Mathcad - CAPE - 2002 - Math Unit 2 - Paper 02

7/29/2019 Mathcad - CAPE - 2002 - Math Unit 2 - Paper 02

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CAPE - 2002Pure Mathematics - Unit 2

Paper 02

Section A (Module 1)

1 a( ) i( ) Solve the simultaneous equations for x > 0, y > 0

xy = 4 and 2 ln x = ln 2 + ln y [6 marks]

ii( ) Show that logx

y1

logy

xlog

xy

yx

for x, y real and positive

Hence if logx

y 2 3 logy

xlogx

yy

x find y as functions of x [8 marks]

b( ) Finddy

dx when

i( ) y x2

tan 3 x.x x [4 marks]

ii( ) yt

1 t2

t

t

x 3 t2

t [7 marks]

a( ) i( ) ln x2. ln 2 y( ).ln x2. y x

22 yx

2y

x x

2

2. 4x x

2

2. x3 8x3 x = 2 y = 2

ii( ) logx

y clogx

y c xc

yxc

y logy

xc

logy

ylogy

xc

yy

c logy

x. 1c logy

x. c1

logy

cy

clog

xy

1

logy

xlog

xy

yx

logx

y 23

logx ylog

xy

x ylog

xy

22 log

xy 3 0log

xy

22 log

xy 3

logx

y 3 logx

y 1 0logx

y 3 logx

y 1 logx

y 3logx

y logx

y 1logx

y

y x3

x y1

xx

1



b( ) i( )dy

dx3 x

2sec

23 x 2 x tan. 3 x.x 3 xx x x x

ii( ) dy

dt

1 t2 1( ) t 2 t( ).

1 t2

2

dy

dt

t t t

t

dy

dt

1 t2

1 t2

2

dy

dt

t

t

dx

dt6 t

dx

dtt

dy

dx

1 t2

6 t 1 t2

2.

dy

dx

t

t t

2 a( ) Find real constants A, B and C such that

3 x2

4 x 1

x 2( ) x2

1

A

x 2

Bx C

x2

1

Bx C

x

Bx

Hence evaluate x3 x

24 x 1

x 2( ) x2

1

d [13 marks]

b( ) The rate at which atoms in a mass of radioactive material are disintegrating isproportional to n the number of atoms present at any time t measured in days.

Initially the number of atoms is m

i( ) Form and solve the differential equation which represents the data

[8 marks]

ii( ) Given that half of the original mass disintegrates in 76 days evaluate theconstant of proportion in the differential equation

[4 marks]

3 x2

4 x 1

x 2( ) x2 1

1

x 2

2 x

x2 1

3 x2

4 x 1

x 2( ) x2 1 x

x

xa( )

ln Kx 2

x2

1

.x

1

x 2

2 x

x2

1

d ln x 2( ). ln x2

1. ln K.ln x2

1. Kx x

2



b( ) i( )dn

dtkn

dn

dtkn

m

n

n1

nd

0

t

tk d

m

n

n1

nd

t

k n m ekt. kt

ii( )1

2m m e

k 76( ).

.1

2m

k

76 k ln1

2.76 k k

1

76ln 2( ).

1

76ln

Section B (Module 2)

3 a( ) A sequence {un} is defined by un 1

un

2nnn

n 1

Prove that un 2

2 unn

un [4 marks]

b( ) Express1

n2 n

33. in the form p

q

n2

where p,q ε R

Hence show that∞n

2 n 3

2 n2

3

lim 2

∞n

2 n 3

2 n2

3

lim [8 marks]

c( ) Given the series1

1 4.

1

4 7.

1

7 10....

i( ) obtain the nth term of the series [3 marks]

ii( ) find the sum of the first n terms of the series [8 marks]

iii( ) find the sum to infinity if it exists [2 marks]

a( ) un 1

2n

un

nn

un 2

2n 1

un 1

n 1n

un 2

2n

2.

2n

un

nn

un 2

2 unn

un

3



b( )1

n2 n

23. 2

3

n2

2

n2

n ∞n

23

n2

lim 2

∞n

23

n2

lim

∞n

2 n 3

2 n2

3

lim

∞n

1

n2 n 3( )

2

lim

∞n

2 n 3

2 n2

3

lim

∞n

2

2

lim 2

∞n

2

2

lim

c( )i( ) T

n

1

3 n 2( ) 3 n 1( ).n 2n

ii( )

1

3 n 2( ) 3 n 1( ).

expands in partial fractions to1

3 3 n. 2( ).( )

1

3 3 n. 1( ).( )

1

31

n

r

1

3 n 2

1

3 n 1=

.

1

3 11

4

1

4

1

7

1

7

1

10 ...1

3 n 5

1

3 n 2

1

3 n 2

1

3 n 1

1

31

1

3 n 1

n

3 n 1n 1

n

n

iii( ) Sn

∞n

1

31

1

3 n 1lim

nnS

n

1

3nn

4



4 The function f is given by f (x) = x4

4 x 1

Show that

a( ) when x > 1 f is strictly increasing [5 marks]

b( ) f (x) = 0 has a root in each of the intervals [0, 1] and 1, 2] 9 marks

c( ) f (x) = 0 has no other roots in the intervals [0, 2] [5 marks]

d( ) If x1

is an approximation to the root of f (x) = 0 in [1, 2] the Newton-Raphsonmethod gives a second approximation

x2

3 x1

41

4 x1

31.

x

x

in [1, 2] [6 marks]

a( )d

dxx

44 x 1 4 x

34

d

dxx

44 x 1 x 4 x

31. 0> for x > 1

b( ) f (0) = 1 f (1) = -4 by Intermediate Value Theorem

root exists in [0, 1]

f (1) = -4 f (2) = 9 by Intermediate Value Theorem

root exists in [0, 1]

f (1), f (2) < 0 f (0), f (1) < 0

f is continuousconsidering 4 x

31. 04 x

31. has solution(s)

1

1

2

1

2i. 3.

1

2

1

2i. 3.

f x( ). x4

4 x 1f x( ). x x has one min turning point at x = 1

t4

4 t 1

t

2 1 0 1 2

10

10

5



c( ) since 1 2,( )min

and no other turning point has real roots

no other roots exist in [0, 2]

d( ) xn 1

xn

xn

44 x

n1

4 xn

34

nx

n

xn 1

xn

4 xn

34. x

n

44 x

n1

4 xn

34

nx

n

xn 1

3 xn

4

1

4 xn

31.

nxn hence x2

3 x1

4

1

4 x1

31.

x

x

in [1, 2]

Section C (Module 3)

5 a( ) A manufacturer of computers is supplied with a particular computer microchip calledMC-40 from three suppliers, Halls Electronics, Smith Sales, and Crawford Sales andSupplies. A small batch of the chips supplied is defective. The information issummarised in the table below.

Supplier of Microchip M-40 % supplied % defective

Halls Electronics 30 3

Smith Sales 20 5

Crawford Sales & Supplies 50 4

When the MC-40 chips arrive at the manufacturer they are carefully stored in aparticular container and not inspected neither is the supplier identified.

i( ) A worker is asked to select one chip at random from the container for installation in a computer. draw an appropriate tree diagram to represent thisselection process.

[8 marks]

6



ii( ) What is the probability that

a( ) it was supplied by Halls Electronics? [1 mark]

b( ) it was not supplied by Crawford Sales & Supplies? [1 mark]

c( ) it was supplied by Smith Sales and it was good? [1 mark]

d( ) it was defective? [4 marks]

e( ) it will work effectively in the computer? [4 marks]

i( ) let H represent P (Halls Electronics)

let S represent P (Smith Sales)

let C represent P (Crawford Sales & Supplies)

let HD

represent P (Halls Sales given defective)

let SD

represent P (Smith Sales given defective)

let CD

represent P (Crawford Sales & Supplies given defective)

let HD

represent P (Halls Electronics given defective)

let SD

represent P (Smith Sales given not defective)

let CD

represent P (Crawford Sales & Supplies given not defective)

7



HD = 0.03

DHH ∩ = (0.30)(0.03)

H = 0.30

DH = 0.97

DHH ∩ = (0.30)(0.97)

SD = 0.05 DSS∩ = (0.20)(0.05)

S = 0.20

DS = 0.95 D

SS∩ = (0.20)(0.95)

CD = 0.04 DCC∩ = (0.50)(0.04)

C = 0.50

DC = 0.96 D

CC∩ = (0.50)(0.96)

ii( ) a( ) P (supplied by Halls Electronics) = P (Halls and D) or P (Halls and not D)

0.30( ) 0.03( ) 0.30( ) 0.97( ) 0.300.30( ) 0.03( ) 0.30( ) 0.97( )

b( ) P (not supplied by Crawford Sales & Supplies) = 1 - P (supplied by Crawford Sales& Supplies)

1 - [P (Crawford Sales & Supplies and defective) or P (Crawford Sales &Supplies and not defective)]

1 0.50( ) 0.04( ) 0.50( ) 0.96( ) 0.501 0.50( ) 0.04( ) 0.50( ) 0.96( )

c( ) P (supplied by Smith Sales and good) = P (Smith Sales and not defective)

0.20( ) 0.95( ) 0.19=

d( ) P (defective) = P (Halls Electronics & defective) or P (Smith Sales & defective) or P(Crawford Sales and Supplies & defective)

0.30( ) 0.97( ) 0.20( ) 0.95( ) 0.50( ) 0.96( ) 0.961=

e( ) P (it will work effectively) = p (not defective) = 1 - 0.961 = 0.039

8



5 b( ) A biology examination includes 4 True or False questions. The probability of astudent guessing the correct answer to the first question is 0.5. Likewise theprobability of a student guessing correctly each of the remaining questions is 0.5.Use the probability model

P r( ).n !

n r( ) ! r !p

rq

n rnn

where n is the number of questions

r is the number of observed successes

p is the probability of guessing correctly

q is the probability of guessing incorrectly

to answer the questions below.

What is the probability of a student

i( ) guessing at least one of the four questions correctly? [3 marks]

ii( ) guessing exactly one of the four questions correctly [3 marks]

i( ) P (at least 1 correctly) = 1 - P (none correctly)

14 !

4 0( ) ! 0 !

1

2

01

2

4

yields15

16 0.9375=

ii( ) P (one correctly) =4 !

4 1( ) ! 1 !

1

2

1

2

3

yields1

4

6 Ms Janis Smith takes out an endowment policy with an insurance company which involvesmaking a fixed payment of $P each year. At the end of n years Janis expects to receivepayment of a sum of money which is equal to her total payments together with interest added

at the rate of α % per annum of the total sum of the fund.

a( ) Show that the total sum in the fund at the end of the second year is

$P (R + R2) where R 1α

100

α[7 marks]

9



b( ) Show by mathematical induction or otherwise that the total sum in the fund at theend of the nth year is

$ =

PR Rn

1.

R 1 [12 marks]

c( ) Find the value of P to the nearest dollar when n = 10, α = 8 and the payout is$100 000.00

[6 marks]

P Pα

100P 1

α

100.P P

α

100P

α

a( ) end of year 1:

end of year 2: P 1α

100. P P 1

α

100. P

α

100

P 1α

100. P 1

α

100.

P 1α

100. P 1

α

100

2

. P R R2.P 1

α

100. P 1

α

100

2

. P R R

b( ) let the statement An

PR Rn

1.

R 1

nPR R

n

be true

at (a) n = 1 A1

PR R 1( ).

R 1

PR R

RA

1PRPR

n = 2 A2

PR R2

1.

R 1

PR R

RA

2PR R 1( ).PR R

A2

P R R2.P R R

for n = k Ak

PR Rk

1.

R 1

PR Rk

Rk

n = k + 1 Ak 1

PR Rk

1.

R 1PR

k 1PR Rk

RPR

k

k

10



Ak 1

PRk 1

PR PRk 1

R 1( )

R 1

PRk

PR PRk

R

Rk A

k 1

PRk 1

PR PRk 2

PRk 2

R 1

PRk

PR PRk

PRk

Rk

Ak 1

PR k 1( ) 1 PR

R 1

PR k PR

Rk A

k 1

PR Rk 1 1.

R 1

PR Rk

Rk

since An

is true for n = k = 1 and true for n = k + 1

fund at end of nth year isPR R

n1.

R 1

c( ) 100000P 1.08( ). 1.08

101

1.08 1

100000P

P 1.08( ). 1.0810

1

1.08 1100000 0

P 1.08( ). 1.0810

1

1.08 1100000 has solution(s) 6391.619323803280322

P = $6 392.00

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Mathcad - CAPE - 2002 - Math Unit 2 - Paper 02