Collins Cambridge International AS & A Level Mathematics Pure Mathematics 2 and 3 Worked SOLUTIONS [First Edition] 9780008257743, 9780008482961

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Cambridge International AS & A Level Mathematics

Pure Mathematics 2 & 3 STUDENT’S BOOK: Worked solutions

Tom Andrews, Helen Ball, Michael Kent, Chris Pearce Series Editor: Dr Adam Boddison

Pure Mathematics 1 International Students Book Title page.indd 1 57736_Pi_viii.indd 1 Mathematics 2 & 3.indd 1 WS TITLE PAGE_Pure

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1

WORKED SOLUTIONS

Worked solutions 1 Algebra Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1

2 a (x + 1)(x + 2) = x2 + 3x + 2 b (x − 3)(x + 4) = x2 + x − 12

a y = 2x + 5

c (x − 2)(x − 3) = x2 − 5x + 6 y

d (x − 3)2 = x2 − 6x + 9

6

e (2x + 1)(x + 2) = 2x2 + 5x + 2

5

f (x + 1)(x + 2)(x + 3) = x3 + 6x2 + 11x + 6

4 3

3

2

 n  n  n −1  n a b +   a n − 2b 2 +   a n − 3b 3 + ...  3 1   2

( a + b )n = a n + 

1 –5

–4

–3

–2

0 –1 –1

 n  n −1  n  n a b +   a n − 2b 2 +   a n − 3b 3 + ...  1  3  2

( a + b )n = a n + 

1

2

3

4

5

6

x

x3 term = 6 ! 23 x 3 = 160x 3 so the coefficient is 160. 3! 3!

Exercise 1.1A

b y = 3x − 2

1

y

a y = 3x + 2 y

3

5

2

4

1

3 –5

–4

–3

–2

0 –1 –1

1

2

3

4

5

6

x

2 1

–2 –3 3

–7

–6

–5

–4

–3

–2

0 –1 –1

c y=3−x

–4

2

3

4

5

x

3

4

5

6

7

8

–2

b y = x−2

y

–5

1

–3

–2

4

y

3

5

2

4

1

3

0 –1 –1 –2

1

2

3

4

5

6

2

x

1 –4

–3

–2

0 –1 –1

1

2

x

–2

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1 ALGEBRA

3

y =3x +2

c

a y = 6−x y

y

7

6

6

5

5

4

4

3

3

2

2

1 –6

–5

–4

–3

–2

1

0

–1 –1

1

2

3

4

5

6

x

–3

–2

–2

0 –1 –1

1

2

3

9

x

4

2

3

1 –2

0 –1 –1

2 1

2

3

4

5

6

x

–2

1 –6

–5

–4

–3

–2

–3

2

8

5

3

–3

7

6

4

–4

6

y

y

–5

5

b y = −x

d y = x −2

–6

4

0 –1 –1

1

2

3

4

5

6

x

1

2

3

4

5

6

x

1

2

3

4

5

6

c y =6− x

Not A as y = 2x − 5 has all positive y values. Not B as y = 2x − 5 has all positive y values.

y

C is the correct graph.

7

y

6 5

8

4

6

3

4

2

2

1 –12 –10 –8

–6

–4

0 –2 –2

2

4

6

8

10 12

x –6

–5

–4

–3

–2

0

–1

–4 –6

d y=−x

Not D as y = 2x − 5 has a positive x-intercept.

y 2 1 –6

–5

–4

–3

–2

0 –1 –1

x

–2 –3 –4 –5

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1

WORKED SOLUTIONS

4

7

a y = f (x )

y y

5

5

4

4

3

3

2

2

1

1 –6

–5

–4

b y = f( x

a f (x) = 3 – 4 (|x + 2|)

–3

–2

0

–1 –1

–6 1

2

3

4

5

6

–5

–4

–3

–2

x

) 8

–2

0

–1 –1

f(x) = 5 − 2x, y = f ( x

3 2 1

2

3

4

5

6

x

)

1 –5

–4

–3

6

–2

25

3

20

2

15

1

10

0 –1 –1

1

2

3

4

5

x

y

4

1

2

3

4

5

6

x

5 0 –25 –20 –15 –10 –5 –5

–2

6

0 –1 –1

b y = |f(x – 2)|

5

–3

–2

–2

y

–4

x

4

1

–5

6

y

2

–6

5

5

3

5

4

x a y = f .  2

4

–3

3

b It is the modulus of a horizontal translation to the left of two units.

5

–4

2

–3

6

–5

1

–2

y

–6

0 –1 –1

f(x) = 3 – |4x|

5

10 15 20

x

–10

y 4 3 2 1 –6

–5

–4

–3

–2

0 –1 –1

1

2

3

4

5

6

x

–2 –3 –4

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1 ALGEBRA

c y = |f(x)| + 3

2

a y = x − 5 and y = 2x

y

y

14

5

12

4

10

3

8

2

6

1

4 2 –10 –8

–6

–4

–4

0 –2 –2

2

4

6

8

10

–2

x

0 –1 –1 1

b x
9 2 − 5x > −4x x 2x or 3 – 7x < –2x 1 3 x < 3 or x > 5 9 9x2 – 30x + 25 = 16 – 16x + 4x2 5x2 – 14x + 9 = 0 14 ± 14 2 − 4 × 5 × 9 or ( x − 1)( 5x − 9) = 0 10 9   x = 1 or 5 x=



10 f(x) = 5 – 4x and g(x) = 3x |5 – 4x| > 3|x| 25 – 40x + 16x2 > 9x2 7x2 – 40x + 25 > 0 40 ± 40 2 − 4 × 7 × 25 14 5 x < or x > 5 7 x=



Exercise 1.2A x2 + 5x + 6 x − 1 x3 + 4x2 + x − 6 x3 − x2 5x2 + x 5x2 − 5x

1



6x – 6 6x – 6

0

2x2 + 17x + 48 2 x − 1 2x3 + 15x2 + 31x + 12 2x3 − 2x2 17x2 + 31x 17x2 − 17x

So (x +1) is not a factor of 3x3 + 2x2 − 7x + 2.

2x3 + x2 − 13x + 6 b x + 1 2x4 + 3x3 − 12x2 − 7x + 6 2x4 + 2x3 x3 − 12x2 x3 + x2

8 a  Use the relation |x – a |  b ⇔ x   a – b or x  a + b.





−13x2 − 7x −13x2 − 13x 6x + 6 6x + 6 0



So (x + 1) is a factor of 2x4 + 3x3 − 12x2 − 7x + 6.

4 a x3 + 2x2 + 4x − 10 x − 2 x4 + 0x3 + 0x2 − 18x + 81 x4 − 2x3 2x3 + 0x2 2x3 − 4x2 4x2 − 18x 4x2 − 8x −10x + 81 −10x + 20 61 8x3 − 6x2 − 47x + 84 b x + 1 8x4 + 2x3 − 53x2 + 37x − 6 8x4 + 8x3 −6x3 − 53x2 −6x3 − 6x2 −47x2 + 37x −47x2 − 47x 84x − 6 84x + 84

−90

48x + 12 48x − 48

60 So the remainder is 60.

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1 Algebra

5

3 2 11 11 x − x+ 2 4 8

2x + 1 3x3 − 4x2 + 0x + 1 3 3x3 + x 2 2 11 2 − x + 0x 2 11 2 11 − x − x 2 4 11 x +1 4 11 11 x+ 4 8 3 − 8

2x2 + 9x + (p + 18) 6 x – 2 2x3 + 5x2 + px – q 2x3 – 4x2 9x2 + px 9x2 – 18x

(p + 18)x – q (p + 18)x – 2(p + 18)



– q + 2(p + 18) = 0 1

2x2 – 3x + (p + 12) x + 4 2x3 + 5x2 + px – q 2x3 + 8x2 – 3x2 + px – 3x2 – 12x

8 x2 + (a – 5)x + (12 – 5a) x + 5 x3 + ax2 – 13x + b x3 + 5x2 (a – 5)x2 – 13x (a – 5)x2 + 5(a – 5)x

2p – q = – 36 1

+ (1 + b)x + (a + 1 + b) x – 1 x3 + bx2 + ax + 10 x3 – x2 (1 + b)x2 + ax (1 + b)x2 – (1 + b)x



25a + b = 60 1

a = 3 and b = –15 9 x–2

8m – 22 = 8 + 2m



p = – 14 and q = 8

m = 5



(10 – 2p)x + p (10 – 2p)x + 5(10 – 2p) p – 5(10 – 2p) = –33 p = 17 11

(4m – 8) x – 6 (4m – 8)x – 2(4m – 8)

2x2 + m x – 2 2x3 – 4x2 + mx + 8 2x3 – 4x2 0 + mx + 8 mx – 2m

6p = – 84



mx2 + 13x + 6 mx3 + x2 – 10x – 6 mx3 – 2mx2 (1 + 2m)x2 – 10x (1 + 2m)x2 – 2(1 + 2m)x



1 + 2

7 – 2x + (10 – 2p) x + 5 x3 + 3x2 – 2px + p x3 + 5x2 –2x2 – 2px –2x2 – 10x

11 + a + b = –1 2

a + b = –12

– q – 4(p + 12) 2

x2

(a + 1 + b)x + 10 (a + 1 + b)x – (a + 1 + b)



4p + q = – 48 2

b – 5(12 – 5a) = 0 1 x2

(p + 12)x – q (p + 12)x + 4(p + 12)



(12 – 5a)x + b (12 – 5a)x + 5(12 – 5a)



8m – 22

8 + 2m

10 5x – 7 x2 + 3x – 4 5x3 + 8x2 – 41x + 28 5x3 + 15x2 – 20x –7x2 – 21x + 28 –7x2 – 21x + 28

0

Exercise 1.3A 1 f(1) = 13 + 4(12) + 1 − 6 = 0 so (x − 1) is a factor of f(x). 2 f(1) = 2(13) + 15(12) + 31 + 12 = 60 so (x − 1) is not a factor of f(x) and the remainder is 60.

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1 

WORKED SOLUTIONS

3 a f(x) = 3x3 + 2x2 − 7x + 2 f(1) = 3 + 2 − 7 + 2 = 0 so (x − 1) is a factor.

8 a f(5) = 375 – 125 – 235 – 15 = 0

f(–3) = –81 – 45 + 141 – 15 = 0 1 1 5 47 f − =− − + − 15 = 0 3 9 9 3 3 2 3x – 5x – 47x – 15 = (x + 3)(3x + 1)(x – 5)

( )



Using algebraic division, 3x3 + 2x2 – 7x + 2 = (x – 1)(3x2 + 5x – 2) and 3x2 + 5x – 2 = (3x – 1)(x + 2)





3x3 + 2x2 − 7x + 2 = (x − 1)(3x − 1)(x + 2)

b 5x3 – 15x2 – 47x – 15 = 2x3 – 10x2

b f(x) = 2x4 + 3x3 − 12x2 − 7x + 6

f(−1) = 2 − 3 − 12 + 7 + 6 = 0 so (x + 1) is a factor.



By algebraic division, 2x4 + 3x3 − 12x2 − 7x + 6 = (x + 1)(2x3 + x2 − 13x + 6)



g(x) = 2x3 + x2 − 13x + 6



g(−3) = −54 + 9 + 39 + 6 = 0 so (x + 3) is a factor.



By algebraic division, 2x3 + x2 – 13x + 6 = (x + 3)(2x2 – 5x + 2) and 2x2 – 5x + 2 = (2x – 1)(x − 2) 2x4 + 3x3 − 12x2 − 7x + 6 = (x + 1)(x + 3)(2x − 1)(x − 2)



4 a f(x) = x4 − 18x + 81

f(2) = 16 − 36 + 81 = 61

b f(x) = 8x4 + 2x3 − 53x2 + 37x − 6

f(−1) = 8 − 2 − 53 − 37 − 6 = −90

5 f(x) = 3x3 − 4x2 + 1 3 3 1 f − = − − 1 + 1 = − so (2x + 1) is not a factor. 2 8 8 6 a f(x) = x4 − 7x3 + 13x2 + 3x − 18 f(2) = 16 − 56 + 52 + 6 − 18 = 0 so (x − 2) is a factor. x4 − 7x3 + 13x2 + 3x − 18 = (x − 2)(x3 − 5x2 + 3x + 9) g(x) = x3 − 5x2 + 3x + 9 g(3) = 27 − 45 + 9 + 9 = 0 so (x − 3) is a factor. g(x) = (x − 3)(x2 − 2x − 3) f(x) = (x + 1)(x − 2)(x − 3)2

( )

b (x + 1)(x − 2)(x − 3)2 = 0 So x = −1, 2 or 3



3 27 117 30 f − =− − + + 21 = 0 2 4 4 2



( )



(x + 3)(3x + 1)(x – 5) = 0 1 So x = −3, − or 5. 3



f(–6) = – 432 + 324 + 120 – 12 = 0

( 12 ) = − 14 + 94 + 10 − 12 = 0

f −

2x3 + 9x2 – 20x – 12 = (x + 6)(2x + 1)(x – 2)

b 2x3 + 9x2 = 20x + 12 2x3 + 9x2 – 20x – 12 = 0

(x + 6)(2x + 1)(x – 2) = 0 1 x = −6, − or 2. 2 10 f(x) = 12x2 + 5x + 7

There are no values of x such that f(x) = 0.

Exercise 1.4A 1 a i

2x − 5 A B = + (x + 2)(x + 3) x + 2 x + 3

2x − 5 ≡ A(x + 3) + B(x + 2)

Substitute x = −3



−11 = −B



B = 11

Substitute x = −2



−9 = A



A = −9 −9 11 2x − 5 = + (x + 2)(x + 3) x + 2 x + 3





2x3 – 13x2 – 10x + 21 = (x – 1)(2x + 3)(x – 7)

c 2x3 – 13x2 – 10x + 21 = 0



9 a f(2) = 16 + 36 – 40 – 12 = 0

b f(1) = 2 – 13 – 10 + 21 = 0 f(7) = 686 – 637 – 70 + 21 = 0

3x3 – 5x2 – 47x – 15 = 0



7 a f(–1) = –2 – 13 + 10 + 21 = 16 so not a factor.



(x – 1)(2x + 3)(x – 7) = 0 3 So x = 1, − or 7. 2



A B ii 2x − 5 = + (x + 2)(x + 3) x + 2 x + 3 2x – 5 = A(x + 3) + B(x + 2)



2 = A + B

1





−5 = 3A + 2B

2







4 = 2A + 2B







A = −9



Substitute into 1 .



B = 11

1 ×2 3

2 – 3

−9 2x − 5 11 = + (x + 2)(x + 3) x + 2 x + 3

7

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1 Algebra

6x − 12 A B b i (x − 1)(x + 5) = x − 1 + x + 5

2

6x − 12 = A(x + 5) + B(x − 1) Substitute x = 1 −6 = 6A A = −1 Substitute x = −5 −42 = −6B B=7



11 − 2x 1 3 = − (x − 2)(x + 5) x − 2 x + 5

−1 6x − 12 7 = + (x − 1)(x + 5) x − 1 x + 5



2 − 3x A B 3 a = + (2x + 1)(3 − x) 2x + 1 3 − x

6x − 12 A B ii = + (x − 1)(x + 5) x − 1 x + 5



6x − 12 = A(x + 5) + B(x − 1)



6 = A + B

1



−12 = 5A − B

2



1 + 2



−6 = 6A

−7 = 7B



B = −1

−12 = −4A



A=3



Substitute x = 7



−32 = 4B



B = −8



3 − 5x = 3 − 8 (x − 3)(x − 7) x − 3 x − 7



22 = 2A



A = 11



Substitute x = −2 18 = −2B



B = −9

4x − 30 = A + B x 2 − 8x + 15 x − 5 x − 3

1



4x – 30 = A(x − 3) + B (x − 5)

2



Substitute x = 3





−18 = −2B





B=9



Substitute x = 5

−12 = −4A





−10 = 2A

A=3





A = −5



4x − 30 −5 9 2 + = x − 8x + 15 x − 5 x − 3

2 + 3





c

−15 = 3A + 3B



Substitute x = 0



1 ×3



2x + 22 A B = + x 2 + 2x x x + 2



3 = −7A − 3B



1 1 − 2x + 1 3 − x

2x + 22 11 9 = − x x+2 x 2 + 2x

−5 = A + B



=

2x + 22 = A (x + 2) + B (x)



3 − 5x = A(x − 7) + B(x − 3)



2 − 3x

( 2x + 1)(3 − x )



3 − 5x A B = + ii (x − 3)(x − 7) x − 3 x − 7

1 2

7 7 = A 2 2 A=1

b

Substitute x = 3



Substitute x = −



3 − 5x = A(x − 7) + B(x − 3)





3 − 5x A B = + (x − 3)(x − 7) x − 3 x − 7



Substitute x = 3



−1 6x − 12 7 = + (x − 1)(x + 5) x − 1 x + 5 c i





Substitute into 1 B = 7



2 − 3x = A(3 − x) + B(2x + 1)



A = −1

11 − 2x A B = + (x − 2)(x + 5) x − 2 x + 5 11 − 2x = A(x + 5) + B(x − 2) Substitute x = −5 21 = −7B B = −3 Substitute x = 2 7 = 7A A=1



Substitute into 1 B = −8





3 − 5x 3 8 = − (x − 3)(x − 7) x − 3 x − 7

3

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1 

WORKED SOLUTIONS

4

A 1 B = + x −9 x−3 x+3 1 = A(x + 3) + B(x − 3) Substitute x = 3 1 = 6A 1 A= 6 Substitute x = −3 1 = −6B 1 B=− 6 1 1 1 = − x 2 − 9 6 ( x − 3) 6 ( x + 3) 2

1 1 1 = − . x − a 2 2a ( x − a ) 2a ( x + a ) 2

 (x − 2)(x − 3) 2x2 − 4x + 8 = A + B(x − 1)(x − 3) + C(x − 1)(x − 2)









A=3





Substitute x = 2





8 − 8 + 8 = B(1)(−1)







B = −8





















A = −2









2 − 3 − 4 = B(1)(−1)





B=5













2 − 4 + 8 = A(−1)(−2)

Substitute x = 3 18 − 12 + 8 = C(2)(1) C=7

6



2

2x − 4x + 8 3 8 7 = − + ( x − 1)( x − 2)( x − 3) x − 1 x − 2 x − 3

2 − 3x − 4x2 = A(x − 1)(1 − 2x) + B(x)(1 − 2x) + C(x)(x − 1) Substitute x = 0

Substitute x = 1

Substitute x =

1 2

( )( )

4 3 2 1 1 − − =C − 2 2 2 2 2 C=2 2

( x )( x − 1)(1 − 2x )





c

( x − 1)( x − 2)( x + 4 )

6 − 6x − 5x 2

=

−2 5 2 + + x x − 1 1 − 2x

=

A B C + + x −1 x − 2 x + 4



Substitute x = 2



  6 − 12 − 20 = B(1)(6)







Substitute x = 1



    6 − 6 − 5 = A(−1)(5)

B  = −

13 3

   A  = 1

Substitute x = −4



   6 + 24 − 80 = C(−5)(−6)





Substitute x = 1

2 − 3x − 4x 2 A B C = + + b ( x )( x − 1)(1 − 2x ) x x − 1 1 − 2x



C=−

5 3

6 − 6x − 5x 2 1 13 5 = − − ( x − 1)( x − 2)( x + 4 ) x − 1 3 ( x − 2) 3 ( x + 4 )









2x 2 − 4x + 8 A B C = + + ( x − 1)( x − 2)( x − 3) x − 1 x − 2 x − 3



2 = A(−1)(1)

6 − 6x − 5x2 = A(x − 2)(x + 4) + B(x − 1)(x + 4) + C(x − 1)(x − 2)

5 a



2 − 3x − 4x

A 1 B = + x 2 − 16 x − 4 x + 4 1 = A(x + 4) + B(x − 4) Substitute x = 4 1 = 8A 1 A= 8 Substitute x = −4 1 = −8B 1 B=− 8 1 1 1 = − x 2 − 16 8 ( x − 4 ) 8 ( x + 4 ) Notice that



A 1 B = + x 2 − a2 x − a x + a 1 = A(x + a) + B(x − a) Substitute x = a 1 = 2aA 1 A= 2a Substitute x = −a 1 = −2aB 1 B=− 2a 1 1 1 = − x 2 − a 2 2a ( x − a ) 2a ( x + a )

7

5 + 3x − x 2 −x + 3x 2 + 4x − 12



Using the factor theorem: Substitute x = 2 into denominator −8 + 12 + 8 − 12 = 0 So (x − 2) is a factor.

3

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1 Algebra



Using the factor theorem: Substitute x = − 2 into denominator 8 + 12 − 8 − 12 = 0 So (x + 2) is a factor.



Using the factor theorem: Substitute x = 3 into denominator −27 + 27 + 12 − 12 = 0 So (3 − x) is a factor as coefficient of x3 is negative. Express the expression as partial fractions. 5 + 3x − x 2 A B C = + + −x 3 + 3x 2 + 4x − 12 x + 2 x − 2 3 − x 5 + 3x − x2 = A(x − 2)(3 − x) + B(x + 2)(3 − x) + C(x + 2)(x − 2) Substitute x = 2 5 + 6 − 4 = B(4)(1) 7 B= 4 Substitute x = 3 5 + 9 − 9 = C(5)(1) C=1 Substitute x = −2 5 − 6 − 4 = A(−4)(5) 1 A= 4



1 7 1 5 + 3x − x 2 = + + −x + 3x 2 + 4x − 12 4 ( x + 2) 4 ( x − 2) 3 − x 3

5 cannot be split into partial fractions x 2 − x + 10 because x2 – x + 10 does not factorise. ( x + 3)( x − 2) x2 + x − 6 = 9 3 2 x − 1)( x + 2)( x + 4 ) ( x + 5x + 2x − 8 8

x2 + x − 6 A B C = + + ( x − 1)( x + 2)( x + 4 ) x − 1 x + 2 x + 4



x2 + x – 6 = A  (x + 2)(x + 4) + B(x – 1) (x + 4) + C(x – 1)(x + 2)



x=1



10



1 1 = 2x 3 − 3x 2 − 32x − 15 ( x + 3)( 2x + 1)( x − 5) 1 A B C = + + ( x + 3)( 2x + 1)( x − 5) x + 3 2x + 1 x − 5



1 = A(2x + 1)(x – 5) + B (x + 3) (x – 5) + C (x + 3)(2x + 1) 1 x=− 2 5 11 1= B − 2 2



B=−





( )( )

The coefficient of



x = –2



4 – 2 – 6 = B(–3)(2) 2 3



B=



x=–4



16 – 4 – 6 = C (–5)(–2) C=



3 5

x2 + x − 6 2 3 4 = + − x + 5x 2 + 2x − 8 3 ( x + 2) 5 ( x + 4 ) 15 ( x − 1) 3

1 4 is − 2x + 1 55

Exercise 1.4B x 2 + 8x + 4 A B C = + 2+ x x x−2 x 2 ( x − 2) x2 + 8x + 4 = A(x)(x − 2) + B(x − 2) + C(x2) Substitute x = 0 4 = −2B B = −2 Substitute x = 2 4 + 16 + 4 = 4C C=6 Substitute x = 1 1 + 8 + 4 = A(1)(−1) + (−2)(−1) + (6)(1) 13 = −A + 2 + 6 A = −5 1

x 2 + 8x + 4 −5 2 6 − 2+ = x x−2 x 2 ( x − 2) x 2

p 3p 5x . 2 ≡ x +3 − ( x + 3) ( x + 3 )2 5x = p(x + 3) −3p Substitute x = −3 −15 = −3p p=5

1 + 1 – 6 = A(3)(5) 4 A=− 15

4 55

5x

( x + 3 )2

≡

5 15 − . x + 3 ( x + 3 )2

3

7x − 3 7x − 3 = x 2 − 8x + 16 ( x − 4 )2







7x − 3 = A(x − 4) + B Substitute x = 4 B = 25 Substitute x = 0 −3 = −4A + 25 A=7

7x − 3

( x − 4 )2

=

A B + x − 4 ( x − 4 )2

7x − 3 7 25 = + x 2 − 8x + 16 x − 4 ( x − 4 )2

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WORKED SOLUTIONS

4 The errors are shown in red. 2x 2 − x − 6 ≡ 2x 2 − x − 6 x 3 + 4x 2 + 4x x(x + 2)2 2x 2 − x − 6 A B C ≡ + + x 3 + 4x 2 + 4x (x) (x + 2) (x + 2)2



2x 2 − x − 6 A(x + 2)2 + B(x)(x + 2) + C(x) ≡ 3 2 x + 4x + 4x x(x + 2)2 2x2 − x − 6 ≡ A(x + 2)2 + B(x)(x + 2) + C(x) Substitute x = 0. −6 = 4A 3 A=− 2 Substitute x = −2. 8 + 2 − 6 = −2C C = −2 Substitute x = 1. 3 2 − 1 − 6 = − (9) + B(1)(3) + (−2)(1) 2



( )







∴

5

−5 = − 27 + 3B − 2 2 21 3B = 2 2x 2 − x − 6 3 7 2 ≡− + − 2(x) 2(x + 2) (x + 2)2 x 3 + 4x 2 + 4x

A 1 B C = + + (x + 1)(x − 2)2 x + 1 x − 2 (x − 2)2 1 = A(x − 2)2 + B(x + 1)(x − 2) + C(x + 1) Substitute x = 2 1 = 3C 1 C= 3 Substitute x = −1 1 = 9A 1 A= 9 Substitute x = 0 1 1 (1) 1= (−2)2 + B(1)(−2) + 9 3

()

B=−

()

1 9







1 1 1 1 = − + (x + 1)(x − 2)2 9(x + 1) 9(x − 2) 3(x − 2)2 Yes,

1 can be split into partial fractions. (x + 1)(x − 2)2



Substitute x = 0 −20 = 4A – 2B Substitute x = 1 −12 = A − B 2 ×2 −24 = 2A − 2B 3 – 1 −4 = −2A A=2



Substitute into 2 : −12 = 2 − B B = 14

2 3

2x 2 + 6x + 5 2 14 25 = + + (x − 2) (x − 2)2 (x − 2)3 (x − 2)3 7 The two different methods to split a rational function with linear factors in its denominator into partial fractions are substitution and equating coefficients. For partial fractions without repeated terms, the substitution method relies only on basic mathematical operations without the need for any algebraic manipulation, whereas the equating coefficients method results in the solving of simultaneous equations. For partial fractions with repeated terms, the substitution method relies on a combination of basic mathematical operations and the solving of simultaneous equations, whereas the equating coefficients method results in the solving of simultaneous equations with three unknowns. 8

2

2x + 6x + 5 A B C 6 = + + x − 2 (x − 2)2 (x − 2)3 (x − 2)3 2x2 + 6x + 5 = A(x − 2)2 +B(x − 2) + C Substitute x = 2 8 + 12 + 5 = C C = 25

1





1 A B C = + + x x − 3 ( x − 3 )2 x 3 − 6x 2 + 9x 1 = A(x – 3)2 + Bx (x – 3) + Cx x=3 1 = 3C 1 C= 3 x=0 1 = 9A 1 A= 9 x=1 1 1= ( 4 ) + B (1)( −2) + 13 (1) 9

()

B=−

()

1 9

1 1 1 1 = − + 9x 9(x − 3) 3 ( x − 3)2 x 3 − 6x 2 + 9x

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1 Algebra

2x − 1 A B C = + + ( 2x + 1)3 2x + 1 ( 2x + 1)2 ( 2x + 1)3

9

2x – 1 = A(2x + 1)2 + B(2x + 1) + C x=−



–2 = C



x=0



1 2

–1 = A + B – 2



x=1



1 = 9A + 3B – 2



A = 0 and B = 1 2x − 1 1 2 = − ( 2x + 1)3 ( 2x + 1)2 ( 2x + 1)3

x 2 + 7x − 1 A B C = + + 3 x + 5 ( x + 5)2 ( x + 5)3 x + 5 ( ) x2 + 7x – 1 = A(x + 5)2 + B(x + 5) + C 10



x = –5



–11 = C



x=0



2

A(x 2 + 1) + (Bx + C )(x + 3) (x + 3)(x 2 + 1) 2 2 x − 7x + 6 ≡ A(x + 1) + (Bx + C)(x + 3) Let x = −3. 9 + 21 + 6 = 10A 18 So A = . 5 Let x = 0. 18 6 = 5 + 3C ≡

( )



–1 = 25A + 5B – 11



x=1



7 = 36A + 6B – 11



A = 1 and B = –3 The coefficient of



1

( x + 5 )2

is –3.

1−7+6 =

3

(185 )(2) + ( B + 45 )(4)

36 16 + 4B + 5 5 −13 So B = . 5 All of the above working is correct. The denominator of A, B and C has been missed from the final statement: State the original fraction as partial fractions. 0=

x 2 − 7x + 6 18 4 − 13x ≡ + (x + 3)(x 2 + 1) 5(x + 3) 5(x 2 + 1)

(x − 3)2 A Bx + C ≡ + 2 x(x 2 − 6) x x −6

A(x 2 − 6) + (Bx + C )x x(x 2 − 6) (x − 3)2 ≡ A(x2 − 6) + (Bx + C)x Let x = 0. 9 = −6A. −3 So A = 2 Let x = 1. 15 4= +B+C 2 −7 B + C = 2 Let x = −1. 15 16 = +B−C 2 17 B – C = 2 5 B = 2 C = −6 ≡

x 2 − 5x + 6 Ax + B C ≡ 2 + (x + 2)(x 2 + 1) x +1 x + 2

( Ax + B)(x + 2) + C(x 2 + 1) (x 2 + 1)(x + 2) 2 x − 5x + 6 ≡ (Ax + B)(x + 2) + C(x2 + 1) Let x = −2. 4 + 10 + 6 = 5C So C = 4. Let x = 0. 6 = 2B + 4 So B = 1. Let x = 1. 1 − 5 + 6 = (A + 1)(3) + 8 So A = −3. ≡

∴

4 . 5 Let x = 1. So C =

∴

Exercise 1.4C 1

x 2 − 7x + 6 A Bx + C ≡ + (x + 3)(x 2 + 1) x + 3 x 2 + 1

x 2 − 5x + 6 1 − 3x 4 ≡ + (x + 2)(x 2 + 1) x 2 + 1 x + 2

∴

5x − 12 3 (x − 3)2 ≡ − x(x 2 − 6) 2(x 2 − 6) 2x

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WORKED SOLUTIONS

4

Let x = 2

8p p x 2 + 8x + 7 ≡ px − 37 − + . 2 2 2 9 ( − 1 ) x 3(x − 1)2 (x − 1) (x + 2) 9(x + 2)

(

5 = 2A +

Put the right-hand side over a common denominator.

)

17 (1) − 14 (7 ) 4

5 A= 4 2 (px − 37)(x − 1) − p(x − 1)(x + 2) + 8p × 3(x + 2) x + 8x + 7 = x 2 + 3x − 5 5x + 17 1 (x − 1)2 (x 2 + 2) 9(x − 1)2 (x 2 + 2) ≡ − ∴ 2 2 − 1) 4 x ( x − x + x + 1 3 4 3 ( ) (px − 37)(x − 1)2 − p(x − 1)(x 2 + 2) + 8p × 3(x 2 + 2) x 2 + 8x + 7 = (x − 1)2 (x 2 + 2) 9(x − 1)2 (x 2 + 2) 3x − 1 A B C ≡ + + 7 2 x + 5 x −1 x +1 x + 5 x − 1 ( ) 9(x2 + 8x + 7) = (px − 37)(x − 1)2 − p(x − 1)(x2 + 2) + 24p(x2 + 2) 3x – 1 ≡ A(x – 1)(x + 1) + B(x + 5)(x + 1) Let x = 1.  + C(x + 5)(x – 1) 9(1 + 8 + 7) = 72p Let x = 1 So p = 2. 2 = B(6)(2) 2

2

2

(

)

(

5

x2 − 5 can be split into partial fractions. x(x 2 − 3)



So

6

x 2 + 3x − 5 Ax + B C ≡ 2 + x + 3 x −1 ( x − 1) x 2 + 3

(

)

x2 + 3x – 5 ≡ (Ax + B)(x – 1) + C(x2 + 3) Let x = 1

B=

17 4

So B =

– 4 = C(4)(– 2) 1 2 Let x = – 5 So C =

– 16 = A(– 6)(– 4) 2 3

So A = ∴ 8

3x − 1

( x + 5 ) ( x 2 − 1)

−2 1 1 + + 3(x + 5) 6(x − 1) 2(x + 1)

3x + 2 ≡ (Ax + B)(2x – 5) + C(x2 + 5) Let x =

5 2

( 38 45 = C( ) 4 4

15 25 + 2 =C +5 2 4

)

38 45 Let x = 0 C=

2 = (B)(– 5) + 4 9 Let x = 1

38 ( 5) 45

B=

A =− 1 ( 3) 4

≡

3x + 2 Ax + B C ≡ + (x 2 + 5)(2x − 5) x 2 + 5 2x − 5

5= A+

1 C=− 4 Let x = 0 – 5 = (B)(– 1) −

)

(

– 1 = C(4)

)

1 6 Let x = –1

x2 − 5 A Bx + C ≡ + 2 x(x 2 − 3) x x −3

x2 − 5 A(x 2 − 3) + (Bx + C )x ≡ 2 x(x 2 − 3) x(x − 3) 2 2 x − 5 ≡ A(x − 3) + (Bx + C)x Let x = 0. −5 = −3A 5 So A = 3 Let x = 1. −10 −4 = +B+C 3 −2 B + C = 3 Let x = −1. −10 −4 = +B−C 3 −2 B – C = 3 −2 B = 3 C = 0 x2 − 5 ≡ 5 − 2x ∴ x(x 2 − 3) 3x 3(x 2 − 3)

(

∴

)

20 ( −3) + 38 (6 ) 45 45

19 45

3x + 2 20 − 19x 38 ≡ + (x 2 + 5)( 2x − 5) 45(x 2 + 5) 45 ( 2x − 5)

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1 Algebra

2x + 1

9

( 2x + 3)( x 2 + 1)

≡

Ax + B C + x 2 + 1 2x + 3

2x + 1 ≡ (Ax + B)(2x + 3) + C(x2 + 1) Let x = − −2 = C

3 2

8 13 Let x = 0 C=−

3 c 1 + 0.1 ≈ 1 +

1 = (B)(3) − 7 13 Let x = 1

8 (1) 13

4 a (1 + x )

B=

(

∴ 10

(

)

7 8 (2) (5) − 13 13

4 13

A=

2x + 1

( 2x + 3)( x 2 + 1)

≡

4x + 7 8 − 13(x 2 + 1) 13(2x + 3)

1 Ax + B C ≡ 2 + x 2 + 2 ( 2x − 1) x + 2 2x − 1

)

1 ≡ (Ax + B)(2x – 1) + C(x2 + 2) Let x = 1 =C

( 14 + 2)

4 9 Let x = 0

(

−1

≈ 1 + ( −1 ) x +

( −1)( −2) x 2 2

( −1)( −2)( −3) x 3 +

3! = 1 – x + x2 – x3 1 b  Replace x with x2: = (1 + x2)−1 1 + x2

4 ( 2) 9



( )( )( )

Exercise 1.5A (−2)(−3) 2 x = 1 − 2x + 3x2 2 b Replace x with –x to get (1 − x)−2 1 a (1 + x)−2 ≈ 1 + (−2)x +

−1 2

is

−1 −3 −5 2 2 2 3 5 5 × ( −2x ) = − × −8 x 3 = x 3 so 3× 2×1 16 ( ) 2 5 the coefficient of x3 is 2

)

−2x − 1 1 ∴ coefficient of 2 is . 9 x +2

1 3 27 2 27 3 − x+ x − x 4 4 16 8 3x < 1 so |x| < 2 3 2

7 The term in x3 in the expansion of (1 − 2x )

1 (1) + 49 (3) 9

2 A =− 9



5 a 1 = (1 + x ) 1+ x

6

1 B=− 9 Let x = 1

1  − 1  − 3 1 −1 2  2  2 2  2   2  3 1 ≈1+ x + x + x 2 2 6 = 1 + 1 x − 1 x2 + 1 x3 2 8 16

≈ 1 – x2 + x4 – x6.

C=

1 = (B)(– 1) +

1 2

0.1 0.01 − = 1.03 to 3 s.f. 3 9

b  Replace x with 2x in part a to get 2 3 1 1 1 1 + 2x ≈ 1 + ( 2x ) − ( 2x ) + ( 2x ) 2 8 16 1 1 = 1 + x − x2 + x3 2 2 c  The expansion is valid if −1 < 2x < 1 so –0.5 < x < 0.5 or x < 0.5.



1 2

1= A−

( )

( )

( 94 + 1)

3= A+

5×3×1 2 a The coefficient of x3 is 2 2 2 = 5 . 3! 16 5 × 3 × 1 × −1 2 2 2 2 5 . b The coefficient of x4 is =− 4! 128 1 −2 1 3 3 2 1 1 1 3 x = 1 + x − x2 3 a (1 + x ) ≈ 1 + x + 3 2 3 9 b −1 < x < 1 or x < 1

8

3

3

3

3

4.01 2 = ( 4 × 1.0025) 2 = 4 2 × (1.0025) 2 3

= 8 × (1 + 0.0025) 2

so you need an expansion of

(1 + x )

3 2

( )

3 × 1 × −1 3×1 2 2 2 3 3 2 2 2 =1+ x + x + x + ... 2 2 6 3 3 1 x 3 ... = 1 + x + x2 − + 2 8 16

= (1 + (−x))−2 ≈ 1 − 2(−x) + 3(−x)2 = 1 + 2x +3x2

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1 

WORKED SOLUTIONS



y

So when x = 0.0025 then 3 1 4.01 2 = 8 + 12x + 3x 2 − x 3 + ... 2 = 8 + 0.03 + 0.000 018 75 – .... which is 8.030 02 to 5 d.p. Three terms are enough for this accuracy.



30 25 20 15 10

−1 1 = (1 − 3x ) 2 1 − 3x

9

1

(1 − 3x )− 2

( ) ( )( ) (− 12 )(− 32 )(− 52 )(3x) + 3! 1 (− 2 )(− 32 )(− 52 )(− 72 )(3x) + 4! 1 3 (− 2 )(− 2 )(− 52 )(− 72 )(− 92 )(3x)

5

−1 −3 2 2 1 = 1 + − (3x) + (3x)2 + 2 2

–8

4

5! 3 27x 135x 3 2835 4 15309 5 =1− x + − + x − x 2 8 16 128 256 1

x x2 x3 − − 2 8 16 1  0.0025 0.00252 0.00253  ∴ 3.99 2  2  1 − − − = 1.997498 2 8 16   1.99750 to 5 d.p. [this gives approximation correct to 9 d.p] 1

(1 − x)2 = 1 −

Exam-style questions 1 a |3x + 5| = 4x

Means 3x + 5 = 4x so x = 5



Or – (3x + 5) = 4x

7x = –5 5  x =− 7 However this is not a valid solution since it makes 4x a negative value, and the modulus cannot be negative. b 2|3x + 5| > |4x| |6x + 10| > |4x|

(6x + 10)2 > 16x2



36x2 + 120x + 100 > 16x2



20x2 + 120x + 100 > 0

x2 + 6x + 5 > 0

4

6

8

x

x < –5 or x > –1



f(3) = 0



2(33) + 32 + 3p + 12 = 0



3p = –75

2x2 + 7x – 4 b x − 3 2x3 + x2 – 25x + 12 2x3 − 6x2 7x2 − 25x 7x2 − 21x

10 3.99 2 = ( 4 − 0.01) 2 = 2 (1 − 0.0025) 2

2

p = –25

2

1

0 –2 –5

2 a f(x) = 2x3 + x2 + px + 12

5

1

–4

–10



3



–6

−4x + 12 −4x + 12

0 2x2 + 7x – 4 = (2x – 1)(x + 4)

2x3 + x2 – 25x + 12 = (x– 3)(2x – 1)(x + 4) 3 a p(x) = x3 + 2ax2 + 3ax + 14 p(7) = 0 73 + 2a(72) + 3a(7) + 14 = 0 119a = –357 a = –3 b i p(x) = x3 – 6x2 – 9x + 14 = 0 x2 + x − 2 x − 7 x3 − 6x2 − 9x + 14 x3 − 7x2 x2 − 9x x2 − 7x −2x + 14 −2x + 14

x2

0 + x − 2 = (x + 2)(x − 1)

x3 − 6x2 − 9x + 14 = (x − 7)(x + 2)(x − 1) = 0

So x = −2, 1 or 7

ii Remainder is −20 4 |3 – 2x| < |4x + 3| (3 – 2x)2 < (4x + 3)2

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1 Algebra

9 – 12x + 4x2 < 16x2 + 24x + 9 12x2 + 36x > 0 x(x + 3) > 0 y

7 a f(x) = 4x4 + 8x3 − 21x2 − 18x + 27 f(1) = 4 + 8 − 21 − 18 + 27 = 0 so (x − 1) is a factor of f(x). 4x3 + 12x2 − 9x − 27 (x − 1) 4x4 + 8x3 − 21x2 − 18x + 27 4x4 − 4x3 12x3 − 21x2 12x3 − 12x2

30 25 20 15

−9x2 − 18x −9x2 + 9x

10 5 –8

–6

0

–2 –5

–4

2

4

6

8

–10



−27x + 27 −27x + 27

x

x < –3 or x > 0



5 a p(x) = x4 – 9x2 – ax + 12

p(3) = 0



34 – 9(32) – 3a + 12 = 0



3a = 12

a = 4 b p(x) = x4 – 9x2 – 4x + 12

p(1) = 1 – 9 – 4 + 12 = 0



Hence (x – 1)is a factor of f(x).

4x2    − 9 (x + 3) 4x3 + 12x2 − 9x − 27 4x3 + 12x2 0 −9x − 27 −9x − 27 0 f(x) = 4x4 + 8x3 − 21x2 − 18x + 27 = (x − 1)(x + 3) (2x − 3)(2x + 3) b 6x4 + 2x3 – 10x = 2x4 – 6x3 + 21x2 + 8x – 27

c f(x) = x4 − 9x2 − 4x + 12 = 0

(x – 3)(x –1)(x + 2)2 = 0



x = –2, 1 or 3

0 x4 − 9x2 − 4x + 12 = (x − 1)(4x3 + 12x2 − 9x − 27) g(x) = 4x3 + 12x2 − 9x − 27 g(−3) = −108 + 108 + 27 − 27 = 0 so (x + 3) is a factor of g(x).



4x4 + 8x3 – 21x2 – 18x + 27 = 0



(x – 1)(x + 3)(2x – 3)(2x + 3) = 0 3 3 so x = 1, – 3, or − 2 2

6 a 7 – 3x < |11x – 5|

either 11x – 5 > 7 – 3x

8 Using remainder theorem



m(3)3 – 3(3)2 + 5(3) – 4m = 3(3)3 – 5(3)2 – m(3) + 2m

⇒ 27m – 27 + 15 – 4m = 81 – 45 – 3m + 2m



14x > 12 6 x> 7 or – (11x – 5) > 7 – 3x



⇒ 23m – 12 = 36 – m



– 11x + 5 > 7 – 3x



⇒ 24m = 48





8x < – 2 x < – 1 4 b 7 – 3(y – 1)2 < |11(y – 1)2 – 5|

7 – 3(y2 – 2y + 1) < |11(y2 – 2y + 1) – 5|



7 – 3y2 + 6y – 3 < |11y2 – 22y + 11 – 5|



4 + 6y – 3y2 < |11y2 – 22y + 6|

c x = (y –

1)2

either (y – y >1±

1)2

6 > 7

6 7



or (y – 1)2 < −



no solutions

1 4

⇒m=2 2x – 1 9 4x2 – 2x + 3 8x3 – 8x2 + 8x – 3 8x3 – 4x2 + 6x –4x2 + 2x – 3 –4x2 + 2x – 3

No remainder.

x2 – 3x + 5 10 a 2 x + 2x + 3 x4 – x3 + 2x2 + px + q x4 + 2x3 + 3x2 –3x3 – x2 + px –3x3 – 6x2 – 9x

16

0

5x2 + (p + q)x + q 5x2 + 10x + 15 (p – 1) + (q – 15) = 0

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1 

WORKED SOLUTIONS

(x2 + 2x + 3)(x2 – 3x + 5) = x4 – x3 + 2x2 + x + 15 So p = 1 and q = 15 b (x2 + 2x + 3)(x2 – 3x + 5) = x4 – x3 + 2x2 + x + 15

x2 + 2x + 3

b2 – 4ac = 4 – 4.1.3 < 0 so no solutions x2 – 3x + 5

14

A 6x + 7 B = + (x + 1)2 (x + 1) (x + 1)2 6x + 7 = A(x + 1) + B Substitute x = −1 B=1 Substitute x = 0 A=6 6x + 7 6 1 = + (x + 1)2 (x + 1) (x + 1)2

b2 – 4ac = 9 – 4.1.5 < 0 so no solutions 11 a 2x – 3 2x2 + 13x + 20 4x3 + 20x2 + px – 60 4x3 + 26x2 + 40x – 6x2 + (p – 40)x – 60 – 6x2 – 39x – 60 b p = 1

(p – 1)x

c 4x3 + 20x2 + x – 60 = (2x2 + 13x + 20)(2x – 3) = 0



(2x + 5)(x + 4)(2x – 3) = 0 3 5 So x = , − or − 4. 2 2 y

12 a

25 20 15 10 5 –8

–6

–4

0 –2 –5

2

4

6

8

x

2x + 4  0 can never be negative so 2 − 12 x 2x + 4 =0 2 − 12 x



⇒ 2x + 4 = 0 ⇒ x = −2 (NB x ≠ 4 or denominator = 0)

13

9 A Bx + C ≡ + (x + 1)(x 2 + 2) x + 1 x 2 + 2



9 A(x 2 + 2) + (Bx + C )(x + 1) ≡ 2 (x + 1)(x + 2) (x + 1)(x 2 + 2) 2 9 ≡ A(x + 2) + (Bx + C) (x + 1) Let x = −1. 9 = 3A So A = 3. Let x = 0. 9=6+C So C = 3. Let x = 1. 9 = 9 + (B + 3)(2) 0 = 2B + 6 So B = −3 9 3 −3x + 3 ∴ ≡ + (x + 1)(x 2 + 2) x + 1 x 2 + 2



–10

b

15

11x − 5 A B = + (x − 3)(3x − 2) (x − 3) (3x − 2) 11x − 5 = A(3x − 2) + B(x − 3) Substitute x = 3 7A = 28 A=4 Substitute x = 0 −5 = (4)(−2) − 3B B = −1 11x − 5 4 1 = − (x − 3)(3x − 2) (x − 3) (3x − 2)

16 a (1 − x)−1 = 1 + (−1)(−x) +







(−1)(−2) (−x)2 2!

(−1)(−2)(−3) (−x)3 + …. 3! = 1 + x + x2 + x3… +

b 1 = 2−x

1

( )

2 1− x 2

=

( ) + ( x2 ) + ...

x x 1 1+ + 2  2 2

2

3

1 x x2 x3 + + + + ... 2 4 8 16 c The expansion is valid if x < 1 which means 2 x < 2. =

17 a

16x 2 + 29x + 7 A B C = + + ( x + 4 )( 2x + 1)2 x + 4 2x + 1 ( 2x + 1)2



16x2 + 29x + 7 = A(2x + 1)2 + B(x + 4)(2x + 1) + C(x + 4)



Let x = –4



256 – 116 + 7 = 49A



A=3



Let x = −

1 2

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1 Algebra

x = 0



29 7 +7= C 2 2 C = –1



Let x = 0



7 = 3 + 4B – 4



B=2



16x 2 + 29x + 7 3 2 1 = + − ( x + 4 )( 2x + 1)2 x + 4 2x + 1 ( 2x + 1)2



b

4−

−1

−1

−1

( ) 



3 x x2  = 1 − + 4 16  4





()

2

 

3 3x 3x 2 = − + −… 4 16 64



−1 2 = 2 ( 2x + 1) 2x + 1



 ( −1)( −2)( 2x )2  2 ( 2x + 1) = 2 1 + ( −1)( 2x ) +  2   2      = 2[1 – 2x + 4x ] −1



     = 2 – 4x + 8x2 – … −



A + B = 5(2)

(1) + (2)



3A = 3





( )

−2 1 2 = −1( 2x + 1) 2 x + 1 ( )

 ( −2)( −3)( 2x )2  −1( 2x + 1) = −1 1 + ( −2)( 2x ) +  2       = – 1[1 – 4x + 12x2] −2

    = – 1 + 4x – 12x2 3 + 2 − 1 x + 4 2x + 1 ( 2x + 1)2

b

 = 7 − 3 x − 253 x 2... … 4 16 64 x2 + 1 A B C = + + ( x − 2)3 x − 2 ( x − 2)2 ( x − 2)3

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

( x2 )

−1



( x − 2)−1 =  −2 1 −



 −2 1 − x   2 



x 1 − 1− 2 2



1 x x2  = − 1 + + 2 4  2      



( )

( )

−1

−1

= ( −2)

=−



4 x 2 = 1− 2 ( x − 2)



=1+ x +

−1

( −2)( −3)

 

2

(− x2 )

2

3x 2 … 4

( )

−3





 ( −3)( −4 ) − x2 5 x − 1 + ( −3) − + 8 2 2  



5 3x 3x 2  = − 1 + + 8 2 2 

( )

=−

2

−2

5 5 x =− 1− 8 2 ( x − 2 )2



( ) 

1 x x2 − − … 2 4 8

( )

( x2 ) +



(1 − x2 )

( )

     

1 + ( −2) −

−1

 ( −1)( −2) − x2 x 1 = − 1 + ( −1) − + 2 2 2  



2 = 3 − 3x + 3x + 2 − 4x + 8x 2 − 1 + 4x − 12x 2 4 16 64

18 a

10 = A + B + 5

B = 4

 ( −1)( −2) x4 3 x = 1 + ( −1) + 4 4 2  

−1

2A – B = –2 (1)

A = 1

( ) 3 x x 3  4 (1 + )  = (1 + ) 4  4 4  3 x 1+ 4 4



x = 3  4 1 +  4  

−1

1 = 4A – 2B + 5

x = 3

−1 3 = 3( x + 4) x+4

3( x + 4)



( )  2

 

5 15x 15x 2 − − … 8 16 16

x2 + 1 = A(x – 2)2 + B(x – 2) + C x = 2

5=C

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1

WORKED SOLUTIONS

1 4 5 + + x − 2 ( x − 2)2 ( x − 2)3 =−

∴

3x 2 5 15x 15x 2 1 x x2 − − +1+ x + − − − 2 4 8 4 8 16 16

b

1 3 5 2 =− − x− x 8 16 16 x c Expansion valid if 2 < 1 which means |x| < 2. 19 a

4 − 8x = 4(1 − 2x) = 2(1 −

( )

1 2x)2

( )( )

1 −1 −3 1 −1     2 2 2 2 2 1 (−2x)3  (−2x)2 + ≈ 2 1 + (−2x) + 6 2 2     2 3 = 2 − 2x − x − x 1 b The expansion is valid if 2x < 1 so x < 2 7x − 5 9 b 20 ≡− + x−a x−3 ( x − a )( x − 3) 7x – 5 = –9(x – 3) + b(x – a) x=3 16 = b(3 – a) x=0 – 5 = 27 – ab – 32 = – ab 32 b= a 32 16 = (3 − a ) a 96 16 = − 32 a a=2 b = 16 3x 2 + 4x + 1 A Bx + C ≡ + 21 a (x − 1)(x 2 + 2) x − 1 x 2 + 2

3x 2 + 4x + 1 8 x + 13 = + ( x − 1) x 2 + 2 3 ( x − 1) 3 x 2 + 2 −1 1 = − (1 − x ) valid for x < 1 ( x − 1)

(

(

)

(

 1 x2  = 21 +  2 2  x +2

−1

)

valid for

)

x2 0, 1 + a > 1 so log (1 + a ) > 0

d 3 = 9 2 1



1

log 9 9 4 =

1 4

3 2 loga 4 = loga 16 loga 8 − loga 2 = loga 4

log a

1 1 − log a = loga 3 2 6

3 loga 2 = loga 8

4 a 7

1 = − log2 32 = −5 32 1 8 d log2 256 = 3 3

b log 2

1 c log2 512 = 4.5 2

5 a log3 (3 × 5) = log3 3 + log3 5 = 1 + c 1 = log3 1 − log3 5 = −c 5

c log3 53 = 3c

1

–1

1

3 = 92 = 94



b log 3



x

1

Exercise 2.2A

11 a The coordinates of P and Q are (a, log a) and ((a + 1), log (a + 1))

0 –1

( )

9 a If y = log a then 10y = a; hence ak = (10y)k = 10yk using the law of indices. Hence log ak = yk and from the definition of y, log ak = k log a log 216 log 63 3log 6 3 = = = = 1.5 b log 36 log 6 2 2log 6 2

y = log x

1

c log 6 = log (2 × 3) = log 2 + log 3

1

x

c

1c  1 d 5 = 3c =  9 2  = 9 2  



log9 5 =

1 c 2

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LOGARITHMS AND EXPONENTIAL FUNCTIONS

6

Exercise 2.3A

a n3 = 216 so n = 6 b

1 n2

= 4 so n = 4² = 16

1

a 7.39 d 6.57

2

a To 5 d.p. the values are: i 2.704 81 ii 2.718 15 iii 2.718 28 b The answers increase and approach the value of e as n increases.

c n0 = 1 so n can be any positive integer. d n1.5 = 27 n × n = 27 n=9 7

a = n3.5 b=

8

n4.5

n1 + 3.5 =

=

a i b=

ax

b In a =

by

n×

n3.5

3

= na

b 1.65 e 0.0498

c The limit is e a y

ii a = by

7 6

from ii above

5

substitute for b using i: a = (ax ) y = a xy This means that xy = 1 and the result follows. 9

a 16 =

24

so log2 16 = 4; 8 =

4

y = e–x

2

4

1

b If x = logc a then c x = a; if y = logc b then c y = b (c y)z =

–4

10 a (x – 1)5 = 1024 = 45; hence x – 1 = 4 and x = 5

–3

–2

–1

4 5

b logx 9x = 3 = 9x(x ≠ 0)

y = ex+1 is A.

= 9 so x = 3

y = (e + 1)x is B.

11 Add the two equations: 2 log4 x = 9; log 4 x = 9 ; 2 9 9 2 x = 4 = 2 = 512

3

4

x

 −1 b y = ex + 1 is A and a translation of   .  0

3 Subtract the two equations: 2 log4 y = 3; log 4 y = 2 ;

 0 y = ex + 1is C and a translation of   . 1 

3

y = 4 2 = 23 = 8 x a2

6

c A is y = ex + 1 = e × ex so it is a stretch of y = ex by a factor of e. a If x = 0, y = ex + 2 = e2. The intercept is (0, e2). 2 b A translation of   will map y = ex + 4 onto 0  

log2 x + log4 x = 12; hence log 2 x + 1 log 2 x = 12 ; 3 log 2 x = 12; 2 2 log2 x = 8; x = 28 = 256 1 c log 8 x = log 3 x = 3 log 2 x; 2 log 2 x + 1 log 2 x + 1 log 2 x = 22 ; 11 log 2 x = 22; 2 3 6 log 2 x = 6 × 22 = 12 ; x = 212 = 4096 11

2

y = ex + 1 is C.

x2

then (a2)y = x; a2y = x; hence 1 2y = loga x and y = log a x 2 b From part a, log 4 x = log 2 x = 1 log 2 x ; 2 2

1

Gradient = ex = y a e0 = 1 b e2 c 2 a One method is to find where each curve crosses the y-axis.

x3

12 a If y = log

0

b Reflection in the y-axis

cy z =

If z = logb a then a; a; a; log c a log c a = which is the yz = logc a; z = y log c b required result.

y = ex

3

23 so log2 8 = 3

4  1 log 216 16 = 24 =  8 3  = 8 3 so log 8 16 = 4 = 3 log 2 8  

b z=

c 0.368

y = e(x − 2) + 4 which is y = ex + 2. 7

a Correct graphs sketched

()

2 x

8 9

b e0.5x = e 4 An x-direction stretch of stretch factor 4. x = −e, e 2 a Using the chain rule, if y = e0.5x + a then dy dy = e0.5x + a × 0.5 = 0.5y ; hence 2 =y dx dx b At A x = 0 and y = ea so A has coordinates (0, ea)

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2

WORKED SOLUTIONS

dy = 0.5y = 1 e a and this is the gradient of 2 dx the tangent. 1 The equation of the tangent is y − e a = 2 e ax ; if y = 0 then −e a = 1 e ax ; x = –2 2 The point is (−2, 0)

c At A,

Exercise 2.4A 1

11 a Using the chain rule f'(x) =

4e2x +1 ×

2

x = 3.51 c 2x − 5 = ln 125 x= d −

5 + ln125 = 4.91 2

1 2 x = ln 0.5 2

x = ± −2 ln 0.5 = ±1.18 3

a 0.5x = e4 x = 109 b 4x + 2 = e3.5

b At P, x = −1 and f(–1) = 4e–2 +1 = 4e–1; the point is (–1, 4e–1)

e3.5 − 2 = 7.78 4 c x = 1 e−4 = 0.009 16 2 120 d x = 5 = 0.809 e 1 ln = ln 1 − ln a = −ln a because ln 1 = 0. a a y = ln 4x = ln 4 + ln x x=

c f'(–1) = 2f (–1) = 2 × 4e–1 = 8e–1 The gradient of the tangent is 8e–1 and the equation is y – 4e–1 = 8e–1 (x + 1)

– 4 = 8x + 8; 8x = –12; x = –1.5 and the point is (–1.5, 0)

a x = ln 2000 = 7.60 b −x = ln 0.03

2 = 2f (x)

Where this meets the x-axis, y = 0 and – 4e–1 = 8e–1 (x + 1); – 4 = 8 (x + 1);

b 4 × ln a = 12 d 1 (ln a + ln b) = 3.75 2

c ln b − 2 × ln a = −1.5

10 a f'(x) = ex = y and this is the gradient of the tangent at (x, y) At P the gradient of the tangent is y so y = y x x and hence x = 1; P is (1, e) b At P the gradient of the tangent is e so the gradient of the normal at P is − 1 e 1 The equation of the normal is y − e = − e ( x − 1); ey – e2 = – x + 1; x + ey = 1 + e2

a ln a + ln b = 7.5

4 5

12

 0  The graph is a translation of y = ln x by  .  ln 4

b

y y = ln 4x

3

A

2 ln 4 1

T

–3

N

–2

y = ln x

0 –1 –1

1

2

3

4

5

6

7

8

x

–2

dy = y so the gradient of the tangent at A is ea dx The equation of the tangent is y – ea = ea(x – a); where it crosses the x-axis y = 0 and so – ea = ea(x – a); –1= x – a; x = a –1 The gradient of the normal is − 1a = −e −a and the e equation is y – ea = – e–a(x – a) where it crosses the x-axis y = 0 and so – ea = – e–a(x – a); e2a = x – a; x = a + e2a The base of the triangle TN = a + e2a – (a – 1) = e2a + 1; the height is ea The area = 1 × e 2a + 1 × e a = 1 e a(e 2a + 1) 2 2

(

)

–3

6

a Either ln 20 + t = ln 100 or t = ln

100 = ln 5. 20

t = 1.609 b −t = ln

35 40

t = 0.134 c 3t = ln

8000 250

t = 1.155 d −0.85t = ln

14.8 32.5

t = ln 0.4554 = 0.925 −0.85

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LOGARITHMS AND EXPONENTIAL FUNCTIONS

7

a

12 a f'(x) = 2ex – e–x. At a stationary point, f'(x) = 0 so 2ex – e–x = 0; 2ex = e–x; 1 1 2e2x = 1; e 2x = 2 ; 2x = ln 2 = − ln 2 ; so x = − 1 ln 2 at a stationary point. 2 1 ln 2 − 1 ln 2 x = − 1 ln 2, y = 2e 2 + e 2 If 21 1 − = 2 e ln 2 2 + e ln 2 2 = 2 × 1 + 2 2 = 2+ 2=2 2

y y = 6e–x

7 6 5

y = 2ex

4 3

( )

2 1 –4

–3

–2

–1

0

1

2

3

4

5

6

x

b f"(x) = 2ex + e–x, which is positive for all values of x and in particular if x = − 1 ln 2 so 2 the value is a minimum.

b Where they cross, 2ex = 6e−x. e2x = 3 2x = ln 3 x = ln 3 = 0.5493; 2 y = 2e0.5493 = 3.464 Coordinates, to 3 s.f., are (0.549, 3.46). 8

a B is (ln 2, 2) b The gradient is the y-coordinate = 2. c The gradient is

9

1 . 2

1 d The gradient of y = ln x at (x, ln x) is or you x dy 1 = could say dx x a y2 + y = 6 and this is a quadratic equation in y.

13 a If 10x ≡ ecx then ln 10x ≡ ln ecx; x ln 10 ≡ cx so c = ln 10 or 2.30 to 3 s.f. dy (ln10)x b If y = 10x = e (ln 10) x then dx = ( ln10 ) e = ( ln10 ) × 10 x c If (ln 10) × 10x = 10 then 10 x = 10 and ln10 10 x = log = log10 − log ( ln10 ) ln10

( )

= 1 – log(ln 10) or 0.638 to 3 d.p. 14 a At Q, 0.2e0.5x = a; e0.5x = 5a; 0.5x = ln 5a; x = 2 ln 5a b

(y + 3)(y − 2) = 0

y – a = 0.5a(x – 2 ln 5a)

y = −3 or 2

Where it crosses the x-axis, y = 0 and – a = 0.5a(x – 2 ln 5a); – 2 = x – 2 ln 5a;

c Either ex = −3 or ex = 2. ex = −3 has no solution because ex is always positive.

x = 2 ln 5a – 2; hence 2 ln 5a – 2 = 6; 4 2 ln 5a = 8; ln 5a = 4; 5a = e4; a = e 5

If ex = 2, then x = ln 2 = 0.693 to 3 s.f.

Exercise 2.5A The logarithms in these solutions can be to any base. 1

a x log 3 = log 11 x=

log11 = 2.183 to 3 d.p. log 3

b x log 4 = log 175

11 a ln (e + e2) = ln (e(1 + e)) = ln e + ln (1 + e) = 1 + ln (1 + e)

x=

log175 = 3.726 to 3 d.p. log 4

c x log 12 < log 6

b ln (e2 − e4) = ln (e2(1 − e2)) = ln e2(1 + e)(1 − e)

x
123 = 1.2947 95

t>

d The multiplier is now 1.2.





1.2t = 8



t log 1.2 = log 8

log 8 t= =11.4 log1.2 11.4 weeks

e Changes in temperature and light conditions can change the growth rate.

Growth will not continue when the pond is fully covered.

7 a 1.65 × 1.05 = 1.7325 million

1.7325 × 1.05 = 1.82 million (to 3 s.f.)

b 1.65 × 1.05t  million c When 1.65 × 1.05t = 2.5

log1.2947 = 6.586 log1.04

4 a If it takes t years to increase by 50%, 1.09t = 1.5.

When 10 × 1.2t = 80





1.05t =



t=

2.5 = 1.51515 1.65 log1.51515 = 8.52 log1.05

8.5 years



log1.5 t= = 4.70 log1.09

d The number cannot increase beyond the capacity of the airport.



4.7 years



b If it takes t years to double, 1.09t = 2. log 2 = 8.04 log1.09



t=



Just over 8 years

5 a (x + 2) log 4 = log 90

log 90 x= − 2 =1.246 log 4

b (2x + 1) log 6 > log 35

log 35 2x + 1 > =1.9843 log 6

1.9843 − 1 x> and hence x > 0.492. 2 c (4x − 3) log 15 = log 8

4x − 3 =





log 8 log15

x = 0.942

6 a y = 10 ×

1.5t

b 10 × 1.53 = 33.75 m² c When 10 × 1.5t = 80

1.5t = 8



t log 1.5 = log 8



5.1 weeks

t=

log 8 = 5.1 log1.5

Changes in travel patterns could make the model incorrect.

8 a log 5x + 3 = log 7x − 1

Hence (x + 3) log 5 = (x − 1) log 7.



Rearrange: 3 log 5 + log 7 = x log 7 − x log 5.

3 log 5 + log 7 = 20.1 to 3 s.f. log 7 − log 5 b log 3x + 2 > log 4x

Hence x =



Hence (x + 2) log 3 > x log 4.



Rearrange: 2 log 3 > x (log 4 − log 3).



Hence x
ln 3; x ln 4.5 > ln 3; x > ln 3 ln 4.5 6 42x – 1 = 53x + 1; (2x – 1) ln 4 = (3x + 1) ln 5; 2x ln 4 – ln 4 = 3x ln 5 + ln 5 3x ln 5 – 2x ln 4 = – ln 4 – ln 5; x (ln 125 – ln 16) = – ln 20; x ln 125 = − ln 20 16 x = − ln 20 = –1.46 to 3 s.f. ln 7.8125 35 + 8x 7 a  = ln 3; ln(35 + 8x) − 2 ln x = ln 3; ln x2 35 + 8x = 3 2 ; 3x – 8x – 35 = 0 x2 (3x + 7) (x – 5) = 0; x = − 7 or 5 3 e 2x +1 = − 7 or 5; it cannot be negative and so b  3 e2x+1 = 5; 2x + 1 = ln 5 x = ln 5 − 1 = 0.305 to 3 s.f. 2 8 a When x = 0, y = 375

375 =

Ae0

=A

b When x = 400, y = 945

945 = 375e400k



e400k =



k=

945 = 2.52 375

ln 2.52 = 0.002 31 400

c If y = 540 then 375e0.002 31x = 540;

2x3 – 15x2 + 31x – 12 = (x – 3) (2x2 – 9x + 4)  = (x – 3) (x – 4) (2x – 1) c If f(x) = 0 then x = 3, 4 or 1 2

1



ln 375 + 0.002 31 x = ln 540 ln 540 − ln 375 x= = 158 to 3 s.f. 0.002 31

So x – 3 is a factor;

11 6 2x − 3 = 3 × 2x +1 ; divide by 6 to get 2x − 3 = 2x

Hence 2x – 3 = ± 2x; the positive sign has no solution hence 2x – 3 = – 2x

2 × 2x = 3; 2x + 1 = 3; (x + 1) ln 2 = ln 3; x = ln 3 − 1 ln 2 12 log y = log k + t log a A graph of log y against t crosses the log y-axis at log k so log k = 3.30 k = 103.3 = 2000 to 3 s.f. The gradient is log a. 4.65 − 3.30 log a = = 0.45 3

a = 100.45 = 2.82 to 3 s.f. 13 a Multiply both sides by ex: e2x = 4 2x = ln 4 x = ln 4 = 0.693 to 3 s.f. 2 b Multiply both sides by ex.

e2x − 3ex − 4 = 0



This is a quadratic in ex.



Factorise.



(ex − 4)(ex + 1) = 0



ex = 4 or ex = −1.



The second has no solution.



If ex = 4 then x = ln 4 or 1.39 to 3 s.f.

14 22x + 8 = 8 × 2x Write y = 2x and then y2 − 8y + 8 = 0. Solve with the quadratic formula: y = 8 ± 64 − 4 × 8 = 8 ± 32 . 2 2 y = 6.828 or 1.172 Either 2x = 6.828 or 2x = 1.172.

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2 

WORKED SOLUTIONS



Take logs: either x = x=

1 b ln s = 2 ln a + 0.2ln t ; in particular,

log 6.828 log 2 = 2.77 or

log1.172 = 0.228. log 2

15 Multiply by 3x: 32x + 4 = 5 × 3x; 32x – 5 × 3x + 4 = 0; this is a quadratic in 3x;

(

)



40 1 − 0.975n > 1599; 1 − 0.975 1 − 0.975n > 1599 × 0.025 = 0.999 375 40



0.975n < 0.000 625; n log 0.975 < log 0.000 625



log0.000 625 = 291.4; log0.975 if S∞ – Sn < 1 then n ⩾ 292



2.364 − 0.2 × 2.303 = 1 ln a ; ln a = 3.807; 2



ln t = 3.3081; t = e3.3081 = 27.3

20 a 2 log2 (x + 4) – log2 x = 5; log 2

(x + 4)2 = 5 ; x

(x + 4)2 = 25 ; x2 + 8x + 16 = 32x; x x2 – 24x + 16 = 0 2 b x = 24 ± 24 − 64 = 24 ± 22.627 = 23.3 or 2 2 0.686 to 3 s.f.

a = 40 = 1600 1 − r 1 − 0.975

c If 1600 – Sn  1599; hence



0.4 ln t = ln 3.7556;

39 17 a  The common ratio is 40 = 0.975; the third term is 39 × 0.975 = 38.025 b S∞ =

2.364 = 1 ln a + 0.2 × 2.303 2

a = e3.807 = 45 132 0.4 c s2 = 45 × t0.4; if s = 13 then t = 45 = 3.7556 ;

(3x – 1)(3x – 4) = 0; either 3x = 1 or 3x = 4; if 3x = 1 then x = 0; log 4 if 3x = 4 then x log 3 = log 4; x = = 1.26 to 3 s.f. log 3

16 3x + 3x + 2 = 3x + 1 + 20; 3x + 9 × 3x = 3 × 3x + 20; 20 ; 7 × 3x = 20; 3x = 7 log 20 20 7 = 0.956 to 3 s.f. x log 3 = log ; x = 7 log 3



Mathematics in life and work 1 The population is multiplied by 1.015 every year. If it increases by 20% in n years then 1.015n = 1.2 Take logarithms: log 1.015n = log 1.2. n log 1.015 = log 1.2 log1.2 n = log1.015 =12.25 so it will be just over 12 years.

n>

18 a At P, 1 e 2x = 4; e2x = 16; 2x = ln 16; 4 x = 1 ln16 = 1 ln 4 2 = ln 4 ; 2 2

2 If the annual rate of growth is r % then you want 20 r 1+ < 1.1. 100





Take logarithms: 20 log  1 +



Hence log 1 +



1+



The coordinates are (ln 4, 4)

b T  he gradient of the tangent is 8; the equation is y – 4 = 8(x – ln 4)

or y = 8x + 4 – 8 ln 4

c Where the tangent meets the y-axis, x = 0 and y = 4 – 8 ln 4 (this is a negative number)



Where the tangent meets the x-axis, y = 0 1 and 0 = 8x + 4 – 8 ln 4 so x = ln 4 – 2 1 The area of the triangle = × ( 8 ln 4 − 4 ) × 2 1 2 ln 4 − 2 = (2 ln 4 – 1) 2 16 = (2 ln 4 – ln e)2 = ln e

(

)

( )

(

)

(

)

(

)

r < log 1.1. 100

log1.1 r < = 0.002 070. 20 100

r < 100.002 070 = 1.004 78 100 Hence r < 100(1.004 78 − 1) = 0.478. The annual growth rate must be less than 0.478%. If ln is used the numbers in the calculation are r ln1.1 < = 0.004 766 and ln 1 + 20 100



(



)

1 + r < e0.004 766 = 1.004 78 which gives the 100 same answer.

1 1 19 a 2 ln s = ln a + n ln t so ln s = 2 ln a + 2 n ln t

1 A graph of ln t against ln s has a gradient of 2 n 1 n = 2.722 − 2.364 = 0.2; n = 0.4 2 4.094 − 2.303

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TRIGONOMETRY

3 Trigonometry Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1

sin 600° = sin (360 + 240)° = sin 240° = – sin 60°

2

=− 3 2 The amplitude is 3 and the period is 360 ÷ 2 = 180°.

6

 1 1  + cos A × = 2  sin A ×   2 2 = sin A + cos A = left-hand side. π + 2 sin x + π = sin x cos π − cos x sin π a sin x − 6 6 6 6

( )

( )

+ 2 sin x cos π + 2 cos x sin π 6 6 π π 3 1 = 3sin x cos + cos x sin = 3sin x × + cos x × 6 6 2 2

y 3 2 1

=

0

x

1 3 3 sin x + cos x 2 2

π then x − π = π and x + π = π + π = 5π 6 12 6 4 6 12 4 –2 1 +1× 1 =3 3 π 5 3 3 π 1 π 3 3 Hence sin 12 + 2 sin 12 = 2 sin 4 + 2 cos 4 = 2 × 2 2 2 2 –3 + π 5 3 3 π 1 π 3 3 1 1 1 3 3 1 = = sin + cos = × + × 3 cos (x + 0.3) = 0.4 hence x + 0.3 = ±1.159sin + 2 sin 12 12 2 4 2 4 2 2 2 2 2 2 Hence x = 0.859 or – 1.46 to 3 s.f. sin A sin B sin A cos B + cos A sin B 7 a Left-hand side = tan A + tan B = cos A + cos B = cos A cos B 2 2 sin x sin x = 1 − cos x sin A sin B sin A cos B + cos A sin B sin( A + B) 4 sin x tan x = sin x × cos x = cos x tan x B= Acos + tan + = = cos A cos B cos A cos B cos A cos B ( 1 + cos x )( 1 − cos x ) = = right-hand side. cos x b tan A − tan B = tan A + tan (−B) because tan (−B) = −tan B Exercise 3.1A 90

–1

1

270

360

b If x =

cos 75° = cos (45° + 30°) = cos 45° cos 30° – sin 45° sin 30° =

2

180

3 −1 1 3 1 1 × − × = 2 2 2 2 2 2

8

π radians = 15° and hence 12 π = tan 15° tan 12

= 2 sin A cos B = right-hand side. b Let A + B = x and A − B = y.

tan 45 − tan 30 = tan (45° – 30°) = 1 + tan 45 tan 30 =

1− 1+

1 3 1 3

3

The expression is sin (2A + 3A) = sin 5A.

4

3 The expression is cos (10° + 20°) = cos 30° = 2 π Right-hand side = 2 sin A + 4 π π = 2 sin A cos + cos A sin 4 4

5

(

30

Add the equations to get 2A = x + y. x+y Hence A = . 2 Subtract the equations to get 2B = x − y. x−y Hence B = . 2 Substitute these expressions into the identity in part a and the result follows.

3 −1 3 +1

=

(

Hence, from part a, tan A + tan (−B) = sin( A − B) = sin( A − B) because cos A cos(−B) cos A cos B cos (−B) = cos B a Left-hand side = sin (A + B) + sin (A − B) = (sin A cos B + cos A sin B) + (sin A cos B − cos A sin B)

)

)

9

a tan (A + B) =

tan A + tan B = tan π = 1 1 − tan A tan B 4

Hence tan A + tan B = 1− tan A tan B.

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3 

WORKED SOLUTIONS



Rearrange: tan A + tan A tan B = 1 − tan B.



Factorise: tan A (1 + tan B) = 1 − tan B.



Hence tan A =

1 − tan B . 1 + tan B

1 − tan A . 1 + tan A 10 a cos (A + B) – cos (A – B) = cos A cos B – sin A sin B – (cos A cos B + sin A sin B ) = cos A cos B – sin A sin B – cos A cos B – sin A sin B = – 2 sin A sin B b Simply swap A and B to get tan B =

b Write A + B = α and A – B = β so that cos (A + B) – cos (A – B) ≡ cos α – cos β;

11 a sin ( A − B ) = sin A cos B − cos A sin B cos A cos B cos A cos B sin A sin B = − = tan A − tan B cos A cos B

(

12 a  sin x +



Rearrange as t2 – 4t + 1 = 0; 4 ± 16 − 4 4 ± 12 t= = =2± 3 2 2 (or completing the square: (t – 2)2 – 4 + 1 = 0; (t – 2)2 = 3; this gives the same result)





So

(



)

π π π = sin x cos + cos x sin = 6 6 6



Hence cos2  ≡



cos15° is positive so take the positive square root: cos15° =

)

π π π = sin x cos + cos x sin 3 3 3

(

)

Hence tan x =

(

)

3 − 1 cos x 3 −1 = 3 +1

3 −1 × 3 +1

3 −1 3 −1







(cos 40° – sin 20°) sin θ = (cos 20 °+ sin 40°) sin θ cos20° + sin 40° cos θ; = = tan θ cosθ cos40° − sin 20° Hence tan θ = 3.732 and θ = 75° or 255° tan x + tan 45° = 2; 1 − tan x tan 45° tan x + 1 = 2; tan x + 1 − tan x

b tan x +

2 sin2  = 1 − cos 2 sin 2 θ =

1 (1 − cos 2θ ) 2

b If θ = π radians then 8

(

(

)

    sin 2 π = 1 1 − cos π = 1  1 − 1  = 1  1 − 2  = 2 − 2 4 8 2 4 2 2  2 2

)

    sin 2 π = 1 1 − cos π = 1  1 − 1  = 1  1 − 2  = 2 − 2 . 8 2 4 2 2 2 4    2    

3− 2 3 +1 = 2− 3 = 3−1 13 a sin θ cos 40° – cosθ sin 40° = cosθ cos 20° + sinθ sin 20°

2+ 3 . 2

3 a cos 2 = 1 − 2 sin2 

  3 1 3 1  1 + 2 − 2  sin x =  2 − 2  cos x or 3 + 1 sin x =

Rearrange: 2cos2  ≡ 1 + cos 2.

1 (1 + cos 2). 2 b If  = 15° then the identity becomes cos2 15° = 1 (1 + cos 30°). 2 1 3 2 + 3 2 . = So cos 15° =  1 + 4 2 2 

= 1 sin x + 3 cos x 2 2 3 1 1 3 Hence sin x + 2 sin x + 2 cos x = 2 sin x + 2 cos x ;

1 x x sin x ≡ sin cos . 2 2 2

2 a cos 2 ≡ 2cos2  − 1

1 3 1 3 sin x + cos x so a = and b = 2 2 2 2 b sin x +

If tan x = 2 + 3 then x = 75° or 255°; if tan x = 2 − 3 then x = 15° or 195°

1 sin 2A ≡ 2 sin A cos A x x x Let x = 2A so = A and sin x ≡ 2sin cos . 2 2 2

sin ( A − B ) sin ( B − C ) sin (C − A ) + + cos A cos B cos B cos C cos C cos A = tan A – tan B + tan B – tan C + tan C – tan A = 0

b

tan x (1 – tan x) + tan x + 1 = 2(1 – tan x); writing t = tan x, t – t2 + t + 1 = 2 – 2t

Exercise 3.2A

then 2A = α + β and 2B = α – β α +β α −β Hence A = and B = ; substitute to 2 2 α +β α −β get cosα − cos β = −2sin  2  sin  2     





2− 2 . Hence, taking the positive root, sin π = 8 2 4 Use the double angle formula: 2 sin  cos  = 1.5 sin . Either sin  = 0 or 2 cos  = 1.5. If sin  = 0 then  = 180°. If 2 cos  = 1.5 then cos  = 0.75 and  = 41.4° or 360° – 41.4° = 318.6°. The three possible solutions in the given domain are 41.4°, 180° and 318.6°.

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TRIGONOMETRY

5

2 cos2 x = 1 + cos 2x so 2 cos3 x = cos x + cos x cos 2x Hence right-hand side =

6

7

1 (sin x sin 2x + 2 cos3 x) 2

1 = (sin x sin 2x + cos x + cos x cos 2x) 2 1 = (cos x + cos (2x − x)) = cos x = left-hand side. 2 Left-hand side = cos 2 + 2 cos  + 1 = 2 cos2  − 1 + 2 cos  + 1 = 2 cos2  + 2 cos  = 2 cos  (cos  + 1) = right-hand side. 1 2× 1 2tan θ 4 = 2 = 1 × 16 = 8 a tan 2θ ≡ 2 = 15 2 15 15 1 − tan θ 1 − 1 16 16 b If tan  = t then 2 =

2t . 1 − t2

Rearrange: 2(1 − t2) = 2t and hence t2 + t − 1 = 0. −1 ± 1 + 4 . 2 t is positive so take the positive root: −1 + 5 tan θ = . 2 The quadratic formula gives t =

8

2t = 4t . 1 − t2

Write tan  = t and then

Rearrange: 2t = 4t(1 − t2). t cannot be 0 for the given range of  and so you can divide by it: 2 = 4(1 − t2).

9

1 and so  = 0.615 to 3 d.p. 2

Hence t 2 = 1 and t = 2 a y 4 3 2 1 –1

0 –1 1

45

90

135

180

x

–2 –3 –4

b 3(2 cos2  − 1) = cos  − 2 Write c = cos  and then 6c2 − 3 = c − 2. Rearrange: 6c2 − c – 1 = 0. Factorise: (3c + 1)(2c − 1) = 0. Hence either 3c + 1 = 0 or 2c − 1 = 0. If 3c + 1 = 0 then c = − If 2c − 1 = 0 then c =

1 and  = 109.5°. 3

1 and  = 60°. 2

10 a If x = π then 2x = π so tan 2x = 1 4 8 tan 2x =

2tan x =1 1 − tan 2 x

Hence 2 tan x = 1 − tan2 x. b tan2 x + 2 tan x − 1 = 0 Use the quadratic formula: −2 ± 22 + 4 −2 ± 8 = . 2 2 Since tan x is positive, take the positive root: tan x =

tan π = −2 + 8 = −2 + 2 2 = 2 − 1 8 2 2 11 left-hand side = sin (2A + A) = sin 2A cos A + cos 2A sin A = 2 sin A cos2 A + (1 − 2 sin2 A) sin A = 2 sin A (1 − sin2 A) + sin A − 2 sin3 A = 2 sin A − 2 sin3 A + sin A − 2 sin3 A = 3 sin A − 4 sin3 A = right-hand side. 12 a sin 2x ≡ 2 sin x cos x so 4 sin x cos x = cos x; cos x (4 sin x – 1) = 0; 1 either cos x = 0 or sin x = ; hence x = 90° or 4 14.5° or 165.5° b 2(1 – 2 sin2 x) = sin x + 1 4 sin2 x + sin x – 1 = 0 x = 22.9˚ or 157.0˚ 13 a cos 4A = cos 2(2A) = 2 cos2 2A – 1 = 2(2 cos2 A – 1)2 – 1 = 2(4 cos4 A – 4 cos2 A + 1) – 1 = 8 cos4 A – 8 cos2 A + 1 π 3π 5π b If x = cos A then cos 4A = 0; 4 A = , , 2 2 2 7π are solutions; and 2 π 3π 5π 7π A= , , or ; x = cos A = 0.924, 0.383, 8 8 8 8 −0.383 or −0.924 (In degrees, 4A = 90°, 270°, 450° or 630° give the same solutions) tan A + tan 2A 14 a tan 3A = tan( A + 2A) = 1 − tan A tan 2A 2 tan A + 2tan A 1 − tan 2 A = tan A 1 − tan A + 2tan A = 2 2 1 − tan A − 2tan 2 A 1 − 2tan 2A 1 − tan A

(

=

)

3tan A − tan 3 A 1 − 3tan 2 A

b Rearrange as 1 – 3 tan2 A = 3 tan A – tan3 A; 3tan A − tan 3 A = 1 ; tan 3A = 1 1 − 3tan 2 A 3A = 45° or 225° or 405° A = 15°, 75°, 135° are the three solutions

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3

WORKED SOLUTIONS

15 a tan ( A + 45° ) = tan A + tan 45° ; tan 45° = 1 so 1 − tan A tan 45° 1 + tan A tan ( A + 45° ) ≡ 1 − tan A

b 5 sin( + 53.1°) = 5 hence sin( + 53.1°) = 1 Hence  + 53.1° = 90° and so  = 36.9°. 3

a cos  + sin  ≡ r sin  cos α + r cos  sin α

2tan x 1 + tan x − 2; b = 1 − tan 2 x 1 − tan x 2 tan x = (1 + tan x)2 – 2(1 – tan2 x );

Therefore r cos α = 1 and r sin α = 1 r = 2 and tan α = 1 so α = π . 4

(

2 tan x = 1 + 2 tan x + tan2 x – 2 + 2 tan2 x; 1 3 tan2 x = 1; tan x = ± ; 3 x = 30° or 150° 2tan A A 2 = 2t 16 a tan A = tan 2 × = 2 2 1 − tan 2 A 1 − t 2

(

b 2t =

=

sin  + cos  = 2 sin θ + π 4

b cos  − 7 sin  ≡ r cos  cos α − r sin  sin α Therefore r cos α = 1 and r sin α = 7.

)

 sin A  2sin A 2 and 1 + t 2 = 1 +  2  cos A  cos A 2 2

r = 12 + 7 2 = 50 and tan α = 7 so α = 81.9°. cos  − 7 sin  = 50 cos ( + 81.9°).

2

cos2 A + sin 2 A 1 2 2 = 2 A 2 A cos cos 2 2

c

4

Therefore r cos α = 8 and r sin α = 6. 6 r = 8 2 + 6 2 = 10 and tan α = 8 so α = 0.644 radians or 36.9° 8 sin x – 6 cos x = 10 sin ( − 0.644) using radians, or 10 sin ( − 36.9°). b The smallest possible value is when the sine is –1 to give a minimum of –10.

sin A 2t 2t 1 − t2 = ÷ = 2 2 tan A 1 + t 1−t 1 + t2 5

Exercise 3.3A 1

a 12 sin  + 5 cos  ≡ r sin  cos α + r cos  sin α so r cos α = 12 and r sin α = 5

a cos  + 3 sin  ≡ r sin  cos α + r cos  sin α

r = 122 + 52 = 13 and tan α =

Therefore r sin α = 1 and r cos α = 3 .

α = 0.395.

1 r = 1 + 3 = 2 and tan α = so α = π 6 3

(

π cos  + 3 sin  = 2 sin θ + 6 b

5 so 12

12 sin  + 5 cos  = 13 sin ( + 0.395)

)

b 13 sin ( + 0.395) = 8 so sin ( + 0.395) = 8 = 0.615 13  + 0.395 = 0.663

y

 = 0.268

2

6 1

0

π

2π

a 10 cos  − 12 sin  ≡ r cos  cos α − r sin  sin α Therefore r cos α = 10 and r sin α = 12. r = 10 2 + 122 = 244 and tan α = 12 = 1.2 so 10 α = 0.876.



–1

10 cos  − 12 sin  = 244 cos( + 0.876) b

–2

2

50 cos ( + 81.9°) = −5 and hence 5 1 =− . cos ( + 81.9°) = − 50 2 Hence  + 81.9° = 135° and  = 53.1°.

a 8 sin x − 6 cos x ≡ r sin x cos α − r cos x sin α

2sin A 2t 1 2 ÷ = A 1 + t2 cos cos2 A 2 2 A A = 2sin cos = sin A 2 2 c cos A =

)

a 3 sin  + 4 cos  ≡ r sin  cos α + r cos  sin α Therefore r cos α = 3 and r sin α = 4. 4 r = 3 + 4 = 5 and tan α = so α = 53.1°. 3 2

244 cos( + 0.876) = 5 so cos( + 0.876) = 0.320  + 0.876 = 1.245 or 2π – 1.245 = 5.038  = 0.369 or 4.162

2

3 sin  + 4 cos  = 5 sin( + 53.1°)

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Trigonometry

7 a  Vertical distance from R to OB is 6 sin  and vertical distance from P to R is 4 cos  so the sum of these is the height of P.



Therefore r cos α = 6 and r sin α = 4.



r = 4 2 + 6 2 = 52



The maximum height is 52 = 7.21 m.



52 sin ( + 0.588) = 7



sin ( + 0.588) = 0.9707



 + 0.588 = 1.328 or 1.813



 = 0.740 or 1.23

8 a  If 0.5 sin  + 0.4 cos  ≡ r sin ( + α) ≡ r sin  cos α + r cos  sin α then r cos α = 0.5 and r sin α = 0.4.

r = 0.52 + 0.4 2 = 0.41 = 0.640



tan α =



The equation is 0.640 sin ( + 0.675) = 0.2 so sin ( + 0.675) = 0.312.



 + 0.675 = 0.318 or π – 0.318 or 2π + 0.318 or...



 = –0.357 or 2.149 or 5.926 or...



The two values in the given range are  = 2.15 or 5.93.

0.4 = 0.8 so α = 0.675 0.5

b If 0.640 sin ( + 0.675) = 0.7 then sin ( + 0.675) = 1.094

This has no solution because –1  sin ( + 0.675)  1.

9 a sin x + sin (x + 10°) = sin x + sin x cos 10° + cos x sin 10°

= sin x + 0.985 sin x + 0.174 cos x = 1.985 sin x + 0.174 cos x

b If 1.985 sin x + 0.174 cos x = r sin (x + α) then r cos α = 1.985 and r sin α = 0.174

tan α =

0.174 so α = 5°; r2 = 1.9852 + 0.1742 so 1.985

r = 1.993

The equation is 2.798 cos (θ + 0.530) = 2; cos (θ + 0.530 ) = 2 = 0.715; 2.798 θ + 0.530 = 0.774 or 5.509; θ = 0.244 or 4.98

Exercise 3.4A 1 a 2 b 3 2 c d 2 3 3 2 

sin 

cos 

π 6

1 2

3 2

π 3

3 2

1 2

tan 

sec 

cosec 

cot 

1 3

2 3

2

3

3

2

2 3

1 3

3 a  sin x = 0.25 therefore x = 0.253 or π – 0.253 = 2.89. b cos x = − 1 therefore x = 1.91 or 2π – 1.91 3 = 4.37. 2 3 c 4 a 2 b 3 5 a tan (x + 30°) = 1 = 2.5 0.4 Hence x + 30° = 68.2° or 180° + 68.2° or 360° + 68.2° or …

Hence x = 38.2° or 218.2° or 398.2° or …



The two solutions in the given domain are 38.2° and 218.2°.

b cos (x − 25°) = 0.5

Hence x − 25° = 60° or 300° or …



Hence x = 85° or 325°.

3 2 = and so 3 cos x = 2 sin x which sin x cos x rearranges to sin x = tan x = 1.5. cos x Hence x = 56.3° or 180° + 56.3° = 236.3°. d sin (2x + 20°) = 1 = 0.625 1.6 Hence 2x + 20° = 38.7° or 180° – 38.7° or 360° + 38.7° or 540° – 38.7° or …

c



The equation is 1.993 sin (x + 5°) = 1; sin (x + 5°) = 0.502; x + 5° = 30.1° or 149.9°; x = 25.1° or 144.9° π π π 10 a  cos θ + = cosθ cos − sin θ sin 4 4 4 1 1 1 cosθ − sin θ ; so a = and b = − 1 = 2 2 2 2

(

)

(

)

b cosθ + 2cos θ + π = cosθ + 2 cosθ − 2 sin θ 4

(

2 and α = 0.530; 1+ 2 r 2 = 2 + (1 + 2)2 so r = 2.798

Hence tan α =



4 c tan α = 6 so α = 0.588.

)

r cosα = 1 + 2 and r sin α = 2

b Write 6 sin  + 4 cos  ≡ r sin ( + α) = r sin  cos α + r cos  sin α.

(

If 1 + 2 cosθ − 2 sin θ ≡ r cos (θ + α ) then

)



2x + 20° = 38.7° or 141.3° or 398.7 or 501.3° or …



Hence x = 9.3° or 60.7° or 189.3° or 240.7°.

= 1 + 2 cosθ − 2 sin θ

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3

WORKED SOLUTIONS

6

a

= y 2

(

=4

1 1 sin 2x 2

cosec2 2x

10 a cotA − cosec 2A =

1

)

2

=

1 1 sin 2 2x 4

= right-hand side.

cos A 1 − sin A sin 2A

cos A 1 2cos2 A − 1 cos2A − = = sin A 2sin A cos A 2sin A cos A sin 2A = cot 2A

= 0

2π

x

b cot x – cot 2x = 4; and cot x – cot 2x = cosec 2x; hence cosec 2x = 4; sin 2x = 0.25;

–1 –2

7

x = 0.126, 1.44, 3.27, 4.59

b f(x)  1 and f(x)  –1 a y

Exercise 3.5A 1

sec2  ≡ 1 + tan2  so 9 = 1 + tan2  Hence tan2  = 8 and tan θ = ± 8 .

6

2 4

2

cosec2  ≡ 1 + cot2  and so 1 + cot2  + cot2  = 9 Hence 2 cot2  = 8 cot2  = 4 cot  = ±2. Hence  = 26.6° or 153.4°.

0

90

180

270

360

x

3

Hence sec2  – tan2  = 1

–2 2

(sec  – tan )(sec  + tan ) = 1.

(sec  − tan )k = 1 and so sec  – tan  = 1 . k b Add the simultaneous equations sec  + tan  = k and sec  – tan  = 1 to get k 1 1 1 2 sec  = k + and hence secθ = k + . 2 k k

–4 4

b

2 cos x = cos x sin x

(

Rearrange: 2 sin x = cos2 x Hence 2 sin x = 1 − sin2x. Write s = sin x and rearrange: s2 + 2s − 1 = 0 Use the quadratic formula: s = = –2.4142 or 0.4142.

4

−2 ± 4 + 4 2

If sin x = 0.4142 then x = 24.5° or 155.5°. 2 If x = 24.5 then y = = 2.20. cos 24.5°

5

The coordinates are (24.5, 2.20) and (155.5, –2.20). a f(x)  0.5 and f(x)  –0.5 b f(x)  1.5 and f(x)  –0.5

6

c f(x)  1 and f(x)  –1 9

Left-hand side = sec2 x + cosec2 x =

)

1 + tan2  + 2 tan  = 4 tan2  + 2 tan  − 3 = 0 Factorise: (tan  − 1)(tan  + 3) = 0. Either tan  = 1 or tan  = −3.  = π = 0.785 or  = π – 1.249 = 1.89. 4

sin x = –2.4142 has no solution.

8

a sec2  ≡ 1 + tan2 

1 1 + cos2 x sin 2 x

2 2 1 = sin x2 + cos2 x = cos x sin x (cos x sin x )2

sec2x − 1 = 6 sec x − 10 sec2 x − 6 sec x + 9 = 0 (sec x − 3)2 = 0 sec x = 3 1 cos x = 3 x = 70.5° 1 sin x 1 + sin x a sec x + tan x = + = cos x cos x cos x 2 2 2 b sec x = 1 + tan x so sec x – tan2 x = 1; (sec x + tan x) (sec x – tan x) = 1; sec x − tan x =

1 cos x = sec x + tan x 1 + sin x

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TRIGONOMETRY

θ θ θ 1 sin θ cosθ sin 2 θ + cos2 θ b θIn the θsame way, sin = 2cos sin cosec + = = = sec 2 4 4 cosθ sin θ cosθ sin θ cosθ sin θ θ θ θ 1 sin θ cosθ sin 2 θ + cos2 θ So sin θ = 2cos 2 × 2cos 4 sin 4 = + = = = secθ cosecθ cosθ sin θ cosθ sin θ cosθ sin θ θ θ θ 1 1 1 = 4cos cos sin . 8 cosec 2x = = = cosec x sec x 2 4 4 sin 2x 2sin x cos x 2 2  − sin  = 0 4 2 sin 1 − tan 2 x 1 9 cot 2x = Factorise: sin  (2 sin  − 1) = 0. = tan 2x 2tan x Either sin  = 0 or 2 sin  − 1 = 0. Divide the numerator and the denominator by If sin  = 0 then  = 0 or π. 1 −1 2 1 2 cot x 1 − . If 2 sin  − 1 = 0 then sin θ = so θ = π or 5π . tan2 x to get tan x = 2 6 6 1 2cot x 2× tan x There are 4 values for .

5° =

7

tan θ + cot θ =

sin x cos x cos 2θ cos2 θ − sin 2 θ sin 2 θ 2 2 + = 3 ; Multiply by sin x cos x to get + tan 2 θ = 1 − 5 cos x sin x 2 + tan θ = 2 2 + tan θ cos θ cos θ cos θ sin2 x + cos2 x = 3 sin x cos x; 2 2 2 3 2 cos 22θ + tan 2 θ = cos θ −2sin θ + tan 2 θ = 1 − sin 2θ + tan 2 θ × 2sin x cos x = 1 ; sin 2x = ; 2xcos = 0.730 or θ cos θ cos θ 2 3 2.412; x = 0.365 or 1.21 = 1 − tan2  + tan2  = 1 7 5 ; multiply by tan x; b tan x + = tan x sin x 6 cos4  = 2 sin2  cos2  2 2 tan x + 7 = 5 sec x; sec x – 1 + 7 = 5 sec x Either cos2  = 0 or cos2  = 2 sin2 . 2 sec x – 5 sec x + 6 = 0; (sec x – 2) (sec x – 3) = 0; If cos2  = 0 then cos  = 0 and  = 90°. 1 1 If cos2  = 2 sin2  then tan2  = 0.5 and tan θ = 0.5 sec x = 2 or 3; cos x = or ; 2 3 or tan θ = − 0.5. π Then  = 35.3° or 144.7°. x = (1.05) or 1.23 3 π π π 7 cos θ + 4 = cos θ cos 4 − sin θ sin 4

10 a

(

Exam-style questions 1

a i

3 2

)

(

b cos 105° = cos (135° − 30°) = cos 135° cos 30° + sin 135° sin 30° 8

)

a

1+ 2 a tan 75° = tan (45° + 30°) = tan 45° + tan 30° = 1 − tan 45° tan 30° 1 − 1 1+ tan 45° + tan 30° 3 = 3 +1 = 1 − tan 45° tan 30° 1 − 1 3 −1 3

3 −1 = 3 +1

2

1 3 = 1 3

1

3 +1 3 −1 –180

θ and then 2 θ θ sin  = 2 cos sin . 2 2

90

180

x

–1

3 −1 = 3 +1

Let  = 2x so x =

0

–90

3 −1 3 −1 3− 2 3 +1 4 − 2 3 = = = 2− 3 × 3−1 2 3 +1 3 −1 1 cos x 3 −1 3 −1 3− 2 3 +1 4 − 2 3 b = × = = = 2− 3 cos x sin x 3 1 2 − 3 +1 3 −1 sin x = cos2 x 3 a sin 2x = 2 sin x cos x b cot 75° =

2

y

=− 1 × 3 + 1 ×1 2 2 2 2 1− 3 = 2 2

()

4 3 cos π = sin π = 1 and sin θ = 1 − = 5 5 4 4 2 So cos θ + π = 4 × 1 − 3 × 1 = 1 = 2 . 4 5 2 5 2 5 2 10

1 ii cos 135° = − cos 45° = − 2

––2

sin x = 1 − sin2 x

sin2 x + sin x − 1 = 0 sin x =

−1 ± 1 + 4 2

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3 

WORKED SOLUTIONS



sin x =

−1 − 5 −1 + 5 or sin x = 2 2



sin x =

−1 − 5 = −1.62 has no solution 2



−1 + 5 = 0.618 has solutions 2 x = 38.2° or 180° − 38.2° = 141.8 These are the two solutions in the interval −180°  x  180°.



Factorise: (2 cos x − 1)(3 cos x + 1) = 0 1 1 so cos x = or − . 2 3



If cos x =

sin x =





α = 0.39 to 2 d.p. So



 + 0.951 = 1.059 or π – 1.059 = 2.083



 = 0.108 or 1.13

5 sin θ + 12 cos θ = 13 cos (θ − 0.39)

10 ; b 13 cos (θ – 0.39) = 10; cos (θ − 0.39) = 13 θ – 0.39 = 0.693 or 2π – 0.693 = 5.590; hence θ = 1.09 or 5.98

296 sin ( + 0.951) = 15 sin ( + 0.951) = 0.872

So 5 = R sin α and 12 = R cos α; R = 52 + 122 = 13 and tan α = 5 so 12

14 tan α = = 1.4 so α = 0.95. 10

b



= 2(4 cos4 A − 4 cos2 A + 1) − 1

15 If 3 sin θ + 4 cos θ ≡ r sin (θ + a) = r sin θ cos α + r cos θ sin α then r cos α = 3 and r sin α = 4; hence r = 32 + 4 2 = 5 4 and tan α = so α = 53.1° 3 Hence 5 sin (θ + 53.1°) = 2; sin (θ + 53.1°) = 0.4;



= 8 cos4 A − 8 cos2 A + 1



10 a cos 2A = 2 cos2 A − 1 so cos2 2A = (2 cos2 A − 1)2 = 4 cos4 A − 4 cos2 A + 1 b cos 4A = 2 cos2 2A − 1

1 sin x 2 sin2 x + sin x = 1 2 sin2 x + sin x – 1 = 0 Factorise this quadratic: (2 sin x − 1)(sin x + 1) = 0. 1 sin x = or –1 2 π x = or 5π or 3π 2 6 6   12 a tan x + cot x = 5

11 2 sin x + 1 =



Multiply by tan x : tan2 x + 1 = 5 tan x



Rearrange: tan2 x − 5 tan x + 1 = 0.



Use the quadratic formula.





r cos α = 10 and r sin α = 14 r = 10 2 + 14 2 = 296 = 17.2





1 If cos x = − 3 then x = 4.37 or 1.91.

14 a 5 sin θ + 12 cos θ ≡ R cos (θ – α) = R cos θ cos α + R sin θ sin α

9 a 10 sin  + 14 cos  = r sin  cos α + r cos  sin α

1 π 5π then x = or . 2 3 3

tan x = 5 ± 25 − 4 = 4.791 or 0.2087 2 Therefore x = 1.37 or 0.206.

b tan x + cot x = k; tan2 x − k tan x + 1 = 0

This only has a solution if k2 − 4  0.



Hence k2  4 so that k  2 or k  −2.

13 6(1 − cos2 x) + cos x = 5 6 − 6 cos2 x + cos x = 5 6 cos2 x − cos x − 1 = 0

θ + 53.1° = 23.6° or 156.4° or 383.6˚ …; values of θ in the interval when 0°  θ  360° are

θ = 103.3° or 330.4˚ to 1 d.p. 16 3 sin θ + 1 – 2 sin2 θ = 2; 2 sin2 θ – 3 sin θ + 1 = 0; (2 sin θ – 1)(sin θ – 1) = 0 sin θ = 1 or 1; θ = 30°, 150° or 90° 2 17 a sin A cos B + cos A sin B = 2(sin A cos B – cos A sin B); sin A cos B + cos A sin B = 2 sin A cos B – 2 cos A sin B; 3 cos A sin B = sin A cos B;

3cos A sin B = sin A cos B ; 3 tan B = tan A cos A cos B cos A cos B

b Replacing A with θ and B with 0.5 you get tan θ = 3 tan 0.5 = 1.639; θ = 1.02 or 4.16 π π 18 a cos + A + cos − A 4 4

(



)

(

)

π π = cos cos A − sin sin A 4 4 π π + cos cos A + sin sin A 4 4 π = 2cos cos A = 2 × 1 cos A = 2 cos A 4 2 π 5π 5π π π +A= − = and also then A = 4 12 12 4 6 π π π π −A= − = 4 4 6 12

b If

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Trigonometry



Hence cos 5π + cos π = 2 cos π 12 12 6

c cos



(

3 1 = 2× = 6 2 2

)

(

5π π π π π π − cos = cos + − cos − 12 4 6 4 6 12

)

= cos π cos π − sin π sin π − cos π cos π 4 6 4 6 4 6 π π − sin sin 4 6 π π = −2sin sin 4 6 = −2 × 1 × 1 = − 1 or − 1 2 2 2 2 2

Mathematics in life and work 1 r sin ( + α) = r sin  cos α + r cos  sin α ≡ 5 sin  + k cos  Hence r cos α = 5 and r sin α = k. Square and add: r2 cos2 α + r2 sin2 α = 52 + k2.

Hence r2 = 25 + k2 and r = 25 + k 2 .

2 Dividing:

r sin α k = r cos α 5

So k = 5 tan α = 5 tan 0.3 = 1.55.

3 r = 25 + 1.5472 = 5.234 5 sin  + 1.547 cos  = −4 so 5.234 sin ( + 0.3) = −4 4 sin ( + 0.3) = − = −0.7643 5.234 Hence ( + 0.3) = −0.8699 or π + 0.8699 or …

The smallest positive solution is  = π + 0.8699 − 0.3 = 3.71 to 3 s.f.

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4 

WORKED SOLUTIONS

4 Differentiation Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1 a y = 2x(x2 − 1) = 2x3 − 2x

2

dy = (ln1.5)e(ln1.5)x dx = (ln 1.5)1.5x.

c y = e(ln 1.5)x so

1 1 x − x −1 2 2

dy 1 1 −2 1 1 x2 + 1 = + x or + 2 or 2 2x dx 2 2 2x 2

c y = x

b 1.5 = eln 1.5 so 1.5x = (eln 1.5)x = e(ln 1.5)x.

dy = 6x 2 − 2 dx

b y =

4 a By definition, if y = ln 1.5 then ey = 1.5 so eln 1.5 = 1.5.

−1 2

dy 1 −3 =− x 2 2 dx

2 2 dy 5 a = 4e −0.5x × −0.5 × 2x = −4xe −0.5x dx dy b When x = –2, = −4 × −2 × e−2 = 8e−2. dx 6 a  If t = 3 then the number of bacteria is N e0.1 × 3 = e0.3 N = 1.35N which is an increase of 35%.

b If y = N e0.1t then

dy = 3x 2 − 12x dx dy If x = 6 then = 108 − 72 = 36. dx The gradient of the normal is − 1 and the 36 1 equation is y − 0 = − (x − 6) which simplifies 36 to 36y + x = 6.

t = 3, c If





Using the chain rule,

1 a 2e2x b −e−x 0.4x c 0.4e d e4x + 2 × 4 = 4e4x + 2

8

2 a 4 × 0.5e0.5x = 2e0.5x b 100e−0.1x × −0.1 = −10e−0.1x 50e2x−10

100e2x−10

c ×2= 3 a f(x) = 4e2x × 2 = 8e2x

b f(x) = 8e2x × 2 = 16e2x

Hence 0.1k = ln 2 and k =

ln 2 = 6.93 hours. 0.1

f(x) = 4e2x − 2e−2x × −2 = 4e2x + 4e−2x = 4(e2x + e−2x) = 4f(x)

b A  t a stationary point f(x) = 0; 2e2x – 2e–2x = 0; e2x = e–2x; e4x = 1; x = 0

−−11

Exercise 4.1A

dy = 0.2N then 0.1 N e0.1k = 0.2N and dt

7 a f(x) = 2e2x − 2e−2x and

1

ff′(′(xx))== 11((11++ xx22))22 ××22xx == xx 22 11++xx22 . 4 0.2y = e2x − 1 and hence ln 0.2y = 2x − 1 and 1 x = (1 + ln 0.2y). 2 1 An equivalent answer is x = (1 + ln y − ln 5). 2

dy = 0.1 N e0.3 = 0.135N  h−1. dt

hence e0.1k = 2.

3 f(x) = (1 + x 2)2

dy = 0.1 N e0.1t and when dt

If x = 0, f(x) = 4f(x) = 4 × 2 = 8 > 0 showing this is a minimum point.

dy = ex − 4e−x. dx At a stationary point ex − 4e−x = 0. Multiply by ex: e2x − 4 = 0 so e2x = 4. Take logarithms: 2x = ln 4 and  1 1 x = ln 4 = ln  4 2  = ln 2 . 2  



d 2y = ex + 4e−x which equals y. d 2x



When x = ln 2 then y = eln 2 + 4e−ln 2 = 2 + 4 ×

1 = 4. 2

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DIFFERENTIATION

This means minimum.

d2y = 4 > 0 so the stationary point is a dx 2

b The gradient of the normal at P is − 1 the equation is y − k = − x . k

The coordinates are (ln 2, 4). 9

a

dy = ae ax ; if the gradient is 1 then aea x= 1; dx 1 1 e ax = ; ax = ln = − ln a. a a ln a ; then a 1 1 = e − ln a = ln a = ; the point P is a e

Hence x = − y = e ax

This meets the x-axis at (100, 0) so 1 0 − k = − × 100; k2 = 100; k = 10. k

Exercise 4.2A 1

(

)

1 ln a or x + y = 1 − ln a . y− =− x+ a a a 10 a

a y = ln 3 + ln x and hence b y = 3 ln x and hence

(− lnaa , a1 ).

b The gradient of the tangent is 1 so the gradient of the normal is −1 and the equation is

dy 1 = . dx x

dy 3 = . dx x

dy 1 3x 2 = 3 × 3x 2 = 3 dx x + 2 x +2 dy 4 1 4 1 a b ×4= c y = 4 ln x so = . x 4x x dx x

c 2 3

a

y

y = ln 2x

dy = e x − a; at a stationary point ex – a = 0; dx ex

1 and k

y = ln x

= a; x = ln a;

then = eln a – a ln a = a – a ln a; coordinates are (ln a, a – a ln a).

0

d 2y = e x > 0 for all x so the point is a minimum. dx 2 dy c If x = 0 then y = 1 and = 1 − a; The dx equation of the tangent is

1

x

b

 0  b ln 2x = ln 2 + ln x so the translation is  .  ln 2 c f(x) = ln kx = ln k + ln x and since ln k is a 1 constant, f ′(x) = . x

y – 1 = (1 – a)x or y = 1 + (1 – a)x. 11 a f(x) = ex + 3e–x; f(x) = ex – 3e–x. If the gradient is 4 then f(x) = ex – 3e–x = 4; e2x – 4ex – 3 = 0. ex = 4.646 (ignore the negative solution). x = ln 4.646 = 1.536 At x = 1.536, f(x) = 5.292. Coordinates are (1.54, 5.29).

4

b At a stationary point, f(x) = 0 so ex – 3e–x = 0; ex = 3e–x; e2x = 3.

It crosses the x-axis at (–4, 0) and there dy 1 = = 1. dx −4 + 5

1 ln 3 = ln 3 ; f (x) = ex + 3e–x 2 which is positive for all values of x and in particular if x = ln 3 so this is a minimum point. 2x = ln 3; x =

The equation of the tangent is y − 0 = 1 (x − −4) or y = x + 4. b It crosses the y-axis at (0, ln 5) and there dy 1 1 = = . dx 0 + 5 5

The minimum value is f(ln 3). 3 = 3+ 3=2 3 3 dy dy 12 a = ke x = y . At P, x = 0, y = k and = k. dx dx = e ln

3

+ 3e − ln

3

= 3+

The equation of the tangent is y – k = kx. Where the tangent meets the x-axis, y = 0 and so –k = kx; x = –1 and the point is (−1, 0).

d The translation is parallel to the y-axis so the gradient for any particular x value is unchanged. dy 1 = . a If y = ln(x + 5) then dx x + 5

5

The equation of the tangent is y − ln 5 = 1 x 5 1 or y = x + ln 5. 5 a If x = e, then y = ln e = 1 so (e, 1) is on the graph. dy 1 dy 1 b dx = x and if x = e then = . dx e

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4 

WORKED SOLUTIONS





The equation of the tangent is 1 1 y − 1 = (x − e) which simplifies to y = x . e e This line passes through the origin.





c The gradient of the normal is –e and the equation is y −1 = −e(x − e)

b At Q, gradient

Where it crosses the x-axis, y = 0 so −1 = −e(x − e)

1 1 = x − e and so x = e + . e e 10 1 = ln 10 − ln x and so f ′(x) = − . 6 a f(x) = ln x x 10 b  f(x) = ln 2 = ln 10 − ln x2 = ln 10 − 2 ln x and x 2 so f ′(x) = − . x

c The equation of the tangent at Q is

Which rearranges to



y + 1 = e(x – e–1);



if x = 0 then y + 1 = e × (–e –1) = –1; y = −2 and the point is (0, −2). dy 1 a ×a = ; if the = dx ax + b ax + b a = 1; gradient is 1 then ax + b

10 a  y = ln (ax + b) so

a = ax + b; ax = a – b; x = 1 − and b are positive, so is

7 a y = log10 x therefore x = 10y = (eln 10)y = e(ln 10)y. Take logs and then (ln 10)y = ln x. 1 y= ln x ln10

b y = log10 ax2 = log10 a + log10 x2



c If y = xn – ln x then

dy 1 = nx n −1 − ; if x = 1 x dx



d If y =

1 a 2 cos 2x b −sin (5x − 2) × 5 = −5 sin (5x − 2) c cos (x2 + 1) × 2x = 2x cos (x2 + 1) 2 a 10 cos 0.5x × 0.5 = 5 cos 0.5x b 3 cos 3x − 6 sin 6x

dy 1 – ln x then = 2x − ; if x dx

dy 1 1 1 1 = 2x − = 0 then 2x = ; x 2 = ; x = . x x 2 dx 2

Then y =

1 1 1 1 1 − ln = + ln 2 = + ln 2 2 2 2 2 2

= 1 (1 + ln 2). 2

9 a Suppose P is the point (a, b). gradient is

dy a 1 1 = = and if the gradient is 2 dx ax + a x + 1 1 1 = so x = 1. then x +1 2 Then y = ln (ax + a) = y = ln(e2 + e2) = ln 2 e2 = ln 2 + ln e2 = 2 + ln 2. The point is ( 1, 2 + ln 2).

Exercise 4.3A

dy then = n − 1. dx x2

a a = 0; = 1; a = b b b

y = ln (ax + a) and 2 = ln (0 + a) = ln a so a = e2.

b y = xn – ln x; if x = 1 then y = 1 – 0 = 1 so S is on the curve.

a . Because a b

a hence x < 1. b

b On the y-axis x = 0 so 1 −

dy 1 = dx x ln10

= log10 a + 2 log10 x; dy 1 2 . so = 2× dx x ln10 = x ln10 dy 1 1 8 a  = 1 − ; at a stationary point, 1 − = 0; x x dx x = 1 and y = 1 – ln 1 = 1 – 0 = 1; S is (1, 1).

1 1 = e; x = = e −1 ; then x e

y = ln x = ln e–1 = –1; Q is (e–1, –1).

c f(x) = ln 10n = ln 10 − ln xn = ln 10 − n ln x and x n so f ′(x) = − . x



1 (x − a); if this passes through (0, 0) a 1 then 0 − b = (0 − a); b = 1. a So ln a = 1; a = e; P is (e, 1). y−b=

c −sin (x2 − 3x − 4) × (2x − 3) = −(2x − 3) sin (x2 − 3x − 4) 3 f'(x) = 4 cos 4x − 4 sin 4x

f"(x) = −16 sin 4x − 16 cos 4x = −16(sin 4x + cos 4x) = −16f(x)



Hence f"(x) + 16f(x) = 0.

4 a y = (sin x)2 dy 1 = ; at P the dx x

1 and the tangent is a



Use the chain rule:



dy = 2 sin x × cos x = 2 sin x cos x = sin 2x dx

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Differentiation

dy = 2 cos x × −sin x. dx = −2sin x cos x = −sin 2x.

b If y = cos2 x then



c sin2 x + cos2 x = 1 so the derivative of

sin2 x + cos2 x is 0.



Hence the two derivatives are identical apart from the sign.

dy = esin x × cos x = cos x esin x dx dy If x = 0 then = e0 × 1 = 1. The gradient is 1. dx b At a stationary point, cos x esin x = 0.



5 a



One solution is x = π and then y = 2 π ,e is a stationary point. so 2

( )

=e

6 a f(x) =  3 cos x − 3 sin2 x × cos x = 3 cos x(1 − sin2 x) = 3 cos x × cos2 x = 3 cos3 x b f"(x) = 3 × 3 cos2 x × −sin x = −9 cos2 x sin x 7 a f(x) = (cos x)−1

Using the chain rule, f(x) = ( −1)(cos x)−2 × (−sin x) = =



sin x cos2 x

b y = (sin x)−1 and dy cos x = (−1)(sin x)−2 × cos x = − dx sin 2 x 1 cos x =− × = −cosec x cot x sin x sin x





dy = 0.1 cos 2.4t × 2.4 dt

= 0.24 cos 2.4t. dy = 0.24 cos 2.4 = −0.177. When t = 1, dt The speed is 0.177 m s−1.

b The speed is zero when cos 2.4t = 0

Then sin 2.4t = ±1 and the displacement is 0.1 m, the largest possible value. d 2y = −0.576 sin 2.4t. dt 2 This is zero when sin 2.4t = 0 and then y is also zero.

c The acceleration is

The bob is in the central position. 1 1 is − 2 x x Using the chain rule,

9 a The derivative of



f ′(x) = − sin

( ) ( )(

)( )

dy = 2cos x + 4sin x ; if x = 0 then dx dy = 2×1+ 4 × 0 = 2 dx dy π then y = 2 – 0 = 2 and = 0 + 4 = 4 ; the 2 dx π tangent is y − 2 = 4 x − ; 2

b If x =

(

)

y – 2 = 4x – 2π; y = 4x + 2 – 2π. dy = 0 then 2 cos x + 4 sin x = 0; dx 4 sin x = –2 cos x; tan x = –0.5; x = 2.68.

c If

11 a f(x) = –2 sin 2x + 2 sin x

1 sin x × = sec x tan x . cos x cos x

8 a  The velocity is

10 a

Since esin x is always positive, cos x = 0. π sin e 2

1 1 sin = 0 . x x2 Since x > 0, this means sin 1 = 0 and hence x 1 = π, 2π, 3π, 4π, … x 1 1 1 1 Hence x = , , , ,… π 2π 3π 4π The turning points are 1 1 1 1 ,1 , , −1 , ,1 , … , −1 , π 2π 3π 4π

b At a stationary point

1 1 1 1 × − 2 = 2 sin . x x x x

b If f(x) = 0 then –2 sin 2 x + 2 sin x = 0; –4 sin x cos x + 2 sin x = 0; –2 sin x cos x + sin x = 0; sin x (–2 cos x + 1) = 0; sin x = 0 or cos x = 1 ; 2 1 π if sin x = 0 then x = 0; if cos x = then x = 3 2 f(0) = 1 – 2 = – 1; f π = cos 2π − 2cos π = −0.5 − 2 = −1.5. 3 3 3 π Stationary points are (0, −1) and , −1.5 . 3

()

(

)

dy dy 12 a  = 1 − 2sin 2x ; if x = 0, y = 1 and = 1 ; the dx dx gradient of the normal is –1; the equation is y – 1 = – 1(x – 0) or y + x = 1 b At a stationary point, 1 – 2 sin 2x = 0; 1 π 5π π 5π sin 2x = ; 2x = or ;x= or ; 12 2 6 6 12 two points in the given interval. c

2 d 2y π d y = −4cos2x –3.46 < 0 so 2 = −4cos2x ; if x = 12 , dx dx 2 this is the maximum point.

The y-coordinate is π π π 3 π+6 3 + cos = + = or 1.13 to 12 6 12 2 12 3 s.f. This is the maximum value.

13 a y = sin 2θ + 2 sin θ so

dy = 2cos2θ + 2cosθ dθ

b At a maximum point, 2 cos 2θ + 2 cos θ = 0; cos 2θ + cos θ = 0;

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4 

WORKED SOLUTIONS





2cos2 θ – 1 + cos θ = 0; (2 cos θ – 1)(cos θ + 1) = 0; 1 cosθ = or −1. 2 θ = π or π are two solutions. 3 d 2y π = – 4 sin 2θ – 2 sin θ; if θ = then 3 dθ 2 d 2y 3 3 = −4 × −2× = −3 3 < 0 so this 2 2 dθ 2 π will be a maximum value. If θ = Then 3 3 3 2 π π 3 y = sin + 2sin = 3 × = 3 3 2 2 and this

is the maximum value. dy π = cos x + 3cos3x; if x = then y = sin π + sin 3π 14 4 4 4 dx 1 1 2 + = = 2 = 2 2 2 dy 3 and = cos π + 3cos π 4 4 dx 1 3 2 − =− = − 2 ; the tangent is 2 2 2 π π y− 2=− 2 x− or y + 2x = 2 1 + 4 4

=

(

)

( )

Exercise 4.4A dy 1 a = sin x + x cos x dx dy = cos x − x sin x dx c dy = 2x sin 2x + x2 × 2 cos 2x dx = 2x sin 2x + 2x2 cos 2x b

2 a ex + xex



= (0.2 sin 10t + 10 cos 10t)e0.2t.



b f(t) = cos t2 e−1.5t

If u =cos t2 and v = e−1.5t then and



du = −2t sin t2 dt

dv = −1.5e −1.5t . dt

Then f(t) = cos t2 × −1.5e−1.5t + (−2t sin t2) × e−1.5t = −(1.5 cos t2 + 2t sin t2)e−1.5t.



c f(t) = sin (2t + 1)e −t

2 2



If u = sin (2t + 1) and v = e −t



2 dv then du = 2 cos (2t + 1) and = −2te −t . dt dt



Then f(x) 2

2

−t = sin (2t + 1) × −2te −t + e × 2 cos (2t + 1)



2

= {−2t sin (2t + 1) + 2 cos (2t + 1)}e −t .

dy 1 6 a = ln x + x × = 1 + ln x x dx b If x = 1,

dy = 1 + 0 = 1. dx

The gradient is 1.

c If

dy = 2 then 1 + ln x = 2. dx



Therefore ln x = 1.



Therefore x = e and y = e ln e = e so the coordinates are (e, e).

b e−x − (x + 1)e−x = e−x − xe−x − e−x = −xe−x

d At a stationary point 1 + ln x = 0 therefore ln x = −1.

c 2xe2x + x2 × 2e2x = 2xe2x + 2x2 e2x = 2x(x + 1)e2x



3 a  f(x) = cos x × cos x + sin x × (−sin x)

Then f(t) = sin 10t × 0.2e0.2t + 10 cos 10t × e0.2t

= cos2 x − sin2 x 1 b sin 2x = 2 sin x cos x so y = sin 2x and 2 dy 1 = × 2 cos 2x = cos 2x. dx 2 c The identity cos 2x ≡ cos2 x − sin2 x means that the two answers are identical.

dy 4 a = ex sin x + ex cos x dx d 2y b = ex sin x + ex cos x + ex cos x − ex sin x dx 2 = 2ex cos x 5 a f(t) = sin 10t e0.2t

du If u = sin 10t and v = e0.2t then dt =10 cos 10t dv = 0.2e0.2t . and dt

So x = e−1 and y = e−1 × −1 = −e−1. The coordinates are (e−1, −e−1).

dy 7 a = e−x − xe−x dx

At a stationary point, e−x − xe−x = 0 so e−x(1 − x) = 0.



e−x cannot be 0 so the only solution is x = 1.



Then y = 1 × e−1 and the coordinates are (1, e−1).

b

dy = 2xe−x − x2 e−x dx



At a stationary point, 2xe−x − x2 e−x = 0 so x(2 − x)e−x = 0.



Either x = 0 or x = 2.



If x = 0, y = 0.



If x = 2, y = 4e−2.



The points are (0, 0) and (2, 4e−2).

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Differentiation

2

8 a If u = x2 and v = e x then

2 2 dv = e −x × ( −2x ) = −2xe −x dx

(



2 2 So dy = 10 × 2xe −x + 10x 2 × −2xe −x dx



= 20xe −x − 20x 3e −x

2

c At a stationary point ex sin x + ex cos x = 0; 3π sin x = – cos x; tan x = –1; x = 4

du = 2x and dx

)



2

2

2

b At a stationary point, 20xe −x − 20x 3e −x = 0; 2

hence 20x (1 − x )e

−x 2

=0

2

e −x cannot be 0 so x(1 – x2) = 0; Hence x(1 – x)(1 + x) = 0.

x = 0, 1 or −1; if x = 0, y = 0; if x = 1, y = 10e–1; if x = −1, y = 10e–1

The stationary points are (0, 0), (1, 10e–1) and (−1, 10e–1).

dy 9 = −0.2e−0.2x sin 2x + 2e−0.2x cos 2x dx At stationary point, −0.2e−0.2x sin 2x + 2e−0.2x cos 2x = 0. 2e−0.2x cos 2x = 0.2e−0.2x sin 2x 2 cos 2x = 0.2 sin 2x



sin 2x 2 = cos 2x 0.2 tan 2x = 10 2x = arctan 10 = 1.4711 or 1.4711 + π or 1.4711 + 2π or... 2x = 1.4711 + nπ where n = 0, 1, 2, 3, ... So x = nπ + 0.736 where n is 0, 1, 2, 3, ... 2

12 a

2



2

−0.05t 2

2 −0.05t 2



15e − 1.5t e = 0. Therefore 15 − 1.5t2 = 0. t2 = 10 t = 10 Then y = 15 10e −0.5 = 28.77.



The maximum speed is 28.8 m s−1

11 a Using the product rule, b

dy = e x sin x + e x cos x dx

d 2y = derivative of ex sin x + derivative of ex cos x dx 2 = ex sin x + ex cos x + ex cos x – ex sin x = 2ex cos x dy 2 − 2y = 2(ex sin x + ex cos x) – 2ex sin x dx d 2y = 2ex cos x = dx 2

e4 = 4 2

3π   4 The stationary point is  3π , e  2  4 

d 2y dy 1 34π 0 − 2 × e y = 2 − 2 = dx 2 dx 2 which is negative so stationary point is a maximum. Also

dy = −e −x sin x + e−x cos x; dx d 2y = e −x sin x − e −x cos x − e −x cos x − e −x sin x dx 2 = −2e −x cos x

b At a stationary point – e–x sin x + e–x cos x = 0;

sin x = cos x; tan x = 1; x =

π 5π 9π , , … 4 4 4

−π −π If x = π , y = e 4 sin π = 1 e 4 ; 4 4 2 −π −π −π d 2y π 1 4 4 cos = −2e × = − 2e 4 < 0 2 = −2e 4 2 dx so the value is a maximum.

c When x =

2 −π 5π d y , = 2e 4 > 0 a minimum 2 4 dx

point.

When x =

2 −π 9π d y , = − 2e 4 < 0 a maximum 4 dx 2

 9π 1 − 94π  point at  , e   4  2

2 2 10 dy = 15e −0.05t + 15te −0.05t × (−0.1t ) dt

= 15e −0.05t − 1.5t 2e −0.05t When the speed is maximum the graph has a stationary point so

Then y =

3π

3π 3π e 4 sin

d The maximum points occur 2π apart when π 9π 17π x= , , ,… 4 4 4

The values of y are

1 − π4 e , 2

1 − 94π 1 − π4 = e × e −2π , e 2 2 1 − 174π 1 − π4 = e × e −4π , … and these form e 2 2 a geometric progression with r = e–2π 1 13 a If u = 1 + x = (1 + x)2 and v = sin πx 2



−1 1 and then du = 1 (1 + x) 2 = dx 2 2 1+ x dv π = cos π x 2 dx 2

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4 

WORKED SOLUTIONS



dy πx 1 × sin Hence dx = 2 2 1+ x



πx π x = sin 2 π π 1+ x + 1 + x × cos cos π x + 2 2 2 2 2 1+ x

3π = −2 and b If x = 3, y = 2sin 2 3 π dy sin 2 2π 3π = − 1 + 0 = − 1 cos = + 4 4 4 2 2 dx

The equation is y + 2 = − 1 (x − 3); 4y + 8 = – x + 3; 4 x + 4y + 5 = 0.

14 If u = x2 and v = 20 − x = (20 −

−1 du dv 1 = 2x and = × (20 − x) 2 × −1 dx dx 2 1 =− 2 20 − x

Then

dy x2 Hence = 2x 20 − x − ; at a stationary dx 2 20 − x point 2x 20 − x −



1 x)2

x2 ; 4x (20 – x) = x2; x ≠ 0 so 2 20 − x 4 (20 – x) = x; 80 – 4x = x; 5x = 80; x = 16. Then y = 16 2 20 − 16 = 512; the stationary point is (16, 512).

Exercise 4.5A dy (x + 1) × 1 − x × 1 x + 1 − x 1 1 a = = = dx (x + 1)2 (x + 1)2 (x + 1)2 dy (x 2 + 1) × 1 − (x + 1) × 2x x 2 + 1 − 2x 2 − 2x = = dx (x 2 + 1)2 (x 2 + 1)2 =

1 − 2x − x 2 (x 2 + 1)2

3 2 = 4x − 3x2 (2x − 1)

2 a If f(x) = b If f(x) =

c If f(x) =

dy = sec2 (4x + 3) × 4 = 4 sec2 (4x + 3) dx

c

dy = 3 sec2 (x2) × 2x = 6x sec2 (x2) dx

dv 2 4 a If u = x and v = tan x then du = 1 and dx = sec x . dx f(x) = x sec2 x + tan x b f(x) = 2x tan 2x + x2 × sec2 2x × 2 = 2x tan 2x + 2x2 sec2 2x c f(x) = e−x × sec2 ax × a − e−x tan ax = e−x (a sec2 ax − tan ax) dy 5 a = −2e−2x sin x + e−2x cos x dx b

dy e 2x cos x − sin x × 2e 2x = dx e 4x



Divide numerator and denominator by e2x to get



x cos x − sin x sin x then f ′(x) = . x x2 x +1 × (− sin x) then f ′(x) = cos x − (x + 1) cos x cos2 x cos x + (x + 1)sin x = . cos2 x 2 x2 then f ′(x) = 2x × sin 2x −2x × 2cos2x sin 2x sin 2x

=

2x sin 2x − 2x 2 cos2x . sin 2 2x

cos x − 2 sin x = cos x − 2 sin x which is the e 2x e 2x e 2x same answer.

6 a f ′(x) =

x × 1 − ln x × 1 1 − ln x x = x2 x2

x 2 ×  − 1  − (1 − ln x) × 2x  x b f ′′(x) = x4 = −x − 2x +4 2x ln x = 2x ln x4 − 3x x x ln − 2 3 x = x3 1 − ln x c At a stationary point = 0. x2 ln e 1 = . Therefore ln x = 1 and x = e and y = e e

dy (2x − 1) × 3x 2 − x 3 × 2 6x 3 − 3x 2 − 2x 3 c = = dx (2x − 1)2 (2x − 1)2

dy = sec2 ax × a = a sec2 ax. dx

b

x2 = 0; 2 20 − x

2x 20 − x =

b

3 a Using the chain rule,



The coordinates of the stationary point are e, 1 . e d2y When x = e then 2 = 2 ln e3 − 3 = − 13 which dx e e is negative so the stationary point is a maximum point.

( )

−x −x −x −x dy 7 a = (1 + e ) × e − (1−x− 2e ) × (−e ) dx (1 + e ) −2x −x −2x −x −x = e + e +−ex 2 − e = 2e −x 2 (1 + e ) (1 + e )

b When x = 0,

dy 2 × 1 1 1 = 2 = so the gradient is . 2 2 dx 2

2e −x = 0 but since (1 + e −x )2 e−x is always positive, this can never happen. Therefore there is no turning point.

c At a stationary point,

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Differentiation

8 a f(x) = cos x sin x

d

sin x × (− sin x) − cos x × cos x So f ′(x) = sin 2 x 2 (sin x + cos2 x) =− = −12 = − cosec2 x . sin 2 x sin x

b Using the product rule, if y = x cot x then dy = cot x − x cosec2 x. dx 9 a If v =



So if y =

dy = dx



2

x x +1

du = 2x and 12 a If u = x2 and v = cos x then dx dv = − sin x dx 2x cos x − x 2 × (− sin x) f′(x ) = cos2 x

x then x2 + 1

x2 + 1 − x × x2 + 1

x 2

x +1 .

dy 1 = = dx (x 2 + 1) x 2 + 1

So then −(x − 1)2 1 1 2  −4 = =−  = 2 2 2  y − 1 dy 2 ( y − 1) dx −2 dx = = ( y − 1)2 dy

2

=

The numerator is x2 + 1 − x2 x x2 + 1 − x × = = x2 + 1 x2 + 1 So



1



y +1− y +1 2 = y −1 y −1

=

x 2 + 1 = (x 2 + 1)2 then

−1 dv 1 2 = (x + 1) 2 × 2x = dx 2

1 −(x − 1)2 from part a. From part b, = 2 dy dx y + 1 − ( y − 1) y +1 −1= x −1 = y −1 y −1

1 2

x +1

1 3

x2 + 1

x2 + 1 then x   x 2   × x − x +1 2 dy x + 1 x 2 − (x 2 + 1) = = 2 dx x x2 x2 + 1 1 =− . 2 x x2 + 1

b If y =

.

12 = 1.851; the equation is cos1 y – 1.851 = 6.584(x – 1) or y = 6.584x – 4.733

c At P, y =

13 a If u = x – 1 and v =

c The structure of the formula is the same so dx −2 = dy ( y − 1)2

(

then



dy = Hence dx =

(

x2 + 1 = x2 + 1

dv 1 2 du = 1 and = x +1 dx 2 dx



(x

+1

)

3 2

and x = −1; y =

−1 2

=

(x

+1

1+ x

(x

2

+1

)

)

1 2

× 2x =

x2 + 1 − x2 + x 2

)

x x2 + 1

x x2 + 1

x2 + 1

x 2 + 1 − x(x − 1) 2

)

x 2 + 1 − (x − 1) ×

b At a stationary point

2 2 = cos x +2sin x = 12 cos x cos x

dy ( x − 1) × 1 − (x + 1) × 1 x − 1 − x − 1 = 11 a dx = ( x − 1 )2 ( x − 1)2 − 2 = 2 x − 1) ( b y(x – 1) = x + 1; so xy – y = x + 1; hence xy – x = y + 1; x (y – 1) = y + 1; y +1 x= y −1

2cos1 + sin1 = 6.584 cos2 1

b If x = 1, f ′ ( x ) =

10 tan x = sin x cos x Use the quotient rule. dy cos x × cos x − sin x × (− sin x) = dx cos2 x

2x cos x + x 2 sin x cos2 x

3 2

3 2

=

1+ x

(x

2

+1

)

3 2

= 0 so 1 + x = 0

−1 − 1 −2 = =− 2 1+1 2

The stationary point is (−1, − 2)

c Where the curve crosses the y-axis, x = 0 and 1+0 the gradient is 3 =1 ( 0 + 1) 2

Where the curve crosses the x-axis, y = 0 and x −1 so = 0; x2 + 1



x = 1 and the gradient is or

1 2 2

1+1

(1 + 1)

3 2

=

2

3 22

=

1 2

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4 

WORKED SOLUTIONS

14 a If u = x and v = (x + 1)2 then

16 a Let u = sin x + cos x and v = ex; then du dv = cos x − sin x and = ex dx dx

du = 1 and dx

dv = 2(x + 1) dx



dy (x + 1) − x × 2(x + 1) = dx (x + 1)4

Hence =

2

x + 1 − 2x 1− x = (x + 1)3 (x + 1)3

At a stationary point, y=

=

dy = 0 so x = 1; then dx

( )

1 1 = ; The point is 1, 1 4 (1 + 1)2 4

dy −2sin x = 0 then = 0 and so sin x = 0; dx ex x = 0, π, 2π, 3π, … Stationary points occur every π units along the axis.



The determinant of the quadratic is 9 – 16 = −7 so it has no solution. The only 0 point is where x = 0 and then y = = 0 ; it is 1 the origin (0, 0). du 15 a Let u = 1 – x2 and v = 1 + x2; then = −2x and dx dv = 2x dx





(

=

(

−2x − 2x 3 − 2x + 2x 2

(1 + x )

2 2

(

)

=

v = (1 + x2)2 Then

−4x

(1 + x )

2 2

So

=

2

d y = dx 2

(

(

)

−4 1 + x 2 + 16x 2

(1 + x )

2 3

Hence

(

x x d 2 y −2 e cos x − e sin x 2 = 2x dx e



(1 + x )



let u = –4x and

=

)

−2( cos x − sin x ) ex

At a stationary point sin x = 0 and becomes

2 2

(

d 2y dx 2

−2cos x e 2x

When x = 0, π, 2π, 3π, … cos x is 1, –1, 1, –1 … d 2y alternates in sign +, −, +, … This means dx 2 and the stationary points are alternately maximum and minimum.

)

2

(

− (−4x) × 4x 1 + x 2

(1 + x )

=

1 a If f(x) = tan−1 ax then f ′(x) =

)

2 4

)

12x 2 − 4

(1 + x )

2 3

d 2y = −4 < 0 so the stationary point dx 2 is a maximum.

c If x = 0,



du dv = cos x and = ex dx dx

Exercise 4.6A

)

−4 1 + x 2

d 2y dy −2sin x , differentiate = . dx dx 2 ex

If u = sin x and v = ex then



du = −4 and dx

(

c To find

−4x

dv = 2 1 + x 2 × 2x = 4x 1 + x 2 dx

)

dy There is a stationary point where dx = 0; –4x = 0; x = 0; the stationary point is (0, 1).

b To differentiate



)

ex is always positive and sin x is positive if −2sin x 0 < x < π which means that is ex negative in that interval; the gradient is

b If

x3 + 3x2 + 4x = 0; x(x2 + 3x + 4) = 0; either x = 0 or x2 + 3x + 4 = 0

2 2 dy −2x 1 + x − 2x 1 − x Hence = 2 dx 1 + x2

cos x − sin x − sin x − cos x −2sin x = ex ex

negative.

1− x = 1; (x + 1)3 1 – x = (x + 1)3; 1 – x = x3 + 3x2 + 3x + 1;

b If the gradient is 1 then



Hence dy e x ( cos x − sin x ) − e x (sin x + cos x ) = dx e 2x

1 a . ×a = 1 + (ax)2 1 + a 2x 2

b If f(x) = tan−1 (0.5x − 1) then 1 0.5 × 0.5 = . f ′(x) = 1 + (0.5x − 1)2 1 + (0.5x − 1)2 c If f(x) = 3tan−1 (x2) then 3 6x f ′(x) = × 2x = . 1 + (x 2)2 1 + x4 2 a

dy 1 10 = 10 × = dx 1 + x2 1 + x2

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Differentiation

b Using the chain rule, dy 1 10 = × 10 = dx 1 + (10x)2 1 + 100x 2 dy c dx = dy 3 a = dx

1 1+ x 10

( )

1 1+ x a

()

2

2

×

1 = 10



0.1 10 2 = + x2 x 100 1+ 100

f(x) =

If the gradient is

a 1 1 then 2 = and so 2a a + x 2 2a

But a > 0, so x = a then y = tan−1 1 = π . 4 π The coordinates are a, . 4

( )

( )

π b At a, the gradient of the normal is −2a. 4

The equation is y − π = −2a(x − a) or 4 y + 2ax = 2a2 + π . 4

4 The derivative of x tan−1 x is tan−1 x +

x . 1 + x2



The derivative of 0.5 ln (1 + x2) is 1 x × 2x = . 0.5 × 1 + x2 1 + x2



x x Hence f ′(x) = tan x + − = tan −1 x . 1 + x2 1 + x2 −1

5 If y = tan−1 2x then

dy 1 2 = ×2= . dx 1 + (2x)2 1 + 4x 2 2 = 1. 1 + 4x 2



If the gradient is 1 then



Therefore 2 = 1 + 4x2. 4x2 = 1 1 x2 = 4 1 x=± 2 1 If x = then y = tan−1 1 = π . 2 4 1 −1 If x = − then y = tan  (−1) = − π . 2 4



The points where the gradient is 1 are  1 , π   2 4 and  − 1 , − π  .  2 4

6 a If y =

1 1+ 1 x

()

2

×  − 12  = − 21 .  x  x +1

7 a Let u = 1 + x2 and v = tan–1 x; then

x2 = a2 and so x = ±a.



( 1x ) then

b If f(x) = tan−1 

1 a × = 2 a a + x2

2a2 = a2 + x2.



Hence if f(x) = tan−1  x − 1 then 1 1 1 f ′(x) = × = 1 + (x − 1) 2 x − 1 2x x − 1 .

1

x − 1 = (x − 1)2 then

−1 dy 1 1 = (x − 1) 2 = . dx 2 2 x −1

du = 2x and dx

1 dv = dx 1 + x 2 dy = 2x tan −1 x + 1 + x 2 × 1 2 = 2x tan −1 x + 1 dx 1+ x 2x tan −1 x − 1 + x 2 × 1 2 −1 1 + x = 2x tan x − 1 b dy = 2 2 dx tan −1 x tan −1 x

( (

(

)

) )

(

8 a The derivative of tan–1 x is

)

1 so 1 + x2

1 ⌠ dx = tan −1 x + c  ⌡ 1 + x2 b f ′ ( x ) = =

1 × a

1 a2 + x 2

1

()

1+ x a

2

×

1 1 × = a a2

1

()

1+ x a

2

1 1 c ⌠ dx = ⌠ dx ; using the   ⌡ 25 + x 2 ⌡ 52 + x 2 answer to part b with a = 5 you get

1 1 x ⌠ dx = tan −1 + c  5 5 ⌡ 25 + x 2

dy 1 a 9 a  = ×a= ; at the origin x = 0 dx 1 + (ax)2 1 + a 2x 2 and the gradient is a a a a then = ; hence 2 1 + a 2x 2 2 1 1 + a2x2 = 2; a2x2 = 1; x = ± a

b If the gradient is



( a1 ) = tan ( ±1) = π4 or − π4 ; 1 π the points are ( , ) and ( − 1 , − π ) a 4 a 4 Then y = tan −1 ±a ×

10 a If y = tan–1(x3) then b If y =

(tan–1 x)3

(

dy = 3 tan −1 x dx

then

)

2

–1

dy = dx

1

( )

1+ x

(

3 2

× 3x 2 =

3 tan −1 x 1 × 2 = 1+ x 1 + x2

)

3x 2 1 + x6

2

c = x3 tan–1 x; let u = x3 and v = tan–1 x; then du dv 1 = 3x 2 and = dx dx 1 + x 2

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4

WORKED SOLUTIONS

So dy = 3x 2 tan −1 x + x 3 × 1 2 dx 1+ x = 3x 2 tan −1 x +

4

x3 1 + x2

When t = 1, x = 3, y = 0 and

a If t = π then x = 4 cos π = 2.828 and 4 4 π y = 3 sin = 2.121 and the coordinates are 4  3 2 (2.83, 2.12) to 3 s.f. or  2 2, 2  

5

( )

( )

dy dy dx 2t = ÷ = = 4t . 1 dx dt dt 2 1 When t = −1, x = − and y = −3 and the gradient of 2 1 1 the normal is − = . 4t 4 1 1 The equation of the normal is y + 3 = x + 4 2

( )

3π 3π b If t = 2 then x = 4 cos 2 = 0 and y = 3 sin 3π = −3 and the point is (0, –3). 2 c

therefore 4y + 12 = x +

3

6

2 1 –3

0 –1 –1

–2

1

2

3

4

a

dy dy dx −5sin t 5 = ÷ = = − tan t 4cos t 4 dx dt dt

b

dy dy dx −6sin 2t 3sin 2t = ÷ = = sin t −2sin t dx dt dt

x

This simplifies to

–2 –3 3

2

a

b

t

c

−3 −2 −1

0

1

2

3

x

9

4

1

0

1

4

9

y

−1

0

1

2

3

4

5

7

y 8

5

3 2 1 0

c

1

2

3

4

5

6

7

dy dy dx 1 = 2t and = 1 so = . dt dx 2t dt

d At (4, 0) t = −2 and the gradient dy 1 1 1 = = =− . 4 dx 2t 2 × −2 3

a

dy dy dx 2t = ÷ = = 2t 1 dx dt dt

dy dy dx 4t 3 b = ÷ = = 4t 3 1 dx dt dt c

8

dy dy dx 2t − 3 = ÷ = or 1 − 3 2t 2t dx dt dt

9

3 × 2sin t cos t = 6 cos t. sin t

dy dy dx 2cos 2t = ÷ = = cos 2t × cos2 2t dx dt dt 2sec2 2t = cos3 2t

dy dy dx 7cos t 7 = ÷ = = − cot t 3 dx dt dt −3sin t 7 π 7 7 π When t = the gradient is − 3 cot 4 = − 3 × 1 = − 3 . 4 dy dy dx −3sin t = ÷ = dx dt dt 4cos 2t π The coordinates are (0, 0) when t = or 3π 2 2 π dy −3 sin 2 = = −3 = 3 . When t = π the gradient dx 4cos π 4 × −1 4 2 3 π When t = the gradient 2 3π dy −3 sin 2 = = −3 × −1 = − 3 . dx 4cos3π 4 × −1 4

4

–1

1 . 2

So 8y + 24 = 2x + 1. So 8y − 2x + 23 = 0.

y

–4

dy = 2. dx

The equation of the tangent is y − 0 = 2(x − 3) or y = 2x − 6 or 2x − y − 6 = 0.

Exercise 4.7A 1

dy dy dx 2t = ÷ = = 2t 1 dx dt dt

x

9

dy dy dx 2sin t = ÷ = . dx dt dt 2 − 2cos t 2× 1 2sin π dy π 2 2 = 4 When t = , = = 4 dx 2 − 2cos π 2 − 2 × 1 2− 2 4 2 =

2 × 2 + 2 = 2 2 + 2 = 2 + 1. 4−2 2− 2 2+ 2

So the gradient of the tangent is 2 + 1.

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Differentiation



10

The gradient of the normal is 1 2 −1 2 −1 1 − =− × =− = 1 − 2. 2−1 2 +1 2 +1 2 −1 dy dy dx b cos t b = ÷ = = − cot t a dx dt dt −a sin t d y = − b cot π = − b When t = π , 4 a a 4 dx a b π and x = a cos  = and y = b sin π = . 4 4 2 2 b = −bx − a  The equation of the tangent is y − . a  2 2 At the y-intercept x = 0 and so b b a y− =− ×− a 2 2 b b y− = 2 2 2 b y= = 2b 2 The y-intercept is (0, 2b ).

11 a

dy dy dx −4sin 2t −0.4sin 2t = ÷ = = 10cos t cos t dx dt dt

b At the highest point the gradient is 0 so −0.4sin 2t = 0. cos t Therefore sin 2t = 0 and t = 0.

c Where it meets the x-axis, y = 0 and x = 20t; the point is (20t, 0).

1 20 × 20t × = 200 2 t which is a constant and does not depend on t.

14 a



This means that the tangent of the angle of slope is 0.720.



So the angle is arctan 0.720 = 0.624 radians. That is 0.624 × 180 = 36° to the nearest degree. π dy dx 12 a = 2sin θ cosθ and = −2sin θ ; dθ dθ dy −2sin θ −1 = = dx 2sin θ cosθ cosθ b When the gradient is 1, cos θ = – 1 and the point is (sin2 θ, 2cos θ) = (0, – 2) and the equation of the tangent is y = x – 2



Then sin θ = ±1; For θ > 0, the point on the curve (sin2 θ, 2 cos θ) is (1, 0) where the tangent is parallel to the y-axis.

dy dy dx 10 10 1 = 10 and = − 2 so = − 2 ÷ 10 = − 2 dt dt dx t t t 10 1 b The equation is y − = − 2 (x − 10t ); t t multiply by t2 : t2y – 10t = –x +10t; hence x + t2y = 20t.

13 a

dy  1 t2 − 1 1 t2 − 1 = 1 − 2  ÷ 2t = × = 2t dx  t  t2 2t 3 If t = 1 the point is (1, 2) and if t = −1 the point is (1, −2)

c Consider points near (1, 2) where t = 1; if t = 0.9 then point is (0.81, 2.01) and dy 0.92 − 1 = < 0; if t = 1.1 then the point dx 2 × 0.93 dy 1.12 − 1 is (1.21, 2.01) then = > 0; this dx 2 × 1.13 means (1, 2) is a minimum point.

Consider points near (1, −2) where t = −1; if t = −0.9 then point is (0.81, −2.01) and dy 0.92 − 1 = > 0; if t = 1.1 then the point dx −2 × 0.93 dy 1.12 − 1 is (1.21, −2.01) then = < 0; this dx −2 × 1.13 means (1, −2) is a maximum point.

15 a

dy dy dx 2 1 = 2t and = 2 so = = dt dt dx 2t t

b The gradient of the normal is –t. The equation is y – 2t = –t(x – t2) or y + tx = 2t + t3 c Where the normal meets the y-axis, x = 0 and y = 2t + t3

Where the normal meets the x-axis, y = 0 and tx = 2t + t3; so x = 2 + t2



The area of the triangle is 1 1 × 2t + t 3 2 + t 2 = t 2 + t 2 2 + t 2 = 2 2 1 2 2 t 2+t 2

c When the tangent is parallel to the y-axis the gradient is infinite so cos θ = 0.

dy 1 dx = 2t and = 1 − 2 so dt dt t

b At a stationary point, t2 – 1 = 0 so t = ±1

dy −0.4sin 2.24 = = −0.720. cos1.12 dx

Then

( )

d The area of the triangle =

c If x = 9, 10 sin t = 9, sin t = 0.9 and t = 1.120.

Where it meets the y-axis, x = 0 and t2 y = 20t 20 20 so y = . ; the point is 0, t t

( (

)(

)

(

)(

)

)

dy dy 3cos3t dx = 3cos3t ; = 16 a  = 2cos2t and dt dt dx 2cos2t

At A the gradient is 0 so cos 3t = 0; one π π solution is 3t = ; t = . 2 6

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4 

WORKED SOLUTIONS



π 3 π = and y = sin = 1; the 3 2 2  3  coordinates of A are  2 , 1 .  

b At B the gradient is infinite so cos 2t = 0; one π π solution is 2t = ; t = 2 4

b Differentiate: 2xy + x2

Then x = sin

Then x = sin

3π 1 π = ; the = 1 and y = sin 4 2 2

Another solution is t = π because sin 2π = sin 3π = 0; then dy = 3cos3π = −3 2 dx 2cos2π



3 3 The gradients are and − 2 2



So 8





dy 2x + y = dx 4y − x y3 2 2 dy 2 2 dy dy 3 − 3 dx ; 3 = − 3 dx ; dx = − 3 x y x y x

d Using the produce rule,



sin x sin y so



dy = 0; dx

dy = cos x cos y ; dx

dy cos x cos y = or cot x cot y dx sin x sin y

2 a 4x + 2y



dy dy dy = 4y ; 2x + y = (4y − x) ; dx dx dx

cos x cos y + sin x (− sin y)

dy dy + 2y =0 dx dx dy If x = 5 and y = 0, 10 + 0 + 20 +0=0 dx dy dy = −10 and = −0.5 Therefore 20 dx dx dy =0 dx

dy = −(2x + 4y) dx dy −(2x + 4y) x + 2y =− = 4x + 2y 2x + y dx

(4x + 2y)

b Differentiate: 2y

b Using the product rule on the left-hand side ( x + y ) + x  1 + ddxy  = 4y ddxy ;

c 0 = −

If x = 0 and y = 5 then x2 + 4xy + y2 = 0 + 0 + 25 = 25

5 a If x = 3 and y = 5 then y2 − x2 = 25 − 9 = 16

dy dy x 1 a x = 0; 2 − 4y = dx dx 2y

x+y+x+x

dy dy 1 = −4 and so =− . 2 dx dx

c 2x + 4y + (4x + 2y)

Exercise 4.8A



dy dy = −4 . dx dx

b Differentiate: 2x + 4y + 4x

dy 3cos0 3 = = dx 2cos0 2



If x = 2 and y = 1 then 4 + 4



c At the origin x = y = 0 so sin 2t = sin 3t = 0 One solution is t = 0; then



4 a If x = 5 and y = 0 then x2 + 4xy + y2 = 25 + 0 + 0 = 25

  1  2 coordinates of B are  1,  or  1, 2     2



dy dy = −4 . dx dx

dy dy = 0 therefore 2x = −y and dx dx

dy 2x =− . y dx

dy 6 3 =− =− . 4 2 dx dy 6 3 c If x = 3 and y = −4 then dx = − −4 = 2 .

b If x = 3 and y = 4 then

3 a  If x = 2 and y = 1 then x2y = 4 and 4(2 − y) = 4 so the point is on the line.



dy − 2x = 0 so dy = x . dx dx y

dy 3 = and the equation of the dx 5 tangent is y − 5 = 3 (x − 3) 5 Or 5y − 25 = 3x − 9 or 3x − 5y + 16 = 0 At (3, 5)

6 Differentiate: dy dy 2(x 2 + y 2) ×  2x + 2y  = 50  2x − 2y  dx  dx    dy = 0 and so dx 2 2 2(x + y ) × (2x) = 50(2x). Therefore 2(x2 + y2) = 50 x2 + y2 = 25 This is a circle of radius 5.

At a stationary point,

7 a Differentiate: 2x − 2y

dy dy x = 0 and so = . dx dx y

dy dy dx sec2 t sec t = ÷ = = dx dt dt sec t tan t tan t x sec t and c If x = sec t and y = tan t then = y tan t the expressions are equivalent. b

sec t d If the gradient is 2, = 2 therefore tan t sec t = 2 tan t. 1 2sin t cos t = cos t So 2 sin t = 1

sin t =

1 so t = π and 5π . 2 6 6

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Differentiation





π If t = 6 then x = sec π = 2 and 6 3 π 1  2 1  y = tan = , . and the point is  6  3 3  3





If t = 5π then x = sec 5π = − 2 and 6 6 3 5 1 π y = tan =− and the point is 6 3  2 1  ,− − .  3 3

c Where the normal and the curve cross, x2 + xy + y2 = 7 and x = –4y – 5

8 a If x = −4 and y = 3 then x2 + xy + y 2 = 16 – 12 + 9 = 13 so the point is on the curve. b Differentiate: 2x + y + x



dy dy + 2y = 0; dx dx

If x = −4 and y = 3 then dy dy dy −8 + 3 − 4 +6 = 0 ; −5 + 2 = 0; dx dx dx dy 5 = dx 2 The equation of the tangent is y − 3 =



Substitute: (–4y – 5)2 + (–4y – 5)y + y2 = 7; 16y2 + 40y + 25 – 4y2 – 5y + y2 = 7



13y2 + 35y + 18 = 0; (y + 2)(13y + 9) = 0; y = −2 9 as given or − ; there are two solutions which 13 shows that the normal meets the curve again.

Exam-style questions 1 2

1 a y = (1 + x 2) and 5 (x + 4); 2



dy dy 9 a Differentiate: 2x + 2y −4+6 = 0; dx dx ( 2y + 6 ) ddxy = 4 − 2x ; ddxy = 42y−+2x6 dy = 0 so 4 – 2x = 0 and At a stationary point dx hence x = 2



Substitute x = 2 into the equation of the curve: 4 + y2 – 8 + 6y = 12



y2 + 6y – 16 = 0; (y – 2)(y + 8) = 0 y = 2 or −8



A and B are (2, 2) and (2, −8)

b

4 and the equation is 3

4 y − 1 = − (x + 1); 3y – 3 = –4x – 4; 3y + 4x + 1 = 0 3

The midpoint of AB is (2, −3) and if x = 2 and y = −3 then 3y + 4x + 1 = –9 + 8 + 1 = 0 which shows the midpoint is on the normal.

10 a If x = 3 and y = −2 then x2 + xy + y2 = 9 – 6 + 4 = 7 so (3, −2) is on the curve. dy dy + 2y = 0 ; at (3, −2) dx dx dy dy dy 6−2+3 −4 = 0; =4 dx dx dx

b 2x + y + x

1

(1 + x 2)2

x 1 + x2

dy x x x = = 1 = y 1 + x 2 (1 + x 2)2 dx

b f(x) = e−x + x × (−e−x) = (1 − x)e−x.  At x = 0, gradient = 1 3 a  y = x2 cos x;

dy π = 2x cos x − x 2 sin x ; if x = then 2 dx

()

2 2 dy = π cos π − π sin π = − π 2 2 2 dx 4 2 cos x dy x ( − sin x ) − 2x cos x ; if x = π 2 ; dx = x x4 dy π 2 ( − sin π ) − 2π cos π 2π 2 then dx = = 4 = 3 π4 π π

b y =

= 1 + 1 + 4 + 6 = 12 so (−1, 1) is on the curve.

of the normal is −

x

2 a f(x) = −2e−2x. At x = 0, gradient = –2

b If x = −1 and y = 1 then x 2 + y 2 − 4x + 6y c If x = −1 and y = 1 then dy 4 − 2x 4 + 2 6 3 = = = = ; the gradient dx 2y + 6 2 + 6 8 4

1 − dy 1 = (1 + x 2) 2 × 2x = dx 2

           =

2y – 6 = 5x + 20; 2y = 5x + 26



1 The gradient of the normal is − and the 4 1 equation is y + 2 = − (x − 3) 4 or 4y + 8 = –x + 3 or x + 4y + 5 = 0

4 a  Using the quotient rule, the derivative is



2x(x 2 + 1) − 2x(x 2 − 1) 2x 3 + 2x − 2x 3 + 2x = (x 2 + 1)2 (x 2 + 1)2 4x = (x 2 + 1)2 . n −1 n n −1 n b The derivative is nx (x + 1n) − nx2 (x − 1) (x + 1)





2n −1 + nx n −1 − nx 2n −1 + nx n −1 = 2nx n −1 . = nx (x n + 1)2 (x n + 1)2

5 y = 2 ln x so

dy 2 = . dx x

If the gradient is 0.5 then 2 = 0.5 so x = 4 and x y = ln 16. The coordinates are (4, ln 16).

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4 

WORKED SOLUTIONS

6 y = e2x – 11ex + 12x;

dy = 2e 2x − 11e x + 12 dx

11 y = x2 tan 2x;

At a stationary point, 2e2x – 11ex + 12 = 0; (2ex – 3)(ex – 4) = 0; ex = 1.5 or 4; x = ln 1.5 or ln 4



d 2y = 4e 2x − 11e x dx 2 d 2y If x = ln 1.5 then = 4e 2 ln1.5 − 11e ln1.5 dx 2 = 4 × 1.52 – 11 × 1.5 = −7.5 which is negative. This is

a maximum point. d 2y = 4e 2 ln 4 − 11e ln 4 If x = In 4 then dx 2 = 4 × 16 – 11 × 4 = 20 which is positive. This is a minimum point. dy 7 a  = 0.3 × 2 cos 2x − 0.4 × (−2sin 2x) dx = 0.6 cos 2x + 0.8 sin 2x d 2y = −1.2 sin 2x +1.6 cos 2x b dx 2 d 2y d 2y + 4y = 0 4y = 1.2 sin 2x − 1.6 cos 2x = − 2 so dx dx 2 dy 8 a = 0.1ex − 0.5e− 0.5x dx b At a minimum point, 0.1ex − 0.5e−0.5x = 0. e0.5x:

0.1e1.5x



Multiply by



Take logs: 1.5x = ln 5.



ln 5 x= = 1.073 1.5



Then y = 0.1e1.073 + e−0.536 = 0.877.



The coordinates are (1.073, 0.877).

= 0.5 and so

e1.5x

dy dy dx sin θ 9 a dθ = dθ ÷ dθ = 1 − cosθ sin θ = 0.5 so sin θ = 0.5(1 − cos θ). b 1 − cosθ

2 sin θ = 1 − cos θ



2sin θ + cos θ = 1

dy 10 a = −e−x sin 10x + 10e−x cos 10x dx b At a stationary point −e−x sin 10x + 10e−x cos 10x = 0.

Therefore −sin 10x + 10 cos 10x = 0.



sin 10x = 10 cos 10x



tan 10x = 10



10x = 1.471 or 1.471 + π or...

x = 0.147 is the smallest value and is the x-coordinate of the first maximum point. c The second value is given by 10x = 4.613.

So x = 0.461 at the first minimum point.

= 5.

dy = 2x tan 2x + x 2 sec2 2x × 2 dx

= 2x tan 2x + 2x2 sec2 2x dy π π 2 2 = tan π + π sec2 π = If x = then 3 36 3 6 dx 3 π π2 π 2 2 × 3+ ×4= + π 2 ; a = and b = 3 3 18 9 3 9

12 (10, 20) is on the curve so 20 = ae10k. dy = kaekx The gradient of the line from (10, 20) to dx 25 = 2.5. (0, −5) is 10 Therefore kae10k = 2.5 From these two equations, 20k = 2.5 and so k = 0.125. Hence 20 = ae10 ×0.125 = ae1.25 a = 20 ÷ e1.25 = 5.73 dy = 3 sin2 x × cos x 13 a y = sin3 x so dx

= 3 cos x sin2 x = 3 cos x (1 − cos2 x)



= 3 cos x − 3 cos3 x

b

d 2y  − 3 sin x − 9 cos2 x × (−sin x) = dx 2 = −3 sin x + 9 sin x cos2 x





= −3 sin x + 9 sin x (1 − sin2 x)





= −3 sin x + 9 sin x − 9sin3 x





= 6 sin x − 9 sin3 x

14 a If y = e

− 1 x2 2

then

−1 x2 e 2

−1 x2 dy × (−x) = −xe 2 … = −xy which is the  = dx result required.

b When x = a, y = e

dy = dx

and

− 1 a2 −ae 2 .

The equation of the tangent at P is y−e



− 1 a2 2

− 1 a2 2

= −ae

− 1 a2 2 (x

− a).

Where this crosses the x-axis, y = 0 and −e

− 1 a2 2

= −ae

− 1 a2 2 (x

− a) .



Hence 1 = a(x − a),



which you can rearrange to x = a +

15 a Differentiate implicitly: 2x − 2y − 2x dy = 0. dx



At a maximum point,



So 2x − 2y = 0 and y = x.



Substitute in the initial equation. x2 − 2x2 + 2x2 = 9.

1 a

dy dy + 4y = 0. dx dx

x2 = 9 x = 3 or −3.

53

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Differentiation



From the sketch it is clear that the maximum point is where x = 3.



Then 9 − 6y + 2y2 = 9.

c At a stationary point, dy = 5e −2x cos ( x + 1.107 ) = 0 ; dx cos(x + 1.107) = 0

y2 − 3y = 0 y(y − 3) = 0

y = 0 or 3.



From the diagram, the maximum point must be (3, 3).

b Where the graph crosses the x-axis, y = 0, and from part a the point is (3, 0). dy dy + 4y = 0. dx dx



Substitute in 2x − 2y − 2x



Hence 6 − 6



The gradient of the normal is –1.



The equation of the normal is y = −1(x − 3) or x + y = 3.

dy dy = 0 and = 1. dx dx

16 a Let u = x and v = x + 1 then du = dv = 1 dx dx Hence, using the quotient rule, dy ( x + 1) − x 1 = = dx (x + 1)2 ( x + 1 )2 b Differentiate to x:

y x + = xy 3 with respect x +1 y +1

dy dy 1 1 + = y 3 + x × 3y 2 dx (x + 1)2 ( y + 1)2 dx

dy dy 1 1 + = y 3 + 3xy 2 ; dx (x + 1)2 ( y + 1)2 dx dy 1 1 dy + =1+3 4 4 dx dx 11 dy −3 dy −3 = = ; 4 dx 4 dx 11

if x = y = 1 then

11 and the c The gradient of the normal is− 3 11 equation is y − 1 = (x − 1) which is 3 3y – 3 = 11x – 11 or 11x – 3y = 8 17 a  y = e–2x sin x; =

e–2x

dy = −2e −2x sin x + e −2x cos x dx

(cos x – 2 sin x)



If cos x – 2 sin x ≡ R cos(x + α) = R cos x cos α – R sin x sin α



Hence R cos α = 1 and R sin α = 2; 2

2

R = 1 + 2 = 5 ; tan α = 2; a = tan–1 2 = 1.107 to 3 d.p.

Hence

b Where the curve crosses the y-axis, x = 0 and dy = 5 cos1.107 = 1.00 dx

dy = 5e −2x cos ( x + 1.107 ) dx



π ; x = 0.464 to 3 d.p. 2 and this is the closest to the y-axis. A solution is x + 1.107 =

Then y = e–2x sin x = 0.177 to 3 d.p. The coordinates are (0.464, 0.177)

dy dx 18 a  = −5sin t and = 5cos t ; hence dt dt dy 5cost = = − cot t dx −5sin t b At a stationary point, − cot t = −

cost =0 sin t

π and sin t = ±1 2



So cos t = 0 and then t = ±



Stationary points are when t =

π at (1, 7) or 2

π at (1, −3) 2 c At the point with parameter t the gradient of the normal is tan t and the equation is when t = −



y – 2 – 5 sin t = tan t (x – 1 – 5 cos t ); so sin t y − 2 − 5 sin t = ( x − 1 − 5cost ) cost y cos t – 2 cos t – 5 sin t cos t = x sin t – sin t – 5 sin t cos t y cos t – 2 cos t = x sin t – sin t; or (y – 2) cos t = (x – 1) sin t d x = 1 and y = 2 satisfies this equation so (1, 2) is on the normal. 19 a If y = 2x then ln y = x ln 2; differentiate both sides 1 dy = ln 2 with respect to x; y dx

so dy = ( ln 2) y = ( ln 2) 2x dx 2

b If y = a x then ln y = x2 ln a; differentiate both sides with respect to x:

dy 1 dy = 2x ( ln a ) so = 2 ( ln a ) xy y dx dx

c If y = xx then ln y = x ln x; differentiate using 1 dy 1 = ln x + x × ; the product rule: y dx x

dy so 1 = ln x + 1 ; y dx dy = ( ln x + 1) y = ( ln x + 1) x x dx

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4 

WORKED SOLUTIONS

20 a  x sec x = and

dv = − sin x dx

(

)



x 2

x +1

  x + 1 x

2

1 x2 + 1 + x = x2 + 1 x2 + 1 dy dx 21 a x = t2 + 2 and y = 2t + 3 so = 2t and =2 dt dt dy 2 1 and = = dx 2t t 1 1 1 If the gradient is − then = − and t = −2; t 2 2 A is (6, −1)

=

1 × x + x2 + 1

The tangent at A is y + 1 = −

1 ( x − 6 ) so 2

Equation of tangent at x = 0: π x x π y − = or y = + 4 2 2 4

Mathematics in life and work dy e x × (e x + 1) − e x × e x ex = = x x 2 dx (e + 1) (e + 1)2 ex = ex + 1 − ex = 1 e +1 ex + 1 ex + 1 x x dy e × x1 = xe 2 = Hence y(1 − y) = x dx . e + 1 e + 1 (e + 1)

2 1 − y = 1 −

2y + 2 = –x + 6 or x + 2y = 4





The tangent at B is y – 5 = x – 3 or y = x + 2



Where these lines cross, x + 2 (x + 2) = 4; x + 2x + 4 = 4; x = 0 and y = 2 so they meet on the y-axis at (0, 2)

c

dy π + 2 π = and y = 2 2 dx −2 Gradient of normal = π+2 Equation of normal: y − π = −2(x − 1) 2 π+2 x dy 23  = e 2x dx 1 + e dy 1 π At x = 0: y = and = 4 dx 2

1

1 b A  t B, = 1 so t = 1 and B is (3, 5) t

y 2x tan −1 x = = 2tan −1x x x f(x) + y = 2x + 2tan −1 x = dy f(x) x 1 + x 2 dx

b At x = 1,

So if y = ln x + x 2 + 1 then  dy 1 = × 1+ dx x + x 2 + 1 

The base of the triangle is 6 and the area is 1 ×6×2=6 2

22 a dy = 2tan −1 x + 2x 2 dx 1+x If f(x) = 1 + x2, then f(x) = 2x

dy cos x + x sin x x sin x 1 = = + and cos x cos2 x dx cos2 x x tan x = sec x + = sec x + x tan x sec x cos x = sec x (1 + xtan x)

b If u = x + x 2 + 1 then du = 1 + 1 × 1 × 2x = 1 + 2 dx x2 + 1



x du ; if u = x and v = cos x then =1 cos x dx

3



y

x

dy kekx × (ekx + 1) − kekx × ekx kekx = = kx 2 kx dx (e + 1) (e + 1)2

=k×

1 ekx × = ky(1 − y) e + 1 ekx + 1 kx

6

B 4

2

–2

0

2

4

6

x4

A –2



The height of the triangle is 2.



x + 2y = 4 meets the x-axis at 4; y = x + 2 meets the x-axis at −2.

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INTEGRATION

5 Integration Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1

a If f(x) = sin (x2 + 1) then by the chain rule f ′(x) = cos(x 2 + 1) × 2x = 2x cos (x2 + 1)

= 11.3889. The area of the ellipse ≈ 4 × 11.3889 = 45.56 to 4 s.f.

b y = uv where u = e–x and v = cos 2x Then

du dv = −e −x and = −2sin 2x. dx dx

dy dv du =u +v dx dx dx = e–x × (–2 sin 2x ) – e–x × cos 2x = –e–x(2 sin 2x + cos 2 x). Hence

4

a

∫ 4x

−1 2dx

3

1

= 4 × 2x 2 + c = = 8 x + c

2

b The integral is a multiple of (2x – 1)6. If f(x) = (2x – 1)6 then

1

f ′(x) = 6 (2x − 1)5 × 2 = 12 (2x − 1)5. Hence ∫ (2x − 1)5 dx = c

3

1 (2x − 1)6 + c 12

0

4x 3 + 3 dx = ∫ 4x + 3x −2 dx = 2x2 – 3x–1 + c x2 3 or 2x 2 − + c x

∫

1

∫0 10e

−2x

x

2

Area =

−x 2

0.5

1

1.5

2

4 3.1152 1.4715 0.4216 0.0733

2

∫0 y dx = 4e

−x 2

dx

≈ 1 × 0.5 {4 + 2(3.1152 + 1.4715 + 0.4216) + 0.0733} 2

1 In this case h = 2. × 0.5 {4 + 2(3.1152 + 1.4715 + 0.4216) + 0.0733} 4 2 1 ∫−2 f(x)dx ≈ 2 × 2 {12.25 + 2(17.89 + 21.04) + 23.38} = 3.5225 = 113.49 h = 0.1 and 0.6

∫0.2

1 − x 2 dx ≈

0.9798 + 2(0.9539  1 × 0.1   2 + 0.9165 + 0.8660) + 0.8  

= 0.3626. 3

x

2

0

y = 4e

1

Exercise 5.1A 1

1

b The width of each strip is 2 ÷ 4 = 0.5.

dx =  −5e −2x  = [–5e–2] – [–5] =  0 5 – 5e–2 = 4.323 to 3 d.p. The area is

b Every chord is below the curve and so every trapezium is smaller than the area it approximates. a y 4

c If f(x) = ln (3x2 + 2) then using the chain rule, 6x 1 f ′(x) = 2 × 6x = 2 . 3x + 2 3x + 2 2

1 × 1 {3 + 2(2.9394 + 2.7495 + 2.4 + 1.8) + 0} 2

≈

a h = 1.

5

c Statement C is correct. a 21 1 1 1 1 1 1 ∫1 x dx ≈ 2 × 0.1 1 + 2 1.1 + 1.2 + 1.3 + ... + 1.9 + 2 = 0.6938 to 4 d.p.

{ (

) }

b It is an overestimate.

The area of a quarter of the ellipse

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5

WORKED SOLUTIONS

6

b

a

15

8

10

5

–1

0

1

2

3

4

5

–5 0 –10 4

c The area between 0 and 4 under y = 4 x is

(

)

b Area = ∫ 12x − 3x 2 dx = 6x 2 − 0

4 x 3  0

≈

= [96 – 64] – [0] = 32 c Each chord will be below the curve so each trapezium will have an area less than the area between the curve and the x-axis.

Percentage error = a h = 1; area ≈

{

(

) }

(

1 2 x is 2

1 0 + 2 ( 0.5 + 2 + 4.5) + 8} = 11 2{ The estimate of the area between them is 20.585 – 11 = 9.585 ≈

4

41

0

0

d The exact area = ∫ 4 x dx −∫

32 − 30 × 100 = 6.25% 32

1 5+2 2

{

1 0 + 2 4 + 4 2 + 4 3 + 8 = 20.585 2

The corresponding area for y =

d Trapezium rule estimate is 1 × 1 × {0 + 2 ( 9 + 12 + 9) + 0} = 30 2

7

4

2

x 2 dx

4

4  2 3 1 = 4 × x 2  −  x 3    3 6 0  0 

) }

24 + 21 + 4 + 3 + 0

2 64 64 1 2 − = 21 − 10 = 10 3 3 6 3 3 a Using the product rule, f(x) = e–x – xe–x . At a stationary point, e–x – xe–x = 0. =

= 18.98 b y = 25 − x 2 → y2 = 25 – x2 → x2 + y2 = 25 This is the equation of a circle of radius 5.

9

e–x (1 – x) = 0; e–x is always positive so 1 – x = 0 and x = 1 f (x) = –e–x – e–x + xe–x; f (1) = –e–1 which is negative. The stationary point is a maximum and the maximum value is f(1) = e–1

5

b

y

0.5 –5

0

5

The area is a quarter of the circle. 1 Area = 4 × π × 52 = 6.25π 8

1 2 x = 4 x ; so x 2 = 8 x ; 2 so x4 = 64x; so x (x3 – 64) = 0

a Where they meet,

–2

0

1

x

c h = 0.5; area 0.5 ≈ 0 + 2 0.5e −0.5 + e −1 + 1.5e −1.5 + 2e −2 2

{

(

)

}

= 0.5706 to 4 d.p.

Either x = 0 or x3 = 64 and x = 4 The graphs cross at (0, 0) and (4, 8)

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INTEGRATION

c If f(x) = ln (5x + 2) then 1 5 . f ′(x) = ×5= 5x + 2 5x + 2

10 a 1

Hence 0.5

5 0

0.5

1

1.5

3

b h = 1 and ∫ 0.5x dx ≈ 0

{

100

∫0

{

100

∫0

2.5

(

0.5x dx ≈

(

{

)

(

b

3

1 1 + 2 0.5 + 0.52 + 0.53 2

= 1.3125 c h = 1 and

2

100

Hence ∫

0

{

}

1

∫

1 Hence ∫ sin(2x − 3)dx = − 2 cos(2x − 3) + c. a –2e–2x + c b 2e0.5x + 2e–0.5x + c 3 c x 2 + e 4x + 5 + c 4 1 1 a − cos 2x − sin 2x + c 2 2 b

d sin(0.1x + 1.3) = 0.1cos(0.1x + 1.3) so the dx integral is 4 sin(0.1x + 1.3) + c = 40sin(0.1x + 1.3) + c. 0.1

4

4 5 = − 5 cos5x − 4 sin 4x + c

c

∫ 4sin 5x − 5cos4x dx

a

∫ x dx = 5 ∫ x dx = 5 ln |x| + c

5

a 4 dx = [ 4 ln x ]1 = 4 ln a x Therefore 4 ln a = 10. ln a = 2.5 a = e2.5 or 12.18 a

a 2 ln | x + 1| + c 1 2 d b dx ln(2x − 1) = 2x − 1 × 2 = 2x − 1

8

1

2 2 1 2 dx = ∫ dx = ln x + c b ∫ 5x 5 x 5

∫

1 x2 − 6 3 1 dx =∫ x − dx = x 2 − 3 ln x + c 2x 2 x 4

Area =

4

4 0.5x − 2

∫2 e

(

1 2x − 3 e + c. 2

c If f(x) = cos(2x – 3) then f ′(x) = − sin(2x − 3) × 2 = –2sin (2x – 3).

x

∫1

c

1 dx = 2x − 3 + c. 2x − 3

Hence ∫ e 2x − 3 dx =

π

3 Therefore integral = 2 ln | 2x − 1| + c.

1 . 2x − 3

b If f(x) = e2x – 3 then f ′(x) = e 2x − 3 × 2 = 2e 2x − 3.

3

6

a If f(x) = (2x − 3)2 then

Hence

}

0

7

−1 1 f ′(x) = (2x − 3) 2 × 2 = 2

2

)

1 1 + 2 × 1 + 0.5100 = 1.5 2

Exercise 5.2A 1

π

2

}

)

0.5x dx ≈

2cos0.5x dx = [ 4sin 0.5x ]0 = [ 4 ] − [ 0 ] = 4

1 1 1 + 2 0.5 + 0.52 + 0.53 + ... + 0.599 + 0.5100 2

)

1

y

}

1 0.5x dx ≈ 1 + 2 0.5 + 0.52 + 0.53 + ... + 0.599 + 0.5100 2 Now 0.5 + 0.52 + 0.53 + ... + 0.599 is the sum of a geometric progression with a = 0.5, r = 0.5 and 0.5 1 − 0.599 n = 99; S99 =1 1 − 0.5

(

π

∫0

a

1

∫ 5x + 2 dx = 5 ln 5x + 2 + c.

dx =  2e0.5x − 2  =  2  −  2e −1  2

)

= 2 1 − e −1 = 1.264. d sin(x 2) = cos(x 2) × 2x = 2x cos(x 2) dx 1 b sin(x 2) + c 2 d c cos(x 2) = − sin(x 2) × 2x = −2x sin(x 2) dx 1 Hence ∫ x sin(x 2)dx = − cos(x 2) + c. 2 10 Both are correct as they have different constants. ln 2x = ln 2 + ln x so Bhaskar’s answer is 1 1 ln 2 + ln x + c which differs from Alice’s only by 2 2 1 the constant ln 2. 2 11 a Using the product rule, f ′(x) = e x(sin x + cos x) + e x(cos x − sin x) 9

a

= 2ex cos x. b

∫e

x

cos x dx =

1 x e (sin x + cos x) + c 2

c Try g(x) = ex (sin x – cos x).

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5 

WORKED SOLUTIONS





Then g′(x) = e x(sin x − cos x) − e x(cos x − sin x) = 2ex sin x.

x 1+ x −1 = = 1+ x − 1 1+ x 1+ x 1+ x 1+ x 1 = 1+ x − ; 1+ x



ii 

1 x Hence ∫ e x sin x dx = 2 e (sin x − cos x ) + c.

12 a Using the product rule with u = x and v = ln x; du dv 1 = 1 and = dx dx x Hence f ′ ( x ) = ln x + x ×

1 = ln x + 1 x

x dx = Hence ⌠  ⌡ 1+ x

3 = 2 (1 + x)2 − 2 1 + x + c 3

1 a tan  x+c

x ln x + c = ∫ ln x dx + x ; rearrange as

b

∫ ln x dx = x ln x − x + c c ∫ ln x dx = [ x ln x − x ]1 = [ a ln a − a ] − [ 0 − 1] a

= a ln a – a + 1 If a ln a – a + 1 = 1 then a ln a = a; hence ln a = 1 and a = e

)

3

13 a Let f(x) = x 2 + 1 2 ; using the chain rule, f ′(x ) =

(

3 2 x +1 2

)

1 2

(

× 2x = 3x x 2 + 1

)

x x +1

f ′(x ) =

(

= x x2 + 1

(

2 2 x +1 3

)

−1 3

)

−1 3

(

)

)

3 2

(

4 x x2 + 1 3

)

2

(

)

3 2

+c

1 = (x − 1)−2 then (x − 1)2 f(x) = –2(x – 1)–3 × 1 = –2(x – 1)–3

14 a i If f ( x ) =

⌠ 1 1 Hence  3 dx = − 2 +c ⌡ (x − 1) 2 ( x − 1) x x −1+1 ii  = (x − 1)3 (x − 1)3 x −1 1 1 1 + = + (x − 1)3 (x − 1)3 (x − 1)2 (x − 1)3

x 1 ⌠ 1 dx +⌠ Hence ⌠ dx = dx   ⌡ (x − 1)3 ⌡ (x − 1)2 ⌡ (x − 1)3 1 =− 1 − +c x − 1 2 ( x − 1)2 1

1

− 1 b i ⌠ dx =∫ (1 + x) 2 = 2(1 + x)2 + c  ⌡ 1+ x

=2 1+ x +c

)dx = 1 tan(x 2) + c. 2

2



=

−(sin 2 x + cos2 x) 1 =− = −cosec2x. sin 2 x sin 2 x

b From part a it follows that 2 ∫ cosec x dx = − cot x + c.

−1 3

⌠ x 3 3 Hence  3 dx = f(x) = x 2 + 1 4 4 ⌡ x2 + 1

=

2

cos x then using the quotient rule, sin x sin x × (− sin x) − cos x × cos x f ′(x) = sin 2 x

+c

so try f ( x ) = x 2 + 1 3 ;

× 2x =

∫ x sec (x

3 a If f(x) =

(

2

d tan(2x − 1) = 2sec2(2x − 1) so the integral dx 1 is tan(2x − 1) + c. 2 2 a Using the chain rule, the derivative of tan (x2) is sec2 (x2) × 2x = 2x sec2 (x2) b Hence

1 2

1 1 2 2 Hence ⌠  x x + 1 dx = 3 f ( x ) + c = 3 x + 1 ⌡ b 3

d tan 2x = 2sec2(2x) so the integral is dx 1 tan 2x + c. 2

c

1

(

1 1 + x dx −⌠ dx  ⌡ 1+ x

Exercise 5.3A

b  ∫f ′(x)dx = ∫(ln x + 1)dx; hence

a

∫

4 a cos  2x = 2 cos2 x – 1 so 2 cos2 x = 1 + cos 2x and 1 1 therefore cos2 x = + cos 2x 2 2 1 1 2 Integrate: ∫ cos x dx = ∫ + cos 2x dx 2 2 1 1 1 1 = x + × sin 2x + c = x + 1 sin 2x + c 2 2 2 2 4 b Similarly to a, from the double angle formula 1 1 cos2 2x = + cos4x 2 2 1 1 2 so ∫ cos 2 x dx =∫ + cos4x dx 2 2 1 1 1 1 1 = x + × sin 4x + c = x + sin 4x + c 2 4 2 8 2 1 d × (− sin x) = − tan x 5 a ln cos x dx = cos x dx b

∫ tan x dx = − ln cos x

c

1 d ln sin x dx = × cos x = cot x ; sin x dx

+c

therefore ∫ cot x dx = ln sin x + c.

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INTEGRATION

6

b If f(x) = cot x then f (x) = –cosec2 x (this was shown in question 3) Try f(x) = cot(3x + 2) then f (x) = – cosec2 (3x + 2) × 3 = – 3 cosec2 (3x + 2) Hence ∫ cosec2(3x + 2)dx = − 1 cot(3x + 2) 3

a sin 2x = 2 sin x cos x hence π ⌠2

π

∫02 sin x cos x dx =⌡

0

1 1 1 =   −  −  = or  4   4  2 π 2 sin x cos x dx 0

∫

π 2

1 1 sin 2x dx =  − cos2x  4 2 0

π

2 =  1 sin 2 x  =  1 × 12  −  1 × 0 2  = 1  2 0  2   2  2 π

b sin 4x = 2 sin 2x cos 2x hence ∫ 8 sin 2x cos2x dx π 8 0

=∫

Exercise 5.4A 1

0

1 sin 4x dx 2

 2 50 dx = 50∫ 2 1 2 dx b ⌠ ⌡ x + 100 x + 10

π

x 1 tan −1 + c = 5tan −1 x + c 10 10 10 6 1 c ∫ 2 dx = 6∫ 2 dx x +6 x2 + 6

8 1 1 π 1 =  − cos4x  =  − cos  −  − cos0  2   8   8 0  8

=0+ 7

= 50 ×

1 1 = 8 8

1 1 a cos 10x = 2 – 1 so cos 5x = + cos10x 2 2 1 1 2 and ∫ cos 5x dx = x + sin10x + c. 2 20 1 1 b cos x = 1 – 2 sin2 0.5x so sin 2 0.5x = − cos x 2 2 1 1 2 and ∫ sin 0.5x dx = 2 x − 2 sin x + c.

( )

2

cos2 5x

The area is

π

∫0 2 cos

2

2

10 1 a ⌠ dx = 10⌠ dx  2  2 ⌡ x + 100 ⌡ x + 10 2 = 10 ×

1 x x tan −1 + c = tan −1 +c 10 10 10

⌠ 10 10 ⌠ b  dx =  dx 2  1 ⌡ 100x + 1 2 ⌡ 100 x + 100

(

= 0.1 ×

0.5x dx.

π

∫0 1 + cos x dx = [ x + sin x ]

π 0

= [π + 0]–

1 x tan −1 +c 0.1 0.1

= tan −1 10x + c 3

a

y

[0 + 0] = π. 9

)

1 dx = 0.1⌠  2 ⌡ x + 0.12

Now cos x = 2 cos2 0.5x – 1 so 2cos2 0.5x = 1 + cos x. The area is

1 x + c = 6 tan −1 x + c tan −1 6 6 6

= 6×

c cos 2ax = 1 – 2 sin2 ax so 1 1 sin 2 ax = − cos 2ax 2 2 1 1 2 and ∫ sin ax dx = x − sin 2ax + c. 2 4a 8

⌠ 1 ⌠ 1 dx = 1 tan −1 x + c dx = a  5 5 ⌡ x 2 + 25 ⌡ x 2 + 52

4

cos x cos 3x – sin x sin 3x = cos (x + 3x) = cos 4x π 4 cos4x 0

∫

π

2

4 1 dx =  sin 4x   4 0

1 1 =  sin π  −  sin 0  = 0 − 0 = 0  4   4  π

Hence ∫ 4 ( cos x cos3x − sin x sin 3x ) dx = 0 ;

b Area =

0

Hence

π

π

π

π

0

Hence ∫ sec2(2x + 1)dx =

1 tan(2x + 1) + c 2

16

4

x

1

∫−4 x 2 + 4 dx = 16∫−4 x 2 + 22 dx =

16 × 1 tan −1 x  2 2  −4  = [8 tan–1 2] – [8 tan–1 (–2)] = [8.8572] – [–8.8572] = 17.71 to 4 s.f.

Hence ∫ 4 cos x cos3x dx =∫ 4 sin x sin 3x dx 10 a The derivative of tan x is sec2 x so try f(x) = tan (2x + 1); then f’(x) = sec2 (2x + 1) × 2 = 2 sec2 (2x + 1)

4

5

4

∫04 cos x cos3x dx − ∫04 sin x sin 3x dx = 0; 0

0

–5

4

a a a dx  1 x ⌠ Area =  = a × tan −1  ⌡0 x 2 + a 2  a a 0 a

x = tan −1  = [tan–1 1] – [tan–1 0] = π . a 0 4 

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5 

WORKED SOLUTIONS

1 −1 x 1 ⌠ 5 a  =  21 dx = ⌠ dx = 5 5 tan 1 ⌡ 5x + 1 2 1  5 x + 5 5 ⌡ 1 −1 tan 5x + c 5

(

b

)

12 12 ∫ 5x 2 + 4 dx = ⌠ 5 x 2 + 4 dx =  5 ⌡

(

)

12 5 5x 12 ⌠ 1 × tan −1 +c dx = 5  2  2 2 5 2 2 x +    5 ⌡ = 6 tan −1 5x + c 2 5 c

1

1  q x2 +   p 

1 dx = q  2 p x +  p  1 p −1 p x+c = = p q tan 2 dx q

b

x2

7 x4 + 1 = (x2)2 + 1 so try f(x) = tan–1 (x2). 2x 1 Then f ′(x) = 4 × 2x = 4 . x +1 x +1 x 1 Hence ∫ 4 dx = tan −1(x 2) + c. 2 x +1 10 1 1 x ⌠ ⌠ 8 a  dx = 10 × dx = 10 × tan −1  2 5 5 ⌡ x + 25 ⌡ x 2 + 25 x +c 5

(

)

2 ⌠ 10x 2 dx = ⌠ 10 x + 25 − 250 dx b   2  x 2 + 25 ⌡ x + 25 ⌡

250  1 x  =⌠ dx = 10x − 250 × tan −1 + c   10 − 2 5 5 x + 25  ⌡ = 10x − 50tan −1

x +c 5

=⌠  ⌡

⌠ x + a − 2a 2a  dx =   1 − 2 dx  x x 2 + a2 + a 2  ⌡ 2

= x − 2a 2 ×

2

−2

= 5tan −1 1 − 5tan −1(−1) = 5 × π + 5 × π = 5π 4 4 2 ∞ 10 x c  Area = ⌠ dx = 5tan −1  =  2  2 0 ⌡0 x + 22 5 × π  − [ 0 ] = 5π   2 2  

Exercise 5.5A

Hence

2

1 x +1.

1

∫ x + 1 dx = ln |x + 1| + c.

b If f(x) = ln (x2 + 1) then f ′(x) = 2x . x2 + 1

Hence



Hence

1 × 2x = x2 + 1

x

∫ x 2 + 1 dx = 12 ln ( x 2 + 1) + c.

c If f(x) = ln |x3 – 3| then f ′(x) =

1 3x 2 × 3x 2 = 3 . x −3 x −3 3

x2

∫ x 3 − 3 dx = 13 ln x 3 − 3 + c.

2 a If f(x) = ln |x2 + 2x + 5| then 1 2(x + 1) f ′(x) = 2 . × ( 2x + 2) = 2 x + 2x + 5 x + 2x + 5

Hence

x +1

∫ x 2 + 2x + 5 dx =

1 ln x 2 + 2x + 5 + c. 2

b If f(x) = ln |2x3 – 3x2 + 12| then 2 1 f ′(x) = 3 × (6x 2 − 6x)= 36(x −2 x) . 2 2x − 3x + 12 2x − 3x + 12 Hence ∫

(x + a)(x − a) x 2 − a2 9 ⌠ dx = ⌠ dx   2 2 2 ⌡ x +a ⌡ x + a2 2

5

2

1 x2 + 1 − 1 dx = dx = ∫ 1 − 2 x +1 x2 + 1

x – tan–1 x + c

= 2tan −1

0

2 10 b  Area = ⌠ dx = 10 × 1 tan −1 x   2 2  2 2  −2 ⌡x +2



x2 + 1 1 dx = ∫ 1 + 2 dx = x − 1 + c x x2 x

∫ x 2 + 1 dx = ∫

–5

1 a If f(x) = ln |x + 1| then f ′(x) =

p 1 tan −1 x+c q pq 6 a ∫

2.5

∞

∫ px 2 + q dx = ∫ 1 p∫

10 a

3 a  ∫

1 x2 − x dx = ln 2x 3 − 3x 2 + 12 + c. 2x − 3x 2 + 12   6 3

1 ex + 1 d x = ∫ 1 + x d x = ∫ 1 + e −x d x e ex

= x – e–x + c

1 x x tan −1 + c = x − 2a tan −1 + c a a a

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Integration

b To find

ex

∫ e x + 4 dx try f(x) = ln (ex + 4).



Then f ′(x) =



Hence Area =

4

1

ex 1 × ex = x . e +4 e +4

1 ⌠  1 ex  ⌠ dx =   1 − x dx  x ⌡0 e + 1 e + 1  ⌡0 

x

ex

∫ e x + 4 dx = ln (ex + 4) + c.

π 4 tan x dx 0

∫

=

π 4 0

∫

sin x dx cos x

= 1 − ln



If f(x) = ln |cos 2x | then 1 f ′(x) = × (−2sin 2x) = –2 tan 2x. cos 2x



1 Hence ∫ tan 2x dx = − ln cos 2x + c. 2 cos0.5x sin 0.5x



If f(x) = ln |sin 0.5x| then 1 f ′(x) = × 0.5cos0.5x ) = 0.5 cot 0.5x. sin 0.5x (



Hence ∫ cot 0.5x dx = 2 ln sin 0.5x + c.

1 × 2x 6 If f(x) = ln (x2 + a2) then f ′(x) = 2 x + a2 2x . = 2 x + a2

Hence

x

1

∫0 x 2 + a 2 dx =  2 ln(x a

2

a

+ a 2) = 0

2

 1 ln 2a 2  −  1 ln a 2  = 1 ln 2a = 1 ln 2 = ln 2. 2 a2   2  2  2 7 If f(x) = ln |ln x| then f ′(x) =

1 1 1 × = . ln x x x ln x

1

∫ x ln x dx = ln ( ln x ) + c.



Hence



Since x > 1, ln x > 0 and the modulus sign is unnecessary. 1

(

)

⌠ e x + 1 dx = 1 1 + e −x dx =  x − e −x 1 8 a   ∫0  0 ⌡0 e x = 1 − e −1  − [ 0 − 1] = 2 − e −1 b  If f ( x ) = e + 1 then f’(x) = x

1

= [ ln e + 1 ] − [ ln 1 + 1 ] = ln (e + 1) – ln 2

( )

= ln e + 1 2

( e 2+ 1 )

cos2x cos2x 9 a  = ; if f(x) = 0.5 sin 2x sin x cos x + 4 0.5sin 2x + 4 + 4 then f (x) = cos 2x cos2x dx = ln|0.5 sin 2x + 4| + c sin x cos x + 4 or ln (0.5 sin 2x + 4) + c since the expression in brackets is always positive.

Hence ∫

sec x × sec x sec2 x sec x = = b  sin x − cos x sec x (sin x − cos x ) tan x − 1 1 cos x If f(x) = tan x – 1 then f(x) = sec2x sec x  dx = ln |tan x – 1| + c Hence ⌠ ⌡ sin x − cos x since sec x =

10 a f(x) = (cos x )–1; hence using the chain rule f(x)= –1 × (cos x)–2 × (–sin x) =

sin x 1 sin x = × = sec x tan x cos2 x cos x cos x

tan x sec x tan x sec x tan x = = b  ; 2 + cos x sec x ( 2 + cos x ) 2sec x + 1 if f(x) = 2 sec x + 1 then f(x) = 2 sec x tan x Hence ∫ =

tan x sec x tan x dx = ∫ dx 2 + cos x 2sec x + 1

1 ln 2sec x + 1 + c 2

Exercise 5.6A 1 a

A B A(x + 5) + Bx x − 10 ≡ + = x(x + 5) x x + 5 x(x + 5)



So A(x + 5) + Bx ≡ x – 10.



If x = 0 then 5A = –10 and A = –2.



If x = –5 then –5B = –15 and B = 3.



So

b

∫ x(x + 5) dx =∫ x + 5 dx −∫ x dx



= 3 ln |x + 5| – 2 ln |x| + c or ln

ex;

1 ex dx =  ln e x + 1  hence ⌠  x 0 ⌡0 e + 1

( )

1

1 1 ex e+1 dx = [ x ]0 − ln = ∫ 1 dx −⌠  x 2 0 ⌡0 e + 1

π  1  = [ − ln cos x ]04 =  − ln − [ − ln1] 2   = ln 2 or 0.347 to 3 s.f. sin 2x 5 a tan 2x = cos 2x

b cot 0.5x =

1 ex + 1 − ex ex c x = =1− x ; hence x e +1 e +1 e +1

x − 10 3 2 = − and x(x + 5) x + 5 x

x − 10

2 a Write

3

2

(x + 5)3 + c. x2

4 A B A(x + 4) + Bx . ≡ + = x(x + 4) x x + 4 x(x + 4)

Then A(x + 4) + Bx ≡ 4.

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WORKED SOLUTIONS



If x = 0 then 4A = 4 and A = 1.



If x = 2 then 4A = 4 and A = 1.



If x = –4 then –4B = 4 and B = –1.



If x = –2 then –4B = 4 and B = –1.



So



Then



= ln |x – 2| – ln |x + 2| + c = ln

4 1 1 and ≡ − x(x + 4) x x + 4

1

4

∫ x(x + 4) dx =

1

∫ x dx − ∫ x + 4

= ln |x| – ln|x + 4| + c or ln

x + c. x+4

1 A B b Write (x − 2)(x − 4) ≡ x − 2 + x − 4 .

Then A(x – 4) + B(x – 2) = 1.



If x = 2 then –2A = 1 and A = −



If x = 4 then 2B = 1 and B =

and −

1 2.

1

=

c

6 6 A B = ≡ + x 2 − 9 (x − 3)(x + 3) x − 3 x + 3



Then A(x + 3) + B(x – 3) = 6.



If x = 3 then 6A = 6 and A = 1.



If x = –3 then –6B = 6 and B = –1. 1 1 and 6 So ≡ − (x − 3)(x + 3) x − 3 x + 3

1 2

1 x−4 x−4 ln + c or ln x−2 2 x−2

6

1

3 Let



x−3 + c. x+3

x4 = 4 ln |x| – ln |x + 1| + c = ln +c x +1

4 a Write







4 dx. x2 + 4 1

x 1 = 4 × tan −1 + c 2 2 2 x so ∫ 2 dx = x − 2tan −1 x + c 2 x +4

3x − 6 A B ≡ + (x + 2)(x − 1) x + 2 x − 1 A(x − 1) + B(x + 2) = (x + 2)(x − 1) Equate the numerators. 3x – 6 = A(x – 1) + B(x + 2) Let x = –2: –12 = –3A so A = 4. Let x = 1: –3 = 3B so B = –1. 3x − 6 4 1 So ≡ − (x + 2)(x − 1) x + 2 x − 1 3x − 6 4 1 and ∫ (x + 2)(x − 1) dx = ∫ x + 2 − x − 1 dx = 4 ln |x + 2| – ln |x – 1| + c (x + 2)4 + c. or ln |x + 2|4 – ln |x –1| + c or ln x −1

4 4 A B = ≡ + . x 2 − 4 (x − 2)(x + 2) x − 2 x + 2

Hence A(x + 2) + B(x – 2) = 4.

4x + 10 A B ≡ + (2x − 3)(2x + 1) 2x − 3 2x + 1 A(2x + 1) + B(2x − 3) . = (2x − 3)(2x + 1) Then A(2x + 1) + B(2x – 3) = 4x + 10. 3 If x = then 4A = 16 and A = 4. 2 1 If x = − then –4B = 8 and B = –2. 2 4 2 4x + 10 . ≡ − Hence (2x − 3)(2x + 1) 2x − 3 2x + 1

6 Write

1

∫ x(x + 1) dx =∫ x dx −∫ x + 1 dx

4 x2 d x =∫ 1 − 2 dx x +4 x +4 2

5 Write

Then A(x + 1) + Bx = 3x + 4. If x = 0 then A = 4. If x = –1 then –B = –3 + 4 and B = –1. 3x + 4 4 1 and ≡ − x(x + 1) x x + 1 4

+c

Hence ∫

+c

3x + 4 A B A(x + 1) + Bx . ≡ + = x(x + 1) x x + 1 x(x + 1)

3x + 4

2



4

1

= ln |x – 3| – ln |x + 3| + c = ln

)

∫ x 2 + 4 d x = 4∫ x 2 + 22 d x

∫ x 2 − 9 dx = ∫ x − 3 dx − ∫ x + 3 dx

(

x2 (x 2 + 4) − 4 4 = =1− 2 x2 + 4 x2 + 4 x +4

=x−∫



)

c

1 1 1 1 dx + ∫ dx 2∫x −2 2 x−4

1 1 = − ln x − 2 + ln x − 4 + c 2 2





(

= 2 ln x 2 + 4 + c = ln x 2 + 4

∫ (x − 2)(x − 4) dx =



x−2 + c. x+2

b If f(x) = ln(x2 + 4) then f ′(x) = 1 × 2x = 2x x2 + 4 x2 + 4 4 2 x Hence ∫ 2 dx = 2∫ 2 x dx x +4 x +4

1 . 2 1 1 1 1 1 ≡− × + × (x − 2)(x − 4) 2 x−2 2 x−4



4 1 1 = − and x2 − 4 x − 2 x + 2 4 1 1 ∫ x 2 − 4 dx = ∫ x − 2 dx − ∫ x + 2 dx



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Integration



Then

4x + 10

2



If x = 0 then 2 – 2B + 4 = 8 and B = –1.

(2x − 3) + c. 2x + 1



Hence

b

∫ (x + 2)(x − 1)2 dx

4

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

= 2 ln |2x – 3| – ln |2x + 1| + c = ln

2

2x 2 + 9x − 11 A B C 7 Write ≡ + + (x + 1)(x − 2)(x + 3) x + 1 x − 2 x + 3 =

A(x − 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x − 2) . (x + 1)(x − 2)(x + 3)

Then A(x – 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x – 2) = 2x2 + 9x – 11. If x = –1 then –6A = 2 – 9 – 11 = –18 and A = 3. If x = 2 then 15B = 8 + 18 – 11 = 15 and B = 1. If x = –3 then 10C = 18 – 27 – 11 = –20 and C = –2. 2x 2 + 9x − 11 3 1 2 So ≡ + − (x + 1)(x − 2)(x + 3) x + 1 x − 2 x + 3 2

2x + 9x − 11

∫ (x + 1)(x − 2)(x + 3) dx = 3 ln x + 1



and



(x + 1)3(x − 2) + ln x − 2 − 2 ln x + 3 + c = ln + c. (x + 3)2

8 a Write

A(x 2 + 1) + x(Bx + C ) . x(x 2 + 1)

A(x2



Then



If x = 0 then A = 3. 3(x2

+ 1) + x(Bx + C) = (x + 1)(x + 3).



So



If x = 1 then 6 + B + C = 8 and hence B + C = 2.



If x = –1 then 6 – (–B + C) = 0 and hence –B + C = 6.



Add the last two equations: 2C = 8 so C = 4 and B = –2. (x + 1)(x + 3) 3 4 − 2x ≡ + 2 . x x +1 x(x 2 + 1)

Hence

b

∫



= 3 ln |x| + 4 tan–1 x – ln (x2 + 1) + c

(x + 1)(x + 3) 3 4 2x dx = ∫ dx + ∫ 2 dx − ∫ 2 dx x x(x 2 + 1) x +1 x +1

= ln

x3 x2 + 1

9 a  Write

+ 4tan −1 x + c

x 2 − 3x + 8 ≡ A + B + C (x + 2)(x − 1)2 x + 2 x − 1 (x − 1)2

2 ) + C(x + 2). = A(x − 1) + B(x + 2)(x − 1 (x + 2)(x − 1)2



Hence A(x – 1)2 + B(x + 2)(x – 1) + C(x + 2) = x2 – 3x + 8.



If x = –2 then 9A = 4 + 6 + 8 = 18 and A = 2.



If x = 1 then 3C = 1 – 3 + 8 = 6 and C = 2.



So 2(x – 1)2 + B(x + 2)(x – 1) + 2(x + 2) = x2 – 3x + 8.

2

1

2

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



2 = 2 ln x + 2 − ln x − 1 − x − 1 + c



= ln

(x + 2)2 2 − +c x −1 x −1

10 a If f(x) = x2 – 2x – 8 then f(x) = 2x – 2 x −1 1 Hence ⌠ dx = ln x 2 − 2x − 8 + c  2 2 ⌡ x − 2x − 8 x − 10 x − 10 b  2 = ; suppose x − 2x − 8 ( x + 2)( x − 4 ) x − 10

≡

A

( x + 2)( x − 4 ) x + 2 = A ( x − 4 ) + B(x + 2)

+

B x−4

(x + 2)(x − 4)

If x = −2, then −12 = −6 A and A = 2; If x = 4 , then –6 = 6B and B = –1 2 x − 10 1 ≡ − ; x 2 − 2x − 8 x + 2 x − 4 x − 10 2 1 ⌠ dx dx =⌠ −  2  ⌡ x − 2x − 8 ⌡ x+2 x−4

Hence

+ 1) + x(Bx + C) = (x + 1)(x + 3).



x 2 − 3x + 8

=

(x + 1)(x + 3) A Bx + C ≡ + 2 x x(x 2 + 1) x +1 =

x 2 − 3x + 8 2 1 2 ≡ − + . (x + 2)(x − 1)2 x + 2 x − 1 (x − 1)2



(

)

= 2 ln x + 2 − ln x − 4 + c

= ln

(x + 2)2 +c x−4

x2 + 3 x2 + 4 − 1 1 11 a  2 = =1− 2 ; hence x +4 x2 + 4 x +4

⌠ x 2 + 3 dx = ⌠  1 − 1  dx  2  x2 + 4 ⌡x +4 ⌡ 1 −1 x = x − 2 tan 2 + c x2 − 3 x2 − 4 + 1 x2 − 3 1 b  2 = =1+ 2 so ⌠ dx  2 2 x −4 x −4 x −4 ⌡x −4 1  1  =⌠ dx = x + ⌠ dx  2  1 + 2 ⌡x −4 x − 4 ⌡ 1 Write 2 in partial fractions: x −4 1 1 A B = ≡ + x 2 − 4 (x − 2)(x + 2) x − 2 x + 2 A ( x + 2) + B(x − 2) x2 − 4

1 ≡ A (x + 2) + B(x – 2); let x = 2 then 1= 4A and 1 A= ; 2

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WORKED SOLUTIONS

let x = −2 then 1 = –4B and B = − 1 ; hence 4 1 1⌠ 1 1 ⌠ dx dx =  −  2 4⌡ x − 2 x + 2 ⌡x −4

(

=

)

1 1 x−2 ln x − 2 − ln x + 2 } + c = ln +c 4 x+2 4{

x2 − 3 1 x−2 Hence ⌠ dx = x + ln +c  2 4 x+2 ⌡x −4 12

x2 x2 − 1 + 1 1 = =1+ 2 x −1 x2 − 1 x −1 2

If

1 1 B A = ≡ + ; x 2 − 1 (x − 1)(x + 1) x − 1 x + 1

so 1 ≡ A(x + 1) + B(x – 1)

{

π 22

 1 π2   1  1 =  − 2 cos 4  −  − 2  =  − cos x  2 0  

1 ; 2

if x = – 1 then 1 = – 2B and B = − 5

1 2

1 1 ⌠ x 2 dx = ⌠  1 + 2(x − 1) − 2(x + 1)  2 ⌡ 2 ⌡2 x − 1

}

5

1 1 dx =  x + ln x − 1 − ln x + 1   2  2 2 5

1 x −1  1 2 1 1 =  x + ln = 5 + ln  −  2 + ln  2 x + 1  2  2 3   2 3   =3+

π

b  Using the answer to a, ∫ 2 x sin x 2 dx 0

If x = 1 then 1 = 2A and A =

5

du = 6x 2 and dx = 1 2 . dx du 6x 4 dx 1 4 2 3 2 4 1 ∫ x (2x + 5) du du = ∫ x u 6x 2 du = 6 ∫ u du = 1 × 1 u 5 + c = 1 (2x 3 + 5)5 + c 6 5 30 du 2 4 a Let u = x then = 2x dx dx 1 ; = du 2x dx 1 1 x sin x 2 du = ∫ x sin u du = ∫ sin u du 2x 2 du ∫ 1 1 2 = − cos u + c = − cos x + c 2 2 3 Let u = 2x3 + 5 then

2 1 1 3 ln = 3 + ln 2 2 2 1 3

= 0.891 to 3 s.f. 5 a ln |x + 2| + c b Let u = x + 2 then x = u – 2 and

Therefore x u−2 2 ∫ x + 2 dx = ∫ u du = ∫ 1 − u du = u − 2 ln u + c = x + 2 − 2 ln x + 2 + c



c As in part b, let u = x + 2 then x = u – 2 and

Therefore x2

∫ x + 2 dx = ∫

Exercise 5.7A dx = 1 so the integral = du 1 5 1 4 3 4 3 ∫ (u − 2)u du =∫ u − 2u du = 5 u − 2 u + c

1 x = u – 2 and

1 1 = (x + 2)5 − (x + 2)4 + c. 2 5 2 a u = x2 + 4 and Then ∫

=

∫2

x 2

x +4

x 1 dx du = ∫ du × 2 x d u u x +4

1

x 2 + 3 dx = ∫ xu 2 ×

1 3 1 2 3 2 = u 2 − 8u 2 + c = (x + 4)2 − 8(x + 4)2 + c 3 3

1 1 1 du = ∫ u 2 du 2x 2

3 1 2 32 1 × u + c = (x 2 + 3)2 + c 2 3 3

b Let u = x + 3 then x = u – 3 and dx = 1. du

x 2 + 4 + c.

x du dx dx = b u = x + 4 so =1= ; dx du ∫ x + 4 1 1 u − 4 dx ∫ u du du = ∫ u 2 − 4u − 2 du

1 (x + 2)2 − 4(x + 2) + 4 ln x + 2 + c 2 du dx 1 6 a Let u = x2 + 3 then = 2x and = dx du 2x

=

x

1 d = 1 − 12 d = 12 + = u ∫ u u u c 2 u

(u − 2)2 du = u 2 − 4u + 4 du ∫ u u

=



∫x

2

dx = 1. du

= ∫ u − 4 + 4 du = 1 u 2 − 4u + 4 ln u + c u 2

du dx 1 . = 2x so = dx du 2x dx = ∫

dx = 1. du

∫x

1

x + 3 dx = ∫ (u − 3)u 2 du

3 1 5 3 = ∫ u 2 − 3u 2 du = 2 u 2 − 3 × 2 u 2 + c 5 3



=

5 3 2 (x + 3)2 − 2(x + 3)2 + c 5

c Let u = x + 3 then x = u – 3 and

dx = 1. du 1



Then ∫ x 2 x + 3 dx = ∫ (u − 3)2u 2 du 1

5

3

1

= ∫ (u 2 − 6u + 9) u 2du = ∫ u 2 − 6u 2 + 9u 2 du.

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Integration

7

5

3

10 Where the graph meets the x-axis, x x + 4 = 0 so x = 0 or −4 0 The integral required is ∫ x x + 4 dx

= 2u2 − 6 × 2u2 + 9 × 2u2 + c 7 5 3



5

7

3

= 2 (x + 3)2 − 12 (x + 3)2 + 6(x + 3)2 + c 7 5

7 a If u = cos x then

du dx 1 = − sin x and =− . sin x dx du

sin x

sin x

dx

∫ tan x dx = ∫ cos x × du du =∫ cos x ×



−4

(− sin1 x )du

1 = −∫ du = − ln u + c u



    

b  To find



∫

π 2 π 4

=∫

= ln sin x + c π

π

 ∫π42 cot x dx = [ ln sin x ]π42 = [ ln1] −  ln

1  2 

= − ln 1 = ln 2 = 1 ln 2 2 2 dx 8 Let x = a sin u and then = a cos u and du 1 dx du = 2 2 du a −x

1 ∫ a 2 − a 2 sin 2 u × a cosu du 1 =∫ × a cos u du a 1 − sin 2 u

∫

= ∫ 1 du

=u+c

x x so u = sin −1 and the integral a a x + c. is u = sin −1 a Now sin u =



9 a If y = 0 then x (x – 2)4 = 0 so x = 0 or 2; the point is (2, 0) 2

b  Area = ∫ x ( x − 2) dx ; let u = x – 2 and then 4

0



∫0 x ( x − 2) ∫u = − 2 (u u=0

4

x=2

∫x = 0 (u + 2)u

dx =

)

4

du 0

1 2 + 2u 4 du =  u 6 + u 5  5  −2  6 2 64 64  = [ 0 ] −  − =2  6 15 5  =

5

4

1 ⌠  3 2 5 8 3  =   u 2 − 4u 2  du =  u 2 − u 2  3    ⌡0 5 0

64 64  128 8 = − = −8 − 0 =− 3  [ ] 15 15  5 8 Hence the area is 8 15 du dθ 1 11 a Let u = 1 − cos θ then = sin θ and = dθ du sin θ 1 1 dθ ∫ sin θ 1 − cosθ dθ = ⌠⌡ sin θ × u 2 du du = ∫ u 2 du 3 2 2 3 = u 2 + c = (1 − cosθ ) 2 + c 3 3 An alternative method is to recognise that 3

the integral is a multiple of (1 − cosθ ) 2 and differentiate that function and find the multiple. b  1 − cos2 2θ = sin 2 2θ so ∫ sin θ 1 − cos2 2θ dθ = ∫ sin θ sin 2θ dx = ∫ 2sin 2 θ cosθ dθ Try f ( x ) = sin 3 x then f(x) = 3 sin2 x cos x and so

∫ sin θ

2 1 − cos2 2θ dθ = sin 3 x + c 3

An alternative method is to find ∫ 2sin 2 θ cosθ dθ

by using the substitution u = sin θ or by spotting that sin3 x differentiates to 3 sin2 xcos x

( = ∫ ( 2sin θ cos θ − sin θ ) dθ

)

2 c sin ∫ θ cos2θ dθ = ∫ sin θ 2cos θ − 1 dθ 2

2 = − cos3 θ + cosθ + c 3

dx =1 du If x = 0 u = −2 and if x = 2, u = 0 x = u + 2 and 2

Using the values of u for the limits, the integral is

4

cos x dx , let u = sin x then sin x

du = cos x and dx = 1 ; dx du cos x cos x dx cos x 1 ∫ cot x dx =⌠⌡ sin x × du du = ∫ sin x × cosu du = ∫ 1 du = ln u + c u

Hence,

When x = −4, u = 0 and when x = 0, u = 4 4 ⌠ (u − 4) u dx du du ⌡0

= – ln |cos x| + c π 2 π cot x d x 4

Make the substitution u = x + 4; then x = u − 4 and dx =1 du

1 −1 1 12 a If f ( x ) = (1 + x)2 then f( x ) = (1 + x) 2 × 1 2 1 1 1 = 2 × 1 + x so ∫ 1 + x dx

= 2 1 + x + c b  Use the substitution u = 1 + x; then x = u − 1 and dx =1 du

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WORKED SOLUTIONS

⌠ 1 −1  x u −1 ⌠ dx = ⌠ du =   u 2 − u 2  du   ⌡ 1+ x ⌡ u ⌡  3 1 3 1 = 2 u 2 − 2u 2 + c = 2 (1 + x ) 2 − 2 (1 + x ) 2 + c 3 3 1 2 This can be written as (1 + x ) 2 {(1 + x ) − 3} + c 3 1 2 = (1 + x ) 2 (x − 2) + c 3 du = 2x and c  Try the substitution u = 1 + x2 ; dx dx 1 = du 2x 1

3 a Let u = x and

2

2



2

1 1  2 32 1⌠  1 −1  2  u 2 − u 2  du = 2  3 u − 2u  2  1 ⌡1

= 2 − 2 = 0.1953 3

So ∫ x 2e xdx = x 2e x − 2∫ xe xdx To find ∫ xe xdx let u = x and Then

dv = ex . dx

x

x

− ∫e xdx = xe x − e x + c

Therefore ∫ x 2e xdx = x 2e x − 2∫ xe xdx

(

)

= x 2e x − 2 xe x − e x + c 1

1

0

0

Hence ∫ x 2e x dx =  x 2e x − 2xe x + 2e x 

dv = sin 2x . dx

du 1 = 1 and v = − cos 2x. Then 2 dx 1 1 ∫ x sin 2x dx = − 2 x cos 2x + ∫ 2 cos 2x dx = − 1 x cos 2x + 1 sin 2x + c 2 4

b To find ∫ x cos 4x dx let u = x and



dv = e x then dx

du = 1 and v = e x and dx

∫ xe dx = xe

du = 1 and v = sin x. dx = (x + 1)sin x + cos x + c

dv = cos 4x . dx

du 1 = 1 and v = sin 4x. 4 dx 1 1 ∫ x cos 4x dx = 4 x sin 4x − ∫ 4 sin 4x dx = 1 x sin 4x + 1 cos 4x + c 4 16

Then

= 2xe0.5x − 4e0.5x + c

= x2 ex – 2xex + 2ex + c

2 a To find ∫ x sin 2x dx let u = x and



dx = 2xe0.5x − ∫ 2e0.5x dx

du = 2x and v = e x dx

∫ (x + 1)cos x dx = (x + 1)sin x − ∫ sin x dx



0.5x

1

dv = cos x . dx



du = 1 and v = 2e0.5x. dx

Then

0

b To find ∫ (x + 1)cos x dx let u = x + 1 and



dv = e0.5x . dx

4 To find ∫ x 2e x dx let u = x2 and

= x sin x + cos x + c

Then

dv = e −x . dx

du = 1 and v = –e–x. dx −x −x −x −x −x ∫ x e d x = −x e + ∫ e d x = −x e − e + c



dv 1 a To find ∫ x cos x dx let u = x and dx = cos x . du = 1 and v = sin x. Then dx ∫ x cos x dx = x sin x − ∫ sin x dx



dx = xe x − ∫ e x dx = xe x − e x + c

Then

∫ xe

Exercise 5.8A



x

du = 1 and v = ex. dx

c Let u = x and

2

2

∫ xe



1⌠ x ⌠ x 1 u −1 du = 1 ⌠ × du =  =  du 2 ⌡1 u 2 ⌡1 u ⌡1 u 2x =





x =1

3

Then

b Let u = x and

⌠ ⌠ x3 x 3 dx du dx =  Then  2 ⌡x = 0 u du ⌡0 1 + x 2



dv = ex . dx

= [ e − 2e + 2e ] − [ 0 − 0 + 2] = e – 2

5 To find ∫ (2x 2 − 1)e −x dx let u = 2x2 – 1 and

dv = e −x . dx

du = 4x and v = –e–x. dx



Then



Therefore ∫ (2x 2 − 1)e −x dx = −(2x 2 − 1)e −x + 4 ∫ xe −x dx. dv = e −x . dx



To find ∫ x e −x dx let u = x and



Then

du = 1 and v = –e–x and dx



∫xe

dx = −xe −x + ∫ e −x dx = −xe −x − e −x + c.



Therefore 2 −x 2 −x −x ∫ (2x − 1)e dx = −(2x − 1)e + 4 ∫ xe dx



−x

= – (2x2 – 1)e–x + 4(–xe–x – e–x) + c = (–2x2 + 1 – 4x – 4)e–x + c = – (2x2 + 4x + 3)e–x + c

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Integration

dv 6 To find I = ∫e x cos x dx let u = cos x and = ex dx du then dx = − sin x and v = e x x x x So I = ∫e cos x dx = e cos x + ∫e sin x dx

To find ∫e x sin x dx let u = sin x and then

∫e

x

dv = ex dx

8 a To find I = ∫ e −x sin 2x dx let u = sin 2x and dv du = e −x then = 2cos 2x dx dx

and v = –e–x.



So I = −e −x sin 2x + 2∫ e −x cos 2x dx.



To find ∫ e −x cos 2x dx let u = cos 2x and

du = cos x and v = e x dx

dv du = e −x then = −2sin 2x and v = –e–x. dx dx

sin x dx = e x sin x − ∫e x cos x dx = e x sin x − I

So I = e x cos x + ∫e x sin x dx = e x cos x + e x sin x − I Rearrange to get 1 x e ( cos x + sin x ) + c 2

2I = e x cos x + e x sin x so I = π 2 x e cos x 0

Hence ∫

π

2 1 dx =  e x ( cos x + sin x )   2 0

 1 π 1 π  1 =  e 2  −   = 2  e 2 − 1   2 2      

∫e

= −e −x cos2x − 2I So I = − e −x sin 2x + 2∫ e −x cos2x dx = − e −x sin 2x − 2e −x cos2x − 4I .



Rearrange to get 5I = –e–x sin 2x – 2e–x cos 2x so I = − 1 e −x(sin 2x + 2cos 2x) + c. 5

 1 −π  2 =  − e 2 × (−2) −  −   5   5 

e

dv =1 dx



du 1 = and v = x; hence dx x

∫ ln x dx

= x ln x − ∫1 dx = x ln x + x + c e

=

2 − π2 2 e + = 0.483 to 3 s.f. 5 5

9 a Let u = x + 1 so x = u – 1 and dx = 1. du

∫x

Hence ∫ ln x dx = [ x ln x − x ]1 = [e – e] – [0 – 1] = 1 e

1

e

∫1 ln x

e

dx = ∫ 2 ln x dx = 2[ x ln x − x ]1 = 2

2

e

1

x d2x+=ln∫ (xln c ∫ ln∫ 2 ( ln ) d2x + ln x ) dx

5



1

3

= 2u2 − 2u2 + c 5 3



3

x + 1 dx = ∫ (u − 1)u 2 du = ∫ u 2 − u 2 du

1

b

π

π

2 1 b Area = ∫ 2 e −x sin 2x dx =  − e −x(sin 2x + 2cos2x) 0 0  5

1

Then

cos2x dx = −e −x cos2x − 2∫ e −x sin 2x dx



7 a  To find ∫ ln x dx use integration by parts with u = ln x and

−x

=

5 3 2 2 (x + 1)2 − (x + 1)2 + c 5 3

1 b Let u = x and dv = (x + 1)2. dx x (xln = ( ln 2) x = + (xln ln2x) x− +x x+ln c x= −x (xln+2c+=ln ) −2 x+ +lncx ) − x + c 3 So du = 1 and v = 2 (x + 1)2 . 3 dx = x ln 2x − x + c e

3

e

1

= [ e ln 2e − e ] − [ ln 2 − 1] = e ln 2e – e – ln 2 + 1

d  To find ∫ ln(x + 1)dx let u = x + 1 and e

e +1

Then ∫ ln(x + 1)dx = ⌠ ⌡2 1 =∫

e +1

2

ln(x + 1)

du dx =1= dx du

dx du du

ln u du = [u ln u − u ]2

e +1

= [(e + 1)ln e + 1 − (e + 1)] − [ 2 ln 2 − 2] = ( e + 1) ln e + 1 − (e + 1) − 2 ln 2 + 2

3

So ∫ x x + 1 dx = x × 2 (x + 1)2 − ∫ 2(x + 1)2 dx 3 3

Hence ∫ ln 2x dx = [ x ln 2x − x ]1

3



5

= 2 x(x + 1)2 − 2 × 2 (x + 1)2 + c 3 5 3 =

3 5 2 4 x(x + 1)2 − (x + 1)2 + c. 3 15

c The answer to part a can be written 5 3 3 2 2 2 2 (x + 1)2 − (x + 1)2 = (x + 1)2 (x + 1) − 5 3 5 3 3 + 1)2

( 52 x − 154 ).

{

}



= (x



The answer to part b can be written 3 5 3 2 4 2 4 x(x + 1)2 − (x + 1)2 = (x + 1)2 x − (x + 1) 3 15 3 15

{

}

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5 

WORKED SOLUTIONS



3

= (x + 1)2

{ 32 x − 154 x − 154 } = (x + 1) ( 52 x − 154 )

Exam-style questions

3 2

1 a = ∫

so the two are the same. 10 ∫ cos3 xdx = ∫ cos2 x × cos x dx ; let u = cos2 x and

b

dv = cos x dx du = −2cos x sin x and v = sin x Then dx

2

Hence ∫ cos3 x dx

2 3 sin x + c 3 dv du 11 Let u = x2 and = sin x ; then = 2x and dx dx v = – cos x

≈

1

1.2

1.4

1.6

1.8

2

0.2 0.3333 + 0.2 + 2 ( 0.3219 + 0.2951 + 0.2625 + 0.2298 )} 2 {

= 0.275 to 3 s.f.

2 2 Then ∫ x sin x dx = −x cos x + 2∫ x cos x dx

3 a If f(x) = cos (4x + 1) then f ′(x) = −4sin(4x + 1)

dv = cos x; then dx



Hence ∫ x cos x dx = x sin x − ∫ sin x dx

Hence ∫ sin(4x + 1)dx = − 1 cos(4x + 1) + c 4

b If f(x) = ln |4x + 1| then f(x) = Hence ∫

= x sin x + cos x + c

3

2

Hence ∫ x 2 sin x dx = −x 2 cos x + 2x sin x + 2cos x + c

1 ×4= 4 4x + 1 4x + 1

3 1 dx =  1 ln 4x + 1  4x + 1  2  4

1 1 1 13 =  ln13  −  ln 9  = ln 9   4  4  4

π

and ∫ 2 x 2 sin x dx =  −x 2 cos x + 2x sin x + 2cos x  2 0 0

4 a If f(x) = tan 0.5x then f ′(x) = sec2 0.5x × 0.5 = 0.5sec2 0.5x.

=π−2 12 a f(x) = e–x – xe–x; at a stationary point, e–x – xe–x = 0; e–x(1 – x) = 0

e

2 x ⌠ dx  3 ⌡1 x + 2

= cos2 x sin x +

π

1.8

dx =  1 e 2x + 1  = 0.5e 4.6  − 0.5e3  1  2 = 39.699 to = 3 d.p.        0.5(e4.6 – e3) 1.8 2x + 1

∫1

x x 3 + 2 0.3333 0.3219 0.2951 0.2625 0.2298 0.2

= cos2 x sin x + 2∫ sin 2 x cos x dx

du = 1 and v = sin x dx

e

x

= cos2 x sin x − ∫ ( −2cos x sin x ) × sin x dx

To find ∫ x cos x dx let u = x and

dx ≈ 1 × 0.2{20.09 + 2(29.96 + 44.70 2 + 66.69) + 99.48} = 40.227

1.8 2x + 1

1



Hence ∫ sec2 0.5x dx = 2tan 0.5x + c

b 1 + tan2 0.5 x = sec2 0.5x or tan2 0.5x = sec2 0.5x – 1

e −x is always positive so the only solution is x = 1.

f  (x) = –e–x – e–x + xe–x and f (1) = –e–1 < 0 so the stationary point is a maximum.



Hence ∫ tan 2 0.5x dx = ∫ (sec2 0.5x − 1) dx = 2tan 0.5x − x + c

b

3

3

5 Area = ∫ e0.5x − e −0.5xdx =  2e0.5x + 2e −0.5x  0 0 0

=  2e1.5 + 2e −1.5  − [2 + 2]

1



= 9.410 – 4 = 5.41 to 3 s.f.

6 a cos 4x = 1 – 2 sin2 2x and so sin 2 2x =

1

The area is ∫ xe −x dx 0

Let u = x and 1

x ∫0 xe dx = 

dv du = e −x then = 1 and v = −e −x dx dx

1  −xe −x   0

1 −x e 0

+∫

=  −2e −1  − [ −1] = 1 − 2e −1  

dx =  −xe −x −

1 e −x  0



1 1 − cos4x . 2 2

Hence ∫ sin 2 2x dx = ∫ 1 − 1 cos4x dx 2 2 1 1 = x − sin 4x + c 2 8 1

1 1 1 b Area = ∫ sin 2 2x dx =  x − sin 4x  8 0 0 2

1 = 0.5 − sin 4  − [0 − 0]= 0.595 to 3 s.f.  8 

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Integration

3

11 a

7 a If f(x) = (2x + 1)2 then

f ′(x) =

1 1 3 (2x + 1)2 × 2 = 3(2x + 1)2. 2 1 2

Hence ∫ (2x + 1) dx =

b Let u = 2x + 1 then

∫ 2x

3 + 1)2

+ c.

×

1 du 2

c

∫ x 2 + 9 dx = ∫

x2



0

3 5 1 1 2 − (2x + 1) 2 2 x 2 x + 1 d x = 2 x + 1 ( )   ∫−0.5 3 5  −0.5 0

=  1 − 1  − [0] = − 2 15  5 3 

12 a

(

)

(

)

= sin 3x – tan x + x + c

(

)

1 2 e −1 2 2 9 a cos 2x – 4 sin2 x cos2 x = cos2 2x – (2 sin x cos x)2 = cos2 2x – sin2 2x =



= cos (2 × 2x) = cos 4x

b  Integral =

π 16 cos4x dx 0

∫

10 a

b 9 cos2 θ – sin2 θ – 6 sin θ cos θ = 9(1 – sin2 θ) – sin2 θ – 3 × 2 sin θ cos θ = 9 – 10 sin2 θ – 3 sin 2θ = 4 + 5(1 – 2 sin2 θ) – 3 sin 2θ = 4 + 5 cos 2θ – 3 sin 2θ c

∫ (9cos θ − sin 2

=

2

)

θ − 6sin θ cosθ dθ

∫ (5cos2θ − 3sin 2θ + 4 ) dθ

5 3 = sin 2θ + cos2θ + 4θ + c 2 2

+ 9) + c

x 2 + 9 − 9 dx = 1 − 9 dx ∫ x2 + 9 x2 + 9 = x − 9∫ 21 dx x +9 = x − 9 × 1 tan −1 x + c 3 3 x = x − 3tan −1 + c 3

)

sin 0.5x



b

π

π

∫02 tan 0.5x dx =  −2 ln cos 0.5x 02 =  −2 ln cos π  − [−2 ln 1]  4  = −2 ln 1 − 0 = −2 ln 1 − ln 2 2

(

π

2 1 2 2+x dx = ∫  2 +  dx = − + ln x + c x x x x2

∫

1 × 2x x2 + 9

∫ tan 0.5x dx = ∫ cos0.5x dx .

16 =  1 sin 4x  0  4

1 1 2 =  1 sin π  − [ 0 ] = × = 4  4 8  4 2

x +c 3

If u = cos 0.5x then

2 2 8 a  ∫ 3cos3x − tan x dx = sin 3x − ∫ sec x − 1 dx 2.5 2.5 1 2 1 4x − 8 1 b  ∫2 2e dx =  2 e4x −8 2 =  2 e  −  2 

−1

du = −0.5sin 0.5x . dx sin 0.5x sin 0.5x dx ∫ cos 0.5x dx = ∫ u du du 1 = ∫ sin 0.5x × du − 0.5 sin 0.5x u = ∫ −2 du = −2 ln u + c u = – 2 ln |cos 0.5x| + c



2 . 15

The area is

2

(

c If 2x 2x + 1 = 0 then x = 0 or –0.5.



1

Hence

5 3 = 1 (2x + 1)2 − 1 (2x + 1)2 + c 5 3



x

∫ x 2 + 9 dx = 2 ln(x



3 1 5 3 = ∫ 1 u 2 − 1 u 2 du = 1 u 2 − 1 u 2 + c 2 2 5 3



1

2x . x2 + 9

=

du dx 1 = 2 and = ; dx du 2

2x + 1 dx = ∫ (u

1

b If f(x) = ln(x 2 + 9) then f ′ ( x ) =

1 1 f(x) = (2x 3 3

1 − 1)u 2

1

∫ x 2 + 9 dx = ∫ x 2 + 32 dx = 3 tan

)

1

= 2 ln 2 2 = 2 × 1 ln 2 = ln 2 2 13 a Using the product rule, f(x) = sin 0.5x + 0.5x cos 0.5x At stationary point, sin 0.5x + 0.5x cos 0.5x = 0; sin 0.5x = – 0.5x cos 0.5x b

tan 0.5x = – 0.5x; tan 0.5x + 0.5x = 0 4

2

0

2π

c  To find ∫ x sin 0.5x dx, let u = x and then

dv = sin 0.5x ; dx

du = 1 and v = – 2 cos 0.5x dx

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5 

WORKED SOLUTIONS



Hence ∫ x sin 0.5x dx = −2x cos 0.5x

If x = 3 then 12 = 4A and A = 3; if x = – 1 then –4 = – 4 B and B = 1

+ ∫ 2cos 0.5x dx

= −2x cos0.5x + 4sin 0.5x + c

So

2π

d  Area = ∫ x sin 0.5x dx = [ – 2x cos 0.5x + 4 sin 0.5x]02π 0

= [4π] – [0] = 4π π 2 sin x cos x dx 0 π 2 1 





π

1

1

∫02 2 sin 2x dx =  − 4 cos2x 0

Alternative method

π 2 sin x cos x dx 0

∫

=

 4 

1



15 a

b

( )

1

Write u = x and dv = sin 2x. dx du 1 = 1 and v = − cos 2x Then 2 dx 1 1 1 ∫ x sin x cos x dx = 2 − 2 x cos 2x − ∫ − 2 cos 2x dx



= [3 ln 2 + ln 2] – [3 ln 3] = 4 ln 2 – 3 ln 3 16 = ln 24 – ln 33 = ln 27

π



{

17 a If f(x) =

x2 e–x

then f(x) = 2xe–x – x2 e–x

At a stationary point 2xe–x –x2 e–x = 0; x (2 – x) e–x = 0; x = 0 or 2; if x > 0 then the only solution is x = 2 f  (x) = 2e–x – 2xe–x – 2xe–x + x2 e–x = (2 – 4x + x2)e–x so f  (2) = – 2e–2 < 0

}

So the maximum value is f(2) = 4e–2

b  I = ∫ x 2e −xdx ; let u = x2 and

dv = e −x then dx

du = 2x and v = –e–x dx

= − 1 x cos 2x + ∫ 1 cos 2x dx 4 4 = − 1 x cos 2x + 1 sin 2x + c 4 8

Then I = −x 2e −x + 2∫ xe −x dx

x+7 = x+7 ≡ A + B x 2 − x − 6 (x − 3)(x + 2) x − 3 x + 2 2 x − 3) = A(x + ) + B(x (x − 3)(x + 2)

Hence ∫ xe −x dx = −xe −x + ∫e −x dx = – xe– x – e– x + c

Equate the numerators: A(x + 2) + B(x – 3) = x + 7. Put x = 3: 5A = 10 so A = 2. Put x = –2: –5B = 5 so B = –1. Therefore 2 x + 7 = 2 − 1 . x −x−6 x−3 x+2 x+7

2

1

∫ x 2 − x − 6 dx = ∫ x − 3 − x + 2 dx

dv To find ∫ xe −x dx let u = x and = e −x then dx du = 1 and v = –e–x dx Hence I = –x2e–x + 2(–xe–x – e–x) + c = – (x2 + 2x + 2)e–x + c 3

18 a If u = (1 − x ) 2 then = −3 1− x 2

= ln

(x − 3)2 +c x+2

16 a If f(x) = x2 – 2x – 3 then f(x) = 2x – 2 and x −1 = 0.5f(x) f(x) x 2 − 2x − 3     x − 1 dx = 0.5ln f(x) + c ⌠ Hence  ⌡ x 2 − 2x − 3

=

A ( x + 1) + B(x − 3) (x − 3)(x + 1)

1

1

b  To find ∫ x 1 − x dx let u = 1 – x; then x = 1 – u 0

dx = −1 du When x = 0, u = 1 and when x = 1, u = 0 and

u=0 1 dx Hence ∫ x 1 − x dx = ⌠ (1 − u) u du du ⌡u =1 0 0

⌠  1 3 =  −u 2 + u 2  du   ⌡1

= 0.5ln x 2 − 2x − 3 + c 4x 4x A B = ≡ + x 2 − 2x − 3 (x − 3)(x + 1) (x − 3) ( x + 1)

1 du 3 = (1 − x)2 × −1 dx 2

3 1 − x dx =  − 2 (1 − x ) 2   3  0 0 2 2  = = [0 ] − −  3  3 1

Hence ∫

= 2 ln x − 3 − ln x + 2 + c

b

)

1

2 =  1 sin 2 x  =  1 × 12  −  1 × 0 2  = 1  2 0  2   2  2

∫ x sin x cos x dx = 2 ∫ x sin 2x dx

(

= [ 3 ln x − 3 + ln x + 1 ]0

1 1 − −  =  4  2

b

4x 3 1 ≡ + x 2 − 2x − 3 x − 3 x + 1

1 4x 3 1 ⌠ dx =⌠ c   2  x − 3 + x + 1 dx ⌡0 x − 2x − 3 ⌡0

14 a sin 2x = 2 sin x cos x so ∫ =

So 4x ≡ A(x + 1) + B(x – 3)

0



 2 3 2 5 =  − u 2 + u 2  = [0 ] −  − 2 + 2  = 4 5   3 5  15  3 1

(This could also be done using integration by parts)

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Integration

c Use the same substitution as part b:

π

1 2 dx 2  ∫0 x 1 − x dx = ⌠⌡u =1 (1 − u) u du du 0

(

)

1

 1 − 2u + u 2 u 2 du = −⌠ ⌡1

b  If f (x) = sin3 x then f(x) = 3 sin2 x cos x; hence

1

π 2 sin 2 x cos x 0

⌠  1 3 5 3 5 =   u 2 − 2u 2 + u 2  du =  2 u 2 − 4 u 2 + 3  5 7  ⌡0   0 2 4 2 16 =  − +  − [0 ] =  3 5 7  105 19 a If f ( x ) = ln 1 + 4x then f( x ) = 1 × 4 = 4 1 + 4x 1 + 4x 1 2 72  u

∫

c h =

8 dx = 2 ln 1 + 4x + c 1 + 4x This can also be solved with the substitution dx 1 1 π π π π = u = 1 + 4x; 0 + sin 2 + sin 2 + sin 2 ≈ du 4 24 2 6 3 2 2 8 8 1 Then ∫ dx = ∫ × du = ∫ du = 2 ln u + c 1 + 4x u 4 u = 2 ln 1 + 4x + c as before. d 2 8 8 ⌠ ⌠ ⌠ d x b  d x = =  d x 2  2  ⌡ 1 + 4x 2  4 x2 + 1 x + 1 4 ⌡ 2 ⌡ –1 –1 = 2 × 2 tan (2x) + c = 4 tan (2x) + c

{ (

)

()

8 8 A B c  = ≡ + 1 − 4x 2 (1 − 2x)(1 + 2x) 1 − 2x 1 + 2x =

A (1 + 2x ) + B(1 − 2x) (1 − 2x)(1 + 2x)

So 8 ≡  A (1 + 2x) + B (1 – 2x)

1 1 Let x = then 8 = 2A and A = 4; let x = − then 2 2 8 = 2B and B = 4 8 4 4 8  2 ≡ + so ⌠ dx  1 − 2x 1 + 2x ⌡ 1 − 4x 2 1 − 4x 4 4 =⌠  1 − 2x + 1 + 2x dx ⌡ 1 × −2 Now if f ( x ) = ln 1 − 2x then f ′ ( x ) = 1 − 2x −2 ; = 1 − 2x

(





(

)

4 4 So ⌠  1 − 2x + 1 + 2x dx ⌡ = −2 ln 1 − 2x + 2 ln 1 + 2x + c

≈

π

π

1 π and ∫ 2 sin 2 x cos2 x dx =∫ 2 sin 2 2x dx 12 0 0 4

{ (

) } π π 1 3 3 1 + +1+ + }≈ = 48 { 4 4 16 4 4 + sin 2

2π 5π + sin 2 +0 3 6

cos 4x = 1 – 2 sin2 2x so sin 2 2x = π 2 0

∫

) }

1 2π 5π π π π π 0 + sin 2 + sin 2 + sin 2 + sin 2 + sin 2 +0 24 2 6 3 2 3 6

π

(

1 1 − cos4x and 2 2

)

2 1 1 1 sin 2 2x dx = ⌠  8 − 8 cos4x dx 4 ⌡0 π

2 =  1 x − 1 sin 4x  =  π  − [ 0 ] = π 0  8  16  8 16

Mathematics in life and work 25 1 h = 5 and ∫ p dv ≈ 1 × 5{5.80 + 2(2.52 + 1.55 2 5 +1.10) + 0.84} = 42.45

2 If pv1.2 = 40 then p = 40v–1.2. 25

Then ∫ p dv =

)

2 Similarly if f ( x ) = ln 1 + 2x then f ′ ( x ) = 1 + 2x

π

2 1 dx =  sin 3 x   3 0

1 1  =  3  − [ 0 ] = 3

Hence ∫

(

π

π

2 1 1 20 a ∫ 2 sin x cos x dx =∫ 2 sin 2x dx =  − cos2x   4 0 0 0 2 1 1 1 =   − −  =  4   4  2

u=0

5

25

∫5

25

40v −1.2dv =  40 v −0.2   −0.2 5

25



=  −200v −0.2  5

= [–105.06] – [–144.96] = 39.9 to 3 s.f.

3 If pv = 24 then p = 24 and v 25 25 24 p d v = d v = [ 24 ln v ]525 = 24 ln 25 − 24 ln 5 ∫5 ∫5 v

25 = 24 ln 5 = 24 ln 5 = 38.6 to 3 s.f. The work done is less by 39.9 – 38.6 = 1.3.

= ln(1 + 2x)2 − ln(1 − 2x)2 + c = ln

(11 +− 22xx ) + c 2

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6

WORKED SOLUTIONS

6 Numerical solution of equations Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. When using numerical methods, it is generally acceptable for intermediate working to be accurate to 6 decimal places. For completeness, some of the intermediate working in this chapter is accurate to 9 decimal places. When using numerical methods, it is generally acceptable for intermediate working to be accurate to 6 decimal places. For completeness, some of the intermediate working in this chapter is accurate to 9 decimal places.

Prerequisite knowledge

Exercise 6.1A

1

1

y

a f(x) = 3x2 + 4x – 11 f(1.2) = 4.32 + 4.8 – 11 = –1.88 f(1.4) = 5.88 + 5.6 – 11 = 0.48

0 –1

There is a sign change between f(1.2) and f(1.4).

x

5

Therefore there is a root of f(x) in the interval 1.2 < x < 1.4.

–5

b f(x) = x3 + 6x2 + 11x + 6 2

f(–1.4) = –2.744 + 11.76 – 15.4 + 6 = –0.384

y

f(–0.8)= –0.512 + 3.84 – 8.8 + 6 = 0.528 There is a sign change between f(–1.4) and f(–0.8), therefore there is a root of f(x) in the interval –1.4 < x < –0.8. –5

–1

0

2

c f(x) = 8x4 + 2x3 – 53x2 + 37x – 6

x

f(1) = 8 + 2 – 53 + 37 – 6 = –12 f(3) = 648 + 54 – 477 + 111 – 6 = 330 3

There is a sign change between f(1) and f(3), therefore there is a root of f(x) in the interval 1 < x < 3.

y

–10

2

f(x) = 1 – ex – ln x f(0.2) = 1.388 f(0.7) = –0.657 There is a sign change between f(0.2) and f(0.7). Therefore there is a root of f(x) in the interval 0.2 < x < 0.7.

3

f(x) = sinx x ; −0.8 < x < −0.7 e f(–0.8) = –1.597 f(–0.7) = –1.297 There is not a sign change between f(–0.8) and f(–0.7). Therefore there is not a root of f(x) in the interval –0.8 < x < –0.7.

x

–3

Asymptotes at x = −1 and y = 0. 4

y 0 1 2 –2

x

Asymptotes at x = 1 and y = 0. 2

73

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NUMERICAL SOLUTION OF EQUATIONS

4

a

y y=

1 x2

b 7+ y = ex

5 13 = x x2

f(x) = 7 +

5 13 − x x2

f(0.95) = –2.141 f(1.1) = 0.802 There is a sign change between f(0.95) and f(1.1). 0

Therefore there is a root of f(x) in the interval 0.95 < x < 1.1.

x

c Rearranging 7 +

One point of intersection of the two graphs 1 therefore 2 = e x has one root. x

Consequently, by solving the quadratic, the roots to the original equation are also found.

1 b f(x) = 2 − e x x

a = 7, b = 5, c = −13

f(0.6) = 0.956

x=

f(0.8) = –0.663 There is a sign change between f(0.6) and f(0.8). Therefore there is a root of f(x) in the interval 0.6 < x < 0.8. sin2 x – cos2 x = 0 f(x) = sin2 x – cos2 x

5

−5 ± 52 − (4)(7)(−13) 14 x = 1.05 or –1.77 7

There is a sign change between f

( π8 ) and f ( 38π ).

π π 3π is in the interval < x < . 4 8 8

Sign change so x = 3 − 1 − 22 has a root in the x x interval –1 < x < 0. f(x) = 3 − 1 − 22 − x x x f(1.4) = –0.134 693 877 6

y

f(1.8) = 0.027 160 493 83

5 y=7+ x y=

13 x2

0

–12

8

Sign change so x = 3x 2 − 1 has a root in the x interval 0 < x < 1. f(x) = 3 − 1 − 22 − x x x f(–1) = 3 f(–0.5) = –2.5

Therefore there is a root near x = π 4 a

f( x ) = 3x 2 − 1 − x x f(10–6) = –1 000 000 f(1) = 1

( π8 ) = −0.707 3π f ( ) = 0.707 8

6

−b ± b 2 − 4ac 2a

x=

f

x=

5 13 gives 7x2 + 5x – 13 = 0. = x x2

x

There are two points of intersection. Therefore the equation 7 +

5 13 = has two roots. x x2

Sign change so x = 3 − 1 − 22 has a root in the x x interval 1.4 < x < 1.8. f(x) = 3 − 1 − 22 − x x x f(1.8) = 0.027 160 493 83 f(2.2) = –0.067 768 595 04 Sign change so x = 3 − 1 − 22 has a root in the x x interval 1.8 < x < 2.2.

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WORKED SOLUTIONS

9

f ( x ) = 1 + 7 − ex 2x

y

f(10–6) = 500 006 f 3 = 2.851 644 263 2

()

No sign change so either ex = 1 + 7 has no roots 2x 3 in the interval 0 < x < or has more than one root 2 in this interval.

–2

0

x

3

10 f(x) = 3x2 – 1 – 2x f(–5) = 73.968 75 f(0) = –2

There is a discontinuity in the interval 2.4 < x < 3.3, so although there is a sign change between f(2.4) and f(3.3) there isn’t a root, as shown on the graph sketch.

Sign change so 2x = 3x2 – 1 has a root in the interval –5 < x < 0. f(x) = 3x2 – 1 – 2x f(0) = –2

3

Functions in the form f(x) = (x ± a)2 do not intersect the x-axis. Instead the curve ‘sits on’ the x-axis (the discriminant equals zero and so the function has one distinct, repeated root). Consequently in the case of f(x) = (x ± a)2, f(a – 1) and f(a + 1) are both positive.

4

f(x) =

f(2) = 7 Sign change so 2x = 3x2 – 1 has a root in the interval 0 < x < 2. f(x) = 3x2 – 1 – 2x f(2) = 7 f(10) = –725

1 +2 x

Sign change so 2x = 3x2 – 1 has a root in the interval 2 < x < 10.

y

Exercise 6.2A 1

2

f(x) = 6x2 – x – 2 Root in the interval –2 < x < 0. f(–2) = 24 f(0) = –2 There is a sign change between f(–2) and f(0). Therefore there is a root of f(x) in the interval –2 < x < 0. Root in the interval 0 < x < 2. f(0) = –2 f(2) = 20 There is a sign change between f(0) and f(2). Therefore there is a root of f(x) in the interval 0 < x < 2. f(x) =

1 ; 2.4 < x < 3.3 (x − 3)

−5 3 10 f(3.3) = 3

f(2.4) =

2

0

–6

a f(x) =

x

1 +2 x

Root in the interval –1 < x < –0.2. f(–1) = 1 f(–0.2) = –3 There is a sign change between f(–1) and f(–0.2). Therefore there is a root of f(x) in the interval –1 < x < –0.2. b f(x) =

1 +2 x

Asymptote in the interval –0.2 < x < 0.2. f(–0.2) = –3 f(0.2) = 7

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Numerical solution of equations



There is a sign change between f(–0.2) and f(0.2).

b & c f( x ) = 12 − 2 x



Therefore there is an asymptote of f(x) in the interval –0.2 < x < 0.2.



( ) f (1 ) = 2 2

5 f(x) = tan x Asymptote in the interval π < x < 3π . 4 4 π f =1 4

() 3π f ( ) = −1 4

There is a sign change between f



Therefore there is an asymptote of f(x) in the π 3π interval < x < . 4 4 Root in the interval 3π < x < 5π . 4 4 3π f = −1 4



( ) 5π f( ) =1 4

No sign change in the interval − 1 < x < 1 2 2 however there is a discontinuity. f(–1) = –1 f −1 = 2 2

( )

( π4 ) and f ( 34π ).



Sign change so there is a root in the interval –1 < x < − 1 . 2 1 f =2 2 f(1) = –1

()

Sign change so there is a root in the interval 1 < x < 1. 2

( 34π ) and f ( 54π ).



There is a sign change between f



Therefore there is a root of f(x) in the interval 3π 5π 2 17

Since a • b is not 0, the lines are not perpendicular.



2 2 Solve (5 − 2k) + (−8) = 2 17 .



(5 – 2k)2 + 64 = 68



4k2 – 20k + 25 + 64 = 68



Find the magnitude of each line segment.



4k2 – 20k + 21 = 0



BA = (–8i – 4j) – (–i + 10j) = (–7i – 14j),



(2k – 3)(2k – 7) = 0

8 a Given three vertices of a rectangle, one must be the right angle.

magnitude AB = (−7)2 + (−14)2 = 245

3 7 k = or 2 2



3 7 For magnitude > 2 17 , k < 2 or k > 2 . 6 a BA = (5i – 2j + 4k) – (8i + 4j + 10k) = –3i – 6j – 6k

BC = (14i + 7j + 4k) – (8i + 4j + 10k) = 6i + 3j – 6k



BA • BC = (–3i – 6j – 6k) • (6i + 3j – 6k)







AB and BC are perpendicular

= –18 – 18 + 36 = 0

b   i OD = OA + AD = OA + BC

OD = (5i – 2j + 4k) + (6i + 3j – 6k) = 11i + j – 2k

1 ii  ((5i – 2j + 4k) + ((14i + 7j + 4k)) 2 19 5 = i + j + 4k) 2 2 iii Magnitude of BA = (−3)2 + (−6)2 + (−6)2 = 9.



Area = 9 × 9 = 81 units2.

 −2  2   −4       7 a AB =  3  –  −5 =  8   2   1   1 

Magnitude of AB = (−4)2 + 8 2 + 12 = 9.

3

s = –2

 5  k   5 − 2k  5 a i   – 2   =  .  −6  1   −8 

b

1

CB = (–i + 10j) – (3i + 8j) = (–4i + 2j), magnitude BC = (−4)2 + 22 = 20



Area = 245 × 20 = 70.

b Fourth vertex = (–8i – 4j) + [(3i + 8j) – (–i + 10j)] = (–4i – 6j) 9 Since ADP is a straight line, DP is parallel to AD, so DP = kd. XP = XC + CD + DP = 6d – 10b + kd. XY = 6d – 4b Also, XP = nXY = n(6d – 4b). Hence 6d – 10b + kd = n(6d – 4b). Equating coefficients for b and d. b: –10 = –4n 5 n = 2 d: 6 + k = 6n 5 6+k=6× 2 6 + k = 15 k = 9 BP = BC + CD + DP = 10d – 10b + 9d = 19d – 10b  −2 − 5λ   −3 + 2µ  10 a At X  −5  =  −17 + 4µ       9 + 7λ   5 − µ 



– 5 = – 17 + 4µ µ = 3 – 2 – 5λ = – 3 + 2(3)  λ = – 1

2 1

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7 

WORKED SOLUTIONS





9 + 7(– 1) = 5 – 3 ✓ Coordinates of X (3, –5, 2)

 −5 b  0     7 

2 • 4  =    −1

( −5)2 + 72

3

22 + 4 2 + ( −1) cosθ





(– 5 × 2) + (0 × 4) + (7 × –1) = 74





– 17 = 74









2

21 cosθ

21 cosθ







XD 2 = (–2 – (–5))2 + (–5 – (–6))2 = 10



CD2 = (–4 – (–5))2 + (1 – (–6))2 = 50



Since XD2 + XC 2 = CD 2, XC is perpendicular to XD.

e Magnitude of DE = magnitude of CF = 3 10 , magnitude of XC = 40 = 2 10 .

−1 

−17  θ = cos  = 115.5  74 21  acute angle 180 – 115.5 = 64.5°



they are parallel and DXGC is a trapezium. 13 a AB = (3i + 7j + k) – (4i + 5j – k) = –i + 2j + 2k

Equation of l1: r = (4i + 5j – k) + t(–i + 2j + 2k).



UV = (–3i + 7j – 11k) – (13i – j + 5k)





Equation of l2: r = (13i – j + 5k) + u(–2i + j – 2k)



 −4  1   −5        AC = 1 − 4 = −3        3   −5  8 



(–i + 2j + 2k) • (–2i + j – 2k) = 2 + 2 – 4 = 0.



Since a • b = 0, the lines are perpendicular.

1   −5 l2 =  4  + µ  −3      −5 8

b Let the point of intersection of l1 and l2 be X.



= –16i + 8j – 16k = 8(–2i + j – 2k)

cosθ =

58 69 98







58 = , θ = 45.1°    7 138



Equating i coefficients: 4 – t = 13 – 2u.

1



Equating j coefficients: 5 + 2t = –1 + u.

2



Equating k coefficients: –1 + 2t = 5 – 2u.

3

θ ( −5)2 + ( −3)2 + 82 cos Subtracting

 2   −5          2 2 2 AB ⋅ AC =  −4 ⋅  −3 = 22 + ( −4 ) + 7 2 ( −5) + ( −3) + 8 2 cosθ     7 8 – 10 + 12 + 56 = 69 98 cosθ

1 AB AC sin θ 2 1        = 2 69 98 sin 45.1                = 29.1 unit2

12 a CF = f – c = (5i + 4j) – (–4i + j) = (9i + 3j)

3 from 2 : 6 = –6 + 3u.

u = 4

Point of intersection: X(5, 3, –3).



Note that the point of intersection is also the mid-point of U and V.



UX = (5i + 3j – 3k) – (13i – j + 5k)



= –8i + 4j – 8k = 4(–2i + j – 2k) Since UVW is isosceles with a right angle at W, XW and UX are of equal length, so OW = OX +XW = (5i + 3j – 3k) ±4(–i + 2j + 2k)

c Area =



Since CG and DX have a common factor,

2  1 l1 =  4  + λ  −4     7  −5

 2   −5       2 b AB ⋅ AC = −4 ⋅ −3 = 22 + ( −4 ) + 7 2     7 8



Area = 3 10 × 2 10 = 60.

f CG = g – c = (2i + 3j) – (–4i + j) = (6i + 2j) = 2(3i + j).

 3  1   2   11 a AB =  0 −  4  =  −4        2  −5  7 

d XC 2 = (–2 – (–4))2 + (–5 – 1)2 = 40



OW = i + 11j + 5k or 9i – 5j – 11k Coordinates of W are (1, 11, 5) or (9, –5, –11).

DE = (10i + 3j – 3k) – (18i + j – 9k) = –8i + 2j + 6k 14 a  = 2(–4i + j + 3k)

92 + 32 = 90 = 3 10 . b CD = d – c = (–5i – 6j) – (–4i + j) = (–i – 7j) FE = e – f = (4i – 3j) – (5i + 4j) = (–i – 7j)



Equation of l1 is (18i + j – 9k) + t(–4i + j + 3k).



FG = (8i – 29j + 31k) – (–6i + 20j – 4k)

c DE = e – d = (4i – 3j) – (–5i – 6j) = (9i + 3j)



Equation of l2 is (–6i + 20j – 4k) + u(2i – 7j + 5k)



(–4i + j + 3k) • (2i – 7j + 5k)



= –4 × 2 + 1 × –7 + 3 × 5 = 0.



DX = 1 DE = 1 (9i + 3j) = (3i + j) 3 3



OX = OD + DX = (–5i – 6j) + (3i + j) = (–2i – 5j)

= 14i – 49j + 35k = 7(2i – 7j + 5k)

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Vectors

b Since FJ is a chord of circle C, F and J are equidistant from l1



AB • AC = (4i – 4j + 6k) • (–4i – 12j + 2k) = –16 + 48 + 12 = 44



Let the point of intersection of the lines be X.



Magnitude of AB = 4 2 + (−4)2 + 6 2 = 68 .



Equating i coefficients: 18 – 4t = –6 + 2u. 1



Equating j coefficients: 1 + t = 20 – 7u.



Magnitude of AC = (−4)2 + (−12)2 + 22 = 164 .



Equating k coefficients: –9 + 3t = –4 + 5u. 3



cos A =



Multiply 2 by 3: 3 + 3t = 60 – 21u.



Subtract from 3 : –12 = –64 + 26u.





2



u = 2

AB • AC = AB AC

44 = 0.4167 68 × 164

A = cos–1(0.4167) = 65.38° 1 Area = × 68 × 164 × sin 65.38° = 48 unit2. 2



Point of intersection = X(–2, 6, 6).



FX = XJ

16 a OR = (28 + 3t)i + (20 – t)j + (4 + 3t)k [(28 + 3t)i + (20 – t)j + (4 + 3t)k] • (3i – j + 3k) = 0



OX – OF = OJ – OX





3(28 + 3t) – (20 – t) + 3(4 + 3t) = 0



OJ = 2OX – OF OJ = 2(–2i + 6j + 6k) – (–6i + 20j – 4k) = 2i – 8j + 16k





84 + 9t – 20 + t + 12 + 9t = 0



19t + 76 = 0



t = –4

J(2, –8, 16) 15 Let the first line be L1, the second L2 and the third L3. Let the intersection of L1 and L2 be A, of L1 and L3 be B and of L2 and L3 be C.

To find A:



Equating i coefficients: 3 + 2t = –1 + 2u.

1



Equating j coefficients: 21 – 2t = –7 + 6u.

2



Equating k coefficients: –7 + 3t = 3 – u.

3



Adding 1 and 2 : 24 = –8 + 8u. Point of intersection: A(7, 17, –1).



To find B:



Equating i coefficients: 3 + 2t = 17 + 2v.

1



Equating j coefficients: 21 – 2t = 19 + 2v.

2



Equating k coefficients: –7 + 3t = 8 + v.

3



Adding 1 and 2 : 24 = 36 + 4v.

v = –3 Point of intersection: B(11, 13, 5).



To find C:



Equating i coefficients: –1 + 2u = 17 + 2v.

1



Equating j coefficients: –7 + 6u = 19 + 2v.

2



Equating k coefficients: 3 – u = 8 + v.

3



Subtracting 1 from 2 : –6 + 4u = 2 Point of intersection: C(3, 5, 1). AB = (11i + 13j + 5k) – (7i + 17j – k) = 4i – 4j + 6k



AC = (3i + 5j + k) – (7i + 17j – k) = –4i – 12j + 2k

OR = 16i + 24j – 8k  b Shortest distance between O and L = OR

= 162 + 24 2 + (−8)2 = 896 = 8 14 = 29.9 17 a QR = (–2j – 6k) – (–2i + 2j) = (2i – 4j – 6k) = 2(i – 2j – 3k) r = (–2i + 2j) + u(i – 2j – 3k)



When t = 9, –2 + 2(9) = 16.



When t = 9, –1 + 2(9) = 17.

c Equating i coefficients: –2 + u = –40 + 7t. 1

Equating j coefficients: 2 – 2u = –2 + 2t.

2



Equating k coefficients: –3u = –1 + 2t.

3



Subtracting 3 from 2 : 2 + u = –1.

u = –3







t = 9





Hence OR = (28 + 3(–4))i + (20 – (–4))j + (4 + 3(–4))k.

b Equating i coefficients: –40 + 7t = 23.

u = 4

u = 2





Point of intersection: P(–5, 8, 9)

d PR = (–2j – 6k) – (–5i + 8j + 9k) = 5i – 10j – 15k

 PS = (23i + 16j + 17k) – (–5i + 8j + 9k) = 28i + 8j + 8k



PR • PS = (5i – 10j – 15k) • (28i + 8j + 8k) = 140 – 80 – 120 = –60



Magnitude of PR = 52 + (−10)2 + (−15)2 = 350 .



Magnitude of PS = 28 2 + 8 2 + 8 2 = 912 .

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7 

WORKED SOLUTIONS



cos  P=

PR • PS = PR PS

−60 = –0.1062 350 × 912

P = cos–1(–0.1062) = 96.10° 1 e Area = × 350 × 912 × sin 96.10° 2 = 281 unit2. 18 a AB = (–4i + 5j + 2k) – (4i + 3j – k) = –8i + 2j + 3k



Magnitude of EG = 18 2 + (−18)2 = 18 2 . cos q =



EF • EG 648 = 36 × 18 2 EF EG

 q = cos–1

 1  648   = cos–1 2  = 45° 36 × 18 2

d FH = (15i + 11j + 4k) – (23i + 9j + 20k) = –8i + 2j – 16k



AC = (6i – j) – (4i + 3j – k) = 2i – 4j + k



AB • AC = (–8i + 2j + 3k) • (2i – 4j + k) = –16 – 8 + 3 = –21





Magnitude of AB = (−8)2 + 22 + 32 = 77 .

e Magnitude of FH = (−8)2 + 22 + (−16)2 = 18.



2 2 2 Magnitude of AC = 2 + (−4) + 1 = 21 .





cos  A = AB • AC = AB AC

−21 = – 33 11 77 × 21 2



  sin  A = – 1 −  33  = 2 22 11  11  Area = 1 × 77 × 21 × 2 22 = 7 6 . 2 11

b EA = (4i + 3j – k) – (i + 2j + k) = 3i + j – 2k

ED = (10i + 11j + 19k) – (i + 2j + k) = 9i + 9j + 18k



EA • ED = (3i + j – 2k) • (9i + 9j + 18k) = 27 + 9 – 36 = 0



Since EA • ED = 0, ∠AED = 90°.

c ED is the height of the tetrahedron.

Magnitude of ED = 92 + 92 + 18 2 = 9 6 . 1 Volume of tetrahedron =  × area of base 3 × height.



=







1 ×7 6 ×9 6 3

= 126 unit3

19 a GH = (15i + 11j + 4k) – (7i – 5j + 6k) = 8i + 16j – 2k

Magnitude of GH = 8 2 + 16 2 + (−2)2 = 18.

b 8i + 16j – 2k = 2(4i + 8j – k)

Equation of l2 is r = (7i – 23j + 24k) + t(4i + 8j – k).



Equating i coefficients: 7 + 4t = 23.

(4i + 8j – k) • (–8i + 2j – 16k) = 4 × –8 + 8 × 2 + –1 × –16 = 0

Area of trapezium =

1 (18 + 2 × 18) × 18 = 486. 2

 8  −7  15   5 20 a AB =  5 –  −4 =  9  = 3  3           3  9   −6  −2  −7  5 r =  −4 + u  3       9   −2  6−t   8 b  11 + 3t  =  5      7 + 2t   3



6–t=8



11 + 3t = 5



t = –2 t = –2

7 + 2t = 3



t = –2

 5   −1 c  3    3  = –5 + 9 – 4 = 0      −2    2  6   −7  13  d AC =  11 –  −4 =  15         7   9   −2



When t = 4, –23 + 8t = –23 + 8(4) = 9.

 13   15  AC  AB =  15    9  = 13 × 15 + 15 × 9 +      −2  −6



When t = 4, 24 – t = 24 – 4 = 20.

(–2) × (–6) = 342

t = 4

c EF = (23i + 9j + 20k) – (7i – 23j + 24k) = 16i + 32j – 4k

EG = (7i – 5j + 6k) – (7i – 23j + 24k) = 18j – 18k EF • EG = 16 × 0 + 32 × 18 + –4 × –18 = 648



Magnitude of EF = 16 2 + 322 + (−4)2 = 36.





| AC | = 132 + 152 + (−2)2 = 398



| AB | = 152 + 92 + (−6)2 = 3 38 342 cosθ = 398 × 3 38 θ = 22.0°



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Vectors

 6   8  −2     e BC = 11 – 5 =  6         7   3  4 

| BC | = (−2)2 + 6 2 + 4 2 = 2 14



Area = 1 × 3 38 × 2 14 = 6 133 = 69.2 2 f CB = BD



OD = 2OB – OC



 8  6  OD = 2  5 –  11      3  7 



Mathematics in life and work 1 MP = (16.4i + 23.1j + 0.5k) – (–2.2i + 21.5j + 0.7k) = (18.6i + 1.6j – 0.2k) km 2 Let a = (18.6i + 1.6j – 0.2k) and b = (13i – 14j – k). a • b = (18.6i + 1.6j – 0.2k) • (13i – 14j – k)



= 18.6 × 13 + 1.6 × –14 – 0.2 × –1





= 219.6



|a| = 18.6 2 + 1.6 2 + 0.22 = 348.56



|b| = 132 + 14 2 + 12 = 366



cos  q =

D(10, –1, –1)

 −19 + t   −2  −17 + t      21 a TF = 14 − 3t – 5 =  9 − 3t         −5 + at   8   −13 + at 

TF  l = 0



 −17 + t   1   9 − 3t    −3 = 0      −13 + at   a 



219.6 = 0.6148 348.56 × 366

q = cos–1(0.6148) = 52.1° 3 Let the foot of the perpendicular be F.

Shortest distance = PF = 348.56  sin 52.1 = 15 km.

4 PF  = [(–2.2i + 21.5j + 0.7k) + t(13i – 14j – k)] – (16.4i + 23.1j + 0.5k)





Hence [(–18.6 + 13t)i + (–1.6 – 14t)j + (0.2 – t)k] • (13i – 14j – k) = 0.

–17 + t – 27 + 9t – 13a + a2t = 0



13(–18.6 + 13t) – 14(–1.6 – 14t) – (0.2 – t) = 0



(a2 + 10)t = 13a + 44



–241.8 + 169t + 22.4 + 196t – 0.2 + t = 0



13a + 44 t= 2 a + 10



366t = 219.6

b 5 =

13a + 44 a 2 + 10



5a2 + 50 = 13a + 44



5a2 – 13a + 6 = 0



(5a – 3)(a – 2) = 0



a=



3 When a = , 5    −19 + 5   −14  14 − 3( 5)    OF =   =  −1   −2   −5 + 3 5  ()  5 



= [(–18.6 + 13t)i + (–1.6 – 14t)j + (0.2 – t)k]

t = 0.6

OF = [(–2.2i + 21.5j + 0.7k) + 0.6(13i – 14j – k)] = (5.6i + 13.1j + 0.1k) km

3 or 2 5



When a = 2,



 −19 + 5   −14  14 − 3 5  ( )  =  −1  OF =   5   −5 + 2( 5)

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8 

WORKED SOLUTIONS

8 Differential equations Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1 a Using the chain rule

2x cos(x2 + 1).

b Using product rule u = e–x, v = cos 2x and du dv = −e −x , = −2sin 2x dx dx

so giving –e–x cos 2x – 2e–x sin 2x.

c

3 3x + 2

2 a

∫ 4x

−0.5

dx = 8 x 0.5 + c = 8 x + c

1 e 4x + c 4 c tan x + c



dy 5 a = 0.005y dt b If y = ae0.005t then dy = a × 0.005e0.005t = 0.005y dt c Substitute t = 0, 50 = a × 1 so a = 50 and y = 50e0.005t.

b

e

f −

When t = 20, y = 50e0.1 = 55.3 so an estimate of the population is 55.3 million.

dy 6 a dx = xy

d ln |3x + 2| + c ln|2x3

dy a − 12 a . = x = dx 2 2 x a This is a solution with k = . 2 1

c If y = a x = ax 2 then

+ 4| + c

1 cos(4x) + c 4

3 a 2 sin 2x

1 x2

b If y = e 2

1 x2 1 x2 dy = e 2 × x = xe 2 = xy. dx

dx c 7 a = dt x

When x = 5,



Therefore

b –cos 2x c e2x

then

dx c = 0.4 so 0.4 = so c = 2. 5 dt

dx 2 dx = or x = 2. dt x dt

Exercise 8.1A

b dx = 2 dt x

dy = 4x 3 + 4x + 2, hence 1 If y = x4 + 2x2 + 2x + 4 then dx d = 2. dy = 1.5 × x 0.5 = 1.5 × x , hence 2 If y = x1.5 + c then dx k = 1.5.



Therefore



When x = 5 the acceleration is 2 − = −0.08  m s−2 25

dy 3 a = kx 2 where k is a constant. dx dy = 3x 2 and this is a b If y = x3 + 4 then dx solution with k = 3.

c The particle will reduce in speed but in theory will never become stationary. dr c 8 a = 2 dt r b 0.5 =

dy k 4 a = dx y

r2

dy 1 − 12 1 = x = then and b If y = x = dx 2 2 x this is a solution with k = 1 . 2 1 x2

d 2x 2 = − 2. dt 2 x

dr 50 c so c = 50 and dt = 2 or 100 r

dr = 50. dt 1



2

dr 1 − 3 = at and dt 3 2 dr 1 −2 1 r2 = a 2t 3 × at 3 = a 3. 3 3 dt

c r = at 3 so

109

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Differential equations



This is a solution if a = 3 150

1 3 a = 50 or a 3 = 150 or 3

1 b y = sin ( 3x ) + c 3

1 9 a dx = kx 3 . dt 1 4 = k ( 8 ) 3 = 2k



(





( )

)

3 2

(

dx = 4 × 3 × 4t + 4 3 dt 3 2

(

 4t dx = 2× +4 dt  3

)



Which gives 4 = 0 + c, so c = 4.



1 Thus y = sin ( 3x ) + 4. 3

c

dy 2 + x 2 − 4x 5 2 x 2 4x 5 = = + − x x x x dx

1 2

1 3 3  2

) 

=

=

1 2x 3

as required

c If x = 0 at the start then the snowball would not start to move. dP k = 10 a Rate of growth  P dt If P = 2kt + c , then −1 c) 2

dP 1 = (2k)(2kt + 2 dt k = 2kt + c k = as required. P dP k = 2 so 2 = b When P = 2,  2 dt k=4 So P = 8t + c When t = 0, P = 2, Then 2 = c c=4 So P = 8t + 4 At t = 2, P = 8(2) + 4 = 2 5 (thousand) So P = 4500 (2s.f.) c P 10

2 + x − 4x 4 x 1 2 4 5 x − x + c. 2 5



So y = 2 ln|x|+



Substitute y = 2 and x = 1.



So 2 = 2 ln|1| +



23 4 Which gives 2 = 0 + 1 − +c, so c = 10 . 2 5



Thus y = 2ln|x| +

2

dy = 8ax 3 dx



y = 2ax4 + c

3

dx = y −0.5 dy x = 2y0.5 + c y 0.5 =

1 4 × (1)2 − × (1)5 + c. 2 5

1 2 4 5 23 . x − x + 10 2 5

( )

x−c x−c or y = 2 2

2



Alternatively separate the variables to get y –0.5 dy = dx.



So ∫ y −0.5 d y = ∫ 1dx and 2y0.5 = x + c.



y=

( x 2+ c ) . Here the constant has changed sign 2

but the solutions are equivalent.

8

4 Rewrite as

6 4

dx 1 = = 10y −2 so dy 0.1y 2

x = ∫ 10y −2 d y = −10y −1 + c or x = −

2 0

π so , 3

1 1 π 4 = 3 sin 3 × 3 + c = 3 sin(π) + c

k=2 1 so dx = 2x 3 dt

4t + 4 b x = 3

Substitute y = 4 and x =

2

4

6

8

10 12 14

t



This can be rewritten as

10 + c. y

10 10 . = c − x or y = c−x y

dy 6 = 8x + 3 = 8x + 6x −3 dx x y = 4x2 – 3x–2 + c

Exercise 8.2A

5

1 a y = 2x − 0.3 x 3 + c = 2x − 0.1 x 3 + c 3 Substitute y = 2 and x = 2, so 2 = 2 × 2 – 0.1 × (2)3 + c.



Substitute x = 1 and y = 9, so 9 = 4 × (1)2 – 3 × (1)–2 + c.



Which gives 9 = 4 − 3 + c, so c = 8.



Thus the equation of the curve is y = 4x2 – 3x–2 + 8.



Which gives 2 = 4 – 0.8 + c, so c = −1.2.



Thus y = 2x – 0.1x3 – 1.2.

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8 

WORKED SOLUTIONS

(t +10100 ) . 7 + 100 107 = b After 1 week, t = 7, so y = ( 10 ) ( 10 )

6 a x˙ = t 3 – 6t 2+ at or dx = t 3 − 6t 2 + at dt x = 1 t 4 − 2t 3 + a t 2 + c 4 2



c At the origin, the displacement is zero, so x = 0. 1 a Hence we solve 0 = 4 t 4 − 2t 3 + 2 t 2 0 = t 4 – 8t 3 + 2at 2 0 = t 2 (t 2 – 8t + 2a)







So t = 0 or t 2 – 8t + 2a = 0



So t =

( −8 )2 − 4 × 1 × 2a

8±

no roots if (–

64 – 8a < 0



64 < 8a



8 0, Re z2 = 7 and Im z2 = –3

(15x ) = –16 2



z2 = 7 – 3i

x4 – 225 = –16x2

d z3 = –7 + 3i

x4 + 16x2 – 225 = 0

| z3| = (−7)2 + 32 = 58

(x2 – 9)(x2 + 25) = 0

3 arg z3 = π – tan–1 7 = 2.74

x2

=9

x = ±3

()

Exercise 9.3A

y = ±5 Square roots are 3 + 5i and –3 – 5i

1 a 4 – 11i

b z = 4 ± 11i 9 362 − 153i × 2 + 3i = 724 + 1086i − 306i+459 = 1183 + 780i = z 91 +–60i 4 = ±11i 2 − 3i 2 + 3i 4+9 13 53i 2 + 3i 724 + 1086i − 306i+459 1183 + 780i (z – 4)2 = –121 × = = = 91 + 60i i 2 + 3i 4 + 9 13 z2 – 8z + 16 = –121 (x + iy)2 = 91 + 60i z2 – 8z + 137 = 0 x2 + 2xyi – y2 = 91 + 60i 2 2 k = 137 x – y = 91 2xy = 60 2 a (–14)2 – 4(1)(58) = 196 – 232 = –36 < 0 y = 30 b (z – 7)2 – 49 + 58 = 0 x x2 –

( 30x ) = 91 2

x4 – 900 = 91x2 x4 – 91x2 – 900 = 0 (x2 – 100)(x2 + 9) = 0 x = ±10 y = ±3 Square roots are 10 + 3i and –10 – 3i. 362 − 153i = 10 2 + 32 = 109 2 − 3i 10 a |z1| = 40 2 + (−42)2 = 58

( )

42 b arg z1 = –tan–1 40 = –0.810



(z – 7)2 + 9 = 0



(z – 7)2 = –9

z – 7 = ±3i z = 7 ± 3i 3 a z = ±5i b z2 = –144 z = ±12i c z2 + 10z + 26 = 0

(z + 5)2 – 25 + 26 = 0



(z + 5)2 = –1

z + 5 = ±i z = –5 ± i

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9 

WORKED SOLUTIONS

d z2 – 14z + 53 = 0

(z –

7)2

– 49 + 53 = 0

(z –

7)2

= –4

z – 7 = ±2i z = 7 ± 2i e

z2



(z + 4)2 – 16 + 80 = 0



(z + 4)2 = –64

+ 8z + 80 = 0

z + 4 = ±8i





(z + 6)2 – 36 + 37 = 0



+ 12z + 37 = 0

(z +

6)2

= –1



z + 6 = ±i

g z2 + 104 = 20z

– 20z + 104 = 0

(z –

10)2

– 100 + 104 = 0

(z –

10)2

= –4

z – 10 = ±2i z = 10 ± 2i h z2 + 18z + 202 = 0

(z + 9)2 – 81 + 202 = 0



(z + 9)2 = –121

z + 9 = ±11i z = –9 ± 11i i z2 + 41 = 10z z2 – 10z + 41 = 0

(z – 5)2 – 25 + 41 = 0



(z – 5)2 = –16

z – 5 = ±4i

5 3 =± i 2 2

9z2 – 48z + 68 = 0

(

9 z2 −

)

16 z + 68 = 0 3

( ) − 649  + 68 = 0 8 9 ( z − ) − 64 + 68 = 0 3 8 9 ( z − ) = −4 3 ( z − 83 ) = −94  8 9 z − 3 

2

j z2 – 12z + 4936 = 0

(z – 6)2 – 36 + 4936 = 0



(z – 6)2 = –4900

z – 6 = ±70i z = 6 ± 70i k 2z2 + 17 = 10z

2z2 – 10z + 17 = 0



2(z2 – 5z) + 17 = 0



 5 2 z − 2 

2

2



2



z –

8 2 =± i 3 3

z =

8 2 ± i 3 3

4 a z = 2 ± 5i z – 2 = ±5i

(z – 2)2 = –25

z2 – 4z + 4 = –25 z2 – 4z + 29 = 0 b z = 7 ± 4i z – 7 = ±4i

(z – 7)2 = –16

z2 – 14z + 49 = –16

z = 5 ± 4i

( )

z–

2

z = –6 ± i z2

2

5 3 ± i 2 2 2 l 9z + 68 = 48z

z = –4 ± 8i f

2

z =



z2

( 52 ) = − 92 ( z − 52 ) = − 94 2 z−

25  −  + 17 = 0 4

z2 – 14z + 65 = 0 c z = –8 ± 20i z + 8 = ±20i

(z + 8)2 = –400

z2 + 16z + 64 = –400 z2 + 16z + 464 = 0 d z = –3 ± 2i z + 3 = ±2i

(z + 3)2 = –4

z2 + 6z + 9 = –4 z2 + 6z + 13 = 0

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Complex numbers

5 a x3 + x2 + 15x – 225 = 0 x3

+

x2

x2 – 2x + 5 = 0

+ 15x – 225 = (x –

5)(x2

+ 6x + 45)

x2 + 6x + 45 = 0

x4 + 3x3 + x2 + 13x + 30 = (x2 – 2x + 5) (x2 + 5x + 6)



(x + 3)2 – 9 + 45 = 0

x2 + 5x + 6 = 0



(x + 3)2 = –36



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



Roots are –2, –3, 1 ± 2i.

x + 3 = ±6i

6 a z3 – 8z2 + 9z – 72 = 0

x = –3 ± 6i

Roots are 5, –3 ± 6i.

b

x3

+

7x2



– 13x + 45 = (x +

9)(x2

– 2x + 5)

x2 – 2x + 5 = 0

f(8) = 0, so (z – 8) is a factor.

z3 – 8z2 + 9z – 72 = (z – 8)(z2 + 9) z2 + 9 = 0



(x – 1)2 – 1 + 5 = 0

z2 = –9



(x – 1)2 = –4

z = ±3i

x – 1 = ±2i



x = 1 ± 2i

b z3 + 4z + 10 = 5z2



Roots are –9, 1 ± 2i.

z3 – 5z2 + 4z + 10 = 0

c

x3



+

10x2

+ 29x + 30 = 0

Roots are 8, ±3i.

f(–1) = 0, so (z + 1) is a factor.

x = –2 ± i

z3 – 5z2 + 4z + 10 = (z + 1)(z2 – 6z + 10)

x + 2 = ±i

z2 – 6z + 10 = 0

(z – 3)2 – 9 + 10 = 0

x2 + 4x + 4 = –1



(z – 3)2 = –1

x2 + 4x + 5 = 0

z – 3 = ±i

x3 + 10x2 + 29x + 30 = (x2 + 4x + 5)(x + 6)

z = 3 ± i







(x +

2)2

= –1

Roots are –6, –2 ± i.

Roots are –1, 3 ± i.

d x = 3 ± 5i

c 2z3 – 8z2 – 13z + 87 = 0

x – 3 = ±5i



f(–3) = 0, so (z + 3) is a factor.



2z3 – 8z2 – 13z + 87 = (z + 3)(2z2 – 14z + 29)

x2 – 6x + 9 = –25



2z2 – 14z + 29 = 0

x2 – 6x + 34 = 0



2(z2 – 7z) + 29 = 0



 7 2 z − 2 



(x –

3)2

= –25

x3 – 10x2 + 58x – 136 = (x2 – 6x + 34)(x – 4)

Roots are 4, 3 ± 5i.

e 3x3 – 38x2 + 135x – 74 = 0

x – 6 = ±i

(x –

x2

2

= –1



2

– 12x + 36 = –1

x2 – 12x + 37 = 0

3x3



Roots are

–

38x2

+ 135x – 74 =

(x2

2 , 6 ± i. 3

f x4 + 3x3 + x2 + 13x + 30 = 0 x = 1 ± 2i x – 1 = ±2i

2

2

x = 6 ± i 6)2

( ) − 494  + 29 = 0 2 ( z − 7 ) − 49 + 29 = 0 2 2 7 9 2( z − ) + = 0 2 2 7 9 2( z − ) = − 2 2 ( z − 72 ) = − 94

(x – 1)2 = –4

2

– 12x + 37)(3x – 2)

3 z – 7 = ± i 2 2 7 3 z = 2 ± i 2

Roots are –3,

7 3 ± i. 2 2

x2 – 2x + 1 = –4

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9

WORKED SOLUTIONS

7

a i z3 – 5z2 – 4z + 60 = 0

b z = 1 ± 7i

f(–3) = 0, hence (z + 3) is a factor.

z – 1 = ±7i

z3

(z – 1)2 = –49

–

5z2

– 4z + 60 = (z +

3)(z2

– 8z + 20)

z2 – 8z + 20 = 0

z2 – 2z + 1 = –49

(z – 4)2 – 16 + 20 = 0

z2 – 2z + 50 = 0

(z –

4)2

2z3 + kz2 + 102z – 50 = (z2 – 2z + 50)(2z – 1)

= –4

z – 4 = ±2i

k = –5

z = 4 ± 2i

Roots are

Roots are –3, 4 ± 2i. 9

ii

Im

×

2

×

0

–3

Re

×

–2

b i z4 – 2z3 + z2 + 2z – 2 = 0 f(1) = 0 and f(–1) = 0, hence (z – 1) and (z + 1) are factors. z4 – 2z3 + z2 + 2z – 2 = (z2 – 1)(z2 – 2z + 2) z2 – 2z + 2 = 0

(49 − 59i)(2 + 13i) z1 = 49 − 59i = (2 − 13i)(2 + 13i) 2 − 13i 98 + 767 + 637i − 118i = 4 + 169 865 + 519i = 173 z1 = 5 + 3i Two roots of the equation are 5 ± 3i z = 5 ± 3i z – 5 = ±3i (z – 5)2 = –9 z2 – 10z + 25 = –9 z2 – 10z + 34 = 0 z3 + 94z = 16z2 + 204 z3 – 16z2 + 94z – 204 = 0

(z – 1)2 – 1 + 2 = 0

(z2 – 10z + 34)(z – 6) = 0

(z – 1)2 = –1

z=6 10 a (x + iy)2 = –27 – 36i

z – 1 = ±i z=1±i

x2 + 2xyi – y2 = –27 – 36i

Roots are ±1 and 1 ± i.

x2 – y2 = –27

ii

2xy = –36

Im 1

×

0 –1 –1

8

1 , 1 ± 7i. 2

()

2 a c=3 3

3

y=

× × 1 ×

()

2 + 10 3

Re

x2 –

( −x18 ) = –27 2

x4 – 324 = –27x2 2

()

2 + 16 3 = 16

(3x3 + 10x2 + 16x – 16) ÷ (3x – 2) = x2 + 4x + 8 x2 + 4x + 8 = 0 (x + 2)2 – 4 + 8 = 0 (x + 2)2 = –4 x + 2 = ±2i x = –2 ± 2i Roots are

−18 x

2 , –2 ± 2i. 3

x4 + 27x2 – 324 = 0 (x2 – 9)(x2 + 36) = 0 x2 = 9 x = ±3 y = 6 Since Im z1 > 0, z1 = –3 + 6i b If –3 + 6i is a root, then so is the complex conjugate, –3 – 6i z = –3 ± 6i z + 3 = ±6i

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Complex numbers

Exercise 9.4A

(z + 3)2 = –36 z2 + 6z + 9 = –36

1 a z = 52 + 52 = 5 2

z2 + 6z + 45 = 0 z4 + 59z2 + 3330 = 2z(2z2 + 3)



z4 + 59z2 + 3330 = 4z3 + 6z

( 55 ) = 34π 5 2 ( cos 3π + isin 3π ) and 5 2e 4 4

arg z = π – tan −1

z4 – 4z3 + 59z2 – 6z + 3330 = 0



(z2

b z =

+ 6z +

45)(z2

– 10z + 74) = 0

z2 – 10z + 74 = 0

(z – 5)2 – 25 + 74 = 0 (z – 5)2 = –49 z – 5 = ±7i



z = 5 ±7i

Other three roots are –3 – 6i and 5 ±7i

11 a z1 = (10 – i) – (2 – 7i) = 10 – i – 2 + 7i = 8 + 6i b z – 8 = 6i (z – z2

8)2

(z2 – 16z + 100)(z – 4) = 0 z = 4 or 8 ± 6i =4

x2

= 8 ± 6i

Square roots of 8 + 6i

(a + ib)2 = 8 + 6i a2 + 2abi – b2 = 8 + 6i

a2 –

( a3 ) = 8 2

a4 – 9 = 8a2 a4 – 8a2 – 9 = 0 (a2 – 9)(a2 + 1) = 0 a2 = 9 a = ±3

π

c z = 7 2 + 32 = 58

arg z = tan −1

( 73 ) = 0.405

58 (cos 0.405 + i sin 0.405) and 58 e0.405i

( )



arg z = –  π − tan −1 10  = –2.11 6  



2 34 (cos 2.11 – i sin 2.11) and 2 34 e–2.11i

(

) ( ) ( 3 3 3 3 3 3 3 3 12 ( cos + i sin ) × 4 ( cos − i sin ) = 12 ( cos + i sin ) × 4  cos ( − ) + i sin ( − ) 4 4 2 2 4 4 2 2   3 3 = 48  cos ( − ) + i sin ( − ) 4 4   πi

)

Roots are 3 + i and –3 – i

Similarly, the square roots of 8 – 6i are 3 – i and –3 + i

πi

8π i

c 5 e 3 × 6 e 5 = 30 e 15

(

d cos 3π + i sin 3π × cos 2π + i sin 2π = cos 7π + i sin 7π 10 10 10 5 5 10

f 4(cos 0.573 + i sin 0.573) × 5(cos 0.228 + i sin 0.228) = 20(cos 0.801 + i sin 0.801) 3 a 3(cos 2 + i sin 2) ÷ 7(cos 3 + i sin 3) = 3 ( cos ( −1) + isin ( −1)) 7

x = ±2, 3 ± i, –3 ± i

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57743_P117_142.indd 128

( )

3 3 3 3 3 3 3 + i sin × 4 cos − i sin = 12 cos + i sin × 4  cos − + 4 4 2 2 4 4 2 

)( ) ( ) (cos 103π + i sin 103π ) × (cos 25π + i sin 25 π ) = (cos 107π + i sin 107π ) π π π π π π e 11 ( cos − i sin ) × 2 ( cos + i sin ) = 11  cos ( − ) + i sin ( − ) × 2 ( cos 6  6 6 2 2 6  11 ( cos π − i sin π ) × 2 ( cos π + i sin π ) = 11  cos ( − π ) + i sin ( − π ) × 2 ( cos π + i sin π ) 6 6 2 2 6 6  2 2  π π = 22 ( cos + i sin ) 3 3

b = ±1

)

b 12 cos

a2 – b2 = 8 2ab = 6 3 b = a

(

− i 4 cos π − i sin π and 4e 6 6 6

2 a 3(cos 2 + i sin 2) × 7(cos 3 + i sin 3) = 21(cos 5 + i sin 5)

c Let z = x2 x = ±2

+ 22 = 4

d z = 6 2 + 10 2 = 2 34

– 16z + 100 = 0

z3 – 20z2 + 164z – 400 = 0

x2

2

π  2  arg z = –tan −1  =–6  2 3 



= –36

(2 3 )

3π i 4

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9 

WORKED SOLUTIONS

(

) ( ) )

) ( ( ) ( )

( () (

)

( ) ) )

( ( )

)

2π + isin 3 3 3 3 3 3 3  2π 18 cos 2π + isin 2π 183 cos + i sin ÷ 4 cos − i sin = 12 cos + i sin ÷ 4 zcos − + i sin 3 3 3 − 2  3 4 4 2 2 4 4 b  1 = 2 = z2 π π 5 5 π 5  3 3 3 3 3 3 3 − isin 3 cos + isin − 5π  3 cos − + i sin ÷ 4 cos − i sin = 12 cos + i sin ÷ 4  cos − + i sin −  12 12 12 12   4 2 2 4 4 2 2   18 cos 2π + isin 2π 18 cos 2π + isin 2π 9 9 z1 3 3 3 3 = 3 cos + i sin = = 4 4 z2 π π 5 5 π 5  − isin 3 cos + isin − 5π  3 cos − 12 12 12 12   2π i πi πi 5 c 5e 3 ÷ 6e 5 = 6 e 15 = 6 cos 13π + isin 13π 12 12 π d cos 3π + i sin 3π ÷ cos 2π + i sin 2π =  cos − π + i sin − π  = cos π11− i sin 10  10 π 1011π 10 10 5 5 10  = 6 cos 12 − i sin 12 π + i sin − π  = cos π − i sin π i sin 3π ÷ cos 2π + i sin 2π =  cos − 10 5 5 10 10  10 10  z r e iθ1 c 1 = 1 iθ2 z 2 r2e π π e 11 cos π − isin π ÷ 2 cos π + isin π == 11  cos − π + isin − π  ÷ 2 cos + isin 6 6 2 2 6 6 = r1 e iθ1 − iθ22 2  r2 π π π π π  π π  isin ÷ 2 cos + isin == 11 cos − + isin − ÷ 2 cos + isin 2 6 2 2 6 6  2  r = 1 e i(θ1 −θ2) r2 = 11  cos − 2π + i sin − 2π  = 11 cos 2π − i sin 2π 3 2 3 3  2 3 π π π π 6 a 8 cos 4 − isin 4 × 8 cos 4 + isin 4 f 4(cos 0.573 + i sin 0.573) ÷ 5(cos 0.228 + i sin 0.228) = 4 π π π π (cos 0.345 + i sin 0.345) = 8  cos − + isin −  × 8 cos + isin 5 4 4 4 4   4 a x = r  cos θ = 9 cos − π = 9 2 = 64 4 2 b 12 cos

) (

)(

) (

) ( (

(

)( ) ) ( ) ( ) ( ( ) ( ) ) ( ) ( ) ( ( ) ( ) (

( ) y = r sin θ = 9 sin ( − π ) = − 9 2 4 2

( (





–3 + 3 i π c x = r cos θ = 3 cos  2 = 0 π y = r sin θ = 3 sin  2 = 3 3i



( ) π y = r sin θ = 14 sin ( − 3 ) = –7 3

π d x = r cos θ = 14 cos − 3 = 7





7 − 7 3i

(

) ( ) = 18 ( cos 2π + i sin 2π ) × 3  cos ( − 5π ) + i sin ( − 5π ) 3 3 12 12   = 54  cos ( 2π − 5π ) + i sin ( 2π − 5π ) 3 12 3 12   = 54 ( cos π + i sin π ) 4 4

5 a z1z2 = 18 cos 2π + i sin 2π × 3 cos 5π − i sin 5π 3 3 12 12



( ( )

)

( ) ) ) )

( ( ) ( ) (( ) ( ) ( ) ( ) ) ) ( ) ( ) ( ) ( ) (

9 2 9 2 – i 2 2 5π b x = r cos θ = 2 3 cos  6 = –3 5π y = r sin θ = 2 3 sin  6 = 3

) )

( )

)

b r(cos θ – i sin θ) × r(cos θ + i sin θ)

= r(cos(–θ) + i sin (–θ)) × r(cos θ + i sin θ)



= r2(cos 0 + i sin 0)



= r2

( ) ( ) 2π 5π π + − = 3 ( 12 ) 4 π π z z = (18 × 3)( cos + isin ) 4 4 π π = 54 ( cos + isin ) 4 4 b 2π − ( − 5π ) = 13π 3 12 12

5π 5π  + isin − 7 a z 2 = 3  cos − 12 12  

1 2

13π 11π − 2π = − 12 12 z1 18  11π 11π  cos − = + isin − z2 3 12 12 

(

= 6 cos

( )

11π 11π − isin 12 12

)

( )

c 2π + 2π = 4π 3 3 3 4π 2π − 2π = − 3 3 2π 2π  z12 = 18 2  cos − + isin − 3 3  

( )

( ) 2π 2π = 324 ( cos − isin ) 3 3

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Complex numbers

(

) (

)

2 2 3 +1 + 3 3 −1 i 3 3 3 +1  +  3 −1  =      8 8 8     2π 2π  2π 2π = 18 2  cos − + isin − × 18 cos + isin 3 3  3 3  3 = 2 4 2π 2π − + = 0. 3 3     arg  3 + 3 3i  = tan −1  3 − 1  = π z13 = 183(cos 0 + i sin 0). 4 + 4i    3 + 1  12     = 5832(1 + 0). 3 2 cos π + i sin π 4 12 12     = 5832, which is real.

d z13 = z12 × z1

( )

( )

3

(

)

(

)

(

)

(

) ii 6 ( cos π + i sin π ) ÷ 4 2 ( cos π + i sin π ) 3 3 4 4 6 π π 3 π = cos + i sin ) = 2 ( cos + i sin π ) ( 12 12 4 12 12 4 2 2π 2π 9 a  z = 8(cos – isin ) = 8  cos ( − 2π ) + isin ( − 2π ) 3 3 3 3  

e Since 5 and 12 are coprime, n = 12.

(

2 8 a z1 = 3 + 3 3

)

2

=6

z2 = 42 + 42 = 4 2



  b arg z1 = tan −1  3 3  = π  3  3 arg z2 = tan −1 4 = π 4 4 π π c z1 = 6 cos 3 + i sin 3

()

(

)

(

z2 = 4 2 cos π + i sin π 4 4



)

Im

Re

–4

d i (3 + 3 3 i)(4 + 4i) = 12 + 12i + 12 3 i + 12 3 i2

(

)

(



(

)

(

)

(

)

(

(

2

)

12 1 − 3  + 12 1 + 3     

)

 12 arg 12 1 − 3 + 12 1 + 3 i = π − tan −1   12  12 3 + 1  7π 12 1 − 3 + 12 1 + 3 i = π − tan −1  =  12 3 − 1  12

(

)

)

( (

))

(

))

(

(

) )

) (

(

)

2

= 1152 = 24 2

( (

( ) 2π y = 8sin( − ) = –4 3 3

2 2 12 1 − 3  + 12 1 + 3b  x = 8cos − 2π = –4 3    

)

12 1 − 3 + 12 1 + 3 i =

12 1 − 3 + 12 1 + 3 i =

–4√3

= 12(1 – 3 ) + 12(1 + 3 )i

2π 3

( (

(a + ib)2 = –4 – 4 3 i 3 + 1  a 7π 2 + 2abi – b2 = –4 – 4 3 i = 12 2 3 − 1  a – b2 = –4

) )

2ab = –4 3 i b =

−2 3 a

2 24 2 cos 7π + i sin 7π  −2 3  12 12 a2 –  a  = –4   ii 6 cos π + i sin π × 4 2 cos π + i sin π = 24 2 cos 7π + i sin 7π 12 3 3 4 4 12 4 a – 12 = –4a2 π π π 7 π 7 π i sin × 4 2 cos + i sin = 24 2 cos + i sin a4 + 4a2 – 12 = 0 3 4 4 12 12 (a2 + 2)2 – 16 = 0 3 + 3 3i 4 − 4i e i × (a2 + 2)2 = 16 4 + 4i 4 − 4i



)



(

(









= = 3



(

=







)

(

)

)

12 − 12i+12 3i+12 3 16 + 16 12 3

(

(

)

3 + 1 + 12 32

) (

3 +1 + 3 8

) (

3 +1 + 3 8

)

3 −1 i

(

)

(

)

a2 + 2 = ±4 a2 = 2 or –6

)

3 −1 i

a = ± 2

)

b = 

3 −1 i

=

3  8

(

)

2

3 3 +1  +   8 

(

)

3 −1  

2

2 3 = 6 2

Roots are 2 – 6  i and – 2 + 6  i

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9 

WORKED SOLUTIONS

10 a | z1 z2| = 12 × 4 = 48

( )

c z1 × i = –2 – 4i; rotation by π anti-clockwise 2 about the origin

( )

b z1 = 12  cos − 5π + isin − 5π  6 6  

d z1 – (2 – 3i) = –6 + 5i; translation by (–2 + 3i)

4cos β = 2 2

e z1 ÷ i = 2 + 4i; rotation by π clockwise about 2 the origin

2 2  2 π β = cos–1 2  = 4

cos β =

( )

( )

z2 = 4(cos β – isin β) = 4  cos − π + isin − π  4 4  

( )

z 5π π 7π – − =– arg  1  = – 4 6 12  z2  c

z1 12 = =3 z2 4 Im

) (

−7π 12



Exercise 9.5A 1 a i rotation by π anti-clockwise about the origin 2 ii rotation by π about the origin iii rotation by π clockwise about the origin 2 iv maps onto itself v rotation by π clockwise about the origin 2 vi rotation by π clockwise about the origin 2 vii maps onto itself viii rotation by π clockwise about the origin 2 ix modulus doubled, rotation by π anti-clockwise 2 about the origin x  modulus multiplied by 0.6, rotation by π about the origin b i rotation by π clockwise about the origin 2 ii maps onto itself iii modulus divided by 3 iv rotation by π clockwise about the origin 2 2 a z1 + (2 – 3i) = –2 – i; translation by (2 – 3i) b z1* = –4 – 2i; reflection in x-axis (real axis)

)

b z1 ÷ 4(cos 2.4 + i sin 2.4) = 1.25(cos 1.61 − i sin 1.61); modulus divided by 4, rotation by 2.4 clockwise about the origin

(

) (

(

)

)

π π π π − isin = 35 cos + isin ; 12 12 3 3 modulus multiplied by 7, rotation by π 12 anti-clockwise about the origin 3π 3π π π d z1 ÷ 5 cos 4 + i sin 4 = cos 2 − i sin 2 ; modulus divided by 5, rotation by 3π 4 clockwise about the origin c z1 × 7 cos

Re 3

(

3 a z1 × 3 cos π + i sin π = 15 cos 9π + i sin 9π ; 5 5 20 20 π modulus multiplied by 3, rotation by 5 anticlockwise about the origin

(

)

e z1* = 5 cos π − i sin π ; reflection in x-axis 4 4 (real axis) 2π 2π 4 a cos 3 − i sin 3 b –1 c 5i d i

()

3π  −1 2  5 a   i −  π − tan 2  = − 4 ii 22 + 22 = 2 2 3π 15π b   i 5 × − 4 = − 4 π arg  z5 = 4

(

ii z 5 = 2 2

)

5

= 128 2

( )

6 a arg z = π − tan −1 10 = 1.86 3 b arg z2 = 1.86 × 2 = 3.72 Since 3.72 > π, arg z2 = 3.72 – 2π = –2.56 c arg  z = 1.86 ÷ 2 = 0.931



d arg 3 z = 1.86 ÷ 3 = 0.621 2π = –3 3 2π y = 6sin =3 3 3 z1 = –3 + 3 3  i

7 a x = 6cos

131

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COMPLEX NUMBERS

b Rotation of 90° anticlockwise about the origin 2π π 7π c + = 2 3 6

9

a

Im B

7

7π 5π – 2π = – 6 6 d z1 = –3 + 3 3 i, z2 = –3 3 – 3i z1 – z2 = (–3 + 3 3 i) – (–3 3 – 3i) = (3 3 – 3) + (3 + 3 3 )i

A

| z1 – z2| = (3 3 − 3)2 + (3 + 3 3)2

0

5

10

Re

= 27 − 18 3 + 9 + 9 + 18 3 + 27 = 72

8

–4

=6 2 −20 + 21i × −2 − 5i a z2 = −2 + 5i −2 − 5i 40 + 100 − + 105 i 42i = 4 + 25 = 145 + 58i 29 = 5 + 2i b

b AB = (10 + 7i) – (–1 + 2i) = 11 + 5i AB = 112 + 52 = 146 AC = (5 – 4i) – (–1 + 2i) = 6 – 6i AC = 6 2 + 6 2 = 72 BC = (5 – 4i) – (10 + 7i) = –5 – 11i BC = 52 + 112 = 146

Im

×

–2

Since AB = BC , the triangle is isosceles.

5

2

0

C

×

5

Re

 8   1  π 10 a arg z = –tan–1 8 3  = –tan–1 3  = – 6 π π arg z2 = 2 × – = – 6 3 π π b Since 2π = 12 × , arg z33 = arg z9 = 9 × – 6 6 3π π =– = 2 2 c |z| = (8 3)2 + (−8)2 = 256 = 16 |z4.5| = 164.5 = 262 144

c Gradient OA = – 5 . 2 2 Gradient OB = 5 . Gradient OA × Gradient OB = –1.

Exercise 9.6A 1

x + iy − 2 + 8i = 13

( x − 2) + i ( y + 8 )

Or i(5 + 2i) = 5i – 2, hence i z2 = z1 and multiplication by i is a rotation of 90° anticlockwise. d Since OAB is a right angle, the centre of the circle is the mid-point of AB.

( −22+ 5 , 2 +2 5 ) = ( 32 , 72 ).

z − 2 + 8i = 13

(x –

2)2

+ (y +

8)2

= 13

= 132

Centre = (2, –8), radius = 13. 2

z −3 = z −2−i x + iy − 3 = x + iy − 2 − i

( x − 3) + iy (x –

3)2

+

y2

= ( x − 2 ) + i ( y − 1) = (x – 2)2 + (y – 1)2

x2 – 6x + 9 + y2 = x2 – 4x + 4 + y2 – 2y + 1 –6x + 9 = –4x + 4 – 2y + 1 –2x + 2y + 4 = 0 –x + y + 2 = 0 or y = x – 2

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9

WORKED SOLUTIONS

3

a x + iy = 10

d x + iy − ( 5 − 6i ) = 2

x2 + y2 = 102 Circle, centre (0, 0), radius 10.

( x − 5) + i ( y + 6 )

=2

(x – 5)2 + (y + 6)2 = 22

Im

Circle, centre (5, –6), radius 2. Im

10

0 0

–10

10

Re

Re

×

–6

–10

b x + iy − 9 = 4

( x − 9) + iy

5

e x + iy + 4 − 3i = 7

=4

( x + 4 ) + i ( y − 3)

=7

(x – 9)2 + y2 = 42 Circle, centre (9, 0), radius 4.

(x +

Im

Circle, centre (–4, 3), radius 7.

5

4)2

+ (y –

3)2

= 72 Im

4 3 2 1 0 –1

1

2

3

4

5

6

7

8

9 10 11 12 13

Re

–2 –3

×

–4

3

–5

c

x + iy + 2i = 8

–4

0

Re

x + i ( y + 2) = 8 x2 + (y + 2)2 = 82 Circle, centre (0, –2), radius 8. 4

Im

a x + iy − 5 = x + iy − 3i

( x − 5) + iy (x –

5)2

+

y2

= x + i ( y − 3) = x2 + (y – 3)2

x2 – 10x + 25 + y2 = x2 + y2 – 6y + 9 0 –2

Re

–10x + 25 = –6y + 9 10x – 6y = 16 5x – 3y = 8

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COMPLEX NUMBERS

d x + iy − 2 − 4i = x + iy + 5 − 2i

Im

( x − 2) + i ( y − 4 ) = ( x + 5) + i ( y − 2) (x – 2)2 + (y – 4)2 = (x + 5)2 + (y – 2)2 x2 – 4x + 4 + y2 – 8y + 16 = x2 + 10x + 25 + y2 – 4y + 4

3 2

–4x + 4 – 8y + 16 = 10x + 25 – 4y + 4

1

14x + 4y + 9 = 0

0 –1

1

2

3

4

Im

Re

5

4

×

–2 2

×

–3

b x + iy + i = x + iy − 4

0

–5

Re

2

x + i ( y + 1) = ( x − 4 ) + iy x2 + (y + 1)2 = (x – 4)2 + y2 x2 + y2 + 2y + 1 = x2 – 8x + 16 + y2 2y + 1 = –8x + 16

e x + iy + 5 + 2i = x + iy − 7 − 6i

8x + 2y = 15

( x + 5) + i ( y + 2)

Im

= ( x − 7) + i ( y − 6)

(x + 5)2 + (y + 2)2 = (x – 7)2 + (y – 6)2 x2 + 10x + 25 + y2 + 4y + 4 = x2 – 14x + 49 + y2 – 12y + 36

×

10x + 25 + 4y + 4 = –14x + 49 – 12y + 36 24x + 16y – 56 = 0 3x + 2y – 7 = 0

Re

×

Im 6

c

x + iy + 3i = x + iy − 7i

×

x + i ( y + 3) = x + i ( y − 7 ) x2 + (y + 3)2 = x2 + (y – 7)2 x2 + y2 + 6y + 9 = x2 + y2 – 14y + 49 6y + 9 = –14y + 49

–5

20y = 40

×

0

7

Re

–2

y=2 Im 4 2

–5

0

2

Re

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9

WORKED SOLUTIONS

5

a

6

Im

a

Im

1 radian Re b

Re

Im

b Im π 4 –3

c

0

8

Re

–8

8

Im

Re

–8

0

c 1

7π 10

–2

d

Im

Re

Im

2 Re

0

Re

–1

3 radian 7

a

x + iy − 3 − 2i = 5

( x − 3) + i ( y − 2) (x –

3)2

+ (y –

2)2

=5

= 52

Circle, centre (3, 2), radius 5. x + iy − 6i = x + iy − 7 + i x + i ( y − 6 ) = ( x − 7 ) + i ( y + 1) x2 + (y – 6)2 = (x – 7)2 + (y + 1)2 x2 + y2 – 12y + 36 = x2 – 14x + 49 + y2 + 2y + 1

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COMPLEX NUMBERS

14x – 14y = 14

(x – 12)(x + 5) = 0

y=x–1

x = 12 because half-line only exists for x  4 and y = 8.

Im

9

Complex number is 12 + 8i a |z – 3| = |z + 2i| |(x – 3) + iy| = |x + i(y + 2)| (x – 3)2 + y2 = x2 + (y + 2)2 x2 – 6x + 9 + y2 = x2 + y2 + 4y + 4 –6x + 9 = 4y + 4

×

2

6x + 4y – 5 = 0 1 |(x + 3) + i(y – 1)| = |(x – 1) + i(y + 5)|

0

–1

1

Re

3

(x + 3)2 + (y – 1)2 = (x – 1)2 + (y + 5)2 x2 + 6x + 9 + y2 – 2y + 1 = x2 – 2x + 1 + y2 + 10y + 25 8x – 12y – 16 = 0 2x – 3y – 4 = 0

b Substitute y = x – 1 into the equation for the circle: (x – 3)2 + (x – 1 – 2)2 = 52

7 13

2x – 3(–

2(x – 3)2 = 52

7 )–4=0 13

31 26 31 7 – i 26 13 b |z – 3| > |z + 2i| is true when z = 0 x=

25 2

x–3=±

1 – 2 13y + 7 = 0 y=–

(x – 3)2 + (x – 3)2 = 52 (x – 3)2 =

6x – 9y – 12 = 0 2

5 2 2

|z + 3 – i| < |z – 1 + 5i| is true when z = 0

5 2 2 5 y=2± 2 2 Complex numbers are  5 2  5 2   3 + 2  +  2 + 2  i  and   x=3±

Im 5 4

 5 2  5 2   3 − 2  +  2 − 2  i    8

5 6

2

Re

Cartesian equation for | z – 3i| = 13 |x + iy – 3i| = 13

–4

|x + i(y – 3)| = 13

3

x2 + (y – 3)2 = 132 Cartesian equation for arg(z – 4) = Gradient = 1, passes through (4, 0) y=x–4 x2 + (x – 4 – 3)2 = 132 x2 + (x – 7)2 = 132 x2 + x2 – 14x + 49 = 169

π 4

10 a i arg(z + 2 – 5i) = π is a half-line starting at (–2, 5) 4 ii arg(z + 2 – 5i) = – π is a half-line starting at 2 (–2, 5) iii |z + 2 – 5i| = 29 is a circle, centre (–2, 5), radius 29 . It passes through the origin.

2x2 – 14x – 120 = 0 x2 – 7x – 60 = 0

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9

WORKED SOLUTIONS

b

2 a 2 The complex numbers use the appropriate combinations of these solutions in the form x + yi 12 a z* = 5 – 2i x = –a 

Im

z 5 + 2i (5 + 2i)(5 + 2i) = = z * 5 − 2i (5 − 2i)(5 + 2i)

5

–2

=

Re

25 + 10i + 10i − 4 25 + 4

21 + 20i 1 = (21 + 20i) 29 29 2 b arg z = tan–1 5 =

() 2 arg z* = –tan ( 5 )

11 a |z + a| = 2a is a circle, centre (–a, 0), radius 2a, equation (x + a)2 + y2 = a2

–1

Im

()

Also arg –3a

–a

()

()

2 2 z = arg z – arg z* = tan–1 5 + tan–1 5 z* 2 = 2tan–1 5 arg

a

( )

20 z = tan–1 21 z*

c z + z* = 5 + 2i + 5 – 2i = 10

Re

Re

b |z – ai| = |z + a(2 + i)|.

A

2

|x + i(y – a)| = |(x + 2a) + i(y + a)|

B

x2 + (y – a)2 = (x + 2a)2 + (y + a)2

0

x2 + y2 – 2ay + a2 = x2 + 4ax + 4a2 + y2 + 2ay + a2

5

10

Im

C

–2

0 = 4ax + 4ay + 4a2 0=x+y+a x + y = –a

Rhombus 13 a w3 = (6 + i)3 = 63 + 3(6)2(i) + 3(6)(i)2 + (i)3 = 216 + 108i – 18 – i

Im

= 198 + 107i b tan w =

1 6

() 1 1 1 arg w = tan ( 6 ) + tan ( 6 ) + tan ( 6 ) 1 = 3tan ( 6 ) 107 From a, arg w = tan ( 198 ) 1 arg w = tan–1 6

–a

3

Re

–1

–1

–1

–a

3

c (x + a)2 + y2 = a2 Let x = –a – y (–a – y + a)2 + y2 = a2 2y2 = a2 y=± 2a 2

–1

–1

Exam-style questions 1

a –1 – 7i b

12 + 7 2 = 5 2 52 + 52 = 5 2

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COMPLEX NUMBERS

()

c arg z = − tan −1 5 = − π 5 4

(

z = 5 2 cos π − i sin π 4 4

4

)

b |w| = 22 + 52 = 29 = 5.39 c zw = (17 – i)(2 + 5i) = 34 + 85i – 2i + 5 = 39 + 83i

d (–1 + 7i)(5 – 5i) = –5 + 5i + 35i + 35 e z + w = (–1 + 7i) + (5 – 5i) = 4 + 2i

2

a

( 42 )= 0.464

14 − 31i 3 + 2i × 3 − 2i 3 + 2i

5

42 + 28i − 93i + 62 = 9+4 =

z 17 − i (17 − i)(2 − 5i) = = w 2 + 5i (2 + 5i)(2 − 5i) 34 − 85i − 2i− 5 = 4 + 25 29 − 87 i = 29 = 1 – 3i a (3 + 2i)(3 + 2i) = 9 + 6i + 6i – 4 = 5 + 12i d

= 30 + 40i arg(z + w) = tan −1

( )

1 a arg z = –tan–1 17 = –0.0589

(5 + 12i)(3 + 2i) = 15 + 10i + 36i – 24 = –9 + 46i b z1 + z2 = (19 – 9i) + (3 + 2i) = 22 – 7i

104 − 65i = 8 – 5i 13

z1 + z 2 = 222 + 7 2 = 533

b w = 8 ± 5i

c

w – 8 = ±5i

19 − 9i 3 − 2i × 3 + 2i 3 − 2i

(w – 8)2 = –25 w2 – 16w + 64 = –25

57 − 38i − 27i − 18 13 39 − 65i = = 3 – 5i 13 =

w2 – 16w + 89 = 0 c = –16, d = 89 c

z1 19 − 9i = 3 + 2i z2

d z − (19 − 9i) = z − (3 + 2i)

Im

x + iy − 19 + 9i = x + iy − 3 − 2i (x − 19) + i( y + 9) = (x − 3) + i( y − 2) (x – 19)2 + (y + 9)2 = (x – 3)2 + (y – 2)2 x2 – 38x + 361 + y2 + 18y + 81 = x2 – 6x + 9 + y2 – 4y + 4

×

5

–38x + 361 + 18y + 81 = –6x + 9 – 4y + 4 –32x + 22y + 429 = 0

0

8

Re

6

a

1 3 + i = 2 2

b tan −1 3

a (3 + 5i)(8 – 7i) = 24 – 21i + 40i + 35 = 59 + 19i w = 59 ± 19i w – 59 = ±19i

2

2

 3 + =1  2 

( 3 ) = π3

   c  1 + 3 i  1 + 3 i = 1 + 3 i + 3 i − 3 = − 1 + 3 i 2 2 2  4 4 4 4 2 2 2 1 3 1 3  1 3 3 3 1 3  2 + 2 i   2 + 2 i  = 4 + 4 i + 4 i − 4 = − 2 + 2 i

(w – 59)2 = –361 w2 – 118w + 3481 = –361 w2 – 118w + 3842 = 0 b tan −1

() 1 2

( 1959 ) = 0.31

1 3  1 3  1 3  2 + 2 i   − 2 + 2 i  = − 4 − 4 = −1 d Roots are –1, 1 ± 3 i 2 2

c w + 2z2 – z1 = (59 + 19i) + 2(8 – 7i) – (3 + 5i) = 59 + 19i + 16 – 14i – 3 – 5i = 72

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9

WORKED SOLUTIONS

e

3 1 × 65 × 65 × 5 2 39 = 2 =

Im √3 2

9

0

–1

a r = 18 ÷ 3 = 6 θ = 2π − − 7π = 11π 5 10 10

( )

11π − 2π = − 9π 10 10  w = 6  cos − 9π + i sin − 9π  10 10  

Re

1 2

( )

( ) w = 6 ( cos 9π − i sin 9π ) 10 10

−√3 2

b Im

7

a Centre of circle = (0, 7) and radius = 7. z − 7i = 7 b r = 72 + 72 = 7 2

()

θ = π − tan −1 7 = 3π 7 4

(

w1 = 7 2 cos 3π + i sin 3π 4 4

(

)

)

8

c w18 = 7 2 (cos 6π + i sin 6π)

c wz = –12

= 92 236 816(1 + 0) = 92 236 816

8

9π 10

6

( )

( )

( )

( )

6  cos − 9π + i sin − 9π  × z = 12(cos(−π) + i sin(−π)) 10 10  

6  cos − 9π + i sin − 9π  × z = 12(cos(−π) + i sin(−π)) 10 10  

a OM = 12 + 8 2 = 65

z=

2

ON = 4 + 7 = 65 2

( )

( )

12(cos(−π) + isin(−π)) = 2  cos − π + isin − π  10 10    6 cos − 9π + isin − 9π  10 10  

( ) ( ) MN = z – z = (4 + 7i) – (1 – 8i) = 3 + 15i z = 12(cos(−π) + isin(−π)) = 2  cos ( − π ) + isin ( − π ) MN = 3 + 15 = 234 10 10   6  cos ( − 9π ) + isin ( − 9π ) 2

Re

1

2

2

10  Two sides are the same, so OMN is isosceles. b Cosine rule: cos O =

(

65

) +( 2

65

) −( 2

2 × 65 × 65 65 + 65 − 234 = 2 × 65 =

−104 4 =− 130 5

c sin2 O + cos2 O = 1

( )

sin2 O + −

4 5

2

( )

)

2

10 |z – 6| = |z – 4 – 6i| |(x – 6) + iy| = | (x – 4) + i(y – 6)| (x – 6)2 + y2 = (x – 4)2 + (y – 6)2 x2 – 12x + 36 + y2 = x2 – 8x + 16 + y2 – 12y + 36 0 = 4x – 12y + 16 0 = x – 3y + 4

=1

sin2 O = 9 25

234

10  −πi z = 2e 10

4 3 When y = 0, x = –4

When x = 0, y = 2

3 sin O = 5 Area = 1 ab sin c 2

Line also passes through (5, 3) |z – 5 – 3i| = 3 is a circle of radius 3 centre (5, 3) For the complex number z = 0, |0 – 6| > |0 – 4 – 6i| is false, so the region is not on the same side of the line as z = 0 and |z – 5 – 3i| < 3 is also false, so the region is inside the circle

139

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COMPLEX NUMBERS

z2 – 4z + 125 = 0

Im

z3

– 2z2 + az + 250 = (z + 2)(z2 – 4z + 125)

(z + 2)(z2 – 4z + 125) = z3 – 2z2 + 117z + 250 a = 117 3

c Real root is –2.

4 3

11 a z = =

2 − 11i 4 − 3i × 4 + 3i 4 − 3i

d w=

–4

5

Re

26 + 29i 6 − i × 6+i 6−i

=

8 − 6i − 44i − 33 16 + 9

=

−25 − 50i 25

= –1 – 2i 14 a x + iy − 8 = x + iy + 2i

156 − 26i + 174i + 29 36 + 1

( x − 8 ) + iy

= x + i ( y + 2)

185 + 148i = 37

(x –

= 5 + 4i

x2 – 16x + 64 + y2 = x2 + y2 + 4y + 4

b iz = i(5 + 4i) = –4 + 5i arg(–4 + 5i) = π – tan −1

8)2

+

y2

= x2 + (y + 2)2

–16x + 64 = 4y + 4

()

4x + y = 15

5 = 2.25 4

b arg (z + 3 – 6i) = − π passes through (–3, 6) 4 with a gradient of –1.

c z = 5 ± 4i z – 5 = ±4i

y – 6 = –1(x + 3)

(z – 5)2 = –16

y=3–x

z2 – 10z + 25 = –16

4x + (3 – x) = 15

z2 – 10z + 41 = 0

3x + 3 = 15

5z3 – 49z2 + 195z + 41 = (5z + 1)(z2 – 10z + 41)

x=4

1 z = − , 5 ± 4i 5

When x = 4, y = –1.

49 1 . Sum of roots = − + (5 + 4i) + (5 – 4i) = 5 5 12 a 53e– 0.557i

4–i c

Im 16

b 148877

15 14

c i 7 – 2i and –7 + 2i ii

13

Im

12

4

11 10

2 –8

–6

–2 0 –2

–4

9 2

4

–4

6

8

Re

8 7 6 5 4

13 a z = (2 –

i)3

= (2 – i)(2 – i)(2 – i)

= (3 – 4i)(2 – i)

3 2 1 –6 –5 –4 –3 –2 –1 0 –1

= 2 – 11i

1 2 3 4 5

Re

–2

b z = 2 ± 11i z – 2 = ±11i (z – 2)2 = –121 z2 – 4z + 4 = –121

–3 –4

d arg(z + 3 – 6i) = π passes through (–3, 6) with 4 a gradient of 1.

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9

WORKED SOLUTIONS

( )

y – 6 = 1(x + 3) y=x+9 4x + (x + 9) = 15 5x = 6 6 x= 5 51 6 When x = , y = . 5 5

( )

6 51 Let the vertices be A(4, –1), B(–3, 6) and C 5 , 5 . Triangle is right-angled with the right angle at B. AB =

( −3 − 4 )2 + (6 − ( −1))

BC =

(

) ( 2

6 51 − ( −3) + −6 5 5

)(

(

)

2

)

=7 2

2

= 12(cos 0.9 + i sin 0.9)

21 = 2 5

b

and 2(cos (π – 0.2) – isin (π – 0.2)) c 3m × 4n = 432

Im P×

w = 4 ( cos0.4 + isin 0.4 ) = 2(cos 0.2 + isin 0.2)

Area = 1 7 2 21 2 = 147 . 2 5 5 15 a

2

sin2 Q = 9 25 3 sin Q = 5 1 Area of PQR = ab sin c 2 1 = × 85 × 85 × 3 5 2 = 51 2 Area of OPQR = 51 × 2 = 51. 2 16 a z2 = 3(cos 0.5 + i sin 0.5) × 4(cos 0.4 + i sin 0.4)

432 = 24 × 33

7

m = 3, n = 2

×Q

5

arg(z3w2) = 3 × 0.5 + 2 × 0.4 = 2.3 17 a z = –2 ± 4i z + 2 = ±4i

0

–6

3

Re

9

×R

–2

(z + 2)2 = –16 z2

+ 4z + 4 = –16

z2 + 4z + 20 = 0 z4 – 6z3 + 14z2 – 64z + 680 = (z2 + 4z + 20) (z2 – 10z + 34)

b | OP | = 6 2 + 7 2 = 85 | OR | = 92 + 22 = 85

z2 – 10z + 34 = 0

RQ = (3 + 5i) – (9 – 2i) = –6 + 7i

(z – 5)2 – 25 + 34 = 0

PQ = (3 + 5i) – (–6 + 7i) = 9 – 2i

(z – 5)2 = –9

Since OP (z1) = RQ and OR (z3) = PQ, sides are parallel and the same length.

z – 5 = ±3i z = 5 ± 3i

Since adjacent sides are also the same length, OPQR is a rhombus.

Other three roots are –2 – 4i, 5 ± 3i. b, e Im

c PR = (9 – 2i) – (–6 + 7i) = 15 – 9i | PR | = 152 + 92 = 306 Cosine rule: cos Q =

(

85

( )

4 Q = cos–1 − 5 = 143.1° d sin2 Q + cos2 Q = 1

( ) 4 5

2

85

=1

) −( 2

2 × 85 × 85

85 + 85 − 306 2 × 85 4 −136 = =− 170 5 =

sin2 Q + −

) +( 2

306

)

×

4

2

× 2

–4

–2

0

2

4

6

Re

–2 × ×

–4

c ABCD is a trapezium. Area = 1 (8 + 6) × 7 = 49. 2

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COMPLEX NUMBERS

d By symmetry, the circle that passes through the roots –2 ± 4i and –5 ± 3i has its centre on the Re axis, say at the point (a, 0) representing the real number a. Then the equation of the circle is (x − a)2 + y2 of the form (x − a)2 + y 2 = b2,where b > 0 is the radius. Since (−2, 4) and (5, 3) lie on the circle,

5 3 – i 2 2 2z = 5 – 3i

d z=

2z – 5 = –3i (2z – 5)2 = –9 4z2 – 20z + 25 = –9

(−2 − a)2 + 4 2 = b 2

4z2 – 20z + 34 = 0

(5 − a)2 + 32 = b 2

2z2 – 10z + 17 = 0

so (−2 − a)2 + 16 = (5 − a)2 + 9

6z3 + 11z + 68 = 22z2 6z3 – 22z2 + 11z + 68 = 0

4 + 4a + a 2 + 16 = 25 − 10a + a 2 + 9 14a = 14 a =1 b=5 Therefore the equation of the circle is |z – 1| = 5

(3z + 4)( 2z2 – 10z + 17) = 0 4 5 3 z=– , + i 3 2 2 20 a (2 + pi)4 = 24 + 4(2)3(pi) + 6(2)2(pi)2 + 4(2)(pi)3 + (pi)4

18 a z* = –8 – 8i

= 16 + 32pi – 24p2 – 8p3i + p4

2

2

|z*| = (−8) + (−8) = 128 = 8 2

b 16 – 24p2 + p4 = 41

()

8  3π b arg z* = –π +  tan –1 =– 8  4 

p4 – 24p2 – 25 = 0 (p2 – 25)(p2 + 1) = 0

c |(–8 + 8i) + (a + 2i)| = 26

p2 = 25

|(a – 8) + 10i| = 26

p = ±5

(a – 8)2 + 102 = 676

c p = –5

(a – 8)2 = 576

Im z = 32(–5) – 8(–5)3

a – 8 = ±24

= –160 + 1000 = 840

a = 32 or –16

Mathematics in life and work

Since a < 0, a = –16 19 a

1

7 + 6i (7 + 6i)(1 − 3i) = 1 + 3i (1 + 3i)(1 − 3i)

29 – 5 = 24 Z = 24 2 + 10 2 = 26 Ω

7 − 21i + 6i + 18 1+9 25 − 15i = 10 5 3 = – i 2 2 3 b arg z = –tan–1 5 = –0.540 =

2

( )

24 tan–1 10 = 1.18 26(cos 1.18 + i sin 1.18)

()

c z + 2i =

5 1 + i 2 2 Im

π 3

1 2 5 2

Re

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WORKED Solutions

Summary Review Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. ii Gradient of AB = 0 − 2 = − 1 3 − −3 3 6−0 =3 Gradient of BC = 5−3

Warm-up Questions 1 A = πr2  ⇒  dA = 2πr dr dr =3 dt dA dA dr = × = 2πr × 3 = 6πr dt dr dt dA = 6π ( 50 ) = 300π m2h −1 When t = 0, r = 50  ⇒  dt 1  2 i OA ⋅ OC =  2p    q

(

(

 − 4p 2 + q 2  ⋅ 2p  q  2

= − 4p + q

(

2

− 4p + q ii AC =

2

2

) + 4p

)−1

2p − 2p

2

=

( −4p

2

− q2 − 1

)

2



  

2

+q =0

0 0

(

= ± −4p 2 − q 2 − 1

AD = 22 + 6 2 = 40

A Level Questions: Pure Mathematics 2

2

)

1 i h = 0.5 (width of strips) x0 = 0 ⇒ y0 = 0.125 x1 = 0.5 ⇒ y1 = 0.107 556… x2 = 1 ⇒ y2 = 0.087 438… 1

1

∫ 6 + 2e x dx ≈

CA > 0   ⇒  CA = AC = 4p 2 + q 2 + 1 iii BA =

 −3  2  −1 OD = OA + AD =   +   =    2   6  8  ⇒  D(–1, 8)

−4p − q − 1

q−q

(Gradient of AB) × (Gradient of BC ) = –1    ⇒ The gradients are perpendicular.

 5 − 3   2 iii AD = BC =  =  6 − 0  6

)

2

=



0

1 2p − q

≈ 0.11 −x ii x + 4e − 2e−2x + c



q − 2p 1 p = 3 and q = 2  ⇒  BA = 4 = 12 + 4 2 + 8 2 8

0.5 0.125 + 2 × 0.107556 + 0.087438 ] 2 [

2 a i ii

⌠ e 2x + 6 dx =  ⌡ e 2x

∫3cos

2

3 cos2x + 1) dx 2∫( 3 1 = sin 2x + x + c 2 2

x dx =

= 81 = 9

1  So the unit vector is 1  4 9   8

1 6−2 = 3 i Gradient of AC = 5 − −3 2     ⇒  Gradient of MB = –2 M −3 + 5 , 2 + 6 ⇒  M(1, 4) 2 2   Equation of line through MB is y – 4 = – 2(x – 1)   ⇒  y = – 2x + 6

(

)

When y = 0, 2x = 6  ⇒  x = 3  ⇒  B(3, 0)

57743_P143_153.indd 143

= b h =

−2x −2x ∫ (1 + 6e ) dx = x − 3e + c

(

)

3 3x sin 2x + +c 4 2

2−1 1 = 2 2

x0 = 1  ⇒ 

y0 = 6 ln3

x1 = 1.5  ⇒ 

6 ln3.5 y2 = 6 ln4

x2 = 2  ⇒

y1 =

  6 1 6 6   6 ⌠  ln(x + 2) dx ≈ 4  ln3 + 2 × ln3.5 + ln4  = 4.84 (2 dp) ⌡ 2

1

143

6/28/18 2:38 PM

SUMMARY REVIEW

16

3

∫

i

6

16

16 6 2 dx = 3 ∫ dx = 3[ ln 2x − 7 ]6 2x − 7 2x − 7

7

0 = − 8 + 4a + b 4a + b = 8 ① Similarly, g(−1) = −18 Remainder theorem −18 = −1 + b − a a − b = 17 ② ① + ②: 5a = 25 a=5 In ②: 5 − b = 17 b = −12 ii g(x) − f(x) = (x3 − 12x2 − 5) − (x3 + 5x2 − 12) = − 17x2 + 7 Maximum occurs when x = 0. Maximum = 7.

6

= 3 ( ln 25 − ln 5) = 3 ln 5 = ln 125

17 − 1 =4 4 x0 = 1 ⇒ y0 = 0 x1 = 5 ⇒ y1 = log 5 x2 = 9 ⇒ y2 = log 9 x3 = 13 ⇒ y3 = log 13 x4 = 17 ⇒ y4 = log 17

ii h =

17

∫ log x dx ≈ 2 (0 + 2( log 5 + log 9 + log13) + log17 ) 4

1

= 13.5 (3 sf )

A Level Questions: Pure Mathematics 2 and Pure Mathematics 3 4

5

5x + 3 = 7x − 1 ln 5x + 3 = ln 7x − 1 (x + 3)ln 5 = (x − 1)ln 7 3 ln 5 + ln 7 = x ln 7 − x ln 5 3 ln 5 + ln 7 = x(ln 7 − ln 5) 3 ln 5 + ln 7 x= ln 7 − ln 5 x = 20.1 (this works equally well using log10) y = 6 sin x − 2 cos 2x dy = 6 cos x + 4 sin 2x dx When x = π , dy = 6 cos π + 4 sin π 6 dx 6 3 dy 3 3 =6 +4 =5 3 2 2 dx ( y − 2) = 5 3 x − π6

i (x + 2) is a factor ⇔ f(−2) = 0 Factor theorem

8

i

y 15

10

y = 15 – x3 5

y = 3lnx

( )

0

y = 8.66x − 2.53 6

i

∫ 4 cos

2

double (θ2 )dθ = ∫ 4( cosθ2 + 1 ) dθ (using angle formulae) = 2∫ (cosθ + 1)dθ

6

ii

1

ii f(x) = x3 + 3 ln x – 15

= 2sin  + 2 + c

f(2.0) =−4.9205…

1

2

∫ 2x + 3 dx = 2 ∫ 2x + 3 dx

−1

−1

=

6 1 ln 2x + 3 ]−1 2[

1 = [In 15 − In 1] 2 =

x

The solution to the equation is the intersection of the graphs. They intersect once, so there is only one root.

= 2(sin  + ) + c 6

5

f(2.5) = 3.373… Change of sign ⇔ Root in the interval [2.0, 2.5]. iii x0 = 2 x1 = 2.346 53…

1 ln15 2

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WORKED Solutions



x2 = 2.317 15 x3 = 2.319 49 x4 = 2.319 31



x5 = 2.319 32 x6 = 2.319 32



x7 = 2.319 32

Therefore, x = 2.319 9 a 2 cosec 2 tan  ≡ 2tan θ sin 2θ 2 sin θ cosθ ≡ 2sin θ cosθ 2sin θ ≡ 2sin θ cos2 θ 1 ≡ cos2 θ ≡ sec2  b i







ii

 = 1.11 rad  or   = 2.03 rad

0

0

∫ 2cosec4x tan 2x dx = ∫ sec











=

3 2

4x2 – 20x + 25 > 9(4x2 + 4x + 1)



4x2 – 20x + 25 > 36x2 + 36x + 9



0 < 32x2 + 56x – 16



0 > 4x2 + 7x – 2 –2 9(2x + 1)2

11 3x + 32x = 33x If u = 3x then: u + u2 = u3 0 = u3 − u2 − u   = u(u2 − u − 1) So u = 0 or u2 − u − 1 = 0 1 ± 1 − 4 (1)( −1) u= 2

  ln  1 + 5   2  x= ln 3

12 i sin 2 sec  ≡ 2sin α cos α cos α ≡ 2 sin 

1 5 π 6



( )

ii 3 cos 2 + 7 cos  = 0

π 6







sec2  = 5 1 cos2 θ = 5 cosθ = ±

1− 5 ⇒ no solutions, since 3x is always 2 positive 1+ 5 3x = 2 1 + 5 ln 3x = ln   2  3x =

1 + 5 x In 3 = In   2 

2 cosec 2 tan  = 5







Converting back into x: 3x = 0 ⇒ no solutions, since 3x is always positive

1 3

1 = (cos t )−3 cos3 t dx 3sin t = −3(cos t)−4 × −sin t = dt cos4 t

13 i x =

y = tan3 t

dy = 3(tan t)2 sec2 t = 3 tan2 t sec2 t dt 3sin 2 t × 1 dy 2 2 dy cos2 t = 3sin t = sin t = dt = cos t 3sin t 3sin t dx dx dt cos4 t



1  ii y − tan3 t = sin t x −   cos3 t  sin t sin 3 t 3 + cos t cos3 t







 sin 3 t − sin t  = x sin t +    cos3 t



 sin t sin 2 t − 1  = x sin t +  ×  cos t cos2 t 



 sin t − cos2 t  = x sin t +  ×  cost cos2 t 





y = x sin t −

y = x sin t − tan t

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Summary REVIEW

14 i 2 tan 2x + 5 tan2 x = 0

2tan x  2  2  + 5     x = 0    1 − tan x 



4t 2 2 + 5t = 0 1 − t 4t + 5t2 (1 − t2) = 0



4t + 5t2 − 5t4 = 0

2−



6 – x = – 6x – 18 5x = – 24 24 x=− 5



t = 0 or 4 + 5t − 5t3 = 0

t3 = t + 0.8



t = 3 t + 0.8



12 −1 5x (5x)2  12 – 5x (5x)2 + 5x – 12  0 (5x + 4)(5x – 3)  0 Therefore: 0  5x  3 5x > 0 for all values of x, so the only condition is   5x  3 log 3 ⇒ log 5x  log 3 ⇒ x log 5  log 3 ⇒ x  log 5 ⇒ x  0.683 (3 sf )

ii Let f(t ) = 3 t + 0.8 − t f(1.2) = 0.059 92…

f(1.3) = −0.0194…



Change of sign ⇒ 1.2 < t < 1.3

iii t0 = 1.3 (or can use 1.2)

( ) ( ) x  3 ln ( x − )  ln   2  8

t1 = 1.280 58

17 ln x − 3  2 ln x − 3 ln 2 2 3 ln x −  ln x 2 − ln 23 2

t2 = 1.276 62 t3 = 1.275 81 t4 = 1.275 64

2

t5 = 1.275 61 t6 = 1.275 60

3 x2  2 8 8x – 12  x2 0  x2 – 8x + 12 0  (x – 6)(x – 2) x  2 or x  6 x−

Therefore t = 1.276

d We know that tan x = 0 or tan x = 1.276

So x = 0, −π, π or x = 0.906, −2.24 y

15 i

18 i Let f (x) = x4 + 2x – 9

20

y = g(x) 15

10

5

–10

–5

ii When f(x) = g(x): x 2 − = 2x + 6 3 6 – x = 6x + 18 7x = – 12

x=−



or

12 7

0

(–3, 0)

y = f (x) 5

(6, 0) 10



f (1.5) = – 0.9375 and f (1.6) = 0.7536



Change of sign ⇒ solution is in the interval [1.5, 1.6]

ii At P, x4 + 2x – 9 = 0  ⇒  x4 = 9 – 2x  9 ⇒  x 3 = 9 − 2   ⇒  x = 3 x − 2 x iii x0 = 1.6

(0, 6) (0, 2)

–15

12 7

24 5

16 5x 

0.8 + t − t3 = 0



Therefore f(x)  g(x)  ⇒  x  − or x  −

t(4 + 5t − 5t3) = 0

x = −2x − 6 3



x

x1 = 1.5362 x2 = 1.5685 x3 = 1.5520 x4 = 1.5604 x5 = 1.5561 x6 = 1.5583 x7 = 1.5572 x8 = 1.5577

Therefore, at P, x = 1.56 (to 2dp)

146

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WORKED Solutions

19 i 5sin 2θ + 2cos 2θ ≡ Rsin (2θ + α)



5sin 2θ + 2cos 2θ ≡ R(sin 2θ cos α + cos 2θ sin α)

5sin 2θ + 2cos 2θ ≡ θ (Rcos α)sin 2θ    + (Rsin α) cos 2θ

5 − 4 = 1 + 5μ

③

From ①: 2μ = 4 − 14

In ②:

Therefore:



1 − 3 = 1 + (4 − 14)

Rcos α = 5 (1)



7 = 14

Rsin α = 2 (2)



=2





(2) ÷ (1)  ⇒ tan θ = 0.4  ⇒  α = 21.80°



In ①:



Square and add  ⇒  R2 = 52 + 22  ⇒  R = 29



4=7+μ

μ = −3

29 sin ( 2θ + 21.80 ) = 4

ii



4 29 2θ + 21.80 = 47.97°, 132.03°, 407.96°, 492.03°



θ = 13.1°, 55.1°, 193.1°, 235.1°



iii

sin ( 2θ + 21.80 ) =

1

2 (10sin 2θ + 4cos 2θ )

  ≡

≡

(2

1

)



To minimise the expression, we need to maximise the denominator  ⇒  1 116



5 − 8 ≠ 1 + 5(−3)

Therefore the lines do not intersect and are skew. 1  ii A vector in the direction of the x-axis is  0    0

2

1 116sin (2θ + 21.80)

  

Check in ③:



29 sin (2θ + 21.80)









Using the scalar product:



1 1  0  2 = 12 + 0 2 + 0 2 12 + 22 + 52 cosθ     0  5

2

20 i 3cosθ + sinθ ≡ Rcos(θ – α)

3cosθ + sinθ ≡ R(cosθ cosα + sinθ sinα)



3cosθ + sinθ ≡ (Rcosα)cosθ + (Rsinα)sinθ



cosθ = 22 i

Therefore: Rsinα = 1 (2)





(2) ÷ (1)  ⇒  tan α = 1   ⇒  α = 18.43° 3 Square and add  ⇒  R2 = 32 + 12  ⇒  R = 10 10cos ( 2x − 18.43) = 2

ii

4 + 12x + x 2 A B C ≡ + + (3 − x)(1 + 2x)2 3 − x 1 + 2x (1 + 2x)2 4 + 12x + x 2 A(1 + 2x)2 + B(3 − x)(1 + 2x) + C(3 − x) ≡ (3 − x)(1 + 2x)2 (3 − x)(1 + 2x)2

Rcosα = 3 (1)



1 = 79.5° 30

2 10



cos ( 2x − 18.43) =



2x – 18.43 = 50.76°, 309.23°, 410.76°, 669.23°



When x = 3



4 + 36 + 9 = 49A







 0  2  7 1 21 i L1: r =  1  + λ  −3 and L2: r =  1  + µ  2          5  −4  1  5 At ‘intersection’:

2 = 7 + μ

①



1 − 3 = 1 + 2μ

②

49 = 49A A=1 1 2



When x = −



1 7C 4−6+ = 4 2   16 − 24 + 1 = 14C







1 C=− 2 When x = 0



  4 = A + 3B + 3C



3 = 1 + 3B − 2       8 = 2 + 6B − 3



6B = 9

θ = 34.6°, 163.8°, 214.6°, 343.8°

A Level Questions: Pure Mathematics 3

4 + 12x + x2 ≡ A(1 + 2x)2 + B(3 − x)(1 + 2x) + C(3 − x)

14C = −7

147

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SUMMARY REVIEW

When θ = π 4 π 1 ln x + 2 = − sin π + ln 2 8 8 ln x + 2 = π + ln 2 8

3 2 Therefore B=

3 −1 4 + 12x + x 2 1 2 + 2 ≡ + (3 − x)(1 + 2x)2 3 − x 1 + 2x (1 + 2x)2

( )

ii (3 − x)−1 = 3−1 1 − x 3

π + ln 2

( )

 (−1)(−2) − x 3 1 x = 1 + (−1) − + 3 3 2!  

( )

=

π + ln 2

x + 2 = e8

−1

2

  +   

x = e8 −2 = 0.962 22 + 4i = 22 + 4i 24 i w = (2 − i)2 4 − 4i + i 2 = 22 + 4i 3 − 4i

 1 x x2 1+ + +   3 3 9 

= 22 + 4i × 3 + 4i 3 − 4i 3 + 4i 2 = 66 + 88i + 12i2 + 16i 9 − 16i

2 = 1 + x + x + 3 9 27

 3 3 (−1)(−2)(2x)2 (1 + 2x)−1 = 1 + (−1)(2x) + +  2 2 2!  3 = 1 − 2x + 4x 2 +  2 3 = − 3x + 6x 2 +  2  1 1 (−2)(−3)(2x)2 − (1 + 2x)−2 = − 1 + (−2)(2x) + +  2 2 2! 

= 50 + 100i 25 = 2 + 4i

Alternative method: Let w = x + iy (x + iy)(2 – i)2 = 22 + 4i (x + iy)(4 – 4i + i2) = 22 + 4i (x + iy)(3 – 4i) = 22 + 4i 1 = − 1 − 4x + 12x 2 +  (3x – 4xi + 3yi – 4yi2) = 22 + 4i 2 (3x + 4y) + ( – 4x + 3y)i = 22 + 4i 1 = − + 2x − 6x 2 +  Equating real and imaginary parts: 2 3x + 4y = 22 ⇒ 12x + 16y = 88 (1) 4 + 12x + x 2 1 3 1 1 1 = + − + −3+2 x + + 6 − 6 x 2 +– 4x + 3y = 4 ⇒ – 12 x + 9y = 12 (2) 3 2 2 9 27 (3 − x)(1 + 2x)2 (1) + (2) 1 12x + x 2 1 3 1 1 2 25y = 100 ⇒ y = 4 2 = 3 + 2 − 2 + 9 − 3 + 2 x + 27 + 6 − 6 x +  )(1 + 2x) Substituting into 3x + 4y = 22 4 8 1 2 3x + 16 = 22 ⇒ x = 2 = − x+ x + 3 9 27 Therefore w = 2 + 4i π  arg((2 + p) + 4i)  3 23 dx = (x + 2)sin 2 2θ ii dθ 4 4 1 2 Im ∫ x + 2 dx = ∫ sin 2θ dθ 1 2 = ∫ (1 − cos 4θ )dθ 4i 2 1 1 ln x + 2 = θ − sin 4θ + C 2 4

(

( ) (

)(

(

)( )

) (

)

)

θ − 1 sin 4θ + C 2 8 When x = 0,  = 0 Therefore C = ln 2 So: θ 1 ln x + 2 = − sin 4θ + ln 2 2 8 ln x + 2 =

π 4

π 4

–4

4

Re

From the Argand diagram: −4  2 + p  4 −6  p  2

148

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WORKED SOLUTIONS

iii S: w = 2 + 4i

I = ∫ e ysin y d y

T: w* = 2 − 4i

Integration by parts:

Im

u = ey ⇒ dv = sin y dy

S

4i

2

Re

a

I =

4a = 20 a=5 Therefore we have a circle with centre (5, 0) and radius 5, so z − 5 = 5 4

1

2I = – ey cos y + ey sin y e y(sin y − cos y) 2 Therefore:

(a − 2)2 + 42 = a2

∫

Integration by parts: du u = ey ⇒ = ey dy dv = cos y ⇒ v = sin y dy

I = – ey cos y + ey sin y – I

Pythagoras’ theorem on the triangle gives:

25 I =

4

−1 ln x dx = ∫ x 2 ln x dx x 1

e y(sin y − cos y) + c = tan −1 x 2  e y(sin y − cos y)  x = tan  + c 2   28 i

Integrate by parts: −1 dv =x 2 u = ln x and dx

4

4

4

4

1

−1  1  =  2x 2 ln x  − ∫ 2x 2 dx  1 1

ii

4

1  1  =  2x 2 ln x − 4x 2   1 = [4 ln 4 − 8] − [0 − 4] I = 4 ln 4 − 4

) ddxy = 1 1

∫ e sin y d y = ⌠⌡ x 2 + 1 dx y

⌠ 1 dx = tan −1 x  ⌡ x2 + 1

3

2

2

4

2

2

2i

(( x − 1) − i 2 )((x − 1) + i 2 )

= (x – 1)2 – 2i2 = (x – 1)2 + 2 = x2 – 2x + 3

)

()

2 2 3 2 x ii ⌠ dx = ⌠ dx = tan −1 +c  2 3 3 3 ⌡x +9 ⌡ x 2 + 32

)(

4

( x − (1 + i 2 ))( x − (1 − i 2 )) =

2x 26 i ⌠ dx = ln x 2 + 9 + c  2 ⌡x +9

(

4

2

 1  2x 2 I =  2x 2 ln x  − ∫ dx  1 1 x

(

(1 + i 2 ) = 1 + 4 × 1 ( i 2 ) + 6 × 1 ( i 2 ) + 4 × 1( i 2 ) + ( i 2 ) = −7 − 4 2i (1 + i 2 ) = 1 + 2 × 1( i 2 ) + ( i 2 ) = −1 + 2 ∴ p ( x ) = ( −7 − 4 2i ) + ( −1 + 2 2i ) + 2 (1 + 2i ) + 6 = 0 A second root of p(x) is (1 − 2i ) 3

1 du 1 = and v = 2x 2 dx x

27 e ysin y x 2 + 1

v = – cos y

I = −e ycos y + e ysin y − ∫e ysin y dy

T

–4i

⇒

I = −e ycos y + ∫e ycos y dy

a

0

du = ey dy

By polynomial long division: x4 + x2 + 2x + 6 = (x2 – 2x + 3)(x2 + 2x + 2) x2 + 2x + 2 = 0 −2 ± 22 − 4(1)(2) 2 −2 ± − 4 x= 2 −2 ± 2i x= 2 x = −1 ± i x=

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Summary REVIEW

∴ – 2a = – 4a3

 7 − − 8   15  5  −1 − − 1 =  0  0 29 i        ⇒       −4 − 8   −12  −4

4a3 – 2a = 0



2a(2a2 – 1) = 0

a = 0  or  2a 2 = 1  ⇒  a = ± 1 2 1 a > 0  ⇒  a = 2

 −8 5 ⇒ l1:  −1 + λ  0      8  −4  

 5 − −1   6  3  −6 − 4 =  −10   ⇒   −5        −7 − 7   −14  −7



5 3     ⇒ l2: −6 + µ  −5      −7  −7

2

ii  1 + 1 x   2 

−2

 1  x  1 + 2 

−2



ii – 8 + 5λ = 5 + 3μ (1)



– 1 = – 6 – 5μ (2)

= 1 − 2x +

32 i u + 2v = 2i 2ui + 2v = 6 (1) – (2) u – 2ui = 2i – 6 u(1 – 2i) = 2i – 6 u = −6 + 2i 1 − 2i



8 – 4λ = – 7 – 7μ (3) From (2)  ⇒ 5μ = – 5  ⇒  μ = – 1



In (1)  ⇒  – 8 + 5λ = 2  ⇒ 5λ = 10  ⇒  λ = 2



Check in (3)





8 – 4(2) = – 7 – 7( – 1)



0 = 0  ⇒ when λ = 2  A(2, –1, 0)

= −10 − 10i = −2 − 2i 1+ 4 Substituting in (1) – 2 – 2i + 2v = 2i 2v = 2 + 4i v = 1 + 2i

3 ⋅  −5 = 52 + 0 2 + 4 2 32 + 52 + 7 2 cos θ .    −7



15 + 0 + 28 = 41 83 cos θ



cos θ =

43 = 0.737… 41 83



θ = 42.5° or 0.742 radians (3sf)

30 i

1

(1 − 4x )− 2 1

(1 − 4x )− 2

( )

=1+ −

1 ( −4x ) + 2

2 u = −6 + 2i × 1 + 2i = −6 −2 10i + 24i 1 − 2i 1 + 2i 1 − (2i)

Im

ii

2i

(− 12 )(− 32 )(−4x) + …

3π 4

2

2

2!

–i

= 1 + 2x + 6x 2 + …

−1 1 + 2x 1 + 2x 1 + 2x ii = × (1 − 4x ) 2 = 2 2 4 − 16x −1 1 + 2x 1 + 2 x = × (1 − 4x ) 2 = 1 + 2x + 6x 2 + … 2 2 The coefficient of x2 is obtained from: 1 2x 6x 2 + ( 2x ) = 5x 2 2 2 The coefficient of x2 is 5.

(



( )

3

(−2)(−3)(ax)2 2!

+ (−2)(−3)(−4)(ax) + … 3! The coefficient of x is – 2a and the coefficient of x3 is – 4a3

Re

–2i

)

31 i (1 + ax)−2 = 1 + ( −2)( ax ) +

3 2 x +… 2

(1) (2)



5 iii  0     −4



 1  x = 1 + ( −2)   2 

(−2)(−3) 1 x   2  +… + 2!



z − w MIN represents the shortest distance between the circle and the line. This can be found by considering the line perpendicular to the half-line (locus of points for w) and passing through the centre of the circle. The shortest distance from the line to the centre of the circle, x, occurs when: 3 x ⇒  x= 3  2 The radius of the circle is 1.  3 −1 ⇒ z − w MIN = 2   cos 45° =

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WORKED Solutions

4

33 i [x 2(2ln x – 1)] 2 = 56 ln 2 – 12



π  24

3 1 ii  sin 4x + sin12x 4 0  16

=

Therefore: x3 7x + 6 ≡ ( x − 3) + (x + 1)(x + 2) (x + 1)(x + 2)

11 96

7x + 6 A B ii (x + 1)(x + 2) ≡ x + 1 + x + 2

Extension Questions 1 i

y



7x + 6 ≡ A(x + 2) + B(x + 1)



When x = – 2  ⇒  –8 = –B  ⇒  B = 8



When x = – 1  ⇒  –1 = A  ⇒  A = –1



x3 1 8 Therefore: (x + 1)(x + 2) ≡ ( x − 3) − x + 1 + x + 2

3 Let y = sin x + cos x  ⇒ 



2 ⇒ d y2 = − sin x − cos x dx We want minimum and maximum values of y so

solve (0, 5)



(– 5, 0)

0

x

( 5, 0)



ii The graphs intersect when 3x − 1 = x − 5 (3x – 1)2 = (x2 – 5)2



9x2 – 6x + 1 = x4 – 10x2 + 25



0 = x4 – 19x2 + 6x + 24



x = – 1 is a root  ⇔ (x + 1) is a factor

By inspection: 0 = (x +

1)(x3 –

x2 –



So the roots of the quartic equation (and x-coordinates of the points of intersection) are: −3 − 33 x = −3 + 33 x = – 1, x = 4, : x = ,: 2 2

f(x) < g(x) occurs when

−3 − 33 < x < −1 2

 −3 + 33  or   < x < 4 2  2 i (x + 1)(x + 2) = x2 + 3x + 2 x−3 x 2 + 3x + 2 x 3 + 0x 2 + 0x + 0 x 3 + 3x 2 + 2x − 3x 2 − 2x − 3x 2 − 9x − 6

7x + 6

2 π d y =− 2− 2 =− 2 2 2 4 , dx 2

 

2 5π d y = 2 + 2 = 2 2 2 4 , dx 2   ⇒  minimum  ⇒  y MIN = − 2

When x =

The cyclic nature of y = sin x + cos x is such that other values of x at minimum and maximum points will yield the same values for yMIN and yMAX.



e−

18x + 24)

By inspection: 0 = (x + 1)(x – 4)(x2 + 3x – 6) Solving the quadratic: x = −3 ± 33 2

When x =



x = 4 is a root  ⇔ (x – 4) is a factor

π 5π , ,… 4 4

⇒ maximum ⇒  y MAX = 2

2



dy =0 dx

0 = cos x – sin x  ⇒ sin x = cos x  ⇒ tan x = 1 x=

(0, 1)

dy = cos x − sin x   dx

2

(x2 – 4)(1 + x2) This is valid since 1 + x2 > 0 1 > x2 + x4 – 4 – 4x2 x4 – 3x2 – 5 < 0 Quadratic formula: x2 =







3 ± 9 − 4(1)(−5) 3 ± 29 = 2 2

π 5π tan 4θ = 1  ⇒  4θ = 4 , 4 , …  ⇒  θ = π , 5π , …  ⇒  θ = (4n + 1)π n ∈ Z 16 16 16   Sometimes true.

7 i cos ((A + B) + C)   ≡ cos (A + B)cos C – sin (A + B) sin C

x=±

3 − 29 ⇒  No real solutions 2  

x=±

3 + 29   ⇒  Two real solutions 2

cos ((A + B) + C) ≡ (cos A cos B – sin A sin B) cos C   – (sin A cos B + cos A sin B) sin C cos ((A + B) + C) ≡ cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C

y

ii Let A = π , B = π , C = π 2 3 4

(1312π ) = cos( π2 )cos( π3 )cos( π4 ) π π π − sin ( )sin ( ) cos ( ) 2 3 4 π π π − sin ( ) cos ( ) sin ( ) 2 3 4 π π π − cos ( ) sin ( ) sin ( ) 2 3 4  13π 3 2  1 − 1× × cos ( = 0 − 1 × × 12 ) 2 2   2  ( 6 + 2) 13π 6 2 cos ( =− − =− 4 12 ) 4 4 cos



x 3 + 29 , 0 – 2

3 + 29 , 0 2 (0, –5)

−

3 + 29