Preliminary Examination: AP Maths
ADVANCED PROGRAMME MATHEMATICS
TIME: 3 HOURS MARKS: 300
SECTION A
CALCULUS AND ALGEBRA
QUESTION 1:
(a) Solve x

R in the following:
(1) x x

2

6
 x (7)
(2) 3 x

4
 x

8 (8)
(b) Solve the following equation, giving your answer in the form a
 bi if appropriate: x
3 
7 x
2 
24

20 x (12)
QUESTION 2:
Prove the following statement by using mathematical induction:
6 r n 

1 r ( r

2 )
 n ( n

1 )( 2 n

7 )
QUESTION 3:
Solve for x (rounded to two decimal places):
(a)
(b) e
2 ln( x
1
 e
 x x
)


3
3 e x

 ln(
2
3
5 )
(c) Given: f ( x )

2 ln( x

1 )
Draw the graphs of f ( x ) and f

1
( x ) on the same Cartesian plane.
QUESTION 4:
(a) Find the following limits if they exist:
(1) lim x

0 sin
(2) lim x

1
2 x tan x x
2 x
2 x 2

8

1

3
(b) Find k if lim x

2 f ( x ) exists and f ( x )



 2 kx
2 x

1 if if x
 x

2
2
27 marks
12 marks
(8)
(6)
(10)
17 marks
24 marks
(5)
(8)
(4)
1

Preliminary Examination: AP Maths
QUESTION 5:
(a) If f ( x )

2 x
3

5
, determine a formula for the n th
derivative of f ( x ).
(10)
(b) Use implicit differentiation to find dy dx
if sin( xy )
 cos x .
19 marks
(9)
QUESTION 6:
(a) A straight line function f ( x ) intersects the x-axis at x = a and the y-axis at y = b.
y
B(0;b) x
0
A(a;0)
(1) Show that f ( x )
 b
 b x .
(2) a
(2) Prove, by using the Riemann Sum, that the area of right-angled triangle AOB is equal to
1
2 ab .
(b) Find the following integrals:
(1)
 cos( 3 x

5 ) sin( 2 x

6 ) dx
(2)
 sec x
2 x dx
QUESTION 7:
(14)
(7)
(11)
34 marks
Given: h ( x )
 x
2
2 x
3

4
(a) Determine the equations of all the asymptotes. (8)
(b) Find the co-ordinates of all the stationary points and determine the nature of the stationary points.
(c) Draw a sketch graph of h ( x ) .
(15)
(8)
(d) Use the graph to determine the value(s) of x if x
2
2 x
3

4
is real. (4)
35 marks
2

Preliminary Examination: AP Maths
QUESTION 8:
(a) The sketch shows the graph of f(x) = x
3
- 4x + 1 with the turning points as shown on the sketch. A and B are two of the x-intercepts.
Kirsten uses Newton’s Method to estimate the smallest positive zero at A.
She uses x = 1 as her initial value and gets the following results:
a
1
= 1; a
2
= -1; a
3
= 3
(1) Explain why Kirsten found the wrong zero. Do this graphically, by using the graph on the diagram sheet. Detach this sheet and put it inside the cover of your answer script. (6)
(2) Use a more appropriate first estimation and determine the coordinates of
A correct to four decimal places. (6)
(b) (1) Use the substitution x

2 sin


4
 x
2 dx
 sin 2
 
2
  c
to show that:
(12)
(2) Part of the ellipse with equation y

16

4 x
2
is shown.
Using the result in (1), find the area enclosed between the curve, the x-axis and the line x = 1, giving your answer correct to two decimal places. (8)
32 marks
3

Preliminary Examination: AP Maths
SECTION B
STATISTICS
QUESTION 1:
At St Mary’s School Form I and Form II pupils are classified as juniors and Form III,
IV and V pupils are classified as seniors. Assume that there are an equal number of pupils per grade. The probability that a junior pupil will do her Mathematics homework is 60%. The probability that a senior pupil will do her Mathematics homework is 80%.
(a) Construct a tree diagram illustrating the situation. (6)
(b) Calculate the probability that a pupil picked at random from all the pupils in the high school will do her Mathematics homework. (3)
(c) If a pupil does her Mathematics homework, calculate the probability that she is a senior. (5)
14 marks
QUESTION 2:
The length of life (in months) of a certain hair drier is approximately normally distributed with a mean 90 months and standard deviation 15 months.
(a) Each hair drier is sold with a 5-year guarantee What percentage of hair driers fail before the guarantee expires? (8)
(b) The manufacturer decides to change the length of the guarantee so that no more than 1% of hair driers fail during the guarantee period. How long should he make the guarantee? (6)
(c) A sample of 36 hair driers is tested and the mean length of life is found to be 88 months. Test at 3% level of significance if the length of life is less than what the manufacturer claims. (8)
22 marks
QUESTION 3: (All answers to be rounded to three decimal palces)
(a) At “Sellitall Supermarket”, 60% of customers pay by credit card. Find the probability that in a randomly selected sample of 12 customers, at least three pay by credit card. (8)
(b) A jar contains 15 white marbles and 45 red marbles.
Standing next to the jar, you close your eyes and draw 10 marbles without replacement.
(1) What is the probability that you will draw exactly 4 white and 6 red marbles?
(6)
(2) What is the probability that you draw at most one white marble? (8)
22 marks
4

Preliminary Examination: AP Maths
QUESTION 4:
(a) In how many ways can the members of two families be lined up for a portrait if one family has 3 members, the other family has 4 members, and all members of each family want to stand together? (6)
(b) In how many arrangements of the letters of the word REVERSE, are the V and S separated? (10)
16 marks
QUESTION 5:
It is believed that only 40% of the youth engage in some kind of sport over the weekend.
(a) You wish to be 95% certain that between 39% and 41% of youth are engaged in some kind of weekend sport. What is the least number of young people you need to interview to achieve this? (8)
(b) If 2500 young people were interviewed to arrive at this estimate, find the 95% confidence interval (correct to four decimal places). (8)
Find m, the maximum lifespan of these insects.
16 marks
QUESTION 6:
The probability density function for the lifespan of a certain insect species is given by f ( x )




16
3 x
2
0

3
4
0
 x
 m elsewhere
where x is the age of the insect in years
10 marks
TOTAL: 300 MARKS
5

Preliminary Examination: AP Maths
Memorandum: AP Maths
QUESTION 1: a) (1) If x ( x x



2 )
1

6
 x

If x < -1
 x ( x

2 )

6
 x
 x
2  x

6

0

(x + 3)(x – 2) = 0
 x
2 x
2 

3 x
3 x


6
6


0
0

x = -3

or x = 2
 x


3

9

4 ( 6 )
2
n/a No solution

(7)
(2) If
3
2 x x x x




4
2
4


4
3 x


8

If

 x x
3 x
 x


4


3

4
12
4
3

 x

8

(8) b) Let f ( x )
 x
3 
7 x
2
f(3) = 0


20 x

1 -7 20 -24

0 3 -12 24

24
1 -4 8 0
( x

3 )( x
2 
4 x

8 )

0

x = 3

or x x x x

4

16

2
4
 
16

2


4

4
2

2
2 i i
4 ( 8 )



(12) [27]
QUESTION 2:
6 r n 

1
18
 r ( r
48


2 )
90
 n ( n

1 )( 2 n


....

6 n ( n

2 )
7 )
 n ( n

1 )( 2 n

7 )

When n = 1
1(1 + 1)(2(1) + 7)=18

Formula is true for n = 1
Assume that the formula is true for a certain value n = k

18

48

90

...

6 k ( k

2 )
 k ( k

1 )( 2 k

7 )

We must show that the formula still holds for n = k + 1
18

48

90

...

6 k ( k

2 )

6 ( k

1 )( k

3 )
 k ( k

1 )( 2 k

( k
= ( k

7 )

6 ( k

1 )( k


1 )( 2 k
1 )( k

2 
13
2 )( 2 k k

18 )

9 )



3 )

6

Preliminary Examination: AP Maths
= ( k

1 )( 2 ( k

1 )

7 )( k

2 )

This is our proposed formula with n = k + 1. Hence since the formula works for n = 1, it will hold for n = 2 and so forth.

(12)
QUESTION 3 a) e
2 x  e x 
3 e x 
2
3
Let e x  k
 b) ln( 1
 x ) ln

3

5 ( 1
 x )
 ln( 5 )
3
 k
2 k
2
 k

2 k

3 k


8

8
 ln( 5

5 x )

3

0

5

5 x
 e
3 
(k – 4)(k + 2) = 0
k = 4


or k = -2
 x


1 e


3 ,
1
5
02
3 

(6) e x 
4

n/a
x = ln 4

1 , 39

(8) c)

  


(10)
[24]
QUESTION 4: a) (1) lim x

0 sin 2 x tan x
2 x
(2) lim x

1 x
2 x
2

8

1

3
 x lim

0
2 sin x x
2 cos cos x .
sin x
 x lim

0
2 sin x x
  lim x

1 x
.
sin x x
  lim x

1 ( x
2
= 2

(5)
 lim x

1 x
2 x
2

8

1

3
 x
2 x
2

8

8
 x
1 )
2

 x
2
8


9
8

3
 

3

3

1 x
2 
8

3


1
6

(8)
7

Preliminary Examination: AP Maths b) f x
( x ) lim

2
 f

(


 2 kx
2 x
 x )

1 lim x

2
 f x x
( x )



2
2
k(2)
2
= 2(2)-1

4k = 3
 k

3
4

(4)
QUESTION 5: a) f f ( x )
( x )

2 x
3


3 ( 2 x
5

5 )

1 f f

( x )
 
3 ( 2 x

5 )

2
( 2 )

( x )

3 .
2 ( 2 x


5 )

3
( 2 )
2  f

( x )
 
3 .
3 .
2 ( 2 x

5 )

4
( 2 )
3 
      f
( n )
( x )

(

1 ) n
3 .
n !
( 2 x

5 )

( n

1 )
( 2 ) n
(10) b) D x sin( xy )

D x cos x  cos xy
 x dy dx
  sin x
 y cos xy
 x cos xy dy dx
  sin x
 x cos dy dx
 xy
 dy
 dx sin x

 sin y x cos x cos xy
 xy y cos
 xy 
(9)
QUESTION 6: a) (1) m =
 b a

c = b
 y
  b a x
 b (2)
(2) A

0 a
  b a x
 b dx
 x i x
0
 a n


0 ; x
1
 a n
;....; x i
 ai
 n f ( x i
)
  b a ai n
 b

[17]
[19]
8

Preliminary Examination: AP Maths
A
 n lim
  i n 

1 a n
 n lim
  i n 

1
 bi n
 b
 abi n
2
 ab n


 n lim
  
 ab n
2 i n 

1 i
 ab n i n 
1

1

 
 n lim
 


 ab n
2
( n
2
2
 n
2
)
 ab n
( n )


 n lim
  
 ab
2
 ab
2 n
 ab

1
= ab
2

(14) b) (1)
 cos( 3 x

5 ) sin( 2 x

6 ) dx
  1  sin( 5 x

1 )
 sin(
 x

11 )
 dx 
2
  
 
1
10 cos( 5 x

1 )

1
2 cos(
 x

11 )
 c (7)
(2)
 sec x
2 x dx
  x cos

2 x dx  Let f(x) = x  g

( x )
 cos 2 x

= x
2 sin 2 x
  1
2 sin 2 xdx
 f

( x )

1
 g ( x )

1
2 sin 2 x

 x
2 sin 2 x

1
4 cos 2 x
 c

(11) [34]
QUESTION 7: a) h ( x )
 x
2
2 x
3

4 b) h

( x )

6 x
2
( x
2
(
 x
2
4 )

2 x (

4 )
2
2 x
3
)
Vertical asymptotes: x
2 x


4

2

0


6 x
4
2 x
4

24 x
2

24 x
2


4 x
4
0

0

Oblique asymptote: y = 2x

(8) 2 x
2
( x
2 
12 )

0

0
 x
2 
4
2 x
2 x
3
x = 0

or x
 
2 3

2 x
3 
8 x

y = 0
 y
 
6
8x x x
 
2 3

2 3

2 3
 x

0 f’(x) +
0 -

0
0
 x

2
0 -
3
3

2 3 x

2
0 +
3
9

Preliminary Examination: AP Maths
(

2 3 ,

6 3 ) local maximum

(0; 0) point of inflection

( 2 3 ; 6 3 ) local minimum  (15)

 


(8) d) x
2 x
2 x


3
4

0
(

2 ; 0 ]


(
 
2 ;

) (4) [35]
QUESTION 8: a) (1) f  x  =
 x 3 -4  x

+1
4

3
-1 a
2
-3 -2.5
-2 -1.5
-0.5

2
1
-2
-1
-3
-4
She used a value too close to a turning point

(6)
0.5
a
1
1
1.5
2 2.5
a
3
3
E
10

Preliminary Examination: AP Maths
(2) Let f(x) = x
3 
4 x

1 f

( x )

3 x
2 
4
 a n

1 a a
1
2
 a n
 a n
3 
3 a n
2
4 a
 n
4

1



0 , 5

0 , 2307
 a
3

0 , 25400 a
4

0 , 2541 x

0 , 2541
 A(0,2541; 0)  (6) b) (1)

4
 x
2 dx (2) A

2
1

16

4 x
2 dx

 x
4


2 sin

2
2
  
1
4 x
2 dx
 dx d


2 cos

4 sin
2 
2
 If x = 1 sin dx

2 cos
 d
 
If x = 2 cos
 d
  
2

2
  sin sin


2



1
2
1 ;

;





6


2
 

6
2







2
4
2 cos
 cos

2
2
.
2 cos cos

 d

2
 d

 
= 2
  sin 2
  c d



(12)


2

 2 (

2
)
2 , 46 units
2
 sin
 

(8)
2 (

6
)
 sin
[32]

3

 
STATISTICS
QUESTION 1: a)
0,6  H
2
5

J
0,4

N
0,8
 H
3
5

S
0,2
 N b) P(H) =
2
5

0 , 6

3
5

0 , 8 
=
18
25

(3)
(6)
11

Preliminary Examination: AP Maths c) P(S/H)=
P ( S

H )

P ( H )

2
=
3
3
5

0
18
25
, 8


(5) [14]
QUESTION 2: a) P(X < 60)

= P ( z

60

15
90
)
 c.)
= P(z < -2)

= 0,5 – 0,47725

= 0,02275

0,03

-1,88
H
0
:
 
90 
= 2,28%

(8) H
1
:
 
90

 z

88

15
90
= -0,8 
b)
6
H is accepted

(8)
0

0,01
0,49
z = -2,33


2 , 33
 x

15
90 x

55 months


(6)
QUESTION 3: a) P(X

3)

=

1 – P(X < 3)

  
= 1 -




12
 0

 0 , 6
0
0 , 4
12 


12
1

 0 , 6
1
0 , 4
11 


12
2

 0 , 6
2
0 , 4
10
= 0,997

(8)
  


 b) (1)


15
4






60
10
45


6



0 , 147  (2)


15
0




45
10





60
10





15
1




45
9



0 , 219 
[22]

(6)

(8) [22]
12

Preliminary Examination: AP Maths
QUESTION 4:
   a) 3 !

4 !

2

288 (6) b) Total number of permutations:

7 !
3 !
2 !

420 
Number of permutations if V and S are together:

6 !
3 !
2 !

2

120

Number of permutations where V and S are separated:
420 – 120 = 300

(10)
QUESTION 5: a) s = 0,4

0 , 4

1 , 96
0 , 4

0 , 6

0 , 39
 n
n = 9219,84

9220 youths need to be interviewed.

(8) b)
 

 0 , 4

1 , 96
0 , 4

0 , 6
; 0 , 4
2500

1 , 96
0 , 4

0 , 6
2500



[16]
 
(0,3808 ; 0,4192)

(8) [16]
QUESTION 6:

0 m

3
16 x
2 
3
4 dx

1




1
16 x
3 
3
4 x

 m
0

1

 m
3 m

16 m
3
3

3 m
4

1



12
12 m m

16

16

0
 m = 2 lifespan is 2 years

(10) [10]
13

Preliminary Examination: AP Maths
ANSWER SHEET
EXAMINATION NUMBER:
8 1 1 5 3 0 6
QUESTION 8:
14