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ANDERSON JUNIOR COLLEGE
JC2 Preliminary Examinations 2015
MATHEMATICS
8864/01
Higher 1
15 September 2015
Paper 1
2.00pm-5.00pm
Additional Materials:
Graph Paper
List of Formulae
Answer all questions.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place
in the case of angles in degrees, unless a different level of accuracy is specified in the
question.
You are expected to use a graphic calculator.
Unsupported answers from a graphic calculator are allowed unless a question
specifically states otherwise.
When unsupported answers from a graphic calculator are not allowed in a question, you
are required to present the mathematical steps using mathematical notations and not
calculator commands.
You are reminded of the need for clear presentation in your answers.
The number of marks is given in [ ] at the end of each question or part question.
1
2
3
4
7
8
9
10
5
11
6
TOTAL
________________________________________________________________________
This Question paper consists of 6 printed pages
1
Pure Mathematics [35 marks]
1.
A straight line has equation x = 3y + k , where k is a constant. Show that the
y-coordinates of the points of intersection of the line and the curve x2  y 2  8 y  9
satisfy the equation 10 y 2  (6k  8) y  (k 2  9)  0 .
[1]
Hence determine the exact range of values of k for which the line intersects the curve.
[3]
2.
(a)
Differentiate ln(5  x 2 )
(b)
Find
[1]
d 2kx2
(e
) , where k is a constant.
dx
Hence or otherwise, find the exact value of
3.

0
2
xe2 x dx .
2
[4]
To store a special liquid chemical, a manufacturer constructs a closed cylindrical
container of negligible thickness with base radius r cm and an internal surface area
of 96  cm2.
(i) Show that the volume of liquid chemical the container can store as r varies, is


given by V   r 48  r 2 cm3 .
[1]
(ii) Using a non-calculator method, find the maximum value of V.
[3]
The manufacturer decides to construct the closed cylindrical container with a fixed
radius of k cm. When liquid chemical is poured into the container at a constant rate
of 2 cm3s1, the height of the liquid chemical in the container increases at a constant
rate of 0.5 cms1. Find k in terms of  .
[3]
4.
The curve C for which
dy
A

has a tangent 6x + y + 6 = 0 at the point where
dx ( x  2)2
the x-coordinate is 1.
(i)
(ii)
(iii)
(iv)
 1

 1 .
Show that the equation of the curve is y = 6 
[3]
 x2 
Sketch the curve, stating its equation(s) of asymptotes and axial intercept(s).
[2]
Write down the equation of the normal to the curve x  1 and draw it on the
[1]
same axes as the curve.
Find the exact area of the finite region bounded by the curve, the normal in (iii)
and the y-axis.
[3]
2
5
(i)
Find the exact value(s) of x for which 10e2 x  8e4 x  3 .
(ii)
Sketch the curve C with equation y  10e2 x  8e4 x  3 , indicating clearly the
turning point(s) , the exact coordinates of any axial intercepts and equation of any
asymptote(s).
State the exact range of values of x for which 10e2 x  8e4 x  3
[3]
[3]
(iii) By inserting a suitable line in your diagram, solve 10e2 x  8e4 x  3  x  5 .
[2]
Write down as an integral an expression for the area of the region bounded by the
curve y  10e2 x  8e4 x  3 and the above line. Evaluate this integral, giving your
answer correct to three decimal places.
[2]
Probability and Statistics [60 marks]
6.
An amateur weather forecaster describes each day as sunny, cloudy or wet. He keeps a
record each day of his forecast and of the actual weather. His results for one particular
year are as given in the table.
Actual
Weather
Sunny
Cloudy
Wet
Total
Sunny
55
15
5
75
Weather Forecast
Cloudy
Wet
14
5
128
31
31
81
173
117
A day is selected at random from that year.
X is the event that the forecast is correct.
S is the event that the forecast is sunny.
W is the event that the actual weather is wet.
264
(i) Show that the P( X ) 
365
(ii) Find the following probabilities:
P( S  W )
(a)
P( X ' | S )
(b)
(iii) Explain why the events X and S are not independent.
Total
74
174
117
365
[1]
[1]
[1]
[1]
The forecaster had forecasted three consecutive sunny days from 10th December to
12th December in that year. Find the probability that in those three days, there were
two sunny days and one wet day, giving your answers to three significant figures. [2]
3
7.
In a company manufacturing uPhones, it is found that 6% of the uPhones manufactured
have a defective touchscreen. The uPhones are randomly packed in boxes of 20.
(i)
An employee inspects a box of uPhones. Find the probability that
(a) only the last uPhone inspected have a defective touchscreen,
(b) exactly one of the uPhones have a defective touchscreen.
Explain why the answer to part (a) is smaller than the answer to part (b).
[1]
[1]
[1]
50% of uPhones that have defective touchscreens actually have cracked touchscreens.
(ii)
Show that there is a probability of 0.120 that a randomly selected box will have
more than one uPhone with a cracked touchscreen.
[2]
Shop A buys a batch of sixty boxes of uPhones.
(iii) Using a suitable approximation, estimate the probability that at least 20% of the
boxes in the batch have more than one uPhone with a cracked touchscreen.
[4]
(iv) The mean number of uPhones with cracked touchscreens per box for the batch is
denoted by A. Write down the distribution for A, stating clearly its mean and
[2]
variance.
8 (a)
A group of students wanting to investigate the electricity consumption pattern of
households in Singapore, decide to obtain a sample of 55 households.
(i) One student suggests using simple random sampling to obtain the sample as
every household would then have the same chance of being selected. Another
student said that they should use stratified sampling.
State one advantage of using stratified sampling for their research.
[1]
(ii) The group finally decide on quota sampling to obtain their sample. Describe how
they can obtain their quota sample.
[2]
State a practical advantage of quota sampling in this context.
[1]
(b) In a certain housing estate, the amount of electricity consumed by a household in a
month, X kilowatt hours (kWh), is normally distributed. The town council manager of
the housing estate claims that the amount of electricity consumed in a month by a
household in the estate is 120 kWh. To test his claim, a group of students took a
random sample of 55 households from the estate, and the result is summarised by
  x  120  725 and   x  120
2
 99801 .
(i) Calculate the unbiased estimates of the population mean and the population
variance, giving your answers to 1 decimal place.
[2]
(ii) The test conducted at the  % significance level shows that there is sufficient
evidence to support the town council manager’s claim. Find the set of values of  ,
giving your answer to one decimal place.
[3]
(iii) In their report, the students concluded that based on the sample studied, the
average amount of electricity consumed in a month by a household in Singapore
is 120kWh. Explain with reasons, whether you agree with their conclusion. [1]
4
9.
In a certain shop, the daily demand of Avonly cheese follow independent normal
distributions. On a weekday, the mean is 3.5 kg and the standard deviation is  kg.
(i)
If there is a 96% chance that the demand of Avonly cheese on a weekday is
[2]
less than 4.55 kg, show that   0.6 .
On weekends, the daily demand for Avonly cheese is 20% more than the daily
demand on a weekday (both the mean and standard deviation are 20% more). The
shop operates seven days in a week all year round. Saturdays and Sundays are
considered weekends. Each week, the shop orders k blocks of Avonly cheese, each
weighing 4 kg, from the producer. No cheese is kept by the shopkeeper from one
week to the next.
Use   0.6 for the following parts of the question.
(ii)
(iii)
If k  7 , find the probability that in a randomly selected week, the shop’s stock
of Avonly cheese will be sold out.
[4]
There is a 65% chance that an employee will make pasta with the leftover
cheese at the end of the week. Estimate the number of weeks in a year that the
[2]
employee will make pasta with the leftover cheese.
The shop sells the Avonly cheese at $5.00 per 100 gm on a weekday and at $5.20 per
100 gm on a weekend. Assume that k is large such that the shop is able to meet the
demand from the customers.
(iv)
10.
Find the probability that the revenue from Avonly cheese sold in the first week
of January is within $100 of the revenue from Avonly cheese sold in the
second week of January.
[4]
A company sells detergent. The population variance of the volume of detergent in the
bottles is 120.1 ml2.
(i) A manager claims that the volume of detergent in the bottles of detergents is at
least 1000ml. A random sample of 100 bottles is taken, and the sample mean in this
sample is 998 ml.
Test at 5% significance level, whether the manager’s claim is valid.
[4]
(ii) For a particular batch, the quality control manager suspects that the average
volume of detergent in the bottles is less than 1000 ml. A random sample of 100
bottles from this batch gives a result of  x  k , where x is the volume of
detergent in a bottle. A test at the 4% significance level indicates that the quality
control manager’s suspicion is not justified. Find the smallest integer value of k.
[3]
(iii) The company introduced a deluxe range of detergent, packed in smaller bottles of
o ml for convenience. The volume of detergent in the bottles for this deluxe range
is normally distributed and the population standard deviation is 10.2 ml. From a
random sample of 60 bottles of this deluxe range, the sample mean is 690 ml. A
test at the 2 % significance level suggests that the company is understating the
average mass. Find largest possible value of 0 to the nearest integer.
[4]
5
11.
A study on the annual household income, x thousand dollars, and the amount the
household spent on insurance in the year, y hundred dollars, is shown below.
x
10
12
14
16
18
20
22
y
2.3
2.8
3.1
3.2
3.3
4.0
5.0
(i) Give a sketch of the scatter diagram for the data as shown on your calculator. [1]
(ii) Calculate the product moment correlation coefficient to 3 significant figures, and
comment on its value in the context of the data.
[2]
(iii) Find the equation of the regression line of y on x, in the form y  mx  c giving
the values of m and c, correct to 4 significant figures.
[1]
[1]
Sketch this line on your scatter diagram.
Explain the meaning of m in this context.
[1]
(iv) Find the equation of the regression line of x on y, giving your answers to 3
[1]
significant figures.
(v) Use an appropriate regression line to estimate the amount that a household with an
annual household income of $21000 would spend on insurance. Comment on the
[2]
reliability of your estimate.
In the annual budget, the government contributes $100 for each household to buy
insurance and each household fully utilises this amount to buy additional insurance on
top of their original insurance.
Without any further calculations on your graphic calculator, state with reasons, how
[1]
this will affect the values of m and c in (iii).
- - - - - - - END OF PAPER - - - - - - - - -
6
1.
Sub. x = 3y + k into x2  y 2  8 y  9 :
 3y  k 
2
 y2  8 y  9  0
9 y 2  6ky  k 2  y 2  8 y  9  0
10 y 2  (6k  8) y  (k 2  9)  0( shown)
For line to intersect curve, D ≥ 0
(6k  8) 2  4(10)(k 2  9)  0
4k 2  96k  424  0
k 2  24k  106  0
To find critical value: Let k  24k  106 = 0
2
24  242  4( 106) 24  1000
k

2
2
24  10 10

 12  5 10
2
For k  24k  106  0 ,
2
12  5 10  k  12  5 10
2(a).
Let y = ln


10 x  x 2 
1
1
ln( x(10  x))  ln x  ln(10  x)
2
2
dy 1  1
1 
  
.
dx 2  x 10  x 
2b.
2
d 2kx2
(e
) = 2kx e2kx .
dx
When k = 1,

2
0
xe2 x dx =    2 xe2 x dx 
0
2
2
2
1
  e 2 x 
0
2
1
1
   e0  e2    e2  1
2
2
3(i).
96  2 rh  2 r 2
h
96  2 r 2 48  r 2

2 r
r
 48  r 2 
V   r 2h   r 2 

 r 
  r (48  r 2 )
V   (48r  r 3 )
1
2
2
2

3(ii).
dV
  (48  3r 2 )
dr
At stationary points,
dV
  (48  3r 2 )  0
dr
48
 16
3
r  4(  0)
r2 
r
4
4
4+
dV
dr
>0
0
<0
tangent
/

\
(V max)
Max V   (48  4  3  42 )  144
V =  k 2 h ( k fixed)
dV
 k2
dh
dV dV dh
dh


 k2
dt
dh dt
dt
1
2  k2  
2
4
2
(k  0)
k2   k 

4(i).

dy
A

dx ( x  2)2
At x = -1, equation of tangent is y  6 x  6 (gradient = 6)
A
= 6  A  6
(1  2) 2
dy
6

dx ( x  2) 2
6
y
dx  6 ( x  2) 2 dx
2
( x  2)
6

C
( x  2)
6
When x = -1 , y = 0, 0 =
 C  C = 6
(1  2)
y=
6
 1

6 = 6 
 1 .
( x  2)
 x2 
(shown)
4(ii)
(iii)
(iv)
Equation of Normal:
1
y  0  ( x  (1))
6
1
y  ( x  1)
6
0 1
 6

 6  dx
Area required =  ( x  1)  
1 6
 ( x  2)

0
 ( x  1)2


  6ln( x  2)  6 x  
 12
 1
1

=   (6ln 2  0)   0   6ln1  6  
12

1
 6  6ln 2
12
5(i).
10e2 x  8e4 x  3
10 8

3 0
e2 x e4 x
Let y = e2 x
10 8
 3 0
y y2
3 y 2  10 y  8  0
(3 y  2)( y  4)  0
2
y
or  4
3
2
or e2 x  4 (rejected since e2 x > 0 for all x)
3
1 2
x  ln
2 3
e2 x 
(ii).
1 2
ln
2 3
10e2 x  8e4 x  3  x  5
10e2 x  8e4 x  3 = x + 5
Insert the line y = x + 5 on the curve in (ii):
Intersection : From GC, x = 0, or 0.52371,
For y > 0, x >
5(iii).
10e 2 x  8e 4 x  3  (x+5) dx


 0.307(3dp.)
Area required = 
0.52371
0
6(i)
(ii)
(iii)
55  128  81 264

365
365
75  31  81 187
P( S  W ) 

365
365
P( X ' S ) 15  5 4
P( X '/ S ) 


75
15
P( S )
X and S are not independent events as
P( X S ) 55 11 264
P( X / S ) 



 P( X )
75 15 365
P( S )
4 101
OR P( X '/ S )  
 P( X ')
15 365
Since X’ and S are independent, X and S are independent
P( X ) 
Required probability
55 54 5
= 3  
75 74 73

7(i)
(a)
(b)
44550
 0.10996  0.110
405150
P(20th uPhone is the only one with defective touchscreen)
19
=  0.94   0.06  0.0185
Let X be the number of uPhones with defective touchscreens,
out of 20. X B(20,0.06)
P( X  1)  0.370
(b) contains the cases in (a) and more ; such as when the
defective uPhone is the 1st phone or 2nd phone.
[or (a) is a proper subset of (b);in (b), the phone that is
defective may be the 1st, or 2nd, or any phone]
Let T be the number of uPhones with cracked touchscreen in a
box of 20
T B(20, 0.5  0.06) i.e B(20,0.03)
P(T  1)  1  P(T  1)
 1  0.88016
(ii)
 0.1198  0.120
Let Y be the number of boxes containing more than one uPhone
with a cracked touchscreen , out of 60 boxes. Y B(60,0.120)
Then Y B(60,0.120)
Since n=60>50 is large such that
np=7.2>5 and n(1-p)=52.8>5
Y N (7.2,6.336)
P(Y  12)  P(Y  11.5)
 0.4379
 0.438
T B(20, 0.03)
Since n  60  50 is large, by CLT
T  T  ...T60
A= 1 2
60
20  0.03  0.97 

T N  20  0.03,
 i.e. N(0.6, 0.0097)
60


8a i
In a stratified sampling, the different housing types will be represented
proportionately, and hence the households chosen will be more
representative of all the households in Singapore. This is important as
different types of households may have different electricity consumption
patterns
Students to survey:
-30 HDB households
-15 private condominium households.
-10 landed properties households
Students to knock on the doors on a weekend until they achieve the
required number in each category of household
Quota sampling saves resources as the students do not need a complete
list of all households in Singapore to calculate the exact number from
each type of households to include in the sample; and they can use any
convenient way to interview the households until they achieve the number
for each category.
8 bi
unbiased estimate of the population mean
725
=
 120  133.181818 =133.2(1dp)
55
unbiased estimate of the population mean
1 
7252 
 1671.188552 =1671.2 (1dp)
= 99801 
54 
55 
Let X be the monthly electricity consumption of a randomly selected
household in the estate, and  the population mean.
Ho:   120 [manager’s claim]
H1:   120
1671.188552 

Under Ho, X N  120,

55



Perform one-tail test at  % significance level. Reject Ho if p 
.
100
p  2P( X  133.181818)  2(0.00839316)  0.01679 .

Support manager’s claim, we do not reject Ho: 0.01679 
100
  1.679
  1.7 (1dp)
No, because the sample is only taken from the households in that
particular housing estate, and is not representative of the households in
Singapore.
9
Let X be the rv “demand of Avonly Cheese on a weekday”
X  3.5
X N (3.5,  2 )  Z 
N (0,1)

4.55  3.5 

P( X  4.55)  0.96  P  Z 
  0.96



0.105
 1.750686

0.105

 0.59976  0.6
1.750686
Let Y  1.2 X N (1.2  3.5, 1.22  0.62 )
or N (4.2, 0.5184)
W  X 1  X 2  ...  X 5  Y1  Y2
N (5  3.5  2  4.2, 5  0.62  2  1.22  0.62 )
N (25.9, 2.8368)
P( X 1  X 2  ...  X 5  Y1  Y2  7  4)
 P(W  28)  0.10623  0.106

Alternative: Let W  X 1  X 2  ...  X 5  1.2 X 6  X 7

52  0.65  1  0.10623  30.2
Estimated number of weeks the employee will make pizza with leftover
cheese is 30.
Let R  50( X 1  X 2  ...  X 5 )  52(Y1  Y2 )
N (50  5  3.5  52  2  4.2,502  5  0.62  522  2  1.22  0.62 )
R
N (1311.8, 7302.5072)
R1  R
2
N (1311.8  1311.8, 7303.5072  7303.5072)
N (0, 14607.0144)
P(100  R1  R2  100)  0.59199  0.592
10 i
ii
Let X be the volume of detergent in a randomly selected bottle, and  the
population mean.
Ho:   1000 [claim   1000 ]
H1:   1000
120.1 

Under Ho, and since n=100 >50 is large, by CLT X N  1000,

100 

Perform one-tail test at 4% significance level.
Reject Ho if p  0.04 .
p  P( X  998)  0.0340 .
Since p< 0.05, we reject Ho.
There is sufficient evidence at 5% significance level to conclude that the
manager’s claim is not valid.
Let Y be the volume of detergent in a randomly selected bottle from this
batch, and  the population mean.
Ho:   1000
H1:   1000 [manager’s suspicion] at 4%
120.1 

Under Ho, by CLT, Y N  1000,

100 

First find critical value C: P(Y  C )  0.04
C  998.081
Suspicion not justified (do not reject Ho)
k
 998.081 .
100
k  99808.1
Smallest integer value of k is 99809
y
iii
Let W be the volume of a randomly selected bottle of deluxe range of
detergent, and  the population mean.
Ho:    0 [manager’s claim]
H1:    0 [understating] at 2%
Under Ho, W
 10.22 
W  0
N  0 ,
and Z 

60 
10.22

60
First find critical value:
P( Z  C1 )  0.02
P( Z  C1 )  0.98
C 1  2.0537489
Manager is understating ( Reject Ho)
z  2.0537489
690  0
10.22
60
 2.0537489
690  0  2.0537489
10.22
60
 0  2.0537489
10.22
 690
60
0  2.0537489
10.22
 690
60
0  2.704  690
0  687.3
Maximum 0  687
N 1,0 
11
(i)
(ii)
r  0.940 .
r is close to 1, suggesting a strong positive linear correlation between
the annual household income and the amount spent on insurance in
year. As household income increases, the spending on insurance
tends to increase.
(iii)
y  0.1911x  0.3286
m  0.1911 means for every additional $1000 increase in annual
household income, the annual expenditure on insurance tends to
increase by an average of $19.11
(iv)
(v)
Regression line : x  4.62 y  0.346
Use y on x.
y=0.191071(15)+0.3286 =4.3411
The estimated amount on insurance is $434 [accept $430or $434.11]
Yes, the estimate is reliable, as we are interpolating for two strongly
linearly correlated variables. x  15 is within the sample data range
of 10  x  22 and r is close to 1.
y increases by 1 unit ($100) regardless of x.
The regression line will shift upwards by 1unit ($100). There will be
no change in the value of m (gradient) but c will increase to 1.3286