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

CAPE - 2002 Pure 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

( ii )

Show that

Hence if 

(b)

dy

Find y

( ii )

y

2

find y as functions of x

3 logy x

[8 marks]

2

x tan . 3 x t

[4 marks] 2

x

2

1

( a)

logx y

for x, y real and positive

logy x

when

dx

( i)

1

logx y

[6 marks] mark s]

[7 marks]

3t

t

2

lln n. x

( i)

2

x.

( ii )

x

4

2

llog ogx y

c

c . logy x

logx y

logx y

3

2

ln. ( 2 y)

2

logx y

2y

x

3

8

x=2 y=2

x

c

y

logy x

1

c

3

1

0

1

2

logx y y

c

logy y 1

logx y

logy c

logx y

logx y

1

x

3

x

logy x

2 logx y

3

3

logx y y

0

1 x

1

(b)

dy

( i)

2

dx

dy

( ii )

t

dt

t. ( 2 t )

dy dt

2

2

t

dy

6t

2

1

t

1

t

2

2

2

1

dx

t

2 2

6 t. 1

t

Find real constants A, B and C such that 2

3x (x

4x

1

2

1

2) x

3x

Hence evaluate

2

(x

(b)

2 x . tan . 3 x

(1) 1

dt

( a)

2

1

dx

2

2

3 x sec 3 x

A x

4x

1

2

1

2) x

2

Bx

C

2

1

x

[13 marks]

dx

The rate at which atoms in a mass of radioactive material are disintegrating is proportional to n the number of atoms present at any time t measured in days. Initially the number of atoms is m Form and solve the differential equation which represents the data

( i)

[8 marks] Given that half of the original mass disintegrates in 76 days evaluate the constant of proportion in the differential equation

( ii )

[4 marks]

2

3x ( a)

(x

1 x

2x 2

2

x

dx

4x

1

2

1

2) x

ln . ( x

2)

1 x

2

ln. x

1

2

2x 2

1

2

x

ln. K

1

ln. K

x x

2

2 1

(b)

n

dn

( i)

1

kn

dt

n

t dn

k dt k 

1 2

m. e

m

m. e

1

ln . ( 2 )

0

m

( ii )

kt

n

k ( 76 ) k  .

76 k  k

ln.

1

k 

2

76

Section B (Module 2)

3

( a)

 A sequence {un} is defined by Prove that

(b)

Express

un 2 1. n

2

n 1

[4 marks]

2 un 3

2n

in the form

3

p

q

where p,q

2

ε

R

n

Hence show that

Given the series

2n

lim n

( c)

n

un 1 un

∞

3 2

2n

[8 marks]

2

3

1

1

1

1. 4

4. 7

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

2

n

un

n

un 2

2

1

un 1

un 2

n 2 .2

2

n

un

un 2

3

2 un

1.

(b)

n

2n

2

3

3

2

lim

2

n

n

1 lim ∞ n

( c)

2n 2n

( i)

3 2

lim

(2 n

∞

n

3

Tn

n

3

2

2

2

n

∞

3)

2

lim ∞ n

2

2

2

1 2).( 3 n

(3 n

1)

1

( ii )

2) . ( 3 n

(3 n

1)

expands in partial fractions to

1

1

( 3. ( 3. n

( 3 . ( 3. n

2) )

n 1. 3

1 3

1

1 r=1

3n

1 2

3n

1

1

1

1

1

4

4

7

7

10

1

...

1 3n

1 5

3n

1 2

3n

1 3

( iii )

Sn

1 lim 1 3 ∞ n

1 3n

4

1

1

1

Sn

2

3n

1 3n

1 3

1

n 1

3n

1

1) )

4

4

The function f is given by f (x) =

x

4x

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-Raphson

method gives a second approximation

x2

d

( a)

4

dx

x

4x

f (0) = 1

(b)

1

3 x1 4.

x1

3

4x

f (1) = -4

4

1

3

1

in [1, 2]

3

4. x

4

[6 marks]

1 >0

for x > 1

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 1 1

3

4. x

considering

1

has solution(s)

0

2 1 2

4

f. ( x )

x

4x

1. . i 3 2 1. . i 3 2

has one min turning point at x = 1

1

10

t

4

4t

1

2

1

0 10 t

5

1

2

f is continuous

( c)

since

and no other turning point has real roots

( 1 , 2 ) min

no other roots exist in [0, 2]

( d)

xn 1

xn 1

xn

xn

4 xn

4 xn

3

3

4

xn. 4 xn

xn 1

4

4 xn

3 xn 4.

xn

4 3

1 4

xn 3

4

4 xn

1

4

1

hence 1

x2

3 x1 4.

x1

4 3

1

in [1, 2] 1

Section C (Module 3) 5

( a)

 A manufacturer of computers is supplied with a particular computer microchip called MC-40 from three suppliers, Halls Electronics, Smith Sales, and Crawford Sales and Supplies. A small batch of the chips supplied is defective. The information is summarised 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 a particular 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 this selection process. [8 marks]

6

( ii )

( i)

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]

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

H

represent P (Halls Electronics given defective)

D

let

represent P (Smith Sales given not defective)

S D

let

represent P (Crawford Sales & Supplies given not defective)

C D

7

HD = 0.03

H ∩ H D = (0.30)(0.03) H = 0.30

H D = 0.97

H ∩ H D = (0.30)(0.97)

SD = 0.05

S ∩ S D = (0.20)(0.05)

S D = 0.95

S ∩ S D = (0.20)(0.95)

S = 0.20

CD = 0.04

C ∩ C D = (0.50)(0.04)

C D = 0.96

C ∩ C D = (0.50)(0.96)

C = 0.50

( ii )

( a)

P (supplied by Halls Electronics) = P (Halls and D) or P (Halls and not D) ( 0.30 ) ( 0.03 )

( b)

( 0.30 ) ( 0.97 )

0.30

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

( c)

( 0.50 ) ( 0.04 )

( 0.50 ) ( 0.96 )

0.50

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 )

( e)

( 0.20 ) ( 0.95 )

( 0.50 ) ( 0.96 ) = 0.961

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 a student guessing the correct answer to the first question is 0.5. Likewise the probability of a student guessing correctly each of the remaining questions is 0.5. Use the probability model n!

P . ( r)

(n

r

r) ! r !

p q

n

r

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)

1

( ii )

6

4! (4

1

0

0) ! 0 ! 2

1

4

15

yields

2

4!

P (one correctly) =

(4

16

1

1

1) ! 1 ! 2

2

= 0.9375

3

yields

1 4

Ms Janis Smith takes out an endowment policy with an insurance company which involves making a fixed payment of $P each year. At the end of n years Janis expects to receive payment 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

9

R

1

α

100

[7 marks]

(b)

Show by mathematical induction or otherwise that the total sum in the fund at the end of the nth year is PR . R

$= ( c)

n

R

1

[12 marks]

1

Find the value of P to the nearest dollar when n = 10, $100 000.00

= 8 and the payout is

α

[6 marks] ( a)

P

end of year 1:

P

α

α

α

α

let the statement

at (a) n = 1

A1

n=2

for n = k

A2

Ak 

n=k+1

PR . R

An

n

R

PR . ( R R PR . R

2

R

PR . R

1

k 

R

Ak  k 1

10

P. R

PR

A2

PR . ( R

A2

P. R

1

1

PR . R R

k 

1

R

be true

A1

1

100

2

100

1

1

100

100

1

1)

α

P

α

α

P. 1

100

α

P. 1

P . 1

100

P. 1

100

P

100

P. 1

( b)

P. 1

100

P. 1

end of year 2:

α

1

PR

k 1 k 

1)

2

R

2

Ak  k 1

PR

k 1 k 

PR

PR

R

Ak  k 1

since

k 1 k 

(R

1)

Ak  k 1

1

PR

( k  k 1)

R

An

1

PR

Ak  k 1

1

PR

10

1.08

1

PR

PR . R

k 1 k 

R

k 2 k 

PR

k 2 k 

1

1

1

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

100000

P . ( 1.08 ) 1.08

PR R

fund at end of nth year is

( c)

k 1 k 

1

P . ( 1.08 ) 1.08 1.08

100000

10

PR . R

n

R

1 1

1

1

has solution(s)

0

P = $6 392.00

11

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