H2 Mathematics 2017 Preliminary Exam Paper 1 Question (9758)

Answer all questions [100 marks]

1
yy
y = f (2x) − 1
C(3, 3)
A(0, 2)
x
O B(2, 0) x

x=1
The diagram shows the curve y = f (2 x ) − 1 with a maximum point at C(3, 3). The curve
crosses the axes at the points A(0, 2) and B(2, 0). The line x = 1 and the x-axis are the
asymptotes of the curve.

On separate diagrams, sketch the graphs of

(i) y = f (x), [2]

(ii) y = f '(x), [2]
stating clearly the equations of the asymptotes and the coordinates of the points
corresponding to A, B and C where appropriate.

x
2 (i) Without using a calculator, solve the inequality ≤ 0 , giving your answer
x −5
2

in exact form. [2]

x
(ii) Hence, find the set of values of x for which ≤ 0. [2]
x −5

→ →
3 Referred to the origin O, the points A and B are such that OA = a and OB = b . The point
P on OA is such that OP : PA = 2 : 3 , and the point Q on OB is such that OQ : QB = 1: 2 .
Given that M is the mid-point of PQ, state the position vector of M in terms of a and b.[1]
Show that the area of triangle OMP can be written as k a × b , where k is a constant to be
determined. [4]

4 Find
(a) ∫ cos(ln x) dx , [3]
⌠ 1 − 2 x dx .
(b)  2 [3]
⌡ 2x +1
5 It is given that z = 3 + i and w = −1 + i .
z2
(i) Without using a calculator, find an exact expression for * . Give your answer in
w
iθ
the form re , where r > 0 and −π < θ ≤ π . [4]

 q π
(ii) Find the exact value of the real number q such that arg 1 −  = . [3]
 z  12

6 It is given that y = ln ( 3 + e x ) .
2
d 2 y  dy  dy
(i) Show that 2
+  = . [3]
dx  dx  dx

(ii) By differentiating the above result, find the first four non-zero terms of the
Maclaurin series for y. Give the coefficients in exact form. [3]

e−2 x
(iii) Hence find the Maclaurin series for , up to and including the term in x 2 .
3 + e−2 x
[2]

ax 2 + bx − 8
7 The curve C has equation y = , where a and b are constants. It is given that C
x−2
has asymptote y = 3 − 2x.

(i) Find the value of a and show that b = 7. [3]

(ii) Sketch C, stating clearly the equations of any asymptotes and the coordinates of
any stationary points and any points of intersection with the axes. [3]

(iii) By drawing another suitable curve on the same diagram, deduce the number of
real roots of the equation ( −2 x 2 + 7 x − 8 ) − 25 ( x − 2 ) = 0 .
2 3
[3]

8 Emily has 1016 toy bricks.

(i) Emily wishes to build a brick structure with one brick in the first row, two bricks
in the second row, three bricks in the third row and so on. What is the maximum
number of rows that she can build and how many bricks will be left
unused? [4]

(ii) Emily keeps all her 1016 bricks in ( 2k − 1) bags of different sizes. She packs m
bricks into the smallest bag. For each subsequent bag, she packs double the
number of bricks she packs in the previous bag. Given that she has 64 bricks in
the kth bag, find the value of m and the number of bags. [5]
 6
⌠ 3x + 1
9 (a) By using the substitution x = 3secθ , evaluate  dx exactly. [5]
⌡3 2 x2 − 9

(b)

x 2 ( y − 2)2
The diagram shows an ellipse with equation + = 1.
16 4

(i) Find the area of the shaded region, giving your answer correct to 3 decimal
places. [2]

(ii) Find the exact volume of the solid generated when the shaded region is
rotated 180° about the y-axis. [4]

dy x − y − 1
10 (a) By using the substitution z = x − y , solve the differential equation = .
dx x − y + 1
Find the particular solution for which y = 1 when x = 1 . [4]

(b) A sky diver jumped out of an aeroplane over a certain mountainous valley with
zero speed and t seconds later, the speed of his descent was v metres per second.
He experienced gravitational force and air resistance which affect v. Gravity would
increase his speed by a constant 10 metres per second2 and the air resistance would
decrease his speed at a rate proportional to the square of his speed. It is given that
when his speed reaches 50 metres per second, the rate of change of his speed is
7.5 metres per second2.

By setting up and solving a differential equation, show that

100 (1 − e− mt )
v= , where m is a constant to be found. [7]
1 + e − mt

Describe briefly what his speed would be after he had descended for a long time
and just before he deployed his parachute. [1]

11
 y cm
r cm

15 cm

3r cm

A plastic water dumbbell consists of a cylinder as a handle and two cylinders as the
weights. The handle has a radius r cm and height 15 cm. Each weight has radius 3r cm
and height y cm. The dumbbell is made of plastic of negligible thickness and the volume
of the dumbbell is a fixed value k cm3.

(i) Given that r = r1 is the value of r which gives the minimum external surface area,
show that r1 satisfies the equation 102π r 3 + 30π r 2 − k = 0 . [6]

(ii) Find the value of r1 if k = 450 . [1]

(iii) It is given instead that r = 2 and y = 7. Water is pumped into an empty dumbbell
through an opening from the top at a rate of 15 cm3s-1. Find the exact rate at
which the depth of the water is increasing after 1 minute. [5]

[Question 12 is printed on the next page.]

12 Laser (Light Amplification by Stimulated Emission of Radiation) has many applications
including medicine, data storage, military and industrial uses. It has the property of spatial
coherence, which allows the laser beam to stay narrow over long distances. When a laser
beam is projected onto a mirror at an angle, it reflects off the mirror at the same angle.

An engineer is designing a device that does industrial cutting using a laser beam. To make
the device compact, the device has a mirror to reflect the beam before it leaves the device.
The laser beam source is located at the origin O. It projects an incident beam with direction
vector i + 2 j + k . The beam hits the mirror at the point P with angle θ. The mirror has an
equation − x + 2 y + 3z = 12 .

O
Incident Reflected
beam beam

θ θ
P mirror

O'

(i) Find the acute angle θ that the beam makes with the mirror. [2]

(ii) By finding O', the image of O in the mirror, find a vector equation of the line
that the reflected beam is on. [7]

(iii) The engineer plans to install a sensor at (3, 1, 0) to monitor the heat produced by
the laser. For the sensor to work properly, the sensor must be less than 2 units
away from either the incident or the reflected beam. Determine if the sensor will
work properly. [4]

~ ~ End of paper ~ ~
 YISHUN JUNIOR COLLEG5E
Mathematics Department

PRELIM Solutions
Subject : JC2 H2 MATHEMATICS 9758/9740 P1 Date :
Qn Solution
1(i)
y

(6,4)

(0,3)

y=1
(4, 1)
O x

x=2
(ii) y

(6, 0)
x
O

x=2

Pg 1
 Qn Solution
2(i) x
2
≤0
x −5
x − + −
≤0 +
( )(
x− 5 x+ 5 ) −√5 0 √5

∴ x < − 5 or 0 ≤ x < 5

x
(ii) ≤0
x −5
x
2
≤0
( x) −5

Replace x by x in the result from (i),

x <− 5 or 0 ≤ x < 5
(Reject ∵ x ≥ 0) or 0 ≤ x < 5

Required set = { x ∈ R : 0 ≤ x < 5}

3 2 1
OP = a OQ = b
5 3
12 1 
OM =  a + b 
25 3 
Area of triangle OMP
1 12 1  2
=   a + b× a
2 25 3  5
1  1 1  2
=   a + b× a
2  5 6  5
1 2 1
= a×a + b×a
2 25 15
1 1
= b×a
2 15
1
= −a × b
30
1
= a×b
30

Pg 2
 Qn Solution
4(a) ⌠ 1
∫ cos(ln x) dx = x cos(ln x) − ⌡ − x sin(ln x). x dx
= x cos(ln x ) + ∫ sin(ln x ) dx
= x cos(ln x ) + x sin(ln x ) − ∫ cos(ln x ) dx

2 ∫ cos(ln x) dx = x cos(ln x) + x sin(ln x ) + constant

1
∫ cos(ln x) dx = 2 x [cos(ln x) + sin(ln x)] + C
(b) ⌠ 1 − 2 x dx
 2
⌡ 2x +1
1⌠ 1
dx − ⌠
1 4x
=   2 dx
2  x2 + 1 2 ⌡ 2x + 1
⌡
2
2 1
= tan −1 2 x − ln ( 2 x 2 + 1) + c
2 2
5(i) z = 3 + i = 3 +1 = 2 ,
w = −1 + i = 1 + 1 = 2
arg( z ) = arg( 3 + i)
1 π
= tan −1=
3 6
arg( w) = arg( −1 + i)
3π
= π − tan −1 1 =
4
2
 i π6 
 2e 
z2
*
=   3π 
w i − 
2e  4 
3 13π
i
2 12
=2 e
3 11π
−i
2 12
=2 e
(ii)  q  z−q
arg 1 −  = arg  
 z  z 
π
= arg ( z − q ) − arg ( z ) =
12
π π π
arg ( z − q ) = + =
12 6 4
π
arg (( 3−q +i =
4
) )
3 − q = 1 ⇒ q = 3 −1

Pg 3
 Qn Solution
6(i) x
y = ln(3 + e )
dy ex
=
dx 3 + e x
dy
(3 + ex ) dx
= ex

d2 y dy
dx
(
2 )
3 + ex + ex
dx
= ex
2 x
d y e dy ex
+ =
dx 2 ( )
3 + e x dx ( 3 + ex )
2
d 2 y  dy  dy
2
+  = (proved)
dx  dx  dx

(ii) d3 y  dy  d y d y
2 2
+ 2   2 =
dx3  dx  dx dx 2
dy 1 d 2 y 3 d3 y 3
When x = 0, y = ln 4, = , = , =
dx 4 dx 2 16 dx3 32
1 x  3  x  3 
2 3
y = ln 4 + x   +   +   + ...
 4  2!  16  3!  32 
1 3 1
= ln 4 + x + x 2 + x3 + ...
4 32 64
(iii) e x
1 3 3 2
x
= + x+ x +…
3+ e 4 16 64
e −2 x 1 3 3
−2 x
= + ( −2 x ) + ( −2 x ) 2 + …
3+ e 4 16 64
1 3 3
= − x + x2 + …
4 8 16

7(i) a=−2

2b − 16
By long division, y = (b − 4) −2x + .
x−2

b − 4 = 3 ⇒ b = 7 (shown)

Pg 4
 Qn Solution
(ii) 2
−2 x + 7 x − 8
The equation is y =
x−2
y

(0, 4)

(1,3)

x
O

(3, −5)
y = 3 − 2x

x=2

(iii)
(−2x2 + 7x − 8)2 − 25(x − 2)3 = 0
(−2x2 + 7x − 8)2 = 25(x − 2)3
2
 −2 x 2 + 7 x − 8 
  = 25( x − 2)
 x−2 

Add graph of y2 = 25(x − 2)

y2 = 25(x − 2)

(0, 4)

(1,3)

x
(2,0)
O

(3,−5)
y = 3 − 2x

x=2

From the graphs, the number of real roots is 2.

Pg 5
 Qn Solution
8(i) This is an AP with a = 1, d =1.

For Sn ≤ 1016
n
(1 + n ) ≤ 1016
2
n 2 + n − 2032 ≤ 0

From GC, −45.58 ≤ n ≤ 44.58

She can complete a maximum of 44 rows.

44
S44 = (1 + 44 ) = 990
2
Number of bricks left = 1016 − 990 = 26

(ii) The sequence is a GP with common ratio 2

S2k −1 = 1016
m  22 k −1 − 1
= 1016
2 −1
m  22 k −1 − 1 = 1016 − − − (1)

Tk = 64 ⇒ m2k − 1 = 64 ---- (2)

(1) ÷ (2):
22 k −1 − 1 1016
=
2k −1 64
From GC, k =4
Sub. into (2): m 24 −1 = 64
⇒m=8

No. of bags = 2(4) − 1 = 7

9(a) dx
= 3sec θ tan θ
dθ
6
⌠ 3x + 1
 dx
⌡3 2 x2 − 9
π
⌠ 3 9sec θ + 1
= ( 3sec θ tan θ ) dθ
⌡π 9sec2 θ − 9
4
π
3 9sec θ + 1
=⌠
π ( 3sec θ tan θ ) dθ
⌡ 3 tan θ
4

Pg 6
 Qn Solution
π
= ∫π3 9sec 2 θ + sec θ dθ
4
π
= 9 tan θ + ln sec θ + tan θ  π3
4

π π π  π π π 
= 9 tan + ln sec + tan −  9 tan + ln sec + tan 
3 3 3  4 4 4 
= 9 3 + ln 2 + 3 − 9 + ln ( 2 +1 )
2+ 3
= 9 3 − 9 + ln
2 +1
(b)(i) x2
Consider y = 2 ± 2 1 −
16
3
⌠ x2
Required area =  2 + 2 1 − dx − 2(6)
⌡− 3 16
= 10.753 (3 dp)

Alternative
x2 y 2 x2
Consider + = 1 ⇒ y = ±2 1 −
16 4 16
3 3
⌠ x2 ⌠ x2
Required area =  2 1 − dx or 4 1 − dx
⌡ −3 16 ⌡0 16
= 10.753 (3 dp)

(ii) 9 1
When x = 3 , y = 2 + 2 1 − =2+ 7
16 2
When x = 0 , y = 4
4
7 ⌠  ( y − 2)2 
Required Volume = π (32 ) + π  16  1 −  dy
2 ⌡2+ 1 7  4 
2
4
9 7 ⌠ ( y − 2) 2
= π + 16π  1− dy
2 ⌡2+ 1 7 4
2
4
9 7  ( y − 2)3 
= π + 16π  y −
2  12  2+ 1 7
2

 2 7 7 7
= 18π + 16π 4 − − 2 − + 
 3 2 96 
1
(
= 64 − 7 7 π
3
)

Pg 7
Qn Solution
Alternative
9 1
When x = 3 , y = 2 1 − = 7
16 2
When x = 0 , y = 2
2
7 2 ⌠  y2 
Required Volume = π (3 ) + π  16  1 −  dy
2 ⌡1 7  4 
2
2
y2
π + 16π ⌠
9 7
= 1 1 − dy
2 ⌡ 7 4
2
2
9 7  y3 
= π + 16π  y −
2  12  1 7
2

9 7 1
= π + 128 − 41 7  π
2 6
1
(
= 64 − 7 7 π
3
)
10(a) dz dy
z = x− y ⇒ = 1−
dx dx
dy dz
⇒ = 1−
dx dx

dy x − y − 1
= .
dx x − y + 1
dz z − 1
⇒ 1− =
dx z + 1
dz z −1
⇒ = 1−
dx z +1
dz 2
⇒ =
dx z + 1
∫ ( z + 1) dz = ∫ 2 dx
z2
+ z = 2x + C
2
2
( x − y ) + x − y = 2x + C
where C is a constant
2
2
( x − y)
−x− y =C
2
When x = 1 , y = 1 ,
⇒ C = −2
2
( x − y) − x − y = −2
Therefore
2

Pg 8
 Qn Solution
(b) dv
= 10 − kv 2 , where k > 0
dt
dv
When v = 50, = 7.5
dt
7.5 = 10 − k (50)2
⇒ k = 0.001
dv
∴ = 10 − 0.001v 2
dt

⌠ 1
dv = ∫ 1 dt

⌡ 10 − 0.001v 2
1 ⌠ 1
 dv = ∫ 1 dt
0.001 ⌡ 10000 − v 2
1000⌠
1
 dv = ∫ 1 dt
⌡ 100 − v 2
2

1000 100 + v
ln = t +C
200 100 − v
100 + v 1 1
ln = t+ C
100 − v 5 5
1 1
100 + v t+ C
= e5 5
100 − v
1 1
100 + v t+ C
= ±e 5 5
100 − v
100 + v
= Ae0.2t where A = ±e0.2C
100 − v

When t = 0 , v = 0 then A = 1
100 + v 100 − v
= e0.2t ⇒ = e −0.2t
100 − v 100 + v
−0.2 t
e (100 + v) = 100 − v
v (1 + e −0.2t ) = 100 (1 − e −0.2t )
100 (1 − e −0.2 t )
v=
1 + e −0.2 t

As t → ∞ , e −0.2t → 0 and v → 100

The sky diver’s speed would increase to a limit of 100 m/s long after he has descended and
before he deployed his parachute.

Pg 9
 Qn Solution
11(i) V = 2 π ( 3r ) y  + (π r × 15 )
2 2
 
k = 18π yr + 15π r 2
2

1
2 (
y= k − 15π r 2 )
18π r
A = 4 π ( 3r )  − 2π r 2 + 15(2π r ) + 2 y  2π ( 3r ) 
2
 
 1 
2 (
= 34π r 2 + 30π r + 2  k − 15π r 2 )  2π ( 3r ) 
18π r 
2
= 34π r 2 + 30π r + ( k − 15π r 2 )
3r
2k
= 34π r 2 + 20π r +
3r
dA 2k
= 68π r + 20π − 2
dr 3r
dA
At minimum area, =0
dr
2k
68π r + 20π − 2 = 0
3r
3 2
204π r + 60π r − 2k = 0
102π r 3 + 30π r 2 − k = 0 (shown)

(ii) 102π r 3 + 30π r 2 − 450 = 0

From GC, r = 1.03 (3 s.f.)

(iii) Volume of water pumped after 1 min = 15 (60)

= 900 cm3
2
Volume of a weight = π ( 3 × 2 ) × 7 = 791.68 cm3
2
Volume of the handle = π ( 2 ) ×15 = 188.50 cm3
Since 900 < 791.68 + 188.50 = 980.18 , the water level is at the handle at 1 min.

Let W = volume of water in the handle and

h = depth of water from the base of the handle
2
W = π ( 2 ) h = 4π h
dW
= 4π
dh
dh dW dh
= ×
dt dt dW
1
= 15 ×
4π
15
Thus the depth of the water is increasing at a rate of cm s −1 .
4π

Pg 10
 Qn Solution
12(i) Let θ be the acute angle between the plane and the incident beam.
 1   −1
  
 2 i 2 
1  3 
  
sin θ =
1+ 4 +1 1+ 4 + 9
6
=
84
Therefore θ = 40.9°
(ii) Let F be the foot of the perpendicular from O to the plane.
 −1
 
OF = λ  2  , for some λ ∈ ℝ
3
 
 −1
 
F is on plane ⇒ OF i 2  = 12
3
 
 −1  −1
  
⇒ λ  2 i 2  = 12
33
  
14λ = 12
6
λ=
7
 −1
12  
OO ' =  2 
7 
3
1
 
OP = µ  2  , for some µ ∈ ℝ
1
 
 −1
 
P is on plane ⇒ OPi 2  = 12
3
 
 1   −1
  
⇒ µ  2 i 2  = 12
1  3 
  
6 µ = 12
µ=2
 2
 
OP =  4 
 2
 

Pg 11
 Qn Solution
 13 
2 
O 'P =  2 
7 
 −11
 2  13 
   
Hence l : r =  4  + γ  2  , γ ∈ ℝ
 2  −11
   
(iii) Let B ≡ ( 3, 1, 0 ) .
Shortest distance of B from incident beam
 3  1  1 
     
 1  ×  2   −3 
0 1  5 
      35
= = = >2
1+ 4 +1 6 6

 3  2  1 
     
PB =  1  −  4  =  −3 
 0   2   −2 
     

Shortest distance of B from reflected beam

 1   13   37 
     
 −3  ×  2   −15 
 −2   −11   41 
      3275
= = = >2
169 + 4 + 121 294 294

Hence sensor will not work properly

Pg 12
H2 Mathematics 2017 Preliminary Exam Paper 2 Question (9758)
Answer all questions [100 marks]

Section A: Pure Mathematics [40 marks]

1 (i) Show that if ar = Tr − Tr −1 for r = 1, 2, 3, …, and T0 = 0 , then

n

∑a
r =1
r = Tn . [1]
n
(ii) Deduce that ∑ π (1 − π ) r
r =1
−r 2
+ 2π r − π  = n 2π − n . [3]
20
(iii) Hence, find the exact value of ∑ π (1 − π ) r
r =4
−r 2
+ 2π r − π  . [2]

2 Functions g and h are defined by

g : x a x 2 + 6 x + 8, x ∈ , x ≤ α ,
h : x a −e x , x ∈ , x > −2.
(i) Given that the function g−1 exists, write down the largest value of α and define
g −1 in similar form. State a transformation which will transform the curve
y = g(x) onto the curve y = g−1(x). [5]

(ii) Given instead that α = −2 , explain why the composite function hg exists and
find the exact range of hg. [2]

3 Do not use a calculator in answering this question.

Given that z = 1 + i is a root of the equation 2 z 4 + az 3 + 7 z 2 + bz + 2 = 0 , find the values
of the real numbers a and b and the other roots. [5]

Deduce the roots of the equation 2 z 4 + bz 3 + 7 z 2 + az + 2 = 0 . [2]

4 A curve C has parametric equations

x = t 2 , y = t − t3 , t ≤ 0 .
(i) The point P on the curve has parameter p. Show that the equation of the tangent
at P is 2 py = x (1 − 3 p 2 ) + p 2 + p 4 . [3]

(ii) If the tangent at P passes through the point (6, 5), find the possible coordinates
of P. [3]

(iii) Find the area of the region bounded by C and the x-axis. [3]
5 The planes p1 and p2 have equations x − 4 y + 8 z = 4 and mx + ny + 2 z = 1 respectively,
where m and n are constants.

(i) If p1 and p2 meet at a line that has equation r = 2i − 0.5 j + λ ( −4i + j + k ), where
λ ∈ , find the values of m and n. [3]

It is given instead that m = 1 and n = 2.

(ii) Find the acute angle between p1 and p2 . [2]
(iii) The point (1, b, 5) is equidistant from p1 and p2 . Calculate the possible value(s)
of b. [6]

Section B: Statistics [60 marks]

6 (a) Find the number of ways to arrange the letters of the word TOTORO such that

(i) all the ‘O’s are together, [1]

(ii) all the ‘O’s are separated, [2]
(iii) the last letter is a consonant. [3]

(b) Tontoro soft toys are sold in four different colours, of which each varies in three
sizes, small, medium and large. Each set of Tontoro soft toys consists of a small,
a medium and a large sized soft toy and exactly two are of the same colour. Find
the number of different possible sets of Tontoro soft toys.
[2]

7 A game is played with a set of 4 cards, each distinctly numbered 1, 2, 3 and 4. A player
randomly picks a pair of cards without replacement. If the sum of the cards’ numbers is
an odd number, the sum is the player’s score. If the sum of the two cards’ number is an
even number, the player randomly picks a third card from the remaining cards. The square
of the third card’s number is the player’s score.

(i) Find the probability that a player obtains a score of 4. [2]

(ii) Find the probability distribution of a player’s score, S. Hence, find the expected
score of a player. [4]

(iii) Find the probability that a player obtains a score lower than 5, given that he draws
three cards. [3]
8 An archaeologist examines rocks to look for fossils. On average, 10% of the rocks selected
from a particular area with a large number of rocks contain fossils. The archaeologist
selects a random sample of 25 rocks from this area. The number of rocks that contain
fossils is denoted by X.

(i) Find the probability that more than 4 but at most 10 rocks contain fossils. [2]

P ( X = k + 1) 25 − k
(ii) Show that = , for k = 0, 1, 2, 3, …, 24. Hence, by
P( X = k ) 9 ( k + 1)
considering P ( X = k + 1) > P ( X = k ) , find the most probable value of X. [4]

The archaeologist explores a new area. On average, p% (p > 10) of the rocks in the new
area contain fossils. A random sample of 20 rocks from the new area is selected. Given
that the probability that there are two rocks that contain fossils is 0.17, find the value of
p, giving your answer correct to 2 decimal places. [3]

9 A researcher investigates the relationship between the population of a particular species

of bacteria in millions (b) and the surrounding temperature in °C (t). The researcher keeps
records so that she can estimate the population of the bacteria at a certain temperature.
Observations at different temperatures give the data as shown in the following table.

t 26.5 27.5 28.5 29.5 30.5 31.5 32.5

b 1.31 2.10 3.65 5.80 α 19.56 31.20

(i) Given that the regression line of b on t is b = - 129.368 + 4.75214t , show that
α = 12.12 , correct to 2 decimal places. [2]

(ii) Sketch a scatter diagram for the data. [1]

(iii) Explain which of b = ct + d or b = kt 3 + l is the more appropriate model for the

relationship between b and t and find the equation of a suitable regression line
for this model. [2]

(iv) Use the model you chose in part (iii) to estimate the population of the bacteria
when the temperature is 33°C. Comment on the reliability of the estimate
obtained. [2]

(v) It is given that the temperature T, in °F, is related to the temperature t, in °C, by
the equation T = 1.8t + 32 . Rewrite your equation from part (iii) so that it can be
used to estimate the population of bacteria when the temperature is given in
°F. [2]
10 In a factory, the average time taken by a machine to assemble a smartphone is 53 minutes.
A new assembly process is trialled and the time taken to assemble a smartphone, x
minutes, is recorded for a random sample of 60 smartphones. The total time taken was
found to be 3129 minutes and the variance of the time was 18.35 minutes2.

The engineer wants to test whether the average time taken by a machine to assemble a
smartphone has decreased, by carrying out a hypothesis test.

(i) Explain why the engineer is able to carry out a hypothesis test without assuming
anything about the distribution of the times taken to assemble a smartphone. [1]

(ii) Find unbiased estimates of the population mean and variance and carry out the
test at the 10% level of significance. [6]

(iii) Explain, in the context of the question, the meaning of ‘at 10% level of
significance’. [1]

After several trials, the engineer claims that the average time taken by a machine to
assemble a smartphone is 45 minutes using the new assembly process. The internal control
manager wishes to test whether the engineer’s claim is valid. The population variance of
the time taken to assemble a smartphone using the new assembly process may be assumed
to be 9 minutes2. A random sample of 50 smartphones is taken.

(iv) Find the range of values of the mean time of this sample for which the engineer’s
claim would be rejected at the 10% significance level. [4]

[Question 11 is printed on the next page.]

11 In the manufacture of child car seats, a resin made up of three ingredients is used. The
ingredients are polymer A, polymer B and an impact modifier. The resin is prepared in
batches and each ingredient is supplied by a separate feeder. The masses, in kg, of polymer
A, polymer B and the impact modifier in each batch of resin are assumed to be normally
distributed with means and standard deviations as shown in the table. The three feeders
are also assumed to operate independently of each other.
Mean Standard deviation
Polymer A 2030 44.8
Polymer B 1563 22.7
Impact modifier µ σ

It is known that 3% of the batches of resin have less than 1350 kg of impact modifier and
30% of the batches of resin have more than 1414 kg of impact modifier.

(i) Show that µ ≈ 1400 and σ ≈ 26.6. [3]

(ii) Given that polymer A costs $2.20 per kg, polymer B costs $2.80 per kg and the
impact modifier costs $1.50 per kg, find the probability that the total cost of 2
batches of resin exceeds $22,000. [4]

(iii) A random sample of n batches of resin is chosen. If the probability that at most 6
batches of resin has more than 1414 kg of impact modifier is less than 0.001, find
the least value of n. [3]

(iv) Each batch of resin is used to make a large number of car seats. It is found that the
tensile strength (N/m2) of resin for a car seat has mean 125 and standard deviation
17. A random sample of 50 car seats is selected. Find the probability that the
average tensile strength of resin for these 50 car seats is less than 130 N/m2. [3]

~ ~ End of Paper ~ ~
 YISHUN JUNIOR COLLEGE
Mathematics Department

PRELIM Solution
Subject : JC2 H2 MATHEMATICS 9758 P2 Date :
Qn Solution
1(i) n n

∑a
r =1
r = ∑ (Tr − Tr −1 )
r =1

= T1 − T0
+ T2 − T1
+ T3 − T2
⋮
+ Tn −1 − Tn − 2
+ Tn − Tn −1
= Tn − T0
= Tn

(ii) Let Tr = r 2π − r
Note T0 = 0
2
Tr − Tr −1 = r 2π − r − ( r − 1) π − r +1
= π − r  r 2 − (r 2 − 2r + 1)π 
=π − r (1 − π ) r 2 + 2π r − π 
= ar

∴ From (i),
n n

∑ π − r (1 − π ) r 2 + 2π r − π  = ∑ ar
r =1 r =1

= Tn = n 2π − n

20
(iii) ∑ π (1 − π ) r
r =4
−r 2
+ 2π r − π 
20 3
= ∑ π − r (1 − π ) r 2 + 2π r − π  − ∑ π − r (1 − π ) r 2 + 2π r − π 
r =1 r =1
−20 −3
= 400π − 9π

2(i) Largest α = −3

Let y = g(x) = x2 + 6x + 8
= (x + 3)2 − 1

(x + 3)2 = y + 1
x +3 = ± y + 1

Pg 1
 Qn Solution
x = −3 ± y + 1
Since x ≤ −3, x = −3 − y + 1

g−1 : x ֏ −3 − x + 1 , x ∈ [−1, ∞)

A reflection about the line y = x will transform the curve y = g(x) onto the curve y = g−1(x).
(ii)
Since R g = [−1, ∞ ) ⊆ (−2, ∞) = D h , the composite function hg exists.
 1
R hg =  −∞, − 
 e

3 By Conjugate Root Theorem, z = 1 − i is also a root.

 z − (1 + i )   z − (1 − i )  = ( z − 1) − i  ( z − 1) + i 
2
= ( z − 1) − i 2
= z2 − 2z + 2
( z 2 − 2 z + 2 )( Az 2 + Bz + C ) = 2 z 4 + az 3 + 7 z 2 + bz + 2
By observation, A = 2 , C = 1 .
i.e. ( z 2 − 2 z + 2 )( 2 z 2 + Bz + 1) = 2 z 4 + az 3 + 7 z 2 + bz + 2
Coeff. of z 2 : 1 − 2 B + 4 = 7 ⇒ B = −1
Coeff. of z 3 : B − 4 = a ⇒ a = −5
Coeff. of z : −2 + 2B = b ⇒ b = −4

2z2 − z +1 = 0
1 ± 1 − 4 ( 2)
z=
2 ( 2)
1 ± 7i
z=
4
1 ± 7i
Hence other roots are 1 − i , .
4
2 z 4 + bz 3 + 7 z 2 + az + 2 = 0
1 1 1 1
2+b +7 2 +a 3 +2 4 = 0
z z z z
Hence
1 1 4 4
z= , , ,
1+ i 1 − i 1 + 7i 1 − 7i
1− i 1+ i 1 − 7i 1 + 7i
z= , , ,
2 2 2 2

Pg 2
 Qn Solution
4(i) x=t ,2
y = t −t . 3

dx dy
= 2t , = 1 − 3t 2
dt dt
dy dy dx 1 − 3t 2
= ÷ =
dx dt dt 2t
2 dy 1 − 3 p 2
At P, x = p , y = p − p3 , =
dt 2p
Equation of tangent at P:
y − (p − p3 ) 1 − 3 p 2
=
x − p2 2p
⇒ 2 py − 2 p(p − p3 ) = ( x − p 2 )(1 − 3 p 2 )
⇒ 2 py − 2 p 2 + 2 p 4 = x (1 − 3 p 2 ) − p 2 + 3 p 4
⇒ 2 py = x (1 − 3 p 2 ) + p 2 + p 4 (shown) -------(1)
(ii) At A, substitute x = 6, y = 5 into eqn (1)
2 p (5) = 6 (1 − 3 p 2 ) + p 2 + p 4
10 p = 6 − 18 p 2 + p 2 + p 4
p 4 − 17 p 2 − 10 p + 6 = 0
From GC, p = 4.35 (rejected) or p = −3.7261 or p = −1 or p = 0.370 (rejected)
Hence coordinates of P: (1,0) and (13.9, 48.0)
(iii) 1
Required area = − ∫ y dx
0
−1
= −∫
0
( t − t ) ( 2t ) dt
3

= 0.267 unit 2
5(i) If p1 and p2 meet at l , then m is perpendicular to n 2 .
 2  m
  
min 2 = 0 ⇒  1 i n  = 0
1  2 
  
2m + n = −2
Since ( 2, − 0.5, 0 ) lies on p2 ,
2 m − 0.5 n = 1
m=0
n= − 2
(ii) Let θ be the acute angle between p1 and p2 .
 1  1
  
 −4 i 2 
 8   2
  
cos θ =
1 + 16 + 64 1 + 4 + 4
1
=
3
Therefore θ = 70.5°

Pg 3
 Qn Solution
(iii) Let B ≡ (1, b, 5 ) .
Observe A1 ( 4, 0, 0 ) lies on p1
 1   4   −3 
     
A1B =  b  −  0  =  b 
5  0  5 
     
 −3   1 
   
 b  •  −4 
 5   8  37 − 4b
   
Shortest distance of B from p1 = =
1 + 16 + 64 9

Observe A2 (1, 0, 0 ) lies on p2

1 1 0
     
A2 B =  b  −  0  =  b 
 5  0  5
     
0 1
   
b• 2
 5   2  10 + 2b
   
Shortest distance of B from p1 = =
1+ 4 + 4 3
37 − 4b 2b + 10 37 − 4b 2b + 10
= or =−
9 3 9 3
7 67
b= b=−
10 2
6(a)(i) 4!
No. of ways = = 12
2!

3! ×4 C3
(ii) No. of ways = = 12
2!

(iii) Case 1: Ending with “T”

5!
No. of ways = = 20
3!
Case 2: Ending with “R”
5!
No. of ways = = 10
3!2!

Total no. of ways = 20 + 10 = 30

3
Choose the two sizes that have the same colour: C2 = 3
(b) 4
Choose colour that is same for two sizes: C1 = 4
3
Choose colour of remaining size: C1 = 3
3 4 3
No. of ways = C2 × C1 × C1 = 36

Pg 4
 Qn Solution
7(i) P(score of 4)
= P(obtain 1 and 3 for the first 2 cards, and obtain 2 for the third card)
1 1  1 1
=  × × 2× =
 4 3  2 12

s 1 3 4 5 7 9 16
(ii)
P(S = s) 1 1 1 4 1 1 1
12 6 12 12 6 12 12
 1  1  1  4 1  1   1
Expected score = 1  + 3   + 4   + 5   + 7   + 9   + 16  
 12   6   12   12   6   12   12 
35
=
6

P(score < 5 | draws three cards)

(iii)
P(score <5 and draws three cards)
=
P(draws three cards)
P(score 4 and 3 cards)+P(score 1 and 3 cards)
=
P(obtain 1,3 or 2,4 for first two cards)
1 1
+
= 12 12
1 1 
 × × 2× 2
4 3 
1
=
2

8(i) Let X be the random variable ‘number of rocks that contain fossils out of 25 rocks’
X ~ B(25, 0.1)
P(4 < X ≤ 10) = P ( X ≤ 10 ) − P ( X ≤ 4 )
≈ 0.0979819403
≈ 0.0980 (3 sig fig)
(ii) P ( X = k + 1) 25Ck +1 (0.1) k +1 (0.9) 25− k −1
= 25
P(X = k) Ck (0.1) k (0.9)25− k
25!
(0.1) k +1 (0.9) 25−k −1
(k + 1)!(25 − k − 1)!
=
25!
(0.1) k (0.9) 25−k
k !(25 − k )!
(25 − k )(0.1) 25 − k
= = for k = 0, 1,2 …, 24 (shown)
(k + 1)(0.9) 9(k + 1)

P ( X = k + 1) > P ( X = k )
P ( X = k + 1) ( 25 − k )
= >1
P( X = k) 9 ( k + 1)
25 − k > 9 k + 9
10 k < 16

Pg 5
 Qn Solution
k < 1.6
⇒ k = 0 or 1 for P ( X = k + 1) > P ( X = k )
Since P ( X = 2 ) > P ( X = 1) > P( X = 0) , most probable value of X = 2
Let Y be the ‘number of rocks that contain fossils out of 20 rocks in the new area’
p
Y ~ B(20, )
100
P(Y = 2) = 0.17
Using g.c.
p p
= 0.045473 or = 0.1815827
100 100
Since p >10, p = 18.16 (2 d.p)
9(i) b = −129.39 + 4.7529t
From GC, t = 29.5

b = −129.39 + 4.7529t
b = −129.39 + 4.7529 (29.5)
= 10.82055

1.31 + 2.1 + 3.65 + 5.8 + α + 19.56 + 31.2

= 10.82055
7
α = 12.124 = 12.12 (2 dp)
(ii)
b
(32.5, 31.2)

(26.5, 1.31)

(iii) From (ii), the scatter diagram shows that as t increases, b increases at an increasing rate which
would not be the case if the data follows a linear model. Hence the model b = kt 3 + l is a better
model.

b = −37.370 + 0.0018516t 3
= −37.4 + 0.00185t 3 ( 3s.f.)

Pg 6
Qn Solution
(iv) When t = 33 ,
3
b = −37.370 + 0.0018516 ( 33)
= 29.171
= 29.2 ( 3s.f.)
The population of the bacteria is 29.2 millions.
Since the estimate is obtained via extrapolation, the estimate is not reliable.
(v)  T − 32 
3

b = −37.370 + 0.0018516  
 1.8 
= −37.370 + ( 3.1749 × 10 −4 ) (T − 32 )
3

= −37.4 + ( 3.17 × 10−4 ) (T − 32 )

3
( 3 s.f.)
10(i) Since n is large, by Central Limit Theorem, the sample mean time for 60 smartphones is
approximately normal. Hence the assumption that the time taken by a machine to assembly a
smartphone is not necessary.

(ii) Unbiased estimate for population mean µ is x

3129
= = 52.15
60
Unbiased estimate for population variance σ 2 is s 2
60
= (18.35 )
59
= 18.661
= 18.7 (3sf)

H 0 : µ = 53
H1 : µ < 53
X −µ
Under H 0 , the test statistic Z = ~ N(0,1) approx. by CLT, where
S/ n
µ = 53, s = 18.661, x = 52.15, n = 60 .

By GC, p-value = 0.0637 ( 3 s.f.) .

Since p-value < 0.1, we reject H 0 and conclude at 10% level that there is sufficient evidence
that average time taken by a machine to assembly a smartphone has reduced.

(iii) There is a probability of 0.1 of concluding that the average time taken by a machine to
assembly a smartphone has decreased when the average time taken by a machine to assembly
a smartphone is 53 minutes.

Pg 7
 Qn Solution
(iv) H 0 : µ = 45
H1 : µ ≠ 45
X −µ
Under H 0 , the test statistic Z = ~ N(0,1) approx. by CLT, where
σ/ n
µ = 45, σ = 9, n = 50 .

0.05

1.6449

Since H0 is rejected,

x − 45 x − 45
< −1.6449 or > 1.6449
9 / 50 9 / 50
x < 44.3021 x > 45.698
x < 44.3 ( 3 s.f.) x > 45.7 ( 3 s.f.)
Range of values of x :
x < 44.3 ( 3 s.f.) or x > 45.7 ( 3 s.f.)
11(i) Let X be the random variable ‘amount (in kg) of impact modifier in a batch of resin’
X ~ N (µ , σ 2 )
P( X < 1350) = 0.03
1350 − µ
P( Z < ) = 0.03
σ
1350 − µ
= −1.88079361
σ
µ − 1.88079361σ = 1350 − −(1)
P( X > 1414) = 0.3
1414 − µ
P( Z < ) = 0.7
σ
1414 − µ
= 0.5244005101
σ
µ + 0.5244005101σ = 1414 − − − (2)
Solve (1) and (2),
µ = 1400.046 =1400 (shown)
σ = 26.609 = 26.6 (shown)

Pg 8
 Qn Solution
(ii) Let Y be the random variable ‘amount (in kg) of Polymer A in a batch of resin’
Let W be the random variable ‘amount (in kg) of Polymer B in a batch of resin’
Y ~ N (2030, 44.82 ) , W ~ N (1563, 22.72 )

Total cost of a batch,

T = 2.20Y + 2.80W + 1.50 X ~ N (10942.4,15345.9572)
Total cost of 2 batches, T1 + T2 ~ N (21884.8,30715.9144)
P(T1 + T2 > 22000) = 0.255 (3.s.f.)
(iii) Let H be the r.v.’ number of batches of resin with more than 1414 kg of impact modifier out
of n batches.’
H ~ B ( n, 0.3)
P( H ≤ 6) < 0.001
Using GC.
When n = 53,
P( H ≤ 6) = 0.00120 > 0.001
When n = 54,
P( H ≤ 6) = 9.44 ×10−4 < 0.001
Therefore, least n = 54
(iv) Let S be the r.v.’ tensile strength (in N/m2) of resin in a car seat’
E(S) = 125, Var(S) =172
S + S + ... + S50
S= 1 2
50
 172 
S ~ N 125,  approx by Central Limit Thm
 50 
P(S < 130) = 0.981 (3 s.f.)

Pg 9