2

1 Without the use of a calculator, solve the inequality

x2 + 2x − 3
≥ 0. [4]
( x 2 − 2 x + 11) ( x + 1)

x( x + 3)
2 The curve C has equation given by y = where x ≠ −2 .
x+2
(i) Show that the gradient of C is always positive for x ∈ ¡ \ {−2} . [3]

(ii) Sketch the graph of C, indicating clearly the asymptotes and intercepts on the
axes whenever applicable. [3]

2
3 (i) Show that e −r
− 2e − r +1
+e −r +2
=
( e − 1)
. [1]
er
( e − 1)
2
N
(ii) Hence find r =1 er +1
in terms of N. [4]

( e − 1)
2
N +1
(iii) Using your result in part (ii), find 
r =9 er +1
in terms of e. [2]

4 It is given that the function y = f ( x ) has the Maclaurin’s series 1 + x + bx 2 + cx 3 + ...

(
and satisfies a − x 2 ) ddyx = y (1 + 2 x − x ) , where a, b and c are real constants.
2

(i) Show that a = 1 and find the value of b and c. [7]

(ii) Given that the first four terms of this series are equal to the first four terms in the
ex
series expansion, in ascending powers of x , of , find the series expansion
1 − x2
e2 x
of y = up to and including the term in x3. [3]
1 − x2

Do not use a graphic calculator in answering this question.

5 (i) By letting z = a + bi, solve z 2 = i , giving your answers in exact form. [3]
(ii) Solve the equation w2 + 2w + (1 − 8i ) = 0 , giving the roots in Cartesian form. [3]
(iii) Hence, solve (1 − 8i ) v 2 + 2iv − 1 = 0 , giving your answers in Cartesian form. [3]

[Turn Over
 3

6 Relative to the origin O, the position vectors of points A and B are a and b respectively,
where a and b are non-zero and non-parallel vectors. The point C with position vector c
lies on the line segment AB such that AC : CB is λ :1 − λ . It is given that b is a unit
4
vector, a = and the angle formed between OA and OB is 120° .
3

(i) Find the value of λ such that the points O, A and C form a right angle AOC. [4]

(ii) Find m, the position vector of M, the midpoint of AC, in terms of a and b. [2]

A circle is drawn with AC as its diameter and O is a point on the circumference of the
circle drawn.

(iii) Determine if OB is a tangent to the circle described above. [3]

(iv) Give a geometrical interpretation of b • m . Hence, explain ( b • m ) b in terms of

its magnitude and direction. [2]

7 A curve is defined by the parametric equations:

x = a cos3 t , y = a sin t , 0 ≤ t < 2π where a is a positive constant.
3

(i) Sketch the curve, showing clearly the coordinates of the points where
π 3π
t = 0, , π and . [2]
2 2
(ii) Show that the equation of the normal at the point where t = p is given by
y sin p = x cos p − a cos 2 p . [4]
π
(iii) Find the value(s) of t when the normal at the point where p = meets the curve
3
again. [3]

x2 y 2 y x
8 (i) Sketch the curve 2
+ 2 = 1 and the line − = 1 , where 0 < b < a on a single
a b b a
diagram, labelling clearly any intersection between the curve and the line. [2]

(ii) Show that the area of the region R, bounded by the curve with equation
a2 y2 y x
x = − a2 − 2
and the line − = 1 , where 0 < b < a is given by
b b a
1 b a2 y2
− ab +  a2 − dy .
2 0 b2

Hence, by substituting y = b cos θ , find the exact value of the area in terms of a
and b. [5]
(iii) Find the exact volume of the solid obtained when R is rotated 4 right angles about
the y-axis, giving your answer in the form of ka 2b , where k is a constant. [3]

[Turn Over
 4

Do not use a graphic calculator in answering this question.

π
i
9 (i) It is given that z = 3e 3 is a root of the equation z 2 − 3z + 9 = 0 . Find in similar
form, the other root of the equation. [1]
iθ −iθ
(ii) Show that e − e = 2i sin θ . [1]
π
i
(iii) Let the root found in (i) to be w1 and w2 = 3e . Hence, find the complex number
9

w2 − w1 , in the form reiθ , where r > 0 and θ ∈ ( − π , π ] . [3]

(iv) Let the point A represent the complex number w2 − w1 on the Argand diagram. A
perpendicular line is drawn from the point A to the real axis. The intersection point
between this line and the real axis is B. Show that the area of the triangle OAB is
 2π   7π 
9sin 2   sin   square units, where O is the origin. [3]
 9   9 

10 Albert and Betty each took a study loan of $100 000 from a bank on 1 January 2014 and
both graduated on 31 December 2017. The bank only starts charging an annual interest
rate of 5% on the outstanding loan at the end of each year from 2018 onwards. Albert pays
the bank $x on the 11th day of every month starting in January 2018. Let n be the number
of years after 2017 that repayment of the study loan has begun.

(i) Find an expression for the amount of outstanding loan at the end of n years. [4]

(ii) Find the minimum value of x if Albert wishes to complete his repayment of the
loan at the end of 2027, giving your answer to the nearest dollar. [3]

An investment fund pays out a constant r % dividend per annum on 31 December every
year based on the amount of funds held to maturity from 1 January till 30 December of the
same year.

On 1 January 2017, Betty decided to invest $50 000 in the fund to finance her repayment
of her study loan using the annual dividend payout. Her repayment is once per year which
she uses the full amount of the annual dividend payout. The schedule of repayment is fixed
on 11 January of each year, starting with effect from 2018.

Find the minimum value of r such that Betty will be able to complete her repayment of
the loan at the end of 2027. [5]

[Note that the principal sum of Betty’s investment remains unchanged at $50 000
throughout the repayment period]

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 5

11 Joe wants to grow vegetables at his house balcony and designs an irrigation device, which
takes the form of a cylindrical water tank, with a fixed cross sectional area A cm2 , and a
small opening at the bottom of it.

It is known that the rate of change of the volume, V cm3 , of water at any time t hours,
dispensed from the tank, is proportional to the square root of the height of the water, h cm,
above the opening.
(i) Show that the rate of change of the height of the water in the tank at any time
dh k
t hours, is given by =− h , where k is a positive constant. [3]
dt A
(ii) Joe fills the water tank to an initial height of 81 cm with the opening closed. The
water is then discharged from the opening and the height of the water level above
the opening is decreasing at a rate of 0.3 cm/hr. By solving the differential
equation in part (i), find h in terms of t. [5]
(iii) Joe is planning for an overseas trip. He wants to make sure that he will be back
before the water tank is empty. What is the maximum number of days, to the
nearest integer, that Joe can be away? [2]

(iv) Given that the tank has a base radius of 20 cm, find the exact rate of change of the
volume of water in the tank at the end of the fourth day. [4]

End of Paper
 2

Section A: Pure Mathematics (40 marks)

b
1 A curve C with equation y = a ( x − 1) + , where a, b and c are real constants,
x+c
undergoes in succession, the following transformations:
A: A reflection in the x-axis
B: A translation of 1 unit in the negative x-direction
The resulting curve with equation y = f ( x ) has the y-axis as one of its asymptotes.

 1 1
Given that 1,  is a turning point of y = , find the values of a, b and c. [5]
 6 f ( x)

2 B

A C
O r

In the figure above, points A and C are fixed points on the circle which form the diameter
passing through the centre O. The circle has a fixed radius r units. The variable point B
moves along the circumference of the circle between A and C in the upper half of the
circle. The chord AB makes an angle of θ radians with the diameter AC.

Use differentiation to find, the exact angle θ , such that S, the area of triangle ABC, is a
maximum. [6]

3 (i) Using the R-formula, express sin θ + m cos θ , for m > 0, in the form R sin(θ + α )
π
where R > 0 and 0 < α < are constants to be determined in terms of m.
2

Hence show that 2(sin θ + m cos θ ) sin(θ − α ) = R ( cos 2α − cos 2θ ) . [3]

π cos 2θ
(ii) Given that α =
2
, evaluate  (sin θ + m cos θ )sin(θ − α ) dθ in terms of m

and θ . [4]

[Turn Over
 3

4 (i) Solve the inequality ln ( x − 1) ≤ 0. [2]

The function f is defined by
f : x a ln ( x − 1) , for x ∈ ¡ , 1 < x ≤ 2.

(ii) Find f −1 ( x) and state the domain of f −1. [3]

−1
(iii) Sketch, on the same diagram, the graphs of y = f ( x) and y = f ( x) , giving the
equations of any asymptotes and the coordinates of any points where the curves
cross the axes. [3]

(iv) Hence solve the inequality f −1 ( x) < f ( x). [1]

Another function g is defined by

4
g : x a 1+ for x ∈ ¡ , x > 0 .
4x + 5
2

(v) Using f −1 , find the exact value of a such that fg ( a ) = 3. [2]

11  −2 
   
5 The line l1 has equation r =  6  + λ  0  , λ ∈ ¡ and the line l2 has equation
0  1 
   
x − 4 y z +1
= = .
2 −3 4

(i) Show that l1 and l2 are skew lines. [3]

The line l1 contains points A and B with coordinates (11, 6, 0) and (1, 6, 5) respectively
with respect to the origin O. The plane p1 which is parallel to l1 has equation
 4
 
r •  3  = 240 .
8 
 

(ii) Find the position vector of the point P on p1 which has the shortest distance to the
line l1 and is equidistant from the points A and B. [5]

A plane p2 contains the point P and is parallel to both l1 and l2 .

(iii) Hence, find the Cartesian equation of p2 . [3]

[Turn Over
 4

Section B: Statistics (60 marks)

6 A committee of eight people is to be chosen from 15 men and 7 women.

Find the number of ways in which the committee can be chosen if it consists of

at least 2 women. [3]

The chosen committee consists of 6 men (Allen, Ben, Calvin, Donald, Edwin and Felix)
and 2 women (Gina and Hazel).
At a meeting, the committee members are seated at a rectangular table as shown in the
diagram below, with seats labelled 1 to 8.

1 2 3 4

TABLE

5 6 7 8

Find the number of possible seating arrangements if Gina and Hazel must be
seated at any two of the corner seats labeled 1, 4, 5 or 8. [2]

7 Oliver is practising for the upcoming target archery competition. During practices,
Oliver shoots from distances ranging from 30m to 90m to the target. The
2
probability, p, that he hits the bullseye is given by p = ( 95 − d ) , where d is the
195
distance between the archer and the target in metres.

Each shot he made is assumed to be independent of any other shots made.

(i) Oliver shoots 18 arrows from a distance of 40 metres from the target.
Find the probability that he hits the bullseye more than 6 times given that
he hits the bullseye at most 10 times. [4]

(ii) Oliver shoots 18 arrows from a distance of x metres from the target. Find x
such that Oliver has a 98% chance of hitting the bullseye at least twice. [3]

[Turn Over
 5

8 For two mutually exclusive events A and B, it is given that P ( A ) = 0.65 and

2
P ( B | A ') = .
7

(i) Show that P ( B ) = 0.1 . [2]

For a third event C, it is given that P ( A ∩ C ) = 0.39 .

(ii) Find P ( C ' | A ) . [2]

It is given that B and C are independent and P ( A '∩ B '∩ C ) = 0.15 .

(iii) Find P ( B ∩ C ) . [2]

(iv) Hence or otherwise, determine whether the events A and C are independent. [1]

9 Connie and Sally play a game using two six-sided dice. One of the dice is fair and each
face is labelled with a digit from ‘1’ to ‘6’ respectively. The other die is biased such that
the score, denoted by Y, has a probability distribution given as follows:

 1
 for y = 1, 3, 5
6

1
P(Y = y ) =  ( y − 1) for y = 2, 4, 6
18
 0 otherwise.



Connie throws the two dice. Sally pays Connie $5 if the difference between the scores
on the fair and biased dice is more than 3. Both players receive nothing if the scores on
the fair and biased dice are identical. Connie pays Sally $3 for all other outcomes. Let X
be Sally’s winnings after one game in dollars.

(i) Find Sally’s expected winnings in one game, leaving your answer in exact form. [4]
(ii) Find the probability that Sally’s total winnings in 50 independent games is at
least $65. [3]

[Turn Over
 6

10 (a) It is given that the regression line y on x for the following bivariate data is
y = 8 + 0.5 x .

x 20 22 24 26 28 30 32 34
y 16 21 a 24 22 24 27 20

Find a. [2]

(b) A botanist conducted an experiment to find out how the age of pine trees, x, in
years, varies with their average height, y, in metres. The data collected were
given below.

x
1 2 3 4 5 6 7 8 9 10
(in years)
y
2.74 3.38 3.75 4.08 4.30 4.48 4.51 4.68 4.72 4.75
(in metres)

(i) Draw a scatter diagram for the given data. [2]

The botanist felt that the data should be modelled by an equation of the form
y = a + bx .

(ii) Give an interpretation, in this context, of the value of b. [1]

(iii) State, with a reason, which of the following models among A, B or C is the
most appropriate for the given data.
b
A: y = a − B: y = a + b x C: y = a + b ln x
x

Write down the equation of the least-squares regression line for the

chosen model, stating clearly the values of a and b. [3]

(iv) Give two reasons why it would be reasonable to use your model to

estimate the age of the pine tree when its height is 4.25 metres. [2]

[Turn Over
 7

11 A road named Spring Avenue has a speed limit of 40 km/h in a housing estate.

The residents were concerned that many vehicles travelled too fast along the road and they
decided to set up a speed tracking device to monitor the speed of vehicles travelling along
this road. The data generated from the device indicated that the mean speed of vehicles
travelling through this road was 44.1 km/hour.
In an attempt to reduce the mean speed of vehicles travelling through Spring Avenue, life-
size photographs of a police officer were put up next to the road. The speed, X km/hour of
a sample of 100 randomly chosen vehicles was then measured and the following data
obtained.

 x = 4327.0, (x − x )
2
= 925.71 .

(i) Calculate the unbiased estimates of the population mean and variance of the speed
of vehicles travelling along Spring Avenue. [2]

(ii) State an assumption that must be made about the sample in order to carry out a
hypothesis test to investigate whether the desired reduction in mean speed had
occurred. [1]

(iii) Given that the assumption that you stated in part (ii) is valid, carry out such a test,
using the 5% level of significance. [5]
(iv) Explain what is meant by “5% level of significance” in the context of this question.
[1]

(v) Subsequently, the residents detected that a measurement error has occurred when
measuring the speed of the 100 randomly selected vehicles. To rectify the error, a
multiplication of a positive constant k to each reading for the 100 randomly
selected cars is recommended. Find the greatest possible value of k, to 3 significant
figures, for the conclusion obtained in (iii) to remain the same. [3]

[Turn Over
 8

12 (i) Aquafresh mineral water is supplied in 1.5-litre bottles. The actual volume in
millilitres, in a bottle may be modelled by a normal distribution with mean 1505 ml
and standard deviation 10.2 ml.

(a) Calculate the probability that the volume of Aquafresh mineral water in a
randomly selected bottle is more than 1480 ml. [1]
(b) The supplier requires that less than 10 per cent of bottles should contain
less than 1480 ml of water.
Assuming that there has been no change in the value of the standard
deviation, calculate the least mean volume in order to satisfy this
requirement. Give your answer to one decimal place. [3]

(ii) Sparkling spring water is supplied in packs of six 0.5-litre bottles. The actual
volume in a bottle may be modelled by a normal distribution with mean 508.5 ml
and standard deviation 3.5 ml.
Find the probability that the volume of water in each of the 6 bottles from a
randomly selected pack is more than 505 ml. [2]

(iii) Calculate the probability that the volume of 6 bottles of Sparkling spring water
in a randomly selected pack differs from twice the volume of one randomly
selected bottle of Aquafresh mineral water by less than 5.5 ml. [3]

(iv) The volume of tap water, V, used by a guest in a bathroom at a small hotel may
be modelled by a random variable with mean 120 litres and standard deviation
65 litres. Give a numerical justification as to why V is unlikely to be normally
distributed. [1]

Explain why V , the mean of a random sample of 30 observations of V, may be

assumed to be approximately normally distributed and state its distribution. [2]

End of Paper
 St Andrew’s Junior College
2018 Preliminary Examination
H2 Mathematics Paper 1 (9758/01) Solutions
1 x2 + 2 x − 3
≥0
( x 2 − 2 x + 11) ( x + 1)
x 2 − 2 x + 11
2 2
2  −2   −2 
= x − 2 x +   −   + 11
 2   2 
= ( x − 1) − 1 + 11
2

= ( x − 1) + 10 > 0 for all real values of x

2

x2 + 2 x − 3
Therefore, ≥0
( x 2 − 2 x + 11) ( x + 1)
x2 + 2x − 3
 ≥0
( x + 1)
( x + 3)( x − 1) ≥ 0, x ≠ −1
( x + 1)
Multiplying both sides by ( x + 1)
2

( x + 3)( x − 1)( x + 1) ≥ 0
Since x ≠ −1 ,

-3 -1 1

Hence,
−3 ≤ x < −1 or x ≥1
2 (i)
x 2 + 3x 2
y= = x +1−
x+2 x+2
dy  2 
= 1+  2 
> 0 for all x ∈ ¡
dx  ( x + 2) 
since (x + 2) 2 > 0 for all x, x ≠ −2.

Alternatively:

Page 1 of 16
 dy ( x + 2)(2 x + 3) − ( x 2 + 3x)(1)
=
dx ( x + 2) 2
x2 + 4x + 6
=
( x + 2) 2
( x + 2) 2 + 2
= >0
( x + 2) 2
Since ( x + 2) 2 ≥ 0 for all x ∈ ¡ ,
( x + 2) 2 + 2 > 0 for all x ∈ ¡
and ( x + 2) 2 > 0 for all x ∈ ¡ \{2},
Hence C has a positive gradient for all x ∈ ¡ .
2 (ii)
y y=x+1

(0,1)
(-1,0) ( 0, 0 ) x

Oblique asymptote: y = x + 1
Vertical asymptote: x = −2
3 (i) e− r − 2e− r +1 + e− r + 2
(
= e− r 1 − 2e + e2 )
2
=
( e − 1)
er
3 (ii) N
( e − 1)
2


r =1 e r +1
( e − 1)
2
N
=
r =1 e r e1
1 N ( e − 1)
2

= 
e r =1 e r
1 N
=  ( e − r − 2e − r +1 + e − r + 2 )
e r =1

Page 2 of 16
 1
= [e −1 − 2e0 + e1
e
+ e −2 − 2e −1 + e0
+ e −3 − 2e −2 + e−1
+ e −4 − 2e −3 + e−2
+ .........................
+ e − N + 2 − 2e − N +3 + e− N + 4
+ e − N +1 − 2e− N + 2 + e− N +3
+ e − N − 2e − N +1 + e− N + 2 ]
1
=
e
( e − 1 + e − N − e − N +1 )

1
= 1 − + e − N −1 − e − N
e

3 N +1
( e − 1)
2
N +1
( e − 1)
2
8
( e − 1)
2

(iii) 
r =9 er +1
=
r =1 e r +1
−
r =1 er +1
1  1 
= 1 − + e − N − 2 − e− N −1 − 1 − + e−9 − e −8 
e  e 
1 1
= e− N − 2 − e − N −1 − 9 + 8
e e
4 (i) y = f ( x)
= 1 + x + bx 2 + cx3 + ...
f ''(0) 2 f '''(0) 3
= f (0) + f '(0)x + x + x +…
2! 3!
Comparing,
f ''(0) f '''(0)
f (0) = 1, f '(0) = 1 , =b, =c
2! 3!
dy d2 y d3 y
When x = 0, y = 1, = 1, 2 = 2b, 3 = 6c.
dx dx dx
dy
(a − x2
dx
) (
= y 1 + 2 x − x2 )
( )
 a − 02 (1) = (1) 1 + 2 ( 0 ) − 02 ( )
 a = 1 (Shown)

When a = 1 , we have
dy
(1 − x2
dx
)
= y 1 + 2 x − x2( )
Differentiate w.r.t. x,

Page 3 of 16
 d2 y dy dy
(1− x 2

dx
)2
− 2x
dx
= y (2 − 2x) +
dx
(
1 + 2x − x2 )
2
d y dy
(1 − x2
dx
)2
= y (2 − 2x) +
dx
(
1 + 4 x − x 2 L (1))
Differentiate w.r.t. x,
3
2 d y d2 y dy
(1− x
dx
)3
− 2 x 2 = −2 y + ( 2 − 2 x )
dx dx
2
d y dy
( )
+ 2 1 + 4x − x2 + ( 4 − 2x )
dx dx
3
2 d y dy d2 y
(1− x
dx
)3
dx dx
(
= −2 y + ( 6 − 4 x ) + 2 1 + 6 x − x 2 L ( 2 ) )
dy
substitute x = 0, y = 1, = 1 into (1) and ( 2 ) ,
dx

d2 y
= 2(1) + 1 = 3
dx 2
d3 y
= −2(1) + 6(1) + 3 = 7
dx 3
Hence, from the given expansion,
f ''(0)
=b
2!
2b = 3
3
b=
2
f '''(0)
=c
3!
f '''(0) = 6c = 7
7
c=
6
3 2 7 3
Hence, the expansion is y = 1 + x + x + x + ...
2 6

Page 4 of 16
4 (ii)
y=
(e ) x 2

1 − x2
ex
= (e x )
(1 − x )
2

Using (i) and Standard Series of e x

 3 7  x 2 x3 
=  1 + x + x 2 + x 3 + ...   1 + x + + L 
 2 6  2 3! 
x 2 x3 1 3 3 7
≈ 1 + x + + + x + x 2 + x3 + x 2 + x3 + x3
2 3! 2 2 2 6
3
10 x
= 1 + 2 x + 3x 2 +
3

5 (i) Let ( a + bi ) = i
2

a 2 − b2 + 2abi = i
Comparing real and imaginary parts,
2ab = 1
a 2 − b2 = 0
and 1
a = b or − b ab =
2
At a = b,

1
a2 =
2
1
a=±
2
1
Hence, b = ±
2
For a = −b ,
1
a 2 = − has no solutions since a ∈ ¡
2
1 1 1 1
Hence the square roots are: + i or − − i
2 2 2 2

5 (ii) Given w2 + 2 w + (1 − 8i ) = 0 ,
−2 ± 4 − 4 (1 − 8i )
w=
2
−2 ± 2 1 − 1 + 8i
=
2
= −1 ± 2 2 ( i)
For w = −1 ± 2 2 ( i ) ,

Page 5 of 16
 2
 1 1 
At i =  ± ± i ,
 2 2 
2
 1 1  1 1
i=  + i = + i
 2 2  2 2
 1 1 
w = −1 ± 2 2  + i 
 2 2 
= −1 ± ( 2 + 2i )
= −1 + 2 + 2i or − 1 − 2 − 2i
= 1 + 2i or − 3 − 2i
5 (1 − 8i ) v 2 + 2iv − 1 = 0
(iii)
2i 1
(1 − 8i ) + − =0
v v2
1 2i
− + + (1 − 8i ) = 0
v2 v
i 2 2i
+ + (1 − 8i ) = 0
v2 v
2
i i
  + 2   + (1 − 8i ) = 0
v v

Comparing with w2 + 2 w + (1 − 8i ) = 0 in (ii),

i
we replace w in (ii) as   in (iii) for both roots in (i),
v
i i
= 1 + 2i = −3 − 2i
v v
i i
v= v=
1 + 2i −3 − 2i
and
(1 − 2i ) i ( −3 + 2i ) i
= 2
=
1+ 2 2
3 + 22
1 1
= (2 + i) = ( −2 − 3i )
5 13
6 (i) Since AC : CB is λ :1 − λ ,

A C B

Page 6 of 16
 By Ratio Theorem,
λb + (1 − λ ) a
c=
λ +1− λ
= λb + (1 − λ ) a
Since OC is perpendicular to OA,
c•a = 0
λb + (1 − λ ) a  • a = 0
λb • a + (1 − λ ) a • a = 0
4  4
λ  ×1× cos120° + (1 − λ ) a = 0, since a =
2

3  3
2
 4  1  4
λ  ×  −   + (1 − λ )   = 0
 3  2  3
 2 16
 −  λ + (1 − λ ) = 0
 3 9
2 16 16
− λ+ − λ =0
3 9 9
22 16
λ=
9 9
16 8
λ= =
22 11
6(ii) 3 8
c = a+ b
11 11
uuuur
By Mid-point Theorem, find OM = m , where M is the mid-
point of AC.
uuuur
OM = m
c+a
=
2
13 8 
=  a + b + a
2 11 11 
1  14 8 
=  a + b
2  11 11 
7 4
= a+ b
11 11

Page 7 of 16
6 7 4 
(iii) m •b =  a + b•b
 11 11 
7 4
= a•b + b•b
11 11
7 4 2
= a•b + b
11 11
7  2 4 2
=  −  + (1)
11  3  11
2
=− ≠0
33
Since the vector b is not perpendicular to m, where OM is the
radius of the circle, OB is not a tangent to the circle.
6 b • m is the length of projection of m on b.
(iv)
(b • m ) b is a vector with magnitude b • m , which is the
length of projection of m on b. Moreover, it is in the opposite
2
direction of b as m • b = − < 0 .
33
7 (i) 3
x = a cos t , y = a sin t
3

t = 0, x = a cos3 0 = a , y = a sin3 0 = 0
π π π
t = , x = a cos3 = 0 , y = a sin 3 = a
2 2 2

t = π , x = a cos3 π = −a , y = a sin3 π = 0

3π 3π 3π
t= , x = a cos3 = 0 , y = a sin 3 = −a
2 2 2

y
t= π/2

x
t=π t=0

t=3π/2

Page 8 of 16
7 (ii) dx
= 3a cos 2 t ( − sin t )
dt
dy
= 3a sin 2 t (cos t )
dt
dy dy 1
= ×
dx dt dx
dt
3a sin 2 t (cos t )
=
3a cos 2 t (− sin t )
dy − sin t
=
dx cos t
cos t
Gradient of normal =
sin t
At t = p, the equation of the normal is:
cos p
y − a sin 3 p = ( x − a cos3 p)
sin p
y sin p − a sin 4 p = x cos p − a cos4 p
y sin p = x cos p − a cos 4 p + a sin 4 p
y sin p = x cos p + a (sin 4 p − cos 4 p )
= x cos p + a (sin 2 p − cos 2 p )(sin 2 p + cos 2 p )
= x cos p + a ( − cos 2 p ) (1)
= x cos p − a cos 2 p (Shown)

7 π
At p = ,
(iii) 3
π π 2π
y sin = x cos − a(cos )
3 3 3
3 1  1
y = x −a− 
2 2  2
3 1 1
y = x+ a
2 2 2
3y = x + a
Since the normal meets the curve again,
3(a sin 3 t ) = (a cos3 t ) + a
3 sin3 t = cos3 t + 1 , a ≠ 0
Using GC, t = 2.32 or 3.14 (3 s.f)

Page 9 of 16
8 (i)

8 (ii) Area of R
= Area of quadrant-Area of triangle
b 1
= −  x dy − ab
0 2
b a2 y2 1
= − 2
− a − 2 dy − ab
0 b 2
1 b a2 y2
= − ab +  a2 − dy
2 0 b2

y = b cos θ
dy
= −b sin θ
dθ

1 b a2 y2
Area= − ab +  a2 − dy
2 0 b2
1 0 a 2b 2 cos 2 θ
= − ab +  π a 2 − (−b sin θ ) dθ
2 2 b2
1 0
= − ab +  π − ab sin 2 θ dθ
2 2

1 0  1 − cos 2θ 
= − ab − ab  π   dθ
2 2  2 
0
1 1 sin 2θ 
= − ab − ab  θ −
2 2 4  π
2

1   π 
= − ab − (ab)  −   
2   4 
π 1 
= ab  −  units 2
 4 2

Page 10 of 16
8 Volume generated
(iii) b 1
= π  x 2 dy − π a 2b
0 3
b a y 
2 2
1
= π   a 2 − 2  dy − π a 2b
0
 b  3
b
 2 a2 y3  1 2
= π a y − 2  − π a b
 3b  0 3
 a 2b3  1 2
= π  a 2b − 2  − 3π a b
 3b 
 a 2b  1 2
= π  a 2b − − πa b
 3  3
2  1
= π  a 2b  − π a 2b
3  3
1
= π a 2b units3
3
9(i) Since z 2 − 3z + 9 = 0 has all real coefficients, given that
π π
i −i
z = 3e 3 is a root of the equation, z = 3e 3
is the other root of
the equation.
9(ii) e iθ − e − iθ
= ( cos θ + i sin θ ) −  cos ( −θ ) + i sin ( −θ ) 
= ( cos θ + i sin θ ) − ( cos θ − i sin θ )
= 2i sin θ

9  π
i −  i
π

(iii) Since w1 = 3e  3
, w2 = 3e 9

Page 11 of 16
 w2 − w1
π  π
i  i − 
= 3e 9
− 3e  3

π  3π 
i  i − 
= 3e  9  − 3e  9 

 2π π   2π π 
i −  i − − 
= 3e  9 9
− 3e  9 9

 π
i −   i 2π9  i − 2π9  
= 3e  9
e   − e   
 
 π
i −    2π  
= 3e  9
 2i sin  9  
  
 π π
 2 π  i − + 
= 6sin   e  9 2 
 9 
 7π 
 2 π  i 
= 6sin   e  18 
 9 

9  2π   7π 
(iv) At point B, OB = 6sin   cos  
 9   18 

Hence,
Area of triangle OAB

Page 12 of 16
 1  7π 
= OB OA sin  
2  18 
1  2π   7π     2π    7 π 
=  6sin   cos    6sin    sin  
2  9   18     9    18 
36 2  2π   7π   7π 
= sin   sin   cos  
2  9   18   18 
  14π  
sin
36 2  2π    18  
= sin    
2  9  2 
 
 2π   7π 
= 9sin 2   sin  
 9   9 

10 Yea n Amount of loan at Amount of loan at the end

(i) r the beginning of nth of nth year
year
201 1 100000 (100000 − 12 x)(1.05)
8
201 2 (100000 − 12 x)(1.05) [ (100000−12x)(1.05) −12x] (1.05)
9
=100000(1.05) −12x(1.05)
2 2

−12x(1.05)
M MM M

n 100000(1.05) n − 12 x (1.05) n
−12 x(1.05) n −1 − ...
−12 x (1.05)

Amount of loan left at the end of n years

= 100000(1.05) n − 12 x(1.05) n − 12 x(1.05) n −1 − ... − 12 x(1.05)
= 100000(1.05) n − 12 x(1.05) 1 + 1.05 + 1.052 + ... + 1.05n −1 
1.05n − 1 
= 100000(1.05) n − 12 x(1.05)  
 1.05 − 1 
= 100000(1.05) n − 252 x(1.05n − 1)

10 (1.05)10 (100000) − 252 x (1.0510 − 1) ≤ 0

(ii)
x  252(1.0510 − 1)  ≥ (1.05)10 (100000)
x ≥ 1027.81

Page 13 of 16
 Hence, minimum value of x = $1028 (to the nearest dollar)
10 Amount of pay-out per year
(iii)  r 
= 50000  
 100 
=500r

Year n Amt left on 1 Jan Amt owed on 31 Dec of

of nth year nth year
2018 1 100000 − 500r [100000 − 500r ] (1.05 )
2019 2 [100000 − 500r ] (1.05 ) [100000 − 500r ] (1.05 )2
−500r −500 r (1.05 )
…
2027 10 [100000−500r] (1.05)
10

−500r (1.05)
9

−500r (1.05) ...

8

−500r (1.05) −500r (1.05)

2

=100000(1.05)
10

−500r (1.05) 1+1.05+L +1.059 

Amount owed on 31 December of 10th year

= 100000 (1.05 ) − 500 r (1.05 ) − 500 r (1.05 )
10 10 9

− 500 r (1.05 ) ... − 500 r (1.05 ) − 500 r (1.05 )

8 2

= 100000 (1.05 ) − 500 r (1.05 ) 1 + 1.05 + L + 1.059 

10

 1 (1.0510 − 1) 
= 100000 (1.05 ) − 500 r (1.05 ) 
10

 1.05 − 1 
 (1.0510 − 1) 
= 100000 (1.05 ) − 500 r (1.05 ) 
10

 0.05 
= 100000 (1.05 ) − 10500 r (1.0510 − 1)
10

For her to have completed paying,

100000 (1.05) − 10500r (1.0510 − 1) ≤ 0
10

r ≥ 24.668 = 24.7 (to 3 sf)

11 V = Ah
(i) dV dh
=A
dt dt

On the other hand,

Page 14 of 16
 dV dVin dVout
= −
dt dt dt
= 0−k h , where k is a positive constant
= −k h

Equating the expressions,

dh
−k h = A
dt
dh −k h
=
dt A

11 dh − k h
(ii) =
dt A
1 k
 h dh = − A  d t
k
2 h = − t + C ---(I)
A
When t = 0, h =81,
∴ C = 18.
k
 2 h = − t + 18
A
dh
When t = 0, = −0.3
dt

dh − k h
= =-0.3 --- (II)
dt A

k
− (9) = −0.3
A
k 1
=
A 30

1
∴2 h = − t + 18
30
1
h = − t + 9 --- (III)
60
1
 h = (9 − t ) 2
60
11 If Joe is to be back before the water tank to be emptied,
(iii) h≥ 0
1
(9 − t ) 2 ≥ 0
60
Equivalently, using (III),

Page 15 of 16
 1
h =− t +9 ≥ 0
60
t
− ≥ −9
60
t
≤9
60
t ≤ 540
540
Number of days he can be away ≤ = 22.5
24
Maximum number of days = 22.
11 At the end of 4th day, 96 hours have lapsed.
(iv) 1
h = (9 − g96) 2 = (7.4) 2
60
Method (1):
A = 400 π given that the radius is 20 cm.
k 1
Using =
A 30
400 π
k=
30
40
k = π
3
dV
= −k h
dt
40π
= − (7.4)
3
296
=− π cm3 /hour
3
296
The water is being delivered at a rate of − π cm3/hour.
3
Method (2):
d V d V dh
= g
dt dh d t
dV
= A = 400 π
dh
dh − k h 1
= = - (7.4)
dt A 30

dV 1 296
= (400π )( − (7.4)) = − π cm3 /hour.
dt 30 3

Page 16 of 16
 St Andrew’s Junior College
2018 Preliminary Examination
H2 Mathematics Paper 2 (9758/02) Solutions
Section A
Qn Solution
1 b
y = a ( x − 1) +
x+c
↓ A (replace y with − y )
b
− y = a ( x − 1) +
x+c
b
y = −a ( x − 1) −
x+c
↓ B (replace x with x + 1)
b
y = −a ( x + 1 − 1) −
x +1+ c
b
y = −ax −
x +1+ c

Since x = 0 is a vertical asymptote, 1 + c = 0

c = −1
b
Therefore, y = − ax − = f ( x) .
x

 1 1
Since 1,  is a turning point on y = , (1, 6 ) is a
 6 f ( x)
turning point on y = f ( x) .

−a − b = 6 --- (1)

dy b
= −a + 2
dx x
dy
Since when x = 1 , = 0,
dx

−a + b = 0 --- (2)

Solving (1), (2) using a GC,

a = −3, b = −3

Therefore, a = −3, b = −3 and c = −1

2 π
Given that AC is the diameter, ∠ABC = .
2

Page 1 of 17
 C

2r

B A

AB = 2r cos θ , BC = 2r sin θ
Let S be the area of triangle ABC.
1
S = ( AB )( BC )
2
1
= ( 2r cos θ )( 2r sin θ )
2
= 2r 2 sin θ cos θ
= r 2 sin 2θ
dS
= r 2 ( 2 cos 2θ )
dθ
dS
For stationary values of S, =0
dθ
Since 2r 2 ≠ 0 ,
cos 2θ = 0
π
Since 0 < θ <  0 < 2θ < π
2
Hence,
π
2θ =
2
π
θ =
4
2
dS
= 2r 2 ( −2sin 2θ ) = −4r 2 sin 2θ
dθ 2

π
At θ = ,
4
2
d S π
= −4r 2 sin 2  
dθ 2
4
= −4r 2 < 0
π
Hence S is maximum when θ = .
4

Page 2 of 17
3 (i)
sin θ + m cos θ = 1 + m 2 sin (θ + tan −1 m )

∴ R = 1 + m 2 , α = tan −1 m

2(sin θ + m cos θ ) sin(θ − α )

= R [ 2sin(θ + α ) sin(θ − α ) ]
= R [ −(cos 2θ − cos 2α ) ]
= R ( cos 2α − cos 2θ )
3 cos 2θ
(ii)  ( sin θ + m cos θ ) sin(θ − α ) dθ

cos 2θ
= 2 dθ
2( sin θ + m cos θ )sin(θ − α )
2 cos 2θ
2  ( cos(2α ) − cos(2θ ))
= dθ
1+ m
−2 cos 2θ π
=
1+ m 2  (cos2θ + 1) dθ since cos(2α ) = −1 when α =
2
−2  2 cos 2 θ − 1 
= 
1 + m 2  2 cos θ 
2  dθ

−2  1 
2   dθ
= 1−
1 + m  2 cos θ 
2

−2  1 
2 
= 1dθ −  sec2 θ dθ 
1+ m  2 
−2 1 
= θ − 2 tan θ  + C ,
2
1+ m
where C is an arbitrary constant

Page 3 of 17
4(i) y

x
(2, 0)

x=1
From the graph, the solution is 1 < x ≤ 2
Alternative
ln ( x − 1) ≤ 0
0 < x −1 ≤ 1
1< x ≤ 2
Since ln ( x − 1) is defined for x > 1 , the solution is 1 < x ≤ 2 .

4 Let y = ln ( x − 1)
(ii)
Since 1 < x ≤ 2 , y = − ln ( x − 1)
− y = ln ( x − 1)
x − 1 = e− y
x = 1 + e− y
f −1 ( x ) = 1 + e− x
Df −1 = R f = [0, ∞)

Alternative:
y = ln ( x − 1)
± y = ln( x − 1)
e± y = x − 1
e − y = x − 1 (since y ≥ 0 and 1 < x ≤ 2)
∴ x = e− y + 1
f −1 ( x ) = 1 + e − x
Df −1 = R f = [0, ∞)

Page 4 of 17
4 y
(iii)
y=x

(0, 2) (1.2785,1.2785)
y = f −1 ( x ) y=1

x
(2,0)

x=1

4 Point of intersection of y = f −1 ( x) and y = f ( x) is

(iv) (1.28, 1.28)
f −1 ( x) < f ( x)
From the graph, 1 < x ≤ 1.27 .
4 fg ( a ) = 3
(v)
g ( a ) = f −1 ( 3)
4
1+ 2
= 1 + e−3
4a + 5
4 1
2
= 3
4a + 5 e
4e3 = 4a 2 + 5
4e3 − 5
a2 =
4
4e3 − 5 4e3 − 5
a= or a = − (reject since Dfg =D g =¡ + )
4 4
5 (i) 11  −2 
   
l1 : r =  6  + λ  0  , λ ∈ ¡
0  1 
   
4  2 
   
l2 : r =  0  + μ  − 3  , μ ∈ ¡
 −1  4 
   

Condition 1:
 −2  2 
   
 0  ≠ k  −3  for any real values of k.
1  4 
   

Page 5 of 17
 Hence lines l1 and l2 are not parallel to each other.

Condition 2:
Suppose both lines intersect,

11 − 2λ   4 + 2μ 
   
6  =  −3μ 
λ   −1 + 4μ 
   
11 − 2λ = 4 + 2μ

6 = −3μ
λ = −1 + 4μ

2μ + 2λ = 7 − − − (1)

 6 = −3μ − − − (2)
−4μ + λ = −1 − − − (3)

Using G.C. to solve (1) and (3),
9 13
μ= , λ=
10 5
Checking with (2),
9
−3μ = −3  
 10 
27
=− ≠6
10
Hence, the two lines do not intersect at any unique points.

Combining both conditions (1) and (2), lines l1 and l2 are

skew lines.

5  11  1 
(ii) uuur   uuur  
OA =  6  , OB =  6 
0  5
   
Since P is on p1 such that it is equidistant from A and B,
ABP forms an isosceles triangle.

Let the mid-point of AB be M.

By mid-point theorem,

Page 6 of 17
 uuur uuur
uuuur OA + OB
OM =
2
 11  1 
   
6  + 6
0  5
=   
2
 
6 
 
= 6 
5
 
2

To find the foot of perpendicular, P, of M on p1 ,

 
6   4
uuru    
OP =  6  + s  3  , for some s ∈ ¡
 5  8 
   
2
 
 6 + 4s 
 
=  6 + 3s 
5 
 + 8s 
2 
 4
Given p1 : r •  3  = 240
8 
 

Since P lies on p1 ,
 
 6 + 4s   4 
   
 6 + 3s  •  3  = 240
5  8 
 + 8s   
2 
5 
4 ( 6 + 4 s ) + 3 ( 6 + 3s ) + 8  + 8s  = 240
2 
24 + 16 s + 18 + 9 s + 20 + 64 s = 240
62 + 89 s = 240
89 s = 178
178
s= =2
89

Page 7 of 17
 Hence,
 
 6 + 4(2) 
uuur  
OP =  6 + 3(2) 
5 
 + 8(2) 
2 
 
 14 
 
=  12 
 37 
 
 2 

5 Normal vector of p2
(iii)
 −2   2 
   
=  0  ×  −3 
1   4 
   
 0 − (−3) 
 
=  −(−8 − 2) 
6−0 
 
3 
 
= 10 
6 
 

Since the plane p2 contains the point P,

 
  14   3 
3
     
r • 10  = 12  • 10 
 6   37   6 
     
 2 
= 273

The Cartesian equation of p2 is:

3x + 10y + 6z = 273

Section B

6 Case 1: Committee only has 1 woman

Page 8 of 17
No. of ways = 15C7 7C1 = 45045
Case 2: Committee has no women
No. of ways = 15C8 = 6435
Hence, total number of ways
= 22C8 − 45045 − 6435 = 268290

ALT

Case 1: Committee only has 2 women

No. of ways = 15C6 7C2 = 105105

Case 2: Committee only has 3 women

No. of ways = 15C5 7C3 = 105105

Case 3: Committee only has 4 women

No. of ways = 15C4 7C4 = 47775

Case 4: Committee only has 5 women

No. of ways = 15C3 7C5 = 9555

Case 5: Committee only has 6 women

No. of ways = 15C2 7C6 = 735

Case 6: Committee only has 7 women

No. of ways = 15C1 7C7 = 15

Hence, total number of ways

= 105105 + 105105 + 47775 + 9555 + 735 + 15
= 268290

Number of ways to choose the 4 corner seats = 4C 2

Number of ways to arrange the remaining 6 committee
members = 6!
Number of ways to arrange the Gina and Hazel = 2!
Hence, total number of ways = 4C2 × 2!× 6! = 8640

Page 9 of 17
7 (i) Let W denote the number of shots that hits the bullseye 40m
away from the target out of 18 shots.

2
W ~ B(18, (95 − 40))
195
22
W ~ B(18, )
39

P (W > 6 W ≤ 10 )
P ([W > 6] ∩ [W ≤ 10])
=
P(W ≤ 10)
P ( 7 ≤ W ≤ 10 )
=
P(W ≤ 10)
P (W ≤ 10 ) − P (W ≤ 6 )
=
P(W ≤ 10)
0.51941
=
0.56103
= 0.926 (to 3 s.f.)

Page 10 of 17
7(ii) Let Y denote the number of shots that hits the bullseye x m
away from the target out of 18 shots.
2
Y ~ B (18, (95 − x))
195
 190 2 
Y ~ B (18,  − x )
 195 195 

P(Y ≥ 2) = 0.98
1− P(Y = 0) − P(Y =1) = 0.98
P(Y = 0) + P(Y =1) = 0.02
18
  190 2 
1−  195 − 195 x 
  
18 190 2 1   190 2 17 
+  C1 × − x  × 1−  − x   = 0.02
 195 195    195 195  

 1 2  18 190 2   1 2  
18 1 17

 + x +
  1 C × − x  × + x   = 0.02
 39 195    195 195   39 195  
Using the GC,
x = 67.3 m (to 3 s.f.)

8 (i)

0.65 B
0.1

0.25

2
P ( B | A ') =
7
P ( B ∩ A ') 2
=
P ( A ') 7
2
P ( B ∩ A ') = (1 − 0.65)
7
= 0.1

P ( B ) =P ( B ∩ A ' )
= 0.1 ( shown )

8
(ii)
C
Page 11 of 17
0.39
 P (C ' | A)
P ( A ∩ C ')
=
P ( A)
P ( A) − P ( A ∩ C )
=
P ( A)
0.65 − 0.39
=
0.65
= 0.4

8
(iii)
C
A
0.15
0.26
B
x
0.1 – x
0.1

Let P ( B ∩ C ) = x
When B and C are independent,
P ( B ∩ C ) = P ( B) × P (C )
x = 0.1× ( 0.39 + 0.15 + x )
x = 0.054 + 0.1x
0.9 x = 0.054
x = 0.06
∴ P ( B ∩ C ) = 0.06

8 P ( C ) = 0.39 + 0.15 + 0.06 = 0.6

(iv)
Since P ( A ) × P ( C ) = ( 0.65 )( 0.6 ) = 0.39 = P ( A ∩ C )
(or P ( C ' ) = 1 − P ( C ) = 0.4 = P ( C ' | A ) )
A and C are independent.

9 (i) Let X be the amount of winnings Sally gets after one game in
dollars.

Page 12 of 17
Let the random variables F be the scores on the fair dice.

P( X = −5)
= P ( F − Y > 3)
= P [ F = 5, Y = 1] + P [ F = 6, Y = 1] +
P [ F = 6, Y = 2] + P [ F = 1, Y = 5]
+ P [ F = 1, Y = 6] + P [ F = 2, Y = 6]
1 1 1 1 1 1 
= × + × + × 
 6 6   6 6   6 18 
1 1 1 5  1 5 
+ × + ×  + × 
 6 6   6 18   6 18 
5
=
27

P( X = 0)
= P [ F = 1, Y = 1] + P [ F = 2, Y = 2] +
P [ F = 3, Y = 3] + P [ F = 4, Y = 4]
+ P [ F = 5, Y = 5] + P [ F = 6, Y = 6]
11 1 1 3 1 5 
=  + + + + + 
6  6 18 6 18 6 18 
1
=
6

P(X = 3) = 1 – P(X=0) – P(X=-5)

5 1 35
= 1− − =
27 6 54
x −5 0 3
5 1 35
P( X = x)
27 6 54

E( X )
=  xP( X = x)
all x

 5  1  35 
= −5   + 0   + ( 3 )  
 27   6   54 
55
=
54

Page 13 of 17
9(ii) E( X 2 )
=  x 2 P( X = x)
all x

2 5  21 2  35 
= ( −5 )   + ( 0 )   + ( 3)  
 27  6  54 
565
=
54

Var(X )
=E( X 2 ) − [ E( X ) ]
2

2
565  55 
= − 
54  54 
27485
=
2916

Since the number of games of 50 is sufficiently large, by

Central Limit Theorem,
  55   27485  
X 1 + X 2 + L + X 50 ~ N  50   ,50    approximately
  54   2916  
P ( X 1 + X 2 + L + X 50 ≥ 65 )
= 0.25839
= 0.258 (to 3 sf)

10 20 + 22 + 24 + ... + 34
(a) x= = 27 and
8
16 + 21 + a + 24 + 22 + 24 + 27 + 20
y=
8
154 + a
= .
8
Since ( x , y ) lies on the regression line,
154 + a
= 8 + 0.5(27)
8
a = 18

(b)
(i)

Page 14 of 17
 y/m

(1,2.74)

0 x/years

(ii) The height of the tree increases by b metres every year.

(iii) Let the product moment correlation coefficient for this
model be r.

For A, rA = 0.958 , For B, rB = 0.972 and for C, rC = 0.996

Since rC = 0.996 is closest to 1, the most appropriate model

is Model C.

Equation required is y = 2.78 + 0.903ln x

a = 2.78 and b = 0.903

(iv) Since y = 4.25 is within the given range of values

[2.74, 4.75], the estimate obtained via interpolation is
reliable.
Since rC = 0.996 is close to 1 which suggests a strong linear
correlation between lnx and y , it would be reasonable to use
Model C.

11 Let X be the random variable “ speed of a vehicle travelling

(i) along Spring Avenue, measured in km/hour”.

The unbiased estimate of population mean = x =

 x = 4327.0 = 43.27
100 100

The unbiased estimate of population variance,

Page 15 of 17
 s2 =
 ( x − x) 2
925.71
= = 9.3506 = 9.35 (to 3 sf)
n −1 99
(ii) The 100 vehicles are selected independently.
(iii) Let μ be the population mean speed
Test H0: μ = 44.1
H1: μ < 44.1
9.3506
Under H0, X ~ N (44.1, ) approximately by Central
100
Limit Theorem, since sample size 100 is sufficiently large.

Carry out one-tailed z-test at 5% significance level.

x = 43.27 gives rise to test statistic z = -2.71 and p-value

0.00332.
Since p-value ≤ 0.05, we reject H0 and conclude that at the
5% significance level, there is sufficient evidence that the
mean speed of vehicles has been reduced after life sized
police photos have been erected along Spring Avenue.

(iv) 5% significance level refers to the probability of 0.05 of

wrongly concluding that the mean speed of vehicles is less
than 44.1 km/hour when it is in fact 44.1 km/hour.

(v) New unbiased estimate of population mean = 43.27k

30857 2
New unbiased estimate of population variance = k
3300
Test H0: μ = 44.1
H1: μ < 44.1
30857
Under H0, X ~ N (44.1, 3300 k 2 ) approximately by
100
Central Limit Theorem, since sample size 100 is sufficiently
large.

To have the same conclusion, i.e. to reject H0,

zcal ≤ −1.64485
(43.27 k ) − 44.1
≤ −1.64485
30857
(k ) 10
3300
30857
1.64485
43.2k − 44.1 ≤ − 3300 k
10
k ≤ 1.01
Since k > 0, 0 < k ≤ 1.01
Greatest possible value of k is 1.01.

Page 16 of 17
12 Let X be the random variable “ volume of Aquafresh mineral
(i)(a water in a bottle in ml”
) X ~ N(1505, 10.22)
P(X > 1480) = 0.993 from G.C.
(i)(b Let μ1 be the new mean volume.
) P( X < 1480) < 0.10
1480 − μ1
P( Z < ) < 0.10
10.2
1480 − μ1
< −1.28155
10.2
μ1 > 1493.07

The least mean volume is 1493.1 ml.

(ii) Let Y be the random variable “ volume of Sparkling spring
water in a bottle in ml”
Y ~ N(508.5, 3.52)
P(the volume of water in each of the 6 bottles from a
randomly selected pack is more than 505 ml)
= [P( Y > 505)]6
= [0.84134]6
= 0.355 (3 sig figures)
(iii) Let T = Y1 + Y2 + ... + Y6 − 2 X
T ~ N (41, 489.66)
P (| T |< 5.5) = P( −5.5 < T < 5.5) = 0.0365

(iv) Let V be the volume of water used by a guest in a bathroom

in litres.
V ~ N(120, 652)
P(V < 0) = 0.0324 which is impossible as the volume of
water used cannot be negative.
Since sample size 30 is sufficiently large, at least 20,
Central Limit Theorem can be applied.
652
∴V ~ N (120, ) approximately.
30

Page 17 of 17