  y = ​− ​ __  ​ ± ​  _______ 2 16 + 3x

√

√

WORKED SOLUTIONS STATISTICS 112. Sampling

(b) y = 1.1(x – 8) _

​​ ​ = 1.1(66.8 – 8) = 64.7 (1 d.p.) y​

sy​​= 1.1 × 18.0 = 19.8 (1 d.p.)

50 (a) 152 × ___ ​​    ​​ = 17.43… 436 17 female employees from Bristol (b) A numbered list of all the employees

119. Box plots and outliers (a) (i) 20 m (ii) Upper quartile (Q3)

113. Mean

(b) It is an outlier. It is a non-typical data value that is usually more than 1.5 × (Q3 − Q1) above Q3.

(a) 251 + 19 + 22 + 15 + 21 + 16 = 344 344 ____ ​   ​ = 17.2 °C 20 (b) The mean will go down because 16 < 17.2.

120. Cumulative frequency diagrams (a) 200

114. Median and quartiles

180 Cumulative frequency

(a) 22 ÷ 2 = 11, so median is half way between the 11th and 12th data values. Median = 1016.5 hPa (b) 22 ÷ 4 = 5.5 so Q1 is the 6th data value Q1 = 1012 hPa (3 × 22) ÷ 4 = 16.5 so Q3 is the 17th data value Q3 = 1026 hPa IQR = Q3 – Q1 = 1026 – 1012 = 14 hPa

160 140 120 100 80 60 40 20

115. Linear interpolation 178 3 ___ ​​   ​​  = 44.5: Q1 = 14.5 + (44.5 − 29) × __ ​​    ​​ = 15.2 (1 d.p.) 4 64 178 3 ​​ ___  ​​  = 89: Q2 = 14.5 + (89 − 29) × __ ​​    ​​ = 17.3 (1 d.p.) 2 64 0.9 × 178 = 160.2 3 90th percentile = 19.5 + (160.2 – 148) × __ ​​    ​​  21 = 21.2 (1 d.p.)

O

2

4

6

8 10 12 14 16 18 20 22 Wind speed (kn)

(b) 90th percentile – 10th percentile = 10.4 – 2.5 = 7.9 kn

121. Histograms (a) 63.5 − 57.5 = 6 57.5 − 54.5 = 3

116. Standard deviation 1

55–57 bar will be 2 cm wide.

(a) n = 31 _ 397 . 3  ​ x   ​   ​   = 12.8    ° C (1 d.p.)​ ​​  = _____ 31 _________________ 5165 . 39 397 . 3 2 ​s = ​  _______ ​     ​ − ​     (_____ ​   ​    ​  ​  =  1.5    ° C (1 d.p.)​ 31 31 ) (b) E.g. No, the mean temperature in Leuchars was higher and Leuchars is further north than Camborne or No, this is not a sufficiently large sample.

√

117. Standard deviation 2 Midpoint

2

5

8.5

15.5

Frequency

48

31

15

6

Mean = 4.7 mins (1 d.p.) Standard deviation = 3.6 mins (1 d.p.)

118. Coding _ 4380 1 ​​y​ = ____ ​   ​ = 23.804 …​ _184  ​​ x So   ​​  = 100 × 23.804… = 2380 m (3 s.f.)

√

______

2727 . 3 sy = ​​  ______ ​       ​ ​​  = 3.849… 184 So ​sx​ = 100 × 3.849… = 385 m (3 s.f.)

(b) Area of 6 × 4 = 24 cm2 represents 30 apples. 24 ​ ___ ​ × 12 = 9.6, so 12 apples represented by area of 9.6 cm2. 30 9.6 ___ ​   ​ = 4.8 so bar is 4.8 cm high. 2

122. Comparing distributions (a) 1987: _ 710 . 9 ​​  = _____ x    ​ ​   ​​   = 22.9 °C (1 d.p.) 31 _________ 16    364 .77 710.9 2 ​sx = ​ _________ ​   ​ ​       − ​(_____ ​   ​     ​ ​= 1.4 °C (1 d.p.) 31 31 ) 2015: _ 727 . 9   = _____  ​ ​   ​​   = 3.5 °C (1 d.p.) ​​ x 31 _________ √ ​  17  255.61 ​  ______ 729 . 9 2 ​s  x = __________ ​   ​ − ​     (​   ​    ​ = 2.3 °C (1 d.p.)​ 31 31 ) (b) E.g. Mean temperatures have increased slightly (increase in mean from 22.9 °C to 23.5 °C) but the temperatures recorded in May 2015 were also more varied (increase in s.d. from 1.4 °C to 2.3 °C). This is also a relatively small sample so it is not sufficient evidence to conclude that average air temperatures have increased between 1987 and 2015.

√ 

_ 668 2 (a) ​​    ​ = ___ x ​   ​​ = 66.8 10_____________ 47870 668 2 ​ ​x​= ​  _____ s ​     ​    − ( ​ ___ ​   ​ )​  ​​= 18.0 (1 d.p.) 10 10

√

200

WORKED SOLUTIONS 123. Correlation and cleaning data

128. Independent events

(a) The group were all young people so it is likely that the age of 81 is an error. Michelle’s decision is valid.

(a) P(A ⋃ B) = P(A) + P(B) − P(A ⋂ B) = P(A) + P(B) − P(A) × P(B) ​ _4 ​ = ​ _5 ​+ P(B) − _​ 5 ​P(B) 3

(b) 20

3

2

P(B) =

16 Time (hours)

2

= ​ _5 ​ + _​ 5 ​P(B)

18 14

_3 _2 ​ 4 ​ − ​ 5 ​ __ 7 _____ ​  _3  ​   = ​  12  ​ ​ 5 ​

(b) P(A9 ⋂ B) = P(A9) × P(B) 7 2 = (​  1 − _​ 5 ​  )​ × __ ​  12  ​

12 10

= ​ __ 20  ​ 7



8

(c) P(B9 | A) = P(B9) = 1 − __ ​ 12  ​ = __ ​ 12  ​

6

7

4

129. Tree diagrams

2

(a)

0

14 15 16 17 18 19 20 21 22 Age (years)

124. Regression (a) For every unit (hPa) increase in pressure, the visibility increases by approximately 0.328    km. (b) E.g. The scatter diagram shows that there is only very weak linear correlation between visibility and mean pressure, so a linear regression model is unlikely to model the data accurately.

125. Using regression lines (a) The data shows strong (positive) linear correlation. (b) (i) This estimate is from within the range of the independent variable (interpolation) so is reliable. (ii) This estimate is from outside the range of the independent variable (extrapolation) so is unreliable.

126. Drawing Venn diagrams R 2 6 9 18

35 13

0.31

P

127. Using Venn diagrams (a) S

15% 63% L

(b) P(L ⋂ T9) = 0.63

0.15

M H

Answer 0.8 0.2 0.5 0.5 0.1 0.9

Correct Incorrect Correct Incorrect Correct Incorrect

(b) P(Correct) = 0.54 × 0.8 + 0.31 × 0.5 + 0.15 × 0.1 = 0.602 P(H 9 ⋂ Correct)    (c) P(H9 | Correct) = _______________ ​      ​ P(Correct) 0.54 × 0.8 + 0.31 × 0.5       = ​ ____________________  ​ 0.602 = 0.97508… = 0.975 (3 s.f.)

130. Random variables 1 (a)

y P(Y = y)

2 1 ​ ___  ​  30

3 2 ___ ​    ​  15

4 3 ___ ​    ​  10

5 8 ___ ​    ​  15

​ 15  ​ = _​ 6 ​  (b) P(Y > 3) = P(Y = 4) + P(Y = 5) = __ ​ 10  ​ + __ 2 (a) 0.1 + a + 0.15 + 2a + 0.15 = 1 3a = 0.6  a = 0.2 5 (b) P(3X + 1 < 6) = P(X < ​ _3 ​ ) 3

8

5

131. The binomial distribution

12

S

E

5

= P(X = −2) + P(X = −1) + P(X = 0) + P(X = 1) = 0.1 + 0.2 + 0.15 + 0.4 = 0.85

N 5

Card

0.54

(c) Negative correlation. The older the person is, the fewer hours of training are likely to be needed to achieve the standard of proficiency.

201

2

17%

5% T

1 (a) There are a fixed number of rolls (50), and each roll is independent. The probabilities are the same on each roll. 1 (b) X ~ B​​(50, __ ​   ​ )​​ 6 40 50 1 10  __ ​   ​ )​  __ ​​   ​​  ​  5 ​​  = 0.1156 (4 d.p.) (i) P(X = 10) = ( ​ __ 10 6 6 (ii) Using calculator with p = 0.1667: P(X < 7) = P(X < 6) = 0.2504 (4 d.p.) 2 Because Emma is selecting without replacement, the probability of success is not the same for each of the trials.

( ) (​ )

132. Hypothesis testing 1 Assume H0, so X ​∼​B(28, 0.7) P(X > 25) = 1 – P(X < 24) = 1 – 0.9843 = 0.0157 1.6% < 2% so there is enough evidence to reject H0.

WORKED SOLUTIONS 2 Let X = the number of students out of 45 who pass. Then X ​∼​ B(45, p) H0: p = 0.493, H1: p ​≠​ 0.493 Assume H0 is true, so X ​∼​B(45, 0.493) P(X > 29) = 1 – P(X < 28) = 1 – 0.9706 = 0.0294 2.9% > 2.5% so there is not enough evidence to reject H0. Conclude that the average pass rate in Edinburgh was no different from the national average.

133. Finding critical regions (a) Assume H0 is true, so X ​∼​B(15, 0.45) P(X < 2) = 0.0107 P(X > 12) = 1 – P(X < 11) = 1 – 0.9937 = 0.0063 The critical regions are X < 2 and X > 12 (b) 0.0107 + 0.0063 = 0.017 Actual significance level = 0.017 or 1.7%

136. Measuring correlation PMCC = −0.5591 Negative linear correlation

137. Hypothesis testing for 0 correlation

140. The normal distribution 1 1 X ​∼​ N(410, 1252) (a) P(X > 500) = 0.2358 (b) P(X < 350) = 0.3156 (c) P(380 < X < 420) = 0.1267 2 P(Y < 2 or Y > 3) = 1 − P(2 < Y < 3) = 1 − 0.5161 = 0.4839

141. The normal distribution 2 Let X be the weight of compost in one bag, so X ​∼​N(20.5, 0.42) P(X < 20) = 0.1056 Let Y be the number of bags in a sample of 10 that have a weight of less than 20 kg, so Y ​∼​B(10, 0.1056) P(Y < 4) = 0.9845

142. The inverse normal function (a) P(432 < W < w) = 0.3 P(W < w) − P(W < 423) = 0.3 P(W < w) − 0.4286 = 0.3 P(W < w) = 0.7286 w = 511 (3 s.f.) (b) P(One courgette between 432 and 511) = 0.3 × 0.7 + 0.7 × 0.3 = 0.42

143. Finding µ and

(a) Linear correlation (b) r = 0.7117 (c) H0: ρ = 0, H1: ρ ​≠​ 0 Sample size = 8 Significance level = 0.05 Probability in each tail = 0.025 From the table, the critical value of r is 0.7067 and the critical region is r > 0.7067. 0.7117 > 0.7067, so the observed value of r is in the critical region. There is sufficient evidence to reject H0 and conclude that there is linear correlation between temperature and mass of product.

( 

)

( 

)

175 − µ P(W < 175) = P​ Z < _______ ​      ​  ​= 0.2 σ 175 − µ ⇒ ​ _______ ​  = −0.8416 σ    µ = 175 + 0.8416σ 230 − µ P(W > 230) = P​ Z > _______ ​  σ    ​  ​= 0.1 230 − µ ⇒ ​ _______ ​  = 1.2816 σ    µ = 230 − 1.2816σ So 175 + 0.8416σ = 230 − 1.2816σ ⇒ 2.1232σ = 55 σ = 25.9042…

= 25.9 (3 s.f.) So µ = 175 + 0.8416 × 25.9042…

138. Conditional probability



= 196.801…

1 + 2 + 11 14 P(Cheese | Avocado or Bacon) = ___________________    ​    ​ = __ ​   ​ 4 + 1 + 2 + 8 + 11 + 5 31



= 197 (3 s.f.)

139. Probability formulae (a) P(C ⋂ V) = P(C) × P(V | C) 7 2 = _​ 3 ​ × __ ​  25  ​ = __ ​  75 ​ 14



(b) P(C ⋃ V) = P(C) + P(V) − P(C ⋂ V) = _​  3 ​ + _​ 5 ​ − __ ​  75 ​ 2





=

2

14

22 ​ __ 25 ​

P(C9 ⋂ V9) = 1 − P(C ⋃ V)

=1−

3 22 __ ​ 25 ​ = __ ​ 25  ​

144. Normal approximations (a) Let X be the number of games won by Deep Thought out of 20, so X ​∼​B(20, 0.56) P(X = 10) = 0.1524 (4 d.p.) (b) Let Y be the number of games won by Deep Thought out of 200, so Y ​∼​B(200, 0.56) µ = 200 × 0.56 = 112

__________________

200 × 0.56(1 − 0.56) ​​ = 7.0199… σ =√ ​​    

Y can be approximated by Z ​∼​N(112, 7.0199…2) P(99.5 < Z < 120.5) = 0.8495 (c) Number of trials is large and probability on each trial is close to 0.5

202

WORKED SOLUTIONS 145. Normal hypothesis testing (a) H0: μ = 25 against H1: μ > 25 Let X represent the diameter of a circle and assume H0 so that X ​∼​N(25, 1.22) So for the sample mean: ​X ∼ N​(25, ​ ___ 15  ​)​​ or ​X ∼ N​(25, 0.096)​​ 1.22

P(​X ​> a) = 0.01 gives a = 25.721 The critical region is ​X ​> 25.721 (b) The observation lies inside the critical region, so there is sufficient evidence to reject H0 at the 1% level, and conclude that the mean diameter of the circles has increased.

203

WORKED SOLUTIONS MECHANICS 148. Modelling in mechanics (a) There is no frictional force between block P and the table top. (b) Air resistance and rotational forces can be ignored. (c) The magnitude of the acceleration is the same for both particles. (d) The weight of the string can be ignored. (e) The tension in the string is the same on both sides of the pulley.

(c) s = ?, u = 4, v = 0, a = −0.4, t = ? v2 = u2 + 2as 02 = 42 + 2 × (−0.4) × s 0.8s = 16 s = 20 m

152. Motion under gravity 3.5 m s21

149. Motion graphs (a) v

29.8 m s22

15 m

(m s21) 18

(a) s = ?, u = 3.5, v = 0, a = −9.8, t = ? v2 = u2 + 2as

V

02 = 3.52 + 2 × (−9.8) × s

O

10 16

30

t (s)

(b) Area under graph = 455 1 18 × 10 + _​ 2 ​ (18 + V) × 6 + 14 × V = 455

180 + 54 + 3V + 14V = 455

17V = 221 V = 13 m s−1

150. Constant acceleration 1 t 5 0s

t 5 12 s

u 5 2 m s21

v 5 23 m s21

19.6s = 12.25 s = 0.625 m Greatest height above water = 15 + 0.625 = 15.625 m (b) s = −15, u = 3.5, v = ?, a = −9.8, t = ? v2 = u2 + 2as = 3.52 + 2 × (−9.8) × (−15) = 306.25 v = ±17.5 The diver hits the water with speed 17.5 m s−1. (c) s = −15, u = 3.5, v = −17.5, a = −9.8, t = ? v = u + at

A

B

(a) s = ?, u = 2, v = 23, a = ?, t = 12 v = u + at 23 = 2 + 12a 12a = 21 a = 1.75 m s−2 (b) s = ?, u = 2, v = 23, a = 1.75, t = 12 1 s = _​ 2 ​ (u + v)t = ​ _2 ​ (2 + 23) × 12 = 150 m 1

153. Forces T

400 kg 200 N

T

1200 kg

1500 N

500 N

1500 − 500 − 200 = (400 + 1200)a

t 5 30 s

u 5 16 m s21

A

9.8t = 21 t = 2.1428… = 2.1 s (2 s.f.)

(a) Using F = ma (whole system):

151. Constant acceleration 2 t 5 0s

−17.5 = 3.5 + (−9.8) × t



800 = 1600a

a = 0.5 m s−2 (b) Using F = ma (trailer): T − 200 = 400 × 0.5

300 m

B

(a) s = 300, u = 16, v = ?, a = ?, t = 30 1 s = ut + _​ 2 ​ at2 1 300 = 16 × 30 + _​ 2 ​ × a × 302 −2   a = −0.4 m s The boat decelerates at 0.4 m s−2. (b) s = 300, u = 16, v = ?, a = −0.4, t = 30 v2 = u2 + 2as = 162 + 2 × (−0.4) × 300 = 16 v = 4 m s−1

T = 400 N

154. Forces as vectors (a) R = F1 + F2 = (2i − 6j) + (3i + kj)

= [5i + (−6 + k)j]

5 = −(−6 + k) k=1 (b) F1 + F2 + F3 = 0 (2i − 6j) + (3i + j) + (pi + qj) = 0 i components: 2 + 3 + p = 0, so p = −5 j components: −6 + 1 + q = 0, so q = 5

204

WORKED SOLUTIONS 155. Motion in 2D

For motion after 1 second:

1 (a) F = ma

s = −0.98, u = 1.96, v = ?, a = −9.8 m s−2, t = ?

24 ​​(​  ​ ) ​ ​​ = 8a − 10 3 1 24 a = ​​ __ ​​ (​  ​ ) ​ ​= ​(​   ​ ) ​ ​​ m  s–2 8 − 10 − 1.25

1.96 m s21 29.8 m s22 0.98 m

_________

(b) Magnitude = √ ​​  32 + 1.252 ​  = 3.25​  m  s–2 N

x

3

s = ut + ​ _2 ​ at2 −0.98 = 1.96t − 4.9t2 4.9t2 − 1.96t − 0.98 = 0 1

________________________ 2

1.25

1.25 x = ​​tan​​  −1​​(____ ​   ​     ​​ = 22.619…° 3 ) Bearing = 90 + 22.619… = 112.6° (1 d.p.) 2 F = F1 + F2 + F3 = (7 + p)i + (q – 3)j F = ma (7 + p)i + (q – 3)j = 0.6(3i + 6j) Equating coefficients of i: 7 + p = 1.8 so p = –5.2 Equating coefficents of j: q – 3 = 3.6 so q = 6.6

−(−1.96) ± √ ​ (−1.96)     − 4 × 4.9 × (−0.98) ​ ____________________________________

t =      ​      ​ 2 × 4.9 = −0.2898… or 0.6898… Particle B returns to its initial position after 0.69 s (2 s.f.).

157. Connected particles 1

T T P Q R (0.2 1 m)g

156. Pulleys

1.5 m s22

0.8g

(a) R(↑): Using F = ma for particle R: T − 0.8g = 0.8 × 1.5

0.2g m s22

T

T

A

B 4g

mg

(a) R(↓): Using F = ma for particle A: 4g − T = 4 × 0.2g T = 4g − 0.8g = 3.2g = 31 N (2 s.f.) (b) Using F = ma for particle B: mg − T = m × (−0.2g) 1.2mg = T 3.2g m = ____ ​   ​  1.2g _8 = ​  3 ​  = 2.7 (2 s.f.) (c) For first 1 second of motion: s = x, u = 0, v = y, a = 0.2g, t = 1 1 s = ut + _​ 2 ​ at2 _1 = 0 + ​ 2 ​ × 0.2g × 1 = 0.98 m v = u + at = 0 + 0.2g × 1 = 1.96 m s−1

T = 0.8g + 1.2 = 9.04 N R(↓): Using F = ma for particles P and Q combined:

(0.2 + m)g − T = (0.2 + m) × 1.5

0.2g + mg − 9.04 = 0.3 + 1.5m 1.96 + 9.8m − 9.04 = 0.3 + 1.5m 8.3m = 7.38 m = 0.88915… = 0.89 kg (2 s.f.) (b) R(↓): Using F = ma on block P:

R

0.2g − R = 0.2 × 1.5 R = 0.2g − 0.3 = 1.66 N So the force exerted on P by Q is 1.7 N (2 s.f.).

P

1.5 m s22

0.2g

158. Combining techniques Consider system before tow-rope breaks: Resultant force = 4000 – (1000 + 1800) = 1200 N 1200 = (m1 + m2) × 0.4 m1 + m2 = 3000 kg Consider trailer after tow-rope breaks: v = u + at 0 = 14 + 2.8a a = −5 m s−2 −1000 = m1 × (−5) m1 = 200 So m2 = 2800 The trailer has a mass of 200 kg and the truck has a mass of 2800 kg.

205

WORKED SOLUTIONS

2

(a) x = t   – 20t + 96t = t2(t2 – 20t + 96) = t2(t – 12)(t – 8) Both linear factors are > 0 for all values in the range 0 < t < 8, and t2 > 0 so x > 0 in this range. dx (b) ​​ ___ ​ ​= 4t3 – 60t2 + 96t dt = 4t(t2 – 15t + 24) dx ___ ​​   ​ ​= 0 ​⇒​ 4t(t2 – 15t + 24) = 0, so either t = 0 or dt t2 – 15t + 24 = 0 ​⇒​ t = 13.17… or t = 1.82… (3 s.f.) Maximum occurs at t = 1.82… At this point, x = 209 m (3 s.f.) Graph is a positive quartic graph and t = 1.82 is the middle of three stationary points, so it must be a maximum.

165. Moments 2 (a) A

R1 = 5g N (b)

R A

1

Taking moments about B: R × 3 = 6g × 2.5 + 18g × BD

36g = 15g + 18g × BD BD = _​ 6 ​  m 7

(a) O

2m

1.6 m D

C

20g

mg

Taking moments about D:

↺

R × 1.6 = 20g × 2

↻

R = 25g R(↑): R − 20g − mg = 0 1 ​​ _2 ​​(V

25g − 20g − mg = 0 m = 5 kg

– U)T 1

U + V = ​​(_____ ​   ​     ​T​as required. 2 )

(b)

4m S A

R

B

2m

1.6 m S 1 75

G

C

20g

D

5g

R(↑): S + S + 75 − 20g − 5g = 0 2S = 25g − 75

Taking moments about C:

↺

40g × 3.5 = 50g × x

= 170 S = 85 Taking moments about C:

↻

40g



S × 2 + 5g × 1.6 = 20g × GC

↺

C

↻

x = 2.8

5

R

B 2m

1

140g = 50gx

7

4m

A

= T(U + _​​  2 ​​V – _​​  2 ​​U)

50g

18g

R = 12g

2 Distance travelled = area under graph

3.5 m

B

R(↑): 2R − 6g − 18g = 0

ds ___ 1 (a) ​ v = ​   ​   = 10 − 2kt dt 2 d  s ___ a = ​   ​ = −    2k​ dt2 So a is constant. (b) v = 0 when t = 4 so 0 = 10 – 2 × k × 4 8k = 10 k = 1.25

xm

D

6g 5m

161. Deriving suvat formulae

164. Moments 1

C 2m

↻

2 (c) When t = 5, x = __ ​​  15  ​  (5​)​​  ​  2 ​ ​  − ​ __ 10  ​ ​(5)​​  ​ + 4​= 2.25 m (3 s.f.)

= UT +

R

166. Centres of mass

1

3 __

R1 × 3 = 6g × 2.5

AD = 5 − _​ 6 ​ = 3​ _6 ​ m or 3.83 m

__ 2 ​  2 ​  So ​x = ​ __ 15  ​​ t​​  ​  − ​  10  ​​ t​​  ​ + 4​ 1

B

Taking moments about B:

+  c​

When t = 0, x = 4 ​⇒​ c = 4 _3

C

6g 5m

_1 dv 1 ___ (a) ​ a = ​   ​  = ​ __ ​  × 0.1​t​​  −​  2 ​ ​ − 0.2 2 dt _1 = 0.05​t​​  −​  2 ​ ​ − 0.2​ _1 When t = 4, a = ​0 . 05(4​)​​ −​  2 ​ ​  − 0 . 2​= 0.025 – 0.2 = –0.175 m s–2 _1 dx (b) ​​  __ ​  = 0.1​t​​  ​  2 ​ ​ − 0.2t dt _3 1 _2 __ ​  2 ​  2 x = ​  3 ​ × 0.1​t​​  ​  − ​  2   ​×  0.2​t​​  ​  +  c

1

R2

2m

160. Variable acceleration 2

1 _​  3 ​  __ 1 2 2 = ​ __ 15  ​​t​​  ​  − ​  10  ​​ t​​  ​ 

R1

↺

3

↻

4

↺

159. Variable acceleration 1

170 + 8g = 20g × GC

GC = 1.2673… = 1.3 m (2 s.f.) DG = 1.6 + 1.3 = 2.9 m (2 s.f.)

206

WORKED SOLUTIONS 167. Resolving forces

170. Projectiles 1 (a) Let speed of projection = u

2.5 m s22

R(↑): uy = u sin 40°

T sin 40°

s = −28, u = uy, a = −9.8, t = 6

T

R

s = ut + _​​  2 ​​at2 1

40° T cos 40°

−28 = 6uy − 4.9(6)2 uy = 24.733… 24.733...  ​​  = 39 m s−1 (2 s.f.) u = ________ ​​  sin  40° (b) R(→): ux = 38.478… cos 40°

500 N 800g

= 29.476…

(a) Using F = ma (car): s = vt

T cos 40° − 500 = 800 × 2.5 T cos 40° = 2500 T = 3263.5… = 3260 N (3 s.f.) (b) R(↑): T sin 40° + R = 800g R = 800g − 3263.5… sin 40° = 5742.2… = 5700 N (2 s.f.)

= 29.476… × 6 = 180 m (2 s.f.) 2 R(↓): v = ?, u = 0, a = 9.8, s = 10 v2 = u2 + 2as v2 = 2 × 9.8 × 10 = 196 v = 14 m s−1 _________

Speed = √ ​​  ​18​​  2​ + ​14​​  2​ ​​  = 23 m s−1 (2 s.f.)

168. Friction P

P sin 15°

171. Projectile formulae (a) R(↑): uy = U sin α

R

uy = U sin α, s = 0, a = −g, t = T 15°

F

s = ut + ​​ _2 ​​at2 1

P cos 15°

0 = (U sin α)T − ​​ _2 ​​ × g × T 2 gT ​⇒​0 = T(U sin α − ​​ ___ ​​)  2 gT either T = 0 (at A) or U sin α − ___ ​​   ​​ = 0 2 2U sin α     ​ so T = ​​ _______ g ​

20g

(a) R(→): P cos 15° − F = 0 F = µR = 0.5R so P cos 15° − 0.5R = 0 R(↑): P sin 15° + R − 20g = 0 P sin 15° + R = 20g 2 × 1 : 2P cos 15° − R = 0 + 2 : P sin 15° + R = 20g

1

(b) R(→), ux = U cos α ux = U cos α, s = R, t = T

2

s = vt R = U cos α × T  2U sin α     ​: Using T = ​​ _______ g ​ 2U sin α _____________ 2 ​U​​  2​  sin  α cos α R = U cos α × _______ ​​  g ​     ​ = ​​    ​  g ​

2P cos 15° + P sin 15° = 20g P(2 cos 15° + sin 15°) = 20g P = 89.4703… = 89 N (2 s.f.) (b) The normal reaction will increase from 20g − P sin 15° to 20g.

169. Sloping planes 25 N

R

F

40° 3g cos 40° 3g

40°

(a) R(↖): R − 3g cos 40° = 0 R = 22.5217… = 23 N (2 s.f.) (b) R(↗): Using F = ma: 25 − 3g sin 40° − F = 3a F = µR = 0.2R so 25 − 3g sin 40° − 0.2 × 22.5217… = 3a 1.5977… = 3a a = 0.53 m s −2 (2 s.f.)

207

Using 2 sin α cos α = sin 2α: ​U​​  2​  sin  2α R = ________ ​​    ​  g ​

172. Static particles 1

3g sin 40°

1

X sin 30°

30°

Y

X X cos 30°

5g

(a) R(↑): X sin 30° − 5g = 0 X = 10g = 98 (b) R(←): Y − X cos 30° = 0 Y = 10g cos 30° __ = 49​√3 ​ = 85 (2 s.f.)

WORKED SOLUTIONS 2 O 50°

175. Connected particles 2

T cos 50°

T

R T

50° T sin 50°

Q (2 kg) 15 N

P

T P (4 kg)

W

(a) R(←): T sin 50° − 15 = 0 15    ​  T = ______ ​  sin 50° = 19.5811… = 19.6 N (3 s.f.) (b) R(↓): W − T cos 50° = 0 W = 19.5811… cos 50° = 12.5864… = 13 N (2 s.f.)

173. Limiting equilibrium R XN F 5 µR 5g

α

R(↖): R − X sin α − 5g cos α = 0 3 4 Using sin α = _​ 5 ​= 0.6 and cos α = _​ 5 ​= 0.8: 1 R − 0.6X = 4g R(↗): X cos α − F − 5g sin α = 0 Using F = μR = 0.5R: 0.8X − 0.5R = 3g 1 : R − 0.6X = 4g + 2 × 2 : 1.6X − R = 6g

4g

(a) R(↗): R − 2g cos θ = 0 4 Using cos θ = _​ 5 ​ : R = 1.6g 1 So F = µR = _​ 4 ​× 1.6g = 0.4g R(↖): Using F = ma on Q: T − F − 2g sin θ = 2a 3 Using sin θ = _​ 5 ​ :       T − 0.4g − 1.2g = 2a T − 1.6g = 2a R(↓): Using F = ma on P: 4g − T = 4a 1 + 2 : 2.4g = 6a a = 0.4g = 3.9 m s−2 (2 s.f.)

2

v 5 0 m s21

174. Static rigid bodies S

C

T = 0.3 s

mg 5 mg

Particle Q reaches its highest point up the slope after a further 0.3 s.

F

1 R(​→​): F = S 2 R(​↑​): R = 5mg + mg = 6mg Taking moments about the bottom of the ladder: 5mg × k cos α + mg × 1.5k cos α = S × 3k sin α ​5mgk​(_​  5 ​)​+ 1.5mgk​(_​  5 ​)​ = 3kS​(_​  5 ​)​​ 4

4

26mg = 9S Ladder in equilibrium so: F < µR S < 6mgµ from 1 and 2 26

0 = 0.24g − 0.8gT

0.8gT = 0.24g

B

​ __ 9  ​mg < 6mgµ

a 5 20.8g m s22

v = u + aT

α

2

For motion after 0.6 seconds: u = 0.24g, v = 0, s = ?, a = −0.8g, t = T (to find)

R

A

1

u 5 0.24g m s21

X = 10g = 98 N (2 s.f.)

(a)

θ

(b) For first 0.6 second of motion: s = ?, u = 0, v = ?, a = 0.4g, t = 0.6 v = u + at = 0.4g × 0.6 = 0.24g m s−1 R(↖): After P hits floor, using F = ma on Q: 2g sin θ + F = 2a 3 Using sin θ = _​ 5 ​: 2a = 1.2g + 0.4g = 1.6g a = 0.8g m s−2

P

α

F 2g

3

176. Vectors in kinematics _____________

(a) |v| = √ ​    (−10)2 + (−5)2 ​ = 11.2 km h−1 (3 s.f.)

(b) r = r0 + vt

= (12i + 25j) + 3(−10i − 5j) 3

= −18i + 10j

from 3

13 __

µ > ​  27 ​ (b) Weight of ladder acts at its midpoint.

208

WORKED SOLUTIONS 177. Vectors and bearings

179. Calculus with vectors

(a)

(a) v = ∫ ​ ​ ​  (​(4t − 1)i − j)​dt​

N

= (2t2 − t)i − tj + c

4j

When t = 0, v = −3i, so c = −3i

θ

v = (2t2 − t − 3)i − tj

θ

When t = 2, v = (3i − 2j) m s−1

2i

(b) i component of velocity = 0 so

tan θ = _​ 4 ​ so θ = 14.03…° 1

Bearing is 360° − 14.03…° = 346° (nearest degree). (b) rA = (3i − 8j) + t(−i + 4j)

1

rB = (−21i + 16j) + t(5i − 2j) = (−21 + 5t)i + (16 − 2t)j 24 = 6t

t = 4 Equating j components when t = 4: A: −8 + 4t = −8 + 16 = 8 B: 16 − 2t = 16 − 8 = 8 So the particles collide when t = 4 at the point with position vector (−i + 8j) m.

178. Variable acceleration 3 (a) v = t e−3t dv ___ ​   ​ = −3t e−3t + e−3t dt = (1 − 3t)e−3t dv ___ ​   ​ = 0 dt ⇒ 1 − 3t = 0 t = _​ 3 ​ 1

When t = _​ 3 ​, v = _​  3 ​ ​e​​  −3(​  3 ​)​= 0.123 m s−1 (3 s.f.) 1

_1

(b) ​∫​ ​  t​  e​​  −3t​  dt​ = − ​ _3 ​ ​e​​  −3t​ + _​  3 ​ ​∫​ ​  ​e​​  −3t​  dt​ 1

1

= − ​ _9 ​  (3t + 1) ​e​​  −3t​ + c 1

So distance travelled is given by

∫​​ 0 ​  ​  t ​e​​  −3t​  dt​​ = [​​ − ​ _19 ​  (3t + 1) ​e​​  −3t​]​​  0​​  2

2

= (​ − ​ _9 ​ ​e​​  −6​)​− (​ − ​ _9 ​ ​e​​  0​)​ 7

= 0.109 m

209

1

Particle is moving parallel to j at t = 3 seconds. (c) r = ∫ ​  ​ ​((2 ​t​​  2​  −t −3)i − tj)​  dt​

= (​ _3 ​t3 − _​ 2 ​t2 − 3t)i − _​ 2 ​t2j + c 2

Equating i components: 3 − t = −21 + 5t

1

(2t + 1)(t − 3) = 0 t = −​ _2 ​ or t = 3

= (3 − t)i + (−8 + 4t)j



2t2 − t − 3 = 0

1

1

When t = 0, r = 0, so c = 0 r = (​ _3 ​t3 − _​ 2 ​t2 − 3t)i − _​ 2 ​t2j 2

1

1

When t = 2, r = −​ _3 ​i − 2j 8

________

√

8 2 10 = __ ​  3  ​ m So distance from origin = ​​  ​​(_​  3 ​ )​​​  ​ + ​2​​  2​ ​​ 

WORKED SOLUTIONS Published by Pearson Education Limited, 80 Strand, London, WC2R 0RL. www.pearsonschoolsandfecolleges.co.uk Copies of official specifications for all Pearson qualifications may be found on the website: qualifications.pearson.com Text © Harry Smith and Pearson Education Ltd 2018 Original illustrations © Pearson Education Limited 2018 Typeset and illustrated by Techset Produced by ProjectOne The right of Harry Smith to be identified as author of this work has been asserted by him in accordance with the Copyright, Designs and Patents Act 1988. First published 2018 21 20 19 18 10 9 8 7 6 5 4 3 2 1 British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library ISBN 978 1 292 32583 5 Copyright notice All rights reserved. No part of this publication may be reproduced in any form or by any means (including photocopying or storing it in any medium by electronic means and whether or not transiently or incidentally to some other use of this publication) without the written permission of the copyright owner, except in accordance with the provisions of the Copyright, Designs and Patents Act 1988 or under the terms of a licence issued by the Copyright Licensing Agency, Barnard’s Inn, 86 Fetter Lane, London EC4A 1EN (www.cla.co.uk). Applications for the copyright owner’s written permission should be addressed to the publisher. Notes from the publisher 1. While the publishers have made every attempt to ensure that advice on the qualification and its assessment is accurate, the official specification and associated assessment guidance materials are the only authoritative source of information and should always be referred to for definitive guidance. Pearson examiners have not contributed to any sections in this resource relevant to examination papers for which they have responsibility. 2. Pearson has robust editorial processes, including answer and fact checks, to ensure the accuracy of the content in this publication, and every effort is made to ensure this publication is free of errors. We are, however, only human, and occasionally errors do occur. Pearson is not liable for any misunderstandings that arise as a result of errors in this publication, but it is our priority to ensure that the content is accurate. If you spot an error, please do contact us at [email protected] so we can make sure it is corrected.

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