Unit 2 Module Tests Answers

8/10/2019 Unit 2 Module Tests Answers

1/28

Page 1of 28

Unit 2 Answers: Module Tests Macmillan Publishers Limited 2013

Module tests numerical answers

Module 1 Test 1

1 (a) (i) 16 cos sin4 4 4

xe x x+

+ (ii)

1

2 1 2x x

(b)1 1

ln16 4

(c) (i)4 2

2 1 3x x+

+

15(ii) 2 ln

4

(d) Proof

2 (a) (i) 21 x c + (ii) 12

(b)4 23 24

16 +

(c) Approx: 0.644, exact: 0.491 (3 dp)3 (a) (ii) y= 2

(b) Proof

(c) (2 2) (3 2)i+ + +

Module 1 Test 2

1 (a) (i) 2 29 tan (3 ) sec (3 ) 8 sin cosx x x x 2

2 2

1 cos( 4)(ii) (iii)

2( 2)1 sin ( 4)

x x x

xx x

+

++ +

(b)2 3/2

2

(1 ) 4x 3)

x

x

+

(c) (i)2

1 3

11 xx+

++ (ii) 3 ln 2

4

+

2 (a) 311 8

81 81e (b) = +

1 1ln 3

3 3y x (c) 5 2 7y x+ =

3 (a) 43 116 16

e + (b) tan , 1, 3, 5, 78n n

=

(c)(i) v = 2 , 2 , 1 , 3 2i i z i i+ = + +

Module 2 Test 1

1 (a) a= 8, b= 1 (b) (i)1 1

2 , 4 , 9,152 2

(c) +2 31 1

2 6x x x

2 (a) 21

1 , 1 12

x x x+ + < < (b) n2+ 3n (c) Proof

3 (a) Proof (b) Proof (c) Proof (d) 1.57


2/28

Page 2of 28


Module 2 Test 2

1 (a) (ii)3 1 1

2 1n n

+ (b) 3, 10, 17, 24, an= 7n4 (c) Proof

2 (a) 2 35 1 1

1 ... ,3 3 3

x x x x

+ < < (b) 0.86 (c) 2 35060

2 23 184 ...3

x x x+ + + +

3 (a) (i)3 5

sin6 120

x xx x= + +

2 4

cos 12 24

x xx= + +

(ii) 2 41 5

12 24

x x+ + +

(iii) 3 51 2

3 15x x x+ + +

(iv) 0.00100

(b)2 3

1 3 1 3

2 2 3 4 3 12 3x x x

+

cos 610.48481

Module 3 Test 1

1 (a) 210 (i) 90 (ii) 60

(b) 4 4 sin 2 3 cos2t tx Ae Be t t= + + 3 4 sin 2 3 cos2tx e t t= +

(c) (i) 514

(ii) 15

28

2 (a) (i) 91 (ii) 88205 5 1

(b) (i) (ii) (iii)18 9 18

(c) ( 1) cosxy e x= +

3 (a) (i)

10 100 50 85 000

15 120 70 119 000

18 105 100 136 250

x y z

x y z

x y z

+ + =

+ + =

+ + =

(ii)

10 100 50 85 000

15 120 70 119 000

18 105 100 136 250

x

y

z

=

(iii)

930 475 201

48 10 1 (iv) TT$ 5000, TT$ 250, TT$ 200135117 75 6

x y z

= = =

(b) x= 1, 4

Module 3 Test 2

1 (a) 720 (b) (i) 30 (ii) 5001 4 3

(c) (i) (ii) (iii)30 5 8


3/28

Page 3of 28


2 (a) Proof

3 6

5 52 1 1 111 7

(b) (i) 3 4 2 7 (ii) Proof (iii)

5 59 7 1 10

x

x

y y

z z

= +

= =

=

(c) (i)

6 0 0

0 6 0

0 0 6

(ii)

3 2 11

3 2 16

3 4 1

3 (a) y= e3x 53

( ) B

b y Axx

= + 2 3 24 6

(c) 25 5

x xy e e x x= +

Full Worked Answers

Module 1 Test 1

1 (a) (i) 1 1 1d

6 sin 6 cos 6 sind 4 4 4 4

x x xe x e x e xx

+ + + = +

16 cos sin4 4 4

xe x x+

= +

1 1/2d 1 2 1(ii) sin (1 2 )d 2 1 2 1 ( 2 1) 2 1 2

xx x x x x

= = +

(b) = 4t

ln = ln 4tln = tln 4

1 dln 4

dt =

d( ln 4) ( ln 4)4

dt

t

= =

When t= 2, 1 2d 1 1

(ln 4 ) (4 ) lnd 16 4t

= =

(c) (i)2

8 10 8 10

(2 1) ( 3)2 5 3

x x

x xx x

+ +

++

8 10

(2 1) ( 3) 2 1 3

x A B

x x x x

+ + + +

8x+ 10 A(x+ 3) +B(2x1)

When1 7

, 14 42 2

x A A= = =

When 3, 14 7 2x B B= = =

8 10 4 2

(2 1) ( 3) 2 1 3

x

x x x x

+ +

+ +

(ii)22 2

1 1 1

8 10 4 2 4d d ln 2 1 2 ln 3

(2 1) ( 3) 2 1 3 2

xx x x x

x x x x

+ = + = + +

+ +

(2 ln 3 2 ln 5) (2 ln1 2 ln 4)= + +


4/28

Page 4of 28


2 ln15 2 ln 4=

152 ln

4

=

(d) y= tan1

xtany=x

2 dsec 1d

yy

x =

2

d 1

d sec

y

x y=

2

1

1 tan y=

+

Since tany=x

2

d 1

d 1

y

x x

=

+

2 (a) (i)2

d1

xx

x

u= 1 x2du= 2xdx

1d d

2u x x =

21 x u =

2

1 1d d

21

xx u

ux

=

1/ 21 d2

u u=

1/2

1

21 2

u c

= +

u c= +

Since 2 22

1 d 11

xu x x x c

x= = +

(ii)

1

10

sin d x x

Let u= sin1x, dv= 1

2

d 1,

d 1

uv x

x x= =

1 111 1

200 0sin ( ) d sin d

1

xx x x x x

x

=

11 2

0

sin 1x x x = +


5/28

Page 5of 28


1sin (1) 1

12

=

=

(b) 20

cos dnnI t t t

=

Letd

, cosd

n vu t tt

= =

1d , sind

nu n t v t t

= =

/2/21

0 0sin sin d n nnI t t n t t t

= /2

1

0sin d

2

nn

nI n t t t

= [1]

12

0sin dnt t t

Let u= tn1,d

d

v

t= sin t

d

d

u

t

2( 1) , cosnn t v t = =

/ 21 1 22 2

00 0sin d cos ( 1) cos d n n nt t t t t n t t t

= +

1/2

20

cos ( 1) , since cos d 2 2

n

nn nn I I t t t

= + =

/2 22

0cos dnnI t t t

=

/21

20

sin d ( 1)n nt t t n I

=

Substituting into [1]:

2( 1)2

n

n nI n n I

=

When n= 4,4

4 24(3)2

I I

=

4

21216

I=

When n= 2,2

2 02(1)2

I I

=

2

024

I

=

[ ]/2

/200 0

0cos d sin sin sin 0 1

2I t t t t

= = = =

2

2 24

I

=


6/28

Page 6of 28


4 2 42

4 12 2 3 2416 4 16

I

= = +

(c)

x 0 1 2e

-2x 1 e-2= 0.135335 e-4= 0.018316Using the trapezium rule:

22

0

1d (1)[(1 0.018316) 2(0.135335)]

2

xe x

+ + = 0.644 (3 dp)For the exact value:

222 2

0 0

1d

2x xe x e

=

41 1 0.4912 2

e= + = (3 dp)

3 (a) (i) x= 4 + 2 cos

d2sind

x =

y= 2 cos 2

d4sin2

d

y

=

d d d

d d d

y y x

x=

4 sin 2 8 sin cos4 cos

2 sin 2 sin

= = =

(ii) Whend

, 4 2 cos , 2 cos , 4 cos2 2 d 2

yx y

x

= = + = = ,

d4, 2, 0

d

yx y

x= = =

Equation of the tangenty+ 2 = 0 (x4)y= 2

(b) y2+ sin (xy) = 2.Differentiate wrtx:

d d2 cos( ) 0

d d

y yy xy y x

x x

+ + =

When , 12

x y

= =

d d2 cos 1 0

2 d 2 2 d

y y

x x

+ + =


7/28

Page 7of 28


d0

d

y

x =

d0

d

y

x=

(c) (i) 2 3 2z i =

(2 3 ) 2z i + =

The locus is a circle centre (2, 3) radius 2

(ii) arg ( 2 3 )4

z i

=

arg ( (2 3 ))4

z i

+ =

The locus is a half-line starting at (2, 3) excluding (2, 3) making an angle of4

radians

with the positive real axis.

Point of intersection is a+ bi2 2a= +

3 2b= +

Point of intersection is (2 2) (3 2)i + + +

Module 1 Test 2

1 (a) (i) 3 2tan (3 ) 4 cosy x x= +

2 2d 3 tan (3 )[3sec (3 )] 8 cos ( sin )d

yx x x x

x= +

2 29 tan (3 ) sec (3 ) 8 sin cosx x x x=


8/28

Page 8of 28


(ii)21

ln2

xy

x

+=

+

21 1

ln2 2

x

x

+

= +

21 1ln(1 ) ln(2 )2 2

y x x= + +

2

d 1

d 2( 2)1

y x

x xx=

++

(iii)1

2 2 2sin( 4) [sin( 4)]y x x= + = +

12 22

d 1[sin( 4)] [2 cos( 4)]

d 2

yx x x

x

= + +

2

2

cos( 4)

sin( 4)

x x

x

+=+

(b) 1sin (2 )y x=

2 1/ 2

2

d 1[1 (2 ) ]

d 1 (2 )

yx

x x

= =

2 1/ 2[1 (4 4 )]x x = + 2 1/2( 4x 3)x = +

22 3/2

2

d 1( 2 4) ( 4x 3)

2d

yx x

x

= + +

2 3/ 2

2

( 4x 3)

x

x

=

+

(c) (i)2

2 2

3 4

1( 1)( 1) 1

x x Ax B C

xx x x

+ + + +

++ + +

2 23 4 ( ) ( 1) ( 1)x x Ax B x C x + + + + + +

Whenx= 1, 6 = 2CC= 3Whenx= 0, 4 =B+ CB= 1Equating coefficients ofx2: 3 =A+ CA= 0

2

2 2

3 4 1 3

1( 1) ( 1) 1

x x

xx x x

+ +

+ ++ + +

(ii)21

20

3 4d

( 1) ( 1)

x xx

x x

+ +

+ + 1

20

1 3d

11x

xx= +

++ 11

0tan ( ) 3 ln 1x x = + +

= tan 11+ 3 ln 2

3 ln 2

4

= +


9/28

Page 9of 28


2 (a)1

3

0dn xnI x e x= 3d,

d

n xvu x e

x

= =

1 3d 1,d 3

n xu nx v ex

= =

1 13 1 3

00

1 1d

3 3

n x n xnI x e n x e x

= 1 1

3 3 1 31 1

0 0

1 1, since d d

3 3

n x n xn n n nI e n I I x e x I x e x

= = =

14 3

40

dxI x e x= 3

4 3

1 4

3 3I e I=

33 2

1

3I e I=

32 1

1 2

3 3I e I=

31 0

1 1

3 3I e I=

Since1

3

0dn xnI x e x=

110 3 3 3

00 0

1 1 1d 3 3 3

x x

I x e x e e

= = =

3 31

1 1 1 1

3 3 3 3I e e

=

32 1

9 9e= +

3 32

1 2 2 1

3 3 9 9I e e

= +

35 2

27 27e=

3 33 1 5 2

3 27 27I e e =

34 2

27 27e= +

3 34

1 4 4 2

3 3 27 27I e e

= +

3 327 16 8

81 81 81e e=

311 8

81 81e=

(b) 2ln(2 1), 1x t y t= + =


10/28

Page 10of 28


d 2 d, 2

d 2 1 d

x yt

t t t= =

+

d d d

d d d

y y x

x t t

=

2(2 1)

2

2 1

tt t

t

= = +

+

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

d(1) (2 1) 3

d

y

x= + =

Gradient of the normal1

3=

Equation of the normal:

1

0 ( ln 3)3y x = 1 1

ln 33 3

y x= +

(c) 2 22 3 .xy x y x+ =

2 2d d4 4 3d d

y yx y x y xy

x x+ + + =

Whenx= 1,y= 1,d d

1 4 4 3d d

y y

x x + + + =

d5 2

d

y

x=

d 2

d 5

y

x=

Equation of the tangent at (1, 1) is 2

5

21 ( 1)

5y x =

5 5 2 2y x = +

5 2 7y x+ =

3 (a) 3

1ln d

e

x x x 3dln ,

d

vu x x

x= =

4d 1 1,d 4

uv x

x x= =

3 4 3

1 11

1 1ln d ln d

4 4

ee e

x x x x x x x

=

4 4

1

1 1ln

4 16

e

x x x

=

4 41 1 1ln4 16 16

e e e

=


11/28

Page 11of 28


43 1

16 16e= +

(b) cos 4+ isin 4= (cos + isin )44 4 3 4 2 2

1 2cos4 sin 4 cos cos ( sin ) cos ( sin )i C i C i + = + +

4 3 43 cos ( sin ) ( sin )C i i+ +

4 2 2 4 3 3(cos 6 cos sin sin ) (4 cos sin 4 cos sini ) = + +

Equating real and imaginary parts4 2 2 4cos 4 cos 6 cos sin sin = +

3 3sin 4 4 cos sin 4 cos sin = 3 3

4 2 2 4

sin 4 4cos sin 4 cos sinNow tan4

cos4 cos 6 cos sin sin

= =

+

Dividing top and bottom by cos4 3 3

44

4 2 2 4

4 4 4

4cos sin 4 cos sincoscostan 4 .

cos 6 cos sin sin

cos cos cos

=

+

3

2 4

4 tan 4 tan

1 6 tan tan

=

+

Let 4 2tan 6 1 0x x x= + = 4 2tan 6 tan 1 0 + =

3 5 7tan 4 , 4 , , ,

2 2 2 2

=

3 5 7, , ,

8 8 8 8

=

tan , 1, 3, 5, 78

nx n

= =

(c) (i) 2 3 4v i= + 2( ) 3 4x iy i+ = +

2 2 (2 ) 3 4x y i xy i + = +

Equating real and imaginary parts:2 2 3x y = [1]

2 4xy= [2]

From [2]2

yx

=

22 2 3x

x

=

4 23 4 0x x = 2 2( 4) ( 1) 0x x + =

2 4 2x x= = sincex

Whenx= 2,

2

12y= =


12/28

Page 12of 28


When2

2, 12

x y= = =

2v i = + , 2 i

(ii)2

(4 3 ) 1 5 0z i z i + + + = 24 3 (4 3 ) 4(1 5 )

2

i i iz

+ + +=

2(4 3 ) 16 24 9 4 20

2

i i i i+ + + =

(4 3 ) 3 4

2

i i +=

Since v = 3 4 2 , 2i i i+ = +

4 3 2 6 43 2

2 2

i i iz i

+ + + += = = +

or4 3 2 2 2

12 2

i i iz i

+ += = = +

Module 2 Test 1

1 (a) 8 8 8 21 2(1 ) 1 ( ) ( )by C by C by+ = + + + 2 21 8 28by b y= + + + 8 2(1 ) (1 ) (1 ) (1 8 28 ...)ay by ay by by+ + = + + + +

2 2 21 8 28 8by b y ay aby= + + + + + 2 21 ( 8 ) (28 8 ) ...y a b y b ab= + + + + +

Now coefficient ofy= 0 and coefficient ofy2= 368 0 8a b a b + = =

228 8 36b ab+ = 228 8( 8 ) 36b b b + =

2 228 64 36b b = 236 36b =

b2= 1

b= 1Since bis positive, b= 1a= 8.Hence a= 8, b = 1

(b) 1 11

2 , 2,2

n nu u n u n+ = + + =

(i) 2 11 1 1

2 2 2 42 2 2

u u= + + = + + =

3 21 1 1

2(2) 4 4 92 2 2

u u= + + = + + =

4 3

1 1 1

2(3) 9 6 152 2 2u u= + + = + + =


13/28

Page 13of 28


First four terms are1 1

2, 4 , 9,152 2

(ii) RTP:22 3

2

nn n

u +

=

Proof:

When n= 1, 12 1 3 4

22 2

u +

= = =

Since u1is given as 2, when n= 122 3

2n

n nu

+=

Assume true for n= k, i.e.22 3

2k

k ku

+=

RTP true for n= k+ 1,

i.e.

2

1

2( 1) ( 1) 3

2kk k

u ++ + +

=

Proof:

Since 11

22

n nu u n+ = + +

11

22

k ku u k+ = + +

Substituting22 3

gives2

kk k

u +

=

2

12 3 1

22 2

kk k

u k+ +

= + +

22 3 4 1

2

k k k + + +=

22 3 4

2

k k +=

22( 1) ( 1) 3

2

k k+ + +=

Hence by PMI22 3

2n

n nu

+=

(c) Letf(x) = ln(1 + sinx)

cos( ) 1 sin

xf x x= +

2

(1 sin ) ( sin ) cos (cos )( )

(1 sin )

x x x xf x

x

+ =

+

2 2

2

sin (sin cos )

(1 sin )

x x x

x

+=

+

2

sin 1 1

1 sin(1 sin )

x

xx

= =

++

2

cos( )

(1 sin )

xf x

x =

+

Whenx= 0,f(0) = ln1 = 0


14/28

Page 14of 28


cos0(0) 1

1 sin0f = =

+

(0) 1f =

2cos0(0) 1(1 sin 0)f = =+

2 3

( ) (0) (0) (0) (0)2! 3!

x xf x f xf f f= + + + +

2 3

ln(1 sin ) 0 (1) ( 1) (1)2! 3!

x xx x+ = + + + +

2 31 1ln(1 sin ) ...2 6

x x x x + = + +

2 (a) (i)1 12 2

1(1 ) (1 )

1

xx x

x

+= +

2 2

1 1 1 3

1 12 2 2 21 1 ( ) ( )

2 2! 2 2!x x x x

= + + + + + +

2 21 1 1 31 12 8 2 8

x x x x

= + + + + +

2 2 21 3 1 1 112 8 2 4 8

x x x x x + + + +

211

2

x x= + +

The expansion is valid for 1


15/28

Page 15of 28


(2)f= (1)f

(3)f+ (2)f

(4)f+ (3)f

+

( )f n+ ( 1)f n

( 1) ( )f n f n+ +

( 1) (1)f n f= +

Since ( 1) ( 1) ( 2) ( 1) ( 1) ( 2)f r r r f n n n+ = + + + = + +

( ) ( 1) (1) (1)(1 1) 2f r r r f= + = + =

1

2( 1) ( 2) ( 1) 2

n

r

r n n

=

+ = + +

2 3 2 2n n= + + 2 3n n= +

(c) RTP2

1

1

2 14 1

n

r

n

nr=

= +

Proof:

When n= 1, LHS2

1 1

34(1) 1= =

RHS 1 12(1) 1 3= =+

LHS = RHS when n= 1

2

1

1

2 14 1

n

r

n

nr=

= +

Assume true for n= k, i.e2

1

1

2 14 1

k

r

k

kr=

= +

RTP true for n= k+ 1, i.e.

1

21

1 1

2( 1) 14 1

k

r

k

kr

+

=

+=

+ +

1

2 2 2

1 1

1 1 1

(4 1) (4 1) 4( 1) 1

k k

r rr r k

+

= =

= + +

2

1

2 1 4 8 3

k

k k k= +

+ + +

1

2 1 (2 1) (2 3)

k

k k k= +

+ + +

1 1

2 1 2 3k

k k

= + + +


16/28

Page 16of 28


21 2 3 1

2 1 2 3

k k

k k

+ +=

+ +

1

2 1k= +

(2 1)k+ ( 1)

2 3

k

k

+

+

1 1

2 3 2( 1) 1

k k

k k

+ += =

+ + +

Hence by PMI

2

1

1

2 14 1

n

r

n

nr=

= +

3 4( ) 4 16 1f x x x= +

(a) 3( ) 16 16f x x =

3( ) 0 16 16 0f x x> > 3 1x >

1x>

Since ( ) 0 when 1f x x> >

f(x) is strictly increasing forx> 1

(b) 2( ) 4 16 1f x x x= +

f(0) = 1(1) 4 16 1 11f = + =

(2) 64 32 1 33f = + =

Sincef(0)f1< 0, by the IMVT

There existsx= such thatf() = 0 there is a root in the interval [0, 1]Sincef1f2< 0, by the IMVTThere existsx= such thatf() = 0 there is a root in the interval [1, 2]

(c) Since the function is strictly increasing forx> 1 and there is a root in the interval [1, 2]there is exactly one root in that intervalHence there is no other root in the interval [1, 2]

(d) 4( ) 4 16 1f x x x= + 3( ) 16 16f x x =

1( )

( )

nn n

n

f xx x

f x+ =

41 3

4 16 116 16

n nn n

n

x xx xx

+ +=

4 4 4 4

13 3 3

16 16 4 16 1 12 1 12 1,

16 16 16 16 16 16

nn n n n n

n

n n n

x x x x x xx

x x x+

+ = = =

(e) Usingx1= 1.54

2 3

12(1.5) 11.572368

16(1.5) 16x

= =

x3= 1.56605.Root is 1.57 to 2 dp

Module 2 Test 2


17/28

Page 17of 28


1 (a) (i)2

2 2

( 1) ( 1)1 r rr

+

2( 1) ( 1) 1 1

A Br r r r + + +

2 ( 1) ( 1)A r B r + +

When 1, 2 2 1r A A= = =

When 1, 2 2 1r B B= = =

2

2 1 1

1 11 r rr =

+

(ii)2

2 2

2 1 1

1 11

n n

r r

r rr= =

= +

11

3=

1 1

2 4

+

1

3

+

1

5

1

4

+

1

6 ...

+

1

3n+

1

1n

1

2n

+

1 1

1n n

+

1

1n

+

1 1 11

2 1n n= +

+

3 1 1

2 1n n=

+

(iii) As 3, sum2

n

(b) 1 17, 3n na a a= + =

2 1 7 3 7 10a a= + = + =

3 2 7 10 7 17a a= + = + =

4 3 7 17 7 24a a= + = + =

Terms are 3, 10, 17, 24an= 7n4RTP 7 4 1na n n=

Proof:

When n= 1, a1= 714 = 3true when n= 1, since a1is given as 3Assume true for n= k, i.e. ak= 7k4RTP true for n= k+ 1, i.e. ak+ 1= 7(k+ 1) 4Proof:Since 1 7n na a = +

1 7k ka a+ = +

7 4 7 (since 7 4)kk a k= + =

7 7 4k= +

7( 1) 4k= +

Hence by PMI7 4na n=


18/28

Page 18of 28


(c) RTP 2 1 2 , 3nn n+ <

Proof:Since 7 < 8 23+ 1 < 23When n= 3, 2n+ 1 < 2n

Assume true for , i.e. 2 1 2kn k k= + <

RTP true for 11, i.e.2( 1) 1 2kn k k += + + + <

Proof:

2 1 2kk+ <

(2 1) 2 2 2kk+ <

Now 4 2 2 2k k k k + = + + +

2 3k + 2( 1) 1k + +

12( 1) 1 2kk + + + <

Hence by PMI 2 1 2 , 3nn n+ <

2 (a)133 1 3 (1 3 )x x =

2 3

1 2 1 2 5

1 3 3 3 3 31 ( 3 ) ( 3 ) ( 3 )

3 2! 3!x x x

= + + + +

2 3513

x x x=

The expansion is valid for1 1

3 3x < <

(b) 3 22 2x x+ = 3 22 2 0x x+ =

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

3 2 3(0.5) 2(0.5) (0.5) 22

f = + =

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

Sincef(0)f1< 0, by IMVT there existsx= in the interval [0.5, 1] such thatf() = 0Hence there is a root in the interval [0.5, 1]Linear interpolation:

1

( ) ( )

( ) ( )

a f b b f a

x f b f a

+

= +

1

3(0.5) (1) (1)

2 420.8

3 5 51

2 2

x

+

= = = =+

3 2(0.8) 2(0.8) (0.8) 2 0.336f = + =

Root is between 0.8 and 1

2(0.8) (1) (1) (0.336)

0.850301 0.336

x +

= =+

3 2

(0.85030) 2(0.85030) (0.85030) 2 0.04739f = + = Root lies between 0.85030 and 1


19/28

Page 19of 28


3(0.85030) (1) (1) (0.04739)

0.857081 0.04739

x +

= =+

(0.85708) 0.006216f =

4 (0.85708) (1) (1) (0.006216) 0.857961 0.006216

x += =+

Root is 0.86 (2 dp)

(c) 2d

4 7d

yy

x= +

2

2

d d8

dd

y yy

xx=

3 2

3 2

d d d d 8 8

d dd d

y y y yy

x xx x

= +

Whenx= 0,y= 2, 2d

4(2) 7 23

d

y

x

= + =

2

2

d(8) (2) (23) 368

d

y

x= =

3

3

d8(2) (368) 8 (23) (23) 10 120

d

y

x= + =

Maclaurins expansion:2 3

( ) (0) (0) (0) (0)2! 3!

x xf x f x f f f= + + +

2 3

2 ( ) (23) (368) (10 120)2! 3!

x xy x = + + + +

2 350602 23 1843

x x x= + + + +

3 (a) (i)3 5

sin6 120

x xx x= + +

2 4

cos 12 24

x xx= + +

(ii)1

2 41

4(cos ) 1

2 2

x xx

= +

22 4 2 4

4

( 1) ( 2)1 ( 1)

2 24 2! 2 2

x x x x = + + + + +

2 4 41 1 112 24 4

x x x= + + +

2 41 512 24

x x= + + +

(iii) 1sin

tan (sin ) (cos )cos

xx x x

x

= =

3 5 2 41 1 1 5... 1 ...6 120 2 24

x x x x x

= + + + + +


20/28

Page 20of 28


3 5 3 5 51 5 1 1 1

2 24 6 12 120x x x x x x= + + + +

3 51 2

3 15x x x= + + +

(iv) 3 51 2

tan(0.001) 0.001 (0.001) (0.001)3 15

+ +

= 0.00100 (5 dp)(b) ( ) cosf x x=

( ) sinf x x=

( ) cosf x x=

( ) sinf x x =

When1

, cos3 3 3 2

x f

= = =

3sin3 3 2

f = =

1cos

3 3 2f

= =

3sin

3 3 2f

= =

Taylors expansion:2 3( ) ( )

( ) ( ) ( ) ( ) ( ) ( )2! 3!

x a x af x f a x a f a f a f a

= + + + + +

( ) cos ,3

f x x a

= =

2 3

1 3 1 33 3cos

2 2 3 2! 2 3! 2

x x

x x

= + + +

2 31 3 1 3

cos2 2 3 4 4 12 3

x x x x

= + +

The Taylors expansion about3

x

= is:

2 31 3 1 3

cos2 2 3 4 3 12 3

x x x x

= + +

Now 61= 60+ 13 180

= +

cos61 cos3 180

= +

2 31 3 1 3

2 2 3 180 3 4 3 180 3 12 3 180 3

= + + + +

+

2 31 3 1 3

2 2 180 4 180 12 180

= +

+

0.4848096= 0.48481 (5 dp)


21/28

Page 21of 28


Module 3 Test 1

1 (a) NUMBERS has 7 different lettersNo. of 3-letter words = 7P3= 210(i) Since we need the S, we choose 2 others from the remaining 6 and then order all three

letters in 3! waysNo. of 3-letter words containing the S = 6C23! = 90

(ii) No. of 3 letter words without the vowels = 5C33! = 60

(b)2

2

d d3 4 50 sin 2

dd

x xx t

tt =

y= CF + PI2

2

ddCF: 3 4 0

dd

x xx

tt =

AQE: m23m4 = 0(m4) (m+ 1) = 0m= 4 or m = 1

4t tx Ae Be = +

PI: Let x= asin 2t+ bcos 2td

2 cos 2 2 sin 2d

xa t b t

t=

2

2

d4 sin 2 4 cos 2

d

xa t b t

t=

Substituting into the differential equation:

4 sin 2 4 cos 2 3(2 cos 2 2 sin 2 ) 4( sin 2 cos 2 ) 50 sin 2a t b t a t b t a t b t t + = Equating coefficients of sin 2t4a+ 6b4a= 508a+ 6b= 50 [1]Equating coefficients of cos 2t: 4b6a4b= 0 6 8a b=

4

3a b

=

Substituting in [1]

48 6 50

3b b

+ =

50 503

b=

3, 4b a= =

General solution is4 4 sin 2 3 cos2t tx Ae Be t t= + +

When 0, 0x t= =

0 3, 3A B A B = + + =

Sincexremains finite as t, A= 0B= 3Required solution isx= 3et4 sin 2t+ 3 cos 2t

(c) (ii) P(two red)5 4 5

8 7 14= =


22/28

Page 22of 28


(ii) P(blue and red) = P(BR) + P(RB)3 5 5 3

8 7 8 7= +

30

56=

15

28=

2 (a) We separate the two Es and the other letters:EE SLCT ION(i)

75

74

73

Number of selections with zero Es C 21

Number of selections with one E C 35

Number of selections with two Es C = 35

= =

= =

=

Total number of selections = 21 + 35 + 35 = 91(ii) We need the number of arrangements of the five letters:Number of arrangements with zero Es 21 5! 2520

Number of arrangements with one E 35 5! 4200

5!Number of arrangements with two Es 35 2100

2!

Total number of arrangements 8820

= =

= =

= =

=

(b) (i) Biased: P(H)1

,3

= P(T)2

3=

Unbiased P(H)1

,2

= P(T)1

2=

P(X) = P(TTT) + P(HHH)2 2 1 1 1 1

3 3 2 3 3 2= +

2 1 5

9 18 18= + =

(ii) P(X Y) = P(X) + P(Y) P(X Y)5 1 4 10 5

18 2 18 18 9= + = =

(iii) P(X Y) = P(X) P(X Y)5 4 1

18 18 18= =

(c)

2d

cos sin cosd

xy

x y x e xx+ =

Dividing by cosx gives:

dtan cos

d

xy y x e xx

+ =

IFtan d ln sec sec

x x xe e x= = =

General solution is

(sec ) cos (sec )d xy x e x x x= sec d xy x e x= (sec ) xy x e c= +

Whenx= 0,y= 2


23/28

Page 23of 28


02

cos0e c= +

2 = 1 + cc= 1

ysecx= ex+ 1y= (ex+ 1) cosx

3 (a) (i) 10 100 50 85000 10 5 8500x y z x y z+ + = + + =

15 120 70 119000 3 24 14 23800x y z x y z+ + = + + =

18 105 100 136250 18 105 100 136250x y z x y z+ + = + + =

(ii)

1 10 5 8500

3 24 14 23800

18 105 100 136250

x

y

z

=

(iii)

1 10 524 14 3 14 3 24

3 24 14 10 5105 100 18 100 18 10518 105 100

= +

= 930 480 585= 135

1 10 5

3 24 14

18 105 100

Matrix of cofactors

930 48 117

475 10 75

20 1 6

=

1

930 475 201

48 10 1135

117 75 6

A

=

(iv)

930 475 20 85001

48 10 1 23800135

117 75 6 136250

x

y

z

=

675000 50001

33750 250135

27000 200

= =

TT$ 5000

TT$ 250

TT$ 200

x

y

z

=

=

=

(b)

2 4 2

1 3 2 0

1 4 1

x

x

x

+

=

3 2 1 2 1 3( 2) ( 4) (2) 0

4 1 1 1 1 4

x xx

x x

+ + =

( 2)[( 3) ( 1) 8] 4[( 1) 2] 2[4 3] 0x x x x x + + + + + = 2( 2) ( 4 5) 4( 1) 2( 1) 0x x x x x + + + + + =


24/28

Page 24of 28


2( 2)( 4 5) 6( 1) 0x x x x + + + =

( 2)( 1)(x 5) 6( 1) 0x x x+ + + + = 2( 1) ( 3 10 6) 0x x x+ + =

( 1) ( 1) ( 4) 0x x x+ + = 1 or 4x x= =

Module 3 Test 2

1 (a) 11 33 2 4 8Treating all the digits as if they are different:For the number to be odd it must end with a 1 or 3.

Number of odd numbers assuming the digits were different = 6543214

Since we have two 1s and two 2s, the number of odd arrangements

6 5 4 3 2 1 4720

2! 2! = =

(b) (i) No. of letters to be chosen = 4EEE SLCTD

54

53

52

51

No. of choices with no Es C 5

No. of choices with one E C 10

No. of choices with two Es C 10

No. of choices with three Es C = 5

Total no.of choices 30

= =

= =

= =

=

=

(ii) No. of arrangements4! 4!

5 4! 10 4! 10 52! 3!

= + + +

120 240 120 20= + + +

= 500(c)

No. drawn = 3

(i) P(all red )4

3

103

C 4 1

120 30C= = =

(ii) P(at least one of each colour) = P(1R 2G) + P(2R 1G)4 6 4 6

1 2 2 110

3

C C C C 60 36 96 4

120 120 5C

+ += = = =

(iii) P(2 R | at least one of each colour)4 6

2 1

103

C C 36P(2 red at least one of each colour) 36 3C 120

96 96P(at least one of each colour) 96 8

120 120

= = = = =

2 (a)2 2 2 2 2 2

2 2 2

1 1 1b c a c a b

a b cb c a c a b

a b c

= +


25/28

Page 25of 28


= bc2 b2c ac2+ a2c+ ab2 a2b= abc+ bc2 b2c ac2+ a2c+ ab2 a2b abc= (a b) (c a) (b c)

(b) 2x+yz= 1

3x+ 4y+ 2z= 79x+ 7yz= 10

(i)

12 1 1

3 4 2 7

9 7 1 10

x

y

z

=

(ii)

12 1 1

3 4 2 7

9 7 110

R22R2 3R1R

32R

3 9R

1

12 1 1

0 5 7 11

0 5 7 11

3 3 2R R R

12 1 1

0 5 7 11

0 0 0 0

Since the three equations reduce to:

2x+yz= 15y+ 7z= 11We have an infinite set of solutions

(iii) 2x+yz= 15y+ 7z= 11Letz= , 5y+ 7= 11

11 7

5y

=

11 7

5 5y=

11 72 1

5 5

x + =

6 122

5 5x

= +

3 6

5 5x

= +

The solution set is:3 6

5 5

11 7

5 5

x

y

z

= +

=

=


26/28

Page 26of 28


(c) (i)

1 1 0 3 2 1 6 0 0

0 1 1 3 2 1 0 6 0

3 1 2 3 4 1 0 0 6

AB

= =

(ii)1 0 0

6 0 1 0

0 0 1

AB =

= 6I1AA I =

16B A= 3 2 1

13 2 1

63 4 1

AB

=

3 (a) 3d 2d

xy y ex

=

32, xP Q e= =

IF =2 d 2x xe e

=

General solution is2 3 2( ) dx x xy e e e x = 2( ) dx xy e e x =

y(e-2x) = ex+ cWhenx= 0,y= 1

1 = 1 + cc= 0y(e2x) = exy= e3x

(b) x= eud d d

d d d

y y u

x u x=

Sincex= eu,d

d

ux eu

=

d 1 1

d uu

x e x = =

d 1 dd dy yx x u

=

d d

d d

y yx

x u=

Differentiating again wrtx:2 2

2 2

d d d d

d dd d

y y y ux

x xx u+ =

2 2

2 2

d d 1 d

dd d

y y yx

x xx u+ =

2 22

2 2

d d d

dd d

y y y

x x xu x = +


27/28

Page 27of 28


Now2

22

d d15 0

dd

y yx x y

xx =

22

2

d d d2 15 0

d dd

y y yx x x y

x xx

+ =

Substituting2 2

22 2

d d d

dd d

y y yx x

xx u+ = and

d d

d d

y yx

x u= gives:

2

2

d d2 15 0

dd

y yy

uu =

AQE: 2 2 15 0m m =

( 5) ( 3) 0m m + =

5 or 3m m= = 5 3u uy Ae Be = +

Since eu=x5 3( ) ( )u uy A e B e = +

5 3y Ax Bx = +

(c)2

22

d d6 6 8 11

dd

y yy x x

xx = + +

y= CF + PI2

2

d dCF : 6 0

dd

y yy

xx =

AQE: 2 6 0m m =

( 3) ( 2) 0m m + =

2 or 3m m= = 2 3x xy Ae Be = +

PI: Let 2y ax bx c= + +

d2

d

yax b

x= +

2

2

d2

d

ya

x=

Substituting into the differential equation:2a (3ax+ b) 6 (ax2+ bx+ c) = 6x2+ 8x+ 11Equating coefficients ofx2: 6a= 6, a= 1

Equating coefficients ofx: 2a 6b= 86

16

b = =

Equating constants: 2a b 6c= 11 2 + 1 6c= 11 c= 2This gives y= x2x 2General solution isy=Ae2x+Be3xx2x 2Whenx= 0,y= 00 =A+B 2 A+B= 2 [1]

2 3d 2 3 2 1d

x xy Ae Be xx

= +

When d0, 1

d

yx

x

= =

1 2 3 1A B = +


28/28

Page 28of 28

2A+ 3B= 2 [2][1] 2 gives:2A+ 2B= 4 [3]

[2] + [3] gives 5B= 6,

6

5B= 6 4

25 5

A A+ = =

Required solution is

2 3 24 6 25 5

x xy e e x x= +

Documents

Unit 2 Module Tests Answers