mathcad - cape - 2007 - math unit 2 - paper 03-2

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CAPE - 2007 - Pure Mathematics Unit 2 - Paper 03-2 Section A (Module 1) 1 The rate of decay of a radioactive substance after t hours was found to be directly proportional to the amount A of the substance remaining a () Form a differential equation to model this situation [3 marks] b ( ) Given that when t = 0 A = A 0 show that A A 0 e kt kt [8 marks] c () If the time taken for the remaining substance to be 1 2 A 0 is T show that A A 0 e t ln . 2 . T t ln . 2 . T A t [4 marks] d ( ) Prove that the time taken for the amount remaining to be reduced to 1 30 A 0 is 4.91T [5 marks] Total 20 marks a () dA dt kA dA dt b ( ) A 0 A A 1 A d 0 t t k d A 0 A A 1 A d t k ln A A 0 . kt ln A A 0 . kt A A 0 e kt A kt c () 1 2 A 0 A 0 e kT 1 2 A 0 A kT kT ln 1 2 . kT k ln 2 . T k ln 2 . T A A 0 e t ln . 2 . T t ln . 2 . T A t d ( ) 1 30 A 0 A 0 e t ln . 2 . T t ln . 2 . T A t ln 1 30 . t ln . 2 . T ln 1 30 . t t ln 30 . ln 2 . T ln 30 . ln 2 . T t = 4.91T 1

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CAPE PURE MATHEMATICS PP QUESTIONS WITH ANSWERS.

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Page 1: Mathcad - CAPE - 2007 - Math Unit 2 - Paper 03-2

CAPE - 2007 - Pure MathematicsUnit 2 - Paper 03-2

Section A (Module 1)

1 The rate of decay of a radioactive substance after t hours was found to be directly proportional to the amount A of the substance remaining

a( ) Form a differential equation to model this situation [3 marks]

b( ) Given that when t = 0 A = A0 show that A A0 e ktkt [8 marks]

c( ) If the time taken for the remaining substance to be12

A0 is T show that

A A0 e

t ln. 2.

Tt ln. 2.

TA

t

[4 marks]

d( ) Prove that the time taken for the amount remaining to be reduced to

130

A0 is 4.91T [5 marks]

Total 20 marks

a( )dAdt

kAdAdt

b( )

A0

AA

1A

d0

ttkd

A0

AA

1A

dt

k lnAA0

. ktlnAA0

. kt

A A0 e ktA kt

c( )12

A0 A0 e kT12

A0 A kT kT ln12

.kT kln 2.

Tk

ln 2.

T

A A0 e

t ln. 2.

Tt ln. 2.

TA

t

d( )130

A0 A0 e

t ln. 2.

Tt ln. 2.

TA

t

ln130

. t ln. 2.

Tln

130

. tt

ln 30.

ln 2.T

ln 30.

ln 2.T t = 4.91T

1

Page 2: Mathcad - CAPE - 2007 - Math Unit 2 - Paper 03-2

Section B - Module 2

2 a( ) The common ratio r of a geometric series is given by r7 x

x2 6

x

x

Find the values of x for which the series is convergent [9 marks]

b( ) An oil well produced 100 000 barrels of oil in its first year but output fell by 7.5% each year thereafter

i( ) Determine to 2 sig fig the number of barrels of oil that could be produced

a( ) from the well as the maximum total output [5 marks]

b( ) by the well in the first 15 years [4 marks]

ii( ) What percentage of the maximum total output of oil would be left after 15 years?

[2 marks]

Total 20 marks

a( )7 x

x2 61< 49 x2 x4 12 x2 36<

x4 37 x2 36 0> x2 1 x2 36 0>

x 6< 1 x< 1< x 6>

b( ) i( ) production is GP with a = 100 000 and r = (1 - 0.075) = 0.925

max output: S∞100000

1 0.925∞ productionmax 1300000 2 sig fig.( )sig fig

ii( ) S15100000 1 0.92515.

1 0.925S15 920000 2 sig fig.( ). sig fig

iii( ) percent =1300000 920000

1300000100 29.231= 29 % 2 sig fig.( ).

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Page 3: Mathcad - CAPE - 2007 - Math Unit 2 - Paper 03-2

Section C - Module 3

3 a( ) Ten students 8 girls and 2 boys are to be seated in a row

i( ) Determine the number of different ways these students can be arranged if there are no restrictions

[2 marks]

ii( ) Find the probability that the two boys do not sit next to each other [6 marks]

b( ) The cost of 2 computers 2 printers and 1 scanner is $5 950

The cost of 4 computers 1 printer and 1 scanner is $11 450

The cost of 5 computers 3 printers and 2 scanners is $14 600

Let x, y and z represent the cost in dollars of a computer a printer and a scanner respectively

i( ) Use the given information to form three equations [3 marks]

ii( ) Write down the system of equations in matrix form [1 mark]

iii( ) Hence calculate the values of x y and z [8 marks]

a( ) i( ) 10 ! 3628800 ways10 ! ways

ii( ) boys next to each other: 9 ! 2 !

P(boys next to each other) = 9 ! 2 !10 !

15

9 ! 2 !10 !

P(boys not next to each other) = 19 ! 2 !10 !

45

19 ! 2 !10 !

b( ) i( ) 2x + 2y + z = 59504x + y + z = 114505x + 3y + 2z = 14600

3

Page 4: Mathcad - CAPE - 2007 - Math Unit 2 - Paper 03-2

ii( )

2

4

5

2

1

3

1

1

2

x

y

z

.5950

11450

14600

2

4

5

2

1

3

1

1

2

x

y

z

.

2

4

5

2

1

3

1

1

2

1

yields

1

3

7

1

1

4

1

2

6

x

y

z

1

3

7

1

1

4

1

2

6

5950

11450

14600

. x 2800 y 100 z 150

4