Mathcad - CAPE - 2007 - Math Unit 2 - Paper 01

7/29/2019 Mathcad - CAPE - 2007 - Math Unit 2 - Paper 01

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CAPE 2007Pure Mathematics Unit 2 - Paper 01

Section A - Module 1

1 a( ) In the diagram below the curves y 2 e2 xx

and y exx

intersect

at P (p, q)

2 e2 x

ex

x

1 0 1 2

1

1

2

Determine the values of p and q [5 marks]

b( ) Solve 22 x 1

622 x 1

for x R [3 marks]

a( ) 2 e 2 x e x2 e 2 x x 2e

2 x1e

x02

e2 x

1e

x2 ex 02 ex x = ln 2

y1

eln 2.

y1

eln 2.

y1

2P ln 2.

1

2,.

b( ) 2 x 1( ) ln 2. ln 6.2 x 1( ) ln 2. x1

2

ln 6.

ln 2.1.x

1

2

ln 6.

ln 2.1. x = 0.792 (3 d.p)

Alternatively: 2 22 x. 62 22 x.

x1

2

lg 3.

lg 2.

lg

lgx = 0.792 (3 dec places)

1


2/9

2 a( ) The parametric equations of a curve are x4

ttand y 2 t

22t

Find the gradient of the curve at the point (4, 4) [5 marks]

b( ) Differentiate with respect to x y tan2

3 x ln x3.x x [4 marks]

a( )dy

dx

4 t

4

t2

dy

dx

t

t

simplifies to t3

x = 4 t = 1dy

dx1

1dy

dx

b( )dy

dx2 tan. 3 x( ). 3 sec

23 x( )..

1

x3

3 x2

3 sec2

3 x( ). 3 x2

x x

x

x

dy

dx6 tan. 3 x( ). sec

2. 3 x( )3

x3 x( )

xx x

x

3 a( ) Express5

3 x( ) 2 x( ).in the form

P

3 x

Q

2 xwhere P and Q are

constants

[3 marks]

b( ) Hence find x5

3 x( ) 2 x( ).d [3 marks]

a( ) 5 P 2 x( ). Q 3 x( ).5 P x Q x P 1 Q 1

5

3 x( ) 2 x( ).

1

3 x

1

2 x

5

3 x( ) 2 x( ). x x

b( ) x1

3 xd

1

2 xln 3 x( ). ln 2 x( ). ln K.ln 2 x( ). Kx x

I ln k 3 x

2 x.. k

x

x

2


3/9

4 Obtain the following

a( ) xx 2 x 5( )4. d by substituting u = 2x - 5 [5 marks]

b( ) xx sec2

x. d using integration by parts [4 marks]

a( )1

2du dx

1

2du dx x

1

2u 5( )u

I u1

2

u 5( ) u4 1

2

. du I1

4

uu5

5 u4

d. u

I1

4

u6

6u

5. Ku

6

6u

5K

uu I

u5

24u 6( ) Ku 6( ) K

uu I

2 x 5( )5

2 x 1( )

24K

24K

x x

b( ) I x tan. x. xtan x. dx x x I x tan. x. ln cos x.. Kln cos x.. Kx x x

5 The cost $c of manufacturing x items may be modelled by the differential equation

dc

dx2 c 10 x

dc

dx2 c x

By using a suitable integrating factor solve the differential equation given that there is a costof $100 when no items are produced

[8 marks]

I e

x2 d x

I e2 xx

e2 x dc

dx. 2 e

2 xc e

2 x10 x.e

2 x dc

dx. 2 e

2 xc

xx

xd

dxc e

2 x. d xe2 x

10 xdxd

dxc e

2 x. d x

c e2 x.

1

2e

2 x10 x x

1

2e

2 x10( ) dc e

2 x. x x x

3


4/9

c e2 x.

1

2e

2 x10 x

5

2e

2 xK

5

2e

2 xK

xx

xx = 0 c = 100

100

1

2 0( )

5

2 1( ) K

5

2 1( ) K K

205

2 c 5 x

5

2

205

2 e2 xx x

Section B - Module 2

6 A sequence {un} is defined by un 1 un 2n

un 1

un

nn 1

a( ) Prove that un 2

2 un

unn

[4 marks]

b( ) If u1 2 find u3 and u5 [4 marks]

a( ) un 1

2n

un

n

nn

un 2

2n 1

un 1

n

nn

un 2

22

n

un 1

.n

nn

2n

un 1

un

2n

un 1

nu

n 22 u

n.

nn

b( ) u1

2 u3

2 2( ). yields 4 u3

2 u1

.

u5

2 4( ). yields 8 u5

2 u3

.

7 The sum of the first and third terms of a GP is 50 and the sum of the second and fourth termsis 150. For this GP find

a( ) the common ratio [4 marks]

b( ) the first term [2 marks]

c( ) the sum of the first five terms [2 marks]

a( ) a 1 r2. 50a 1 r2. ar 1 r

2. 150ar 1 r2. r = 3

b( ) a = 5 c( ) S5

5 35

1.

3 1yields 605

4


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8 Find the term independent of x in the expansion of 2 x2 5

x3

10

[7 marks]

10( ) 9( ) 8( ) 7( )

242( )

65( )

4yields 8400000

9 a( ) Use the fact that nC

k

n !

k! n k( ) !.Ck

n

k n kkto express in terms of factorials

i( ) the coefficient u of xn

in the expansion of 1 x( )2 n

[2 marks]

ii( ) the coefficient v of xn

in the expansion of 1 x( )2 n 1

[3 marks]

b( ) Hence show that u = 2v [4 marks]

a( ) i( )2 n( ) !

n ! 2 n n( ) !.x

n 2 n( ) !

n !( ) n !( )x

n.2 n( ) !

n ! 2 n n( ) !.x

n n

n nx

nu

2 n( ) !

n !( ) n !( ).

n

n n

ii( )2 n 1( ) !

n ! 2 n 1 n( ) !.x

n 2 n 1( ) !

n ! n 1( ) !.x

n2 n 1( ) !

n ! 2 n 1 n( ) !.x

n n

n nx

nv

2 n 1( ) !

n ! n 1( ) !.

n

n n

iii( )2 n( ) 2 n 1( ) !

n !( ) n. n 1( ) !.u

2 n( ) 2 n 1( ) !

n !( ) n. n 1( ) !.u

2 2 n 1( ) !.

n ! n 1( ) !.

n

n nu 2 vv

10 a( ) Given that f r( ).2 r 1

r 1( ) r 2( ).f r( ).

r

r rprove that

f r( ). f r 1( ).2 r 3( ).

r r 1( ). r 2( ).f r( ). f r 1( ).

r

r r r[4 marks]

b( ) Hence find

3

n

r

r 3

r r 1( ). r 2( ).=

[4 marks]

a( )2 r 1

r 1( ) r 2( ).

2 r 3

r r 1( ).

r 2 r 1( ). r 2( ) 2 r 3( )

r r 1( ). r 2( ).

2 r 1

r 1( ) r 2( ).

2 r 3

r r 1( ).

r r r r

r r r

f r( ). f r 1( ).2 r 3( ).

r r 1( ). r 2( ).f r( ). f r 1( ).

r

r r r

5


6/9

b( )

2 Sn

7

2

9

6

9

6

11

12...

2 n 1

n 1( ) n 2( ).

2 n 3

n n 1( )....

Sn

1

2

7

2

2 n 3

n n 1( ).

n

n nnS

n

7 n2

11 n 6

4 n n 1( ).

n n

n nn

Section C - Module 3

11 a( ) Determine the number of ways in which the letters of the word S T A T I S T I C S

may be arranged so that the vowels are placed together

[3 marks]

b( ) A team of five is chosen at random from 4 boys and 6 girls. Calculate the numberof ways that this team can be chosen to include at least 3 girls

[5 marks]

a( ) vowels are A I I considered as one item

1 7( ) ! 3 !

3 ! 3 !. 2 !.3360 ways

1 7( ) ! 3 !

3 ! 3 !. 2 !.ways

b( )

(3 girls from 6 and 2 boys from 4) or (4 girls from 6 and 1 boy from 4)or (5 girls from 6 and no boys)

6C

34

C2

. 6C

44

C1

. 6C

54

C0

.

120( ) 60( ) 6( ) 186120( ) 60( ) 6( )

6


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12 Two unbiased dice each with six faces are tossed randomly one after another

a( ) Determine the set of possible outcomes [2 marks]

b( ) Find the probability that

i( ) the product of the numbers on the two dice is a multiple of 5 [2 marks]

ii( ) the second die shows the number 2 [1 mark]

iii( ) the product of the numbers on the two dice is a multiple of 5 or the seconddie shows the number 2

[2 marks]

a( )

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

b( ) i( )11

36ii( )

1

6iii( )

11

36

6

36

1

36

4

9

11

36

6

36

1

36

13 The matrices X and Y are given by X =

4

5

7

2

6

9

1

8

3

and Y =

1

5

4

6

2

6

Calculate

a( ) the determinant of X [4 marks]

b( ) YT

X [3 mark]

a( ) X 4 18( ) 72( )( ). 2 15( ) 56( )( ). 45( ) 42( )X yields 365

4

5

7

2

6

9

1

8

3

yields 365

b( ) 1

6

5

2

4

6

4

5

7

2

6

9

1

8

3

. yields7

56

68

78

29

4

7


8/9

14 Y and X are 3 x 1 matrices and are related by the equation Y = AX where

A

1

7

3

0

5

2

3

0

1

is a non-singular matrix

a( ) Find A1

[6 marks]

b( ) X, when Y =

10

12

8

[3 marks]

a( )

A1

1

7

3

0

5

2

3

0

1

1

A1

yields

5

2

7

2

1

2

3

4

1

15

2

21

2

5

2

b( ) 10

12

8

1

7

3

0

5

2

3

0

1

X.

10

12

8

X

X

5

2

7

2

1

2

3

4

1

15

2

21

2

5

2

10

12

8

. yields

1

1

3

8


9/9

15 Air is pumped into a spherical balloon of radius r cm at the rate of 275 metres cube persecond. When r = 10 calculate the rate of increase of

a( ) the radius r [5 marks]

b( ) the surface area S [4 marks]

[The volume V and surface area S of a sphere of radius r are given by

V4 r

3

3

rand S 4 r

2r respectively]

a( )dr

dt

dr

dV

dV

dt.

dr

dt

dr

dV

dV dr

dt

1

4 r2

275( )dr

dt r

dr

dt10

275

4 10( )2

dr

dtyields

11

16 .( )cm

b( ) dSdt

dSdr

drdt

.dSdt

dSdr

dr dSdt

8 r 1116

.dSdt

r

dS

dt10

8 . 10( ).11

16

dS

dtyields 55 cm

2

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Mathcad - CAPE - 2007 - Math Unit 2 - Paper 01

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