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Transcript of ACJC Math Prelim 10
ANGLO-CHINESE JUNIOR COLLEGE
MATHEMATICS DEPARTMENT MATHEMATICS Higher 2 Paper 1 19 August 2010
JC 2 PRELIMINARY EXAMINATION
Time allowed: 3 hours
Additional Materials: List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your Index number, Form Class, graphic and/or scientific calculator model/s on the cover page. Write your Index number and full name on all the work you hand in. Write in dark blue or black pen on your answer scripts. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the 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. Where unsupported answers from a graphic calculator are not allowed in the 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 brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 6 printed pages.
[Turn Over
9740 / 01
A-PDF Merger DEMO : Purchase from www.A-PDF.com to remove the watermark
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1
Page 2 of 6
ANGLO-CHINESE JUNIOR COLLEGE
MATHEMATICS DEPARTMENT JC2 Preliminary Examination 2010
MATHEMATICS 9740 Higher 2 Paper 1
Calculator model: _____________________
Arrange your answers in the same numerical order.
Place this cover sheet on top of them and tie them together with the string provided.
Question no. Marks 1
2
3
4
5
6
7
8
9
10
11
12
13
14
/ 100 Index No: Form Class: ___________ Name: _________________________
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1
Page 3 of 6
1 The nth term of a sequence is given by � � 1 21 n
nu n� � , for 1n t . The sum of the first n terms is denoted by Sn. Use the method of mathematical induction to show that
� � � �1 11
2n
nn n
S � � � for all positive integers n. [4]
2 The 3 flavours of puddings produced by a dessert shop are mango, durian and strawberry. A
mango pudding requires 5g of sugar and 36ml of water. A durian pudding requires 6g of sugar and 38ml of water. A strawberry pudding requires 4g of sugar and 40ml of water. The puddings are sold in pairs of the same type at $1.60, $2.20 and $1.80 for mango, durian and strawberry respectively. On a particular day, 754g of sugar and 5972ml of water were used to make the puddings and all the puddings made were sold except for a pair of strawberry puddings. The collection from the sale of puddings was $142.40. Formulate the equations required to determine the number of each type of pudding made on that day. [4]
3 The diagram below shows the graph of f( )y x . The curve passes through the origin and
has a maximum point at � �4 , 4A and asymptotes 2x � and 2y . Sketch on separate diagrams, the graphs of
(i) 1f( )y x [3]
(ii) f '( )y x [3] showing clearly asymptotes, intercepts and coordinates of turning points where possible.
4 The complex number w has modulus 3 and argument 23S . Find the modulus and argument of
*i
w� , where w* is the complex conjugate of w. Hence express
*i
w� in the form a ib� , where
a and b are real, giving the exact values of a and b in non-trigonometrical form. [4]
Find the possible values of n such that *
niw�§ ·
¨ ¸© ¹
is purely imaginary. [2]
[Turn Over
y
x
f( )y x
2
2�
� �4 , 4 A
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1
Page 4 of 6
5 Solve the equation 4 0z i� , giving the roots in the form ire D , where 0r ! and S D S� � d . [3]
The roots represented by 1z and 2z are such that � � � �1 2arg arg 0z z! ! . Show 1z , 2z and
1 2z z� on an Argand diagram. Deduce the exact value of � �1 2arg z z� . [3]
6 An economist is studying how the annual economic growth of 2 countries varies with time. The annual economic growth of a country is measured in percentage and is denoted by G and the time in years after 1980 is denoted by t. Both G and t are taken to be continuous variables.
(i) Country A is a developing country and the economist found that G and t are can be modeled
by the differential equation 12
dG Gdt
� . Given that, when 0t , 0G , find G in terms of t.
[4] (ii) Comment on the suitability of the above differential equation model to forecast the future
economic growth of Country A. [1] (iii) Country B is a developed country and the economist found that G and t can be modeled by
the differential equation 12
dG Gdt
�§ · �¨ ¸© ¹
.
Given that Country B has been experiencing decreasing economic growth during the period of study, sketch a member of the family of solution curves of the differential equation model for Country B. Hence, comment on the economic growth of Country B in the long term. [2]
7 (i) Given that � � 1cosf xx e
�
, where 1 1x� d d , find � �f 0 , � �f 0c and � �f 0cc . Hence write down
the first three non-zero terms in the Maclaurin series for � �f x . Give the coefficients in terms
of ke S , where k�� . [4] (ii) Given that � �g tan secx x x � , where x is sufficiently small for 3x and higher powers of x
to be neglected. Deduce the first three non-zero terms in the series expansion of � �g x .
Hence, show that � � � � 22 2 2f g 2x e x e x eS S S
� | � . [3]
(iii) Explain clearly why it is inappropriate to state that � � � � 22 2 2f g 2a a
a ax e x dx e x e dx
S S S
� �� | �³ ³ ,
where a�� . [1]
8 (i) Show that � � � � � �21 2 1
! 1 ! 2 ! 2 !Ar Br C
r r r r� �� �
� � �, where A, B and C are constants to be
found. [2]
(ii) Hence find � �2
1
3 3 32 !
n
r
r rr
� ��¦ . [3]
(iii) Give a reason why the series � �2
0
3 3 32 !r
r rr
f
� ��¦ converges, and write down its value. [2]
y
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1
Page 5 of 6
9 Relative to the origin O, two points A and B have position vectors given by 3 3a i j k � � and 5 4 3b i j k � � respectively.
(i) Find the length of the projection of OA)))&
on OB)))&
. [2] (ii) Hence, or otherwise, find the position vector of the point C on OB such that AC is
perpendicular to OB. [2] (iii) Find a vector equation of the reflection of the line AB in the line AC. [3] 10(a) The first 2 terms of a geometric progression are a and b ( b < a ). If the sum of the first n
terms is equal to twice the sum to infinity of the remaining terms, prove that 3n na b . [3]
(b) The terms 1 2 3, , ,...u u u form an arithmetic sequence with first term a and having non-zero common difference d. (i) Given that the sum of the first 10 terms of the sequence is 105 more than 510u , find
the common difference. [3] (ii) If 26u is the first term in the sequence which is greater than 542, find the range of
values of a. [3] 11 The region R in the first quadrant is bounded by the curve � �22 2y a a x � , where 0a ! ,
and the line joining � �2 ,0a and � �0,a . The region S, lying in the first quadrant, is bounded
by the curve � �22 2y a a x � and the lines 2x a and y a . (i) Draw a sketch showing the regions R and S. [1] (ii) Find, in terms of a, the volume of the solid formed when S is rotated completely about the x-axis. [4]
(iii) By using a suitable translation, find, in terms of a, the volume of the solid formed when R is rotated completely about the line 2x a . [4]
12 The curve C has the equation 23 2x axy x a� � � where a is a constant.
(i) Find dydx and the set of values of a if the curve has 2 stationary points. [4]
(ii) Sketch the curve C for a = 1, stating clearly the exact coordinates of any points of intersection with the axes and the equations of any asymptotes. [3]
Hence, find the range of values of k such that the equation 23 2 ( 1)x x k x� � � has exactly 2 real roots. [2]
13 The curve has the parametric equations
25
1x
t
� , 1tany t�
(i) Sketch the curve for 2 2t� d d . [1] (ii) Find the cartesian equations of the tangent and the normal to the curve at the point
where 1t . [5] (iii) Find the area enclosed by the x-axis, the tangent and the normal at the point where t = 1. [3]
[Turn Over
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1
Page 6 of 6
14 The functions f and g are defined as follows:
f : sin�x x , ,2 2
x S Sª º� �« »¬ ¼,
� �� �g : 1 38S
� ��x x x , x�� .
(i) Sketch the graph of the function g, labeling clearly the exact values of the coordinates of turning point(s) and intersections with the axes, if any. [1] State the range of the function g in exact values. [1]
(ii) Given that gf exists as a function. By considering the graphs of f and g, explain why
gf ( ) gf ( )D Ez if 2 2S SD E� d � d . [2]
Hence what can be said about the function gf ? [1] Without sketching the graph of gf , find the range of gf in the form > @,a b , giving the exact values of a and b. [1]
(iii) (a) Give a reason why fg does not exist as a function. [1]
(b) Find the greatest exact value of k for which fg is a function if the domain of g is restricted to the interval > @1, k . [2]
- End of Paper -
ANGLO-CHINESE JUNIOR COLLEGE
MATHEMATICS DEPARTMENT MATHEMATICS Higher 2 Paper 2 23 August 2010
JC 2 PRELIMINARY EXAMINATION
Time allowed: 3 hours
Additional Materials: List of Formulae (MF15)
READ THESE INSTRUCTIONS FIRST Write your Index number, Form Class, graphic and/or scientific calculator model/s on the cover page. Write your Index number and full name on all the work you hand in. Write in dark blue or black pen on your answer scripts. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the 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. Where unsupported answers from a graphic calculator are not allowed in the 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 brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.
This document consists of 5 printed pages.
[Turn Over
9740 / 02
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2
Page 2 of 5
ANGLO-CHINESE JUNIOR COLLEGE
MATHEMATICS DEPARTMENT JC2 Preliminary Examination 2010
MATHEMATICS 9740 Higher 2 Paper 2
Calculator model: _____________________
Arrange your answers in the same numerical order.
Place this cover sheet on top of them and tie them together with the string provided.
Question no. Marks
1
2
3
4
5
6
7
8
9
10
11
12
13
14
/ 100 Index No: Form Class: ___________ Name: _________________________
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2
Page 3 of 5
Section A: Pure Mathematics [40 marks]
1 Find the exact value of � �
12
2 11
1 .xxe dx
e ��
�³ [4]
2 The variable complex numbers z and w are such that 2 3z i� � and � �arg 5 3w i S� � .
(i) Illustrate both of these relations on a single Argand diagram. [2] (ii) State the least value of z w� . [1]
(iii) Find the greatest and least possible values of � �arg 3z � , giving your answers in radians correct to 3 decimal places. [4]
3 Find in terms of a, the range of values of x that satisfy the inequality ln 2 0axx
§ ·� t¨ ¸© ¹
,
where 1a! . [4] 4 (a) State the derivative of 3cos x . Hence, find 5 3sin .x x dx³ [4]
(b) Find the exact value of � �2 5 3
2 2
10
5x dx�
�³ , in the form 2 3a b� , using the
substitution 5 secx T . [6]
5 The plane p1 has equation 2 3x y z� � . The line l1 has equation 1 12 4
x zy� � .
(i) Show that the line 1l is parallel to, but not contained in the plane p1. [2] (ii) Find the cartesian equation of the plane p2 which contains the line l1 and is perpendicular to the plane p1. [3]
(iii) Find, in scalar product form, the vector equation of the plane p3 which contains the point � �4,1, 1� and is perpendicular to both p1 and p2. [2]
Another line l2 which is parallel to the vector 203
§ ·¨ ¸¨ ¸¨ ¸�© ¹
intersects the line l1 at the
point � �1,0,1�A . (iv) Given that the line l2 meets the plane p1 at the point B, find the coordinates of B. [4] (v) Find the sine of the acute angle between the line l2 and the plane p1, and hence, find the
length of the projection of the line segment AB on the plane p1, giving your answer in surd form. [4]
Section B: Statistics [60 marks]
6 Mr Raju, who owns a supermarket wishes to find out what customers think about the goods
that he sells. He has been advised that he should take a random sample of his customers for this purpose. State, with reasons, which of the following sampling procedures is preferable. A. Select every 10th customer on each day in a typical week. B. Select the first 20 customers on each day in a typical week [2]
[Turn Over
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2
Page 4 of 5
7 One day, Pinocchio went shopping and bought a pair of size 30 Wood Shoes. The right
shoes have lengths which are normally distributed with mean 20 cm and standard deviation 0.14 cm. The left shoes have lengths which are normally distributed with mean 20.1 cm and standard deviation 0.11 cm. The length of the right shoe is independent of the length of the left shoe. When wearing the pair of shoes, Pinocchio takes six steps, heel to toe as shown in the diagram. Calculate the probability that the distance AB, from the back of the first step to the front of the sixth step, exceeds 120 cm. [3]
8 On Ulu Island the weights of adult men and women may both be taken to be independent
normal random variables with means 75kg and 65 kg and standard deviations 4 kg and 3 kg respectively. Find the probability that the weight of a randomly chosen man and the weight of a randomly chosen woman differ by more than 1 kg. [3] Explain if this is equal to the probability that the difference in weight between a randomly chosen married woman and her husband is more than 1kg. [1]
9 Research has shown that before using an Internet service, the mean monthly family
telephone costs is $72. A random sample of families which had started to use an Internet service was taken and their monthly telephone costs were :
$70, $84, $89, $96, $74 Stating a necessary assumption about the population, carry out a test at the 5% significance level, whether there is an increase in the mean monthly telephone costs. [5] If the assumption stated above still holds, and if the standard deviation of the monthly telephone costs is $9.89, find the range of values of the mean monthly family telephone costs 0P that would lead to a reverse in the decision to the above test. [3]
10 Six overweight men registered at a slimming centre for a slimming programme. The
following table records x, the height (to the nearest cm) and y, the weight (to the nearest 0.1 kg) of these six men.
(i) Given that the least square regression line of x on y line is 103.6 0.726x y � , show that the value of k to the nearest 0.1 kg is 80.3. Hence or otherwise, find the least square regression line of y on x in the form y ax b � , giving the values of a and b to the nearest 3 decimal places. [5] (ii) Based on the data given, use an appropriate regression line to predict the weight of an overweight man who is 165 cm tall. [2] (iii) Find the value of the product moment correlation coefficient between x and y and sketch the scatter diagram of y against x. A particular man among the 6 men who registered for the slimming programme is unusually overweight. Indicate who this man is. [3]
Man A B C D E F x (height in cm) 150 157 160 162 167 170 y (weight in kg) 65.1 73.2 85 k 80.9 89.9
A B
Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 2
Page 5 of 5
11 In a hotel, large number of cups and saucers are washed each day. The number of cups that
are broken each day while washing averages 2.1. State in context, a condition under which a Poisson distribution would be a suitable probability model. [1] Assume that the number of broken cups and saucers follow a Poisson distribution.
(i) Show that on any randomly chosen day, the probability that at least 3 cups are broken is 0.350 correct to 3 significant figures. [1] The probability that there will be at least two days in n days with at least 3 broken cups is more than 0.999. Find the least value of n. [3]
(ii) The number of saucers broken each day averages 1.6, independently of the number of cups broken. The total number of cups broken and saucers broken during a week of 7 days is denoted by T. State a possible model for the distribution of T. [2]
A random sample of 100 weeks is chosen. Using a suitable approximation, find the probability that the average weekly total number of broken cups and saucers does not exceed 26. [3]
12 Fish are bred in large batches and allowed to grow until they are caught at random for sale.
When caught, only 20% of the fish measure less than 8 cm long. (i) What is the probability that the 10th fish caught is the sixth fish that is less than 8 cm long? [2] (ii) A large number, n, of fish are caught and the probability of there being 10 or fewer fish in
the catch which measures less than 8 cm long is at most 0.0227 . Using a suitable approximation, derive the approximate inequality 10.5 0.2 0.8n n� d � . [4] Hence find the least possible number of fish to be caught. [2]
13 An automated blood pressure machine is being tested. Members of the public, %p of whom
have high blood pressure (hypertension), try it out and are then seen by a doctor. She finds that 80% of those with hypertension and 10% of those with normal blood pressure have been diagnosed as hypertensive by the machine. The probability that a randomly chosen patient who was diagnosed as hypertensive by the machine actually has hypertension is
32 .
(i) Find the value of p [3] (ii) Hence, find the probability that a randomly chosen patient does not have hypertension,
given that the machine diagnosed him as having normal blood pressure. [2] Comment on the usefulness of the machine. [1]
14 Ten balls are identical in size and shape of which 2 are red, 3 are blue and 5 are green. The two red balls are labeled ‘1’ and ‘2’, the three blue balls are labeled ‘1’, ‘2’ and ‘3’, and the five green balls are labeled ‘1’, ‘2’, ‘3’, ‘4’ and ‘5’.
(i) Find the number of ways of choosing 2 balls of identical colour. [2] (ii) Find the number of ways of choosing 6 balls if it includes at least one ball of each colour. [4] (iii) A person arranged 3 balls in a row with the numbered sides facing him forming a 3-digit
number. Among these 3 balls, none of them are green. Find the number of possible 3-digit numbers facing that person.
[The number formed is independent of the colours of the balls used. i.e. the number 112 is counted as one number whether the colour of the ball labeled ‘2’ is red or blue.] [3]
- End of Paper -
1
Ang
lo-C
hine
se J
unio
r Col
lege
H
2 M
athe
mat
ics 9
740
2010
JC
2 P
RE
LIM
Mar
king
Sch
eme
Pape
r 1:
1 Le
t Pn d
enot
e th
e st
atem
ent
()
()
11
12
nn
nn
S+
+=−
For n
= 1
, LH
S =
()2
21
11
11
Su
==−
= R
HS
= (
)()
21
21
12
−=
LHS
= R
HS
∴ P
1 is t
rue.
Ass
ume
P k tr
ue fo
r som
e k
+∈ℤ
, i.e
. (
)(
)1
11
2k
kk
kS
++
=−
Prov
e th
at P
k+1 i
s tru
e,
i.e.
()
()(
)2
11
21
2k
kk
kS
++
++
=−
()
()
()
()
()
()
()
()
() [
]
()
()(
)
11
12
2
1 1 2
LHS
11
11
2 11
21
21
12
21
21
RH
S2
kk
k
kk
k k k
SS
uk
kk
kk
k
kk
kk
++
++
+ + +
==
+
+=−
+−
+
+=−
−+
+=−
−−
++
=−
=
Sinc
e P 1
is tr
ue, a
nd P
k is t
rue ⇒
Pk+
1 is t
rue,
by
the
prin
cipl
e of
mat
hem
atic
al in
duct
ion,
Pn i
s tru
e n
+∀
∈ℤ
. 2
Let x
be
the
no o
f man
go p
uddi
ngs p
rodu
ce.
Let y
be
the
no o
f dur
ian
pudd
ings
pro
duce
. Le
t z b
e th
e no
of s
traw
berr
y pu
ddin
gs p
rodu
ce.
56
475
4x
yz
++
=…
……
……
……
……
……
(1)
3638
4059
72x
yz
++
=…
……
……
……
……
(2)
0.8
1.1
0.9(
2)14
2.4
xy
z+
+−
=…
……
……
….(3
) So
lvin
g (1
), (2
) & (3
) usi
ng G
C,
46,
42
an
d
68.
xy
z=
==
3 (i)
1 f()
yx
=
2−
y
x A’
(4, 0
.25)
0.5
y=
0x=
2
3 (ii)
f'
()
yx
=
4
1*
3i
w−=
2ar
gar
g()
arg(
*)*
23
6i
iw
wπ
ππ
−
=
−−
=−
−−
=
11
31
31
cos
sin
*3
66
32
26
6i
ii
iw
ππ
−
=
+=
+=
+
1co
ssi
n*
36
6
nn
in
ni
wπ
π−
=
+
*
ni
w−
is
pur
ely
imag
inar
y, c
os6nπ
=0
()
21
,
62
nk
kπ
π=
+∈ℤ
,
()
32
1,
nk
k∴
=+
∈ℤ
.
5 4
44
20
1
iz
iz
iz
eπ
−=
⇒=
⇒=
24
2,
2,1,
0,1
ik
ze
kπ
π
+
==−
−
12
42 7
35
88
88
, 2,
1,0,
1
,,
,
ik
ii
ii
ze
k
ze
ee
e
ππ
ππ
ππ
+
−−
==−
−
=
z 1
z 2
z 2 +
z 2
x
y
8π
5 8π
O y
x 2−
A’
(4,0
)
3
5 8
1
82
i i
ze
ze
π π
= =
()
12
14
3ar
g8
28
8z
zπ
ππ
+=
+=
6
12
dGG
dt+
=
11
12
ln1
0.5
dGdt
G Gt
C
=+ +
=+
∫∫
0.5 0.
5
11
, w
here
tC t
C
Ge
GAe
Ae
++
=±
=−+
=±
Whe
n 0
t=,
0G=
, 1
A=
0.
51
tG
e∴
=−+
Ex
ampl
es o
f pos
sibl
e co
mm
ents
: Th
e m
odel
is n
ot su
itabl
e be
caus
e …
Th
e ec
onom
ist i
s as
sum
ing
that
that
ther
e ar
e no
fluc
tuat
ions
in th
e ec
onom
ic
grow
th in
the
futu
re.
The
econ
omis
t is
ass
umin
g th
at t
he c
ount
ry w
ill e
njoy
per
petu
al e
cono
mic
gr
owth
in th
e lo
ng te
rm.
The
econ
omis
t is
ass
umin
g C
ount
ry A
is
alw
ays
expe
rienc
ing
posi
tive
and
incr
easi
ng e
cono
mic
gro
wth
in th
e fu
ture
.
Fa
ctor
s aff
ectin
g ec
onom
ic g
row
th re
mai
ns u
ncha
nged
. In
the
lon
g te
rm, C
ount
ry B
is
expe
cted
to
be s
till
in r
eces
sion
with
an
econ
omic
gr
owth
dec
reas
ing
tow
ards
-1%
. 7(
i)
()
1co
sf
xx
e−
=
()
2f
0eπ
⇒=
()
1co
s
2
1f
1x
xe
x
−−
′=
−
()
2f
0eπ
′⇒
=−
t
G 1−
0.5
1,
1t
GBe
B−
=−+
=
0
4
(ii)
(iii)
()
()
()
()
()(
)()
11
11
3co
sco
s2
22
2
3co
sco
s2
22
211
1f
12
21
11
1
f
01
xx
xx
xe
ex
xx
x
exe
xe
x
π
−−
−−
−
−
−
−−
′′=
+−
−−
−
−
′′=
−−
⇒=
−
()
22
22
f...
2ex
ee
xx
ππ
π
∴=
−+
+
()
()
12
2
2
gta
nse
c 1ta
n1
11
...co
s2
2
12
−
=+
=+
≈+
−=
++−
−+
≈+
+
xx
x
xx
xx
xx
xx
()
()
22
22
22
2
22
2
fg
12
2
2 (
Show
n)
ex
xe
xe
ex
xe
x
ee
x
ππ
ππ
π
ππ
+≈
−+
++
+
=+
The
stat
emen
t()
()
22
22
fg
2a
a
aa
xe
xdx
ex
edx
ππ
π
−−
+≈
+∫
∫ is
inap
prop
riate
as
()
()
22
22
fg
2x
ex
ex
eπ
ππ
+≈
+ o
nly
whe
n a
is su
ffic
ient
ly sm
all.
8 (i) (ii)
()
()
()(
)(
)(
)
()
2
12
1!
1!
2!
12
22
12
!
12
!
rr
r
rr
rr
rr
r−+
++
++
−+
+=
+
+−
=+
1,1,
1.A
BC
==
=−
()
() (
)(
)
2
1
2
1 133
32
!
13
2!
12
13
!1
!2
!
n rn r n r
rr
r rr
r rr
r
=
= =
+−
+ +−
=+
=−
+
+
+
∑
∑ ∑
5
(iii)
()
()
()
()
()
()
12
11!
2!3!
12
12!
3!4!
12
13!
4!5!
12
14!
5!6!
3
12
12
!1
!!
12
11
!!
1!
12
1!
1!
2!
nn
n
nn
n
nn
n
−+
+−
+
+
−+
+−
+
=
+−
+
−
−
+
−+
−+
+−
+
+
+
⋮
(
)(
)1
11
32
1!
2!
nn
=−
+
+
+
()
()
()
2
0
2
1
33
32
!
33
33
2!2
!
n rn r
rr
r
rr
r
=
=
+−
+
+−
=−
++
∑
∑
()
()
()
31
11
32
21
!2
!n
n
=−
+−
+
+
+
As
()
()
11
, 0
and
0.1
!n+
2!
nn
→∞
→→
+
the
serie
s con
verg
es to
0.
∴
9(i) (ii)
Leng
th o
f the
pro
ject
ion
of O
A
on
OB
= 3
51
201
42
250
503
3
⋅−
==
or
4 2
Met
hod
1:
From
(i),
22
OC=
55
12
22
44
550
33
OC
=−
=−
Met
hod
2:
Line
OB
: 0
50
40
3λ
=+
−
r
53
53
41
41
33
33
ACO
CO
Aλ
λλ λ
−
=−
=−
−=
−−
−
Sinc
e AC
OB
⊥
,
A
C
B O
6
(iii)
53
54
1.
40
33
32 5λ λ λ
λ
−
−
−−
=
−
−
=
5
24
53
OC
=−
Sinc
e 2 5
OC
OB
=
, :
2:3
OC
CB=
5
11
'4
55
3O
BO
B
=−
=−
−
Or u
se m
idpo
int t
heor
em,
()
1'
2O
CO
BO
B=
+
55
52
1'
22
44
45
53
33
OB
OC
OB
=−
=−
−−
=−
−
5
320
11
''
41
15
53
318
ABO
BO
A
=−
=−
−−
=−
Vec
tor e
quat
ion
of li
ne'
AB is
3
201
1,
318
r
αα
=+
∈
ℝɶ
10
(a)
(b) (
i)
GP
: a =
a, r
= b a
The
sum
to in
finity
of t
he re
mai
ning
term
s, 1
nar
Sr
∞=
−
()
()
2
12
11
12
31 1 3
1 3
3nn
n
nn
n
n
n
nn
SS
ar
arr
rr
r
r r b a ba∞
= −=
−−
−=
= =
= =
7
(ii
)
[]
[]
105
105
1010
29
105
104
2 1045
105
1040
21
Su
ad
ad
ad
ad
d
=+
+=
++
+=
++
=
2554
224
(21)
542
38
u a a
≤
+≤
≤
a
nd
2654
225
(21)
542
17
u a a
>
+>
>
17
38a
∴<
≤
11
(i) (ii)
(iii)
Whe
n S
is ro
tate
d co
mpl
etel
y ab
out t
he x
-axi
s,
()
()
()
()
22
0
22
3
0
32
3
Req
uire
d vo
lum
e2
22
22
22
22
2 c
u. u
nits
a
a
aa
aa
xdx
ax
aa
aa
a
a
ππ
ππ
ππ π
=−
−
−=
−
−
=−
=
∫
Afte
r a tr
ansl
atio
n of
2a
units
in th
e ne
gativ
e x-
dire
ctio
n,
New
equ
atio
n is
(
)2
22
22
(2
)y
ya
ax
ax
a=
−+
⇒=−
Whe
n R
is ro
tate
d co
mpl
etel
y ab
out t
he li
ne
2x
a=
,
()()
22
2
0
53
20
33
3
12
Req
uire
d vo
lum
e2
3 44
35
44
8 c
u. u
nits
35
15
a
a
ya
ady
a
ya
a
aa
a
ππ
ππ
ππ
π
=−
−
=−
=−
=
∫
12
x
y a
2a
O
R S
()
22
2y
aa
x=
−
8
(i) (ii)
()(
)(
)(
)
()
()
2
2
2
22
2
32 6
32
36
2
xax
yx
ax
ax
ax
axdy dx
xa
xax
a
xa
++
=+ +
+−
++
=+
++
−=
+
For s
tatio
nary
poi
nts,
()
22
0
36
20
dy dx xax
a
= ++
−=
For 2
stat
iona
ry p
oint
s, (
)() (
)2
2
2
264
32
0
2424
1
aa
a
a
−−
>
>−
>−
2 is a
lway
s pos
itive
.a
.
a ∴∈ℝ
2
32
43
21
1x
xy
xx
x+
+=
=−
++
+
Po
ints
of i
nter
sect
ion
with
the
axes
: (0
,2)
Asy
mpt
otes
: 3
2
and
1y
xx
=−
=−
. Th
e ra
nge
of v
alue
s of
k s
uch
that
the
equ
atio
n 2
32
(1)
xx
kx
++
=+
has
exa
ctly
2
real
root
s :
1.93
or
11.9
kk
><−
13
(i)
9
(ii)
(iii)
()
()
()
()2
22
22
2
2
-52t
1
and
=
11
. 1+t
1.
101
1=
10
dydx
dtdt
tt
dydy
dtdx
dtdx
tt
t t
=+
+
=
=−
+
+−
Whe
n 1,
1 5tdy dx
=
=−
Equa
tion
of ta
ngen
t :
()
14 5
52
11
52
4
πy x
πy
x
−=−
− =−
++
Gra
dien
t of n
orm
al is
5
Equa
tion
of n
orm
al:
()
45
5 225
52
4
πy x
πy
x
−=
− =−
−
()
()
()
2
11
125
Are
a =
52
24
52
44
131.
60 (3
s.f )
80
ππ
π
π
×+
−−
×
==
14
Func
tions
G
iven
:
sin
,f
xx
֏
,2
2x
ππ
∈−
5
10
(i) (ii)
and
:(
1)(3
),8
gx
xx
π+
−֏
x∈ℝ
.
(,
] 2gR
π=−∞
Fr
om th
e gr
aph
of
()
yf
x=
,
()
()
ff
αβ
≠ if
2
2π
πα
β−
≤<
≤.
In fa
ct,
1(
)(
)1
ff
αβ
−≤
<≤
.
From
the
grap
h of
(
)y
gx
=,
()
()
gfgf
αβ
<
or
()
()
gfgf
αβ
≠ 0
x
y
-1
1 3
(1,) 2
π
2π
38
π
(
)y
gx
=
0 x
y
2π
2π−
-1
1
α
β
()
fα
()
fβ
()
yf
x=
0 x
y
1 -1
2π
38
π
()
yg
x=
11
(iii)
(a)
(iii)
(b)
if 1
()
()
1f
fα
β−≤
<≤
. gf
is o
ne-o
ne (o
r inc
reas
ing)
.
[0,
] 2gfR
π=
.
(,
][
.]
22
2g
fR
Dπ
ππ
=−∞
⊄−
=.
Alte
rnat
ive:
5
(4)
[(4
)][
]8
fgf
gf
π=
=−
is u
ndef
ined
bec
ause
5
[,
]8
22
fD
ππ
π−
∉−
=.
Hen
ce, f
g do
es n
ot e
xist
as a
func
tion.
()
2g
xπ
=
()(
)1
38
2x
xπ
π+
−=−
()(
)1
34
xx
+−
=−
2
23
4x
x−
++
=−
2
27
0x
x−
−=
2
322
x±
=
18
x=
±
Sinc
e 1,
x≥
1
8x=
+.
11
8k
≤≤
+
Gre
ates
t val
ue o
f k is
18
+.
0 x
y
-1
1 3
(1,
) 2π
()
yg
x=
2y
π=−
12
1
Ang
lo-C
hine
se J
unio
r Col
lege
H
2 M
athe
mat
ics 9
740
2010
JC
2 P
RE
LIM
Mar
king
Sch
eme
Pape
r 2:
1
() (
)(
)
()
()
()
12
21
1
11
22
22
12
11
12
11
22
12
12
2
11
2
42
2
1
11
11
11
22
22
14
12
xx
xx
xx
xx
xx
edx
e
edx
edx
ee
ee
ee
ee
ee
−−
−−
−
−−
−−
−
−
−
=−
−+
−
=−
++
+
=+
−+
+
∫
∫∫
2
(i)
2
(ii)
2 (ii
i)
Leas
t val
ue o
f z
w−
= 1
G
reat
est
()
arg
3z+
=
11
13
tan
sin
0.82
6 (3
dp)
526
−−
+=
Leas
t (
)ar
g3
z+ =
1
11
3ta
nsi
n0.
432
(3 d
p)5
26−
−
−=−
3
ln2
0a
xx
−≥
whe
re
1a>
.
2
21
20
11
8)
11
8)
24
40
11
81
18
04
4
ax
xx
xa
xa
ax
x
xa
ax
orx
−≥
−−
≥
+
+−
+−
−
≥
−+
++
≤<
≥
4 (a
) (
)(
)3
23
cos
3si
nd
xx
xdx
=−
x
y
i(2,
1)
O
−3
(5,−
3)
o
2
3
263
3
z
w
2
5
3
32
3
33
32
33
23
33
3
sin (
sin
)
11
cos
cos
33
3
cos
cos
31
cos
sin
33
xx
dx
xx
xdx
xx
xx
dx
xx
xx
dx
xx
xc
=
=−
−−
=−
+
=−
++
∫ ∫
∫
∫
4
(b)
5se
c10
,4
5se
cta
n2
5,3
xw
hen
x
dxw
hen
xd
πθ
θ
πθ
θθ
θ==
=
==
=
()
()
[]
25
32
2
10
33
22
4
3
2
4 33
2
44
33 4
4
5
5sec
55
sec
tan
1se
c5
tan
1co
s1
cot
cos
5si
n5
11
1co
s5
sin
5
12
12
23
..5
155
15
xdx
d
d dor
ecd
ec
iea
b
π π
π π ππ
ππ
ππ π
π
θθ
θθ
θθ
θ θθ
θθ
θθ
θθ
−
−
−
=−
= =
=−
=−
=−
==−
∫ ∫
∫ ∫∫
5 (i) (ii)
1
1:
23
1p
r
⋅
=
−
ɶ
1
12
:0
1,
14
l
rλ
λ−
=+
∈
ℝɶ
21
1.
22
24
04
1
=+
−=
−
11
0.
21
01
31
1
−
=−+
−≠
−
⇒ l 1
is p
aral
lel t
o p 1
.
⇒
l 1 is
not
con
tain
ed in
p1.
Alte
rnat
ive
met
hod:
1
21 2
23
14
1
λλλ
−+
=−
≠
+
−
i
Sinc
e no
solu
tion
for λ
, ∴ l 1
is p
aral
lel a
nd n
ot c
onta
ined
to p
1 1
23
21
32
14
1−
×=−
−
3
(ii
i) (iv
) (v)
2
31
3:
20
23
01
41
11
pr
−−
−
⋅
==
++
=
iɶ
2
:3
24
px
yz
−+
+=
3
24
2:
11
18
14
54
14
pr
⋅=
⋅=
+−
=
−
ɶ
3
2:
15
4p
r
⋅=
ɶ
2
12
:0
0,
13
l
rµ
µ−
=+
∈
−
ℝɶ
1
1:
23
1p
r
⋅
=
−
ɶ
()
12
10
.2
3
25
3
1 po
int
is
1,0,
21
31
µµ
µµ
−+
=⇒
−+
=⇒
=∴
−
−
−
B
2
11
15
sin
02
136
783
1θ
=⋅
=
−
−
Leng
th o
f the
pro
ject
ion
of A
B on
p1 =
co
sAB
θ
= 2
2553
531
01
1331
878
786
63
−=
==
−
6
Proc
edur
e A
is p
refe
rabl
e as
it is
unb
iase
d.
Early
cus
tom
ers m
ay n
ot b
e ty
pica
l cus
tom
ers i
n ge
nera
l.
7 Le
t R b
e th
e r.v
for t
he le
ngth
of a
righ
t sho
e an
d L
for t
he le
ft sh
oe
R
N(2
0, 0
.142 ) a
nd L
N
(20.
1 , 0
.112 )
Met
hod
1
X =
R +
L
N(4
0.1,
0.0
317)
3X
N
(40.
1 x
3, 0
.031
7 x
32 ) P(
X >
120
) = 0
.713
M
etho
d 2
R +
L
N(4
0.1,
0.0
317)
P(
R +
L >
3120
) =
0.71
3 8
Let M
and
W b
e th
e rv
for t
he w
eigh
t of a
n ad
ult m
an a
nd w
oman
resp
ectiv
ely.
M
N
(75,
42 ) an
d W
N
(65,
32 ) W
M
~ N
(-10
, 52 )
P(1
>−
MW
) = P
(W –
M >
1) +
P(W
– M
< -1
) = 0
.978
O
r P(
1>
−M
W) =
1 -
P(1
<−
MW
) = 1
– P
(-1<
W –
M<1
) = 0
.978
4
No
, (i)
The
wei
ght o
f a h
usba
nd a
nd w
ife m
ay n
ot b
e in
depe
nden
t O
r (ii)
Ran
dom
ness
is n
ot th
ere
( a ra
ndom
ly c
hose
n w
omen
but
spou
se is
not
ra
ndom
ly c
hose
n)
Or (
iii) D
istri
butio
n of
wei
ght o
f mar
ried
wom
an is
diff
eren
t fro
m d
istri
butio
n of
ad
ult w
oman
. Et
c 9
Tele
phon
e co
sts a
re a
ssum
ed to
be
norm
ally
dis
tribu
ted.
To
test
H0 :
µ=
72
agai
nst H
1: µ
> 72
at 5
% le
vel o
f sig
nific
ance
Und
er H
0 , T
= n
sx0
_µ
−t(5
-1)
Test
stat
istic
s : T
= n
sx0
_µ
−=
82.6
7210
.667
7/5
−=
2.22
19
p va
lue
= P(
T >
2.22
19) =
0.0
452
< 0.
05
Rej
ect H
0 at t
he 5
% le
vel o
f sig
nific
ance
. We
conc
lude
that
ther
e is
suff
icie
nt
evid
ence
at t
he 5
% le
vel o
f sig
nific
ance
that
ther
e is
evi
denc
e of
an
incr
ease
in
mea
n m
onth
ly c
osts
. To
test
H0 :
µ=
0µ
agai
nst H
1: µ
> 0µ a
t 5%
leve
l of s
igni
fican
ce
Und
er H
0 , Z
=
nx 89.9
0
_µ
− N
(0,1
)
Test
stat
istic
s : Z
=
nx σ
µ 0_−
= 0
82.6
9.89
/5µ
−=
(82.
6 -
0µ)
89.95
Do
not r
ejec
t H0 i
f P(Z
> (8
2.6
- 0µ)
89.95
) > 0
.05
(82.
6 -
0µ)
89.95
< 1.
6448
5....
......
.(1)
µ 0 >
75.
3 10
(i)
16
1x=
(fro
m c
alcu
lato
r or c
ompu
tatio
n)
5
(ii)
(iii)
whe
n 16
1x=
, 10
3.6
0.72
6x
y=
+
(1
6110
3.6)
/0.7
26y=
−
79.0
6336
088
=
usin
g y
yn
=∑
1
79.0
6336
088
(65.
173
.285
80.9
89.9
)6
k=
++
++
+
80.3
k=
U
se G
.C. t
o fin
d re
gres
sion
line
of y
on
x:
97
.593
1.09
7y
x=−
+
Use
y o
n x
line
to p
redi
ct w
eigh
t. W
hen
165
x=
, 97
.593
1.09
7(16
5)y=−
+
83.4
y=
(1 d
.p.)
– us
ing
3 d.
p. o
f a a
nd b
to c
ompu
te.
or
83.5
y=
- us
ing
full
accu
racy
of a
and
b to
com
pute
. U
sing
G.C
., 0.
893
r=
C
is u
nusu
ally
ove
rwei
ght.
11
(i)
Bre
akag
es o
ccur
rand
omly
or
Bre
akag
es o
ccur
inde
pend
ently
or
Mea
n nu
mbe
r of b
reak
ages
is a
con
stan
t Le
t A b
e th
e r.v
for t
he n
umbe
r of b
roke
n cu
ps p
er d
ay
A
Po(2
.1)
P(A≥
3) =
1 –
P(A
≤2)
= 0
.350
369
= 0.
350
(3 si
g fig
s)
Let X
be
the
r.v fo
r the
num
ber o
f day
s with
a le
ast 3
bro
ken
cups
out
of n
cup
s. X
y
65.1
73.2
80.3
80
.9
85.0
89.9
150
157
160 1
62
167
170
x
6
(ii)
P(X≥
2) >
0.9
99
1 –
P(X≤
1) >
0.9
99...
......
...(1
) P(
X≤
1) <
0.0
01
Usi
ng G
C,
n P(
X≤
1)
21
0.00
145
22
0.00
0984
le
ast n
is 2
2 T
Po(2
.1x7
+ 1
.6 x
7)
TPo
(25.
9)
Met
hod
1:
Sinc
e n
is la
rge,
_ T
(2
5.9,
1009.
25) a
ppro
x by
cen
tral l
imit
theo
rem
P(_ T≤
26)
= 0
.578
(to
3 si
g fig
) M
etho
d 2:
10
0 w
eeks
, Y
Po(2
590)
λ
= 2
590
> 10
. N
orm
al a
ppro
x to
Poi
sson
Y
N
(259
0,25
90) a
ppro
x P(
Y≤
2600
) = P
( Y <
260
0.5)
(With
cc)
=
0.57
8 (to
3 si
g fig
)
12 (i) (ii)
Let X
be
the
r.v fo
r the
num
ber o
f fis
h w
hich
mea
sure
s les
s tha
n 8
cm lo
ng.
95
45
14
1(
)(
)(
)5
55
C =
0.0
0330
( to
3 sf
)
X
B(n
,0.2
) Si
nce
n la
rge
and
p= 0
.2 ,
X
N(0
.2n,
(0.2
)(0.
8)n)
app
rox
P(X≤
10) ≤
0.02
27
P(X
< 1
0.5)
≤0.
0227
P(Z
< n
n4.0
2.05.
10−
) ≤
0.02
27
nn
4.02.0
5.10
−≤
-2.0
0092
9....
......
(1)
10.5
– 0
.2n ≤
-0.8
0037
2n
H
ence
10.
5 –
0.2n
≤-0
.8n
app
rox
Met
hod
1
Usi
ng G
C Y
1 = 1
0.5
-0.2
x +
0.8
x
→
Tabl
e
A
ns :
91
Met
hod
2 : U
se G
C a
nd g
raph
7
Met
hod
3:
Hen
ce (1
0.5
– 0.
2n)2 ≥
(-0.
8n
)2 H
ence
4n2 -
484n
+ 1
1025
≥0
appr
ox…
…(2
) Fr
om G
C :
n≥
90.6
or n
≤ 3
0.4
n ≥91
or n
≤30
(N
A b
ecau
se d
oes n
ot sa
tisfy
(1) )
Le
ast n
= 9
1
13 (i) (ii)
Let H
be
the
even
t tha
t the
mem
ber o
f pub
lic h
as h
yper
tens
ion
Let D
be
the
even
t tha
t the
mac
hine
dia
gnos
ed h
yper
tens
ion
(i)
P(H
/D) =
32=
)(
)(
DP
DH
P∩
32=
)1.0)(
1(8.0
8.0k
kk −
+
(
1)
k =
0.2
p
% =
20
%
p =
20
P(H
’/D’)
=
)(
)(
DP
DH
P′′
∩′=
)9.0)(
1(2.0
)9.0)(
1(k
kk −
+−=
0.94
7
Exam
ples
of p
ossi
ble
com
men
ts:
(1)
If it
doe
s not
find
you
hyp
erte
nsiv
e th
en y
ou c
an b
e re
ason
ably
con
fiden
t th
at y
our b
lood
pre
ssur
e is
nor
mal
. (2
) If
it d
iagn
oses
hyp
erte
nsio
n, th
en y
ou sh
ould
con
sult
your
doc
tor f
or fu
rther
te
sts.
(3
) A
ny o
ther
logi
cal c
omm
ents
with
refe
renc
e to
the
cont
ext o
f the
que
stio
n
14
(i) (ii)
Cas
e 1:
2 re
d ba
lls -
21
2 =
Cas
e 2:
2 b
lue
balls
- 3
32 =
Cas
e 3:
2 g
reen
bal
ls -
5
102 =
No.
of w
ays =
13
1014
++
=
Cas
e 1:
No
red
ball.
C
hoos
e 6
balls
from
a to
tal o
f 8 (b
lue
and
gree
n)
bal
ls:
8 6
.
Cas
e 2:
No
blue
bal
l.
C
hoos
e 6
balls
rom
a to
tal o
f 7 (r
ed a
nd g
reen
)
H
H’
D
D
D ’ D’
0.8 0.2
0.1
0.9
k 1-k
8
(ii
i)
bal
ls:
7 6
[N
ote:
We
can’
t exc
lude
gre
en b
alls
bec
ause
tota
l
n
umbe
r of r
ed a
nd b
lue
balls
is o
nly
5.]
No.
of w
ays =
10
87
210
(28
7)17
56
66
−+
=−
+=
Alte
rnat
ive
Met
hod:
C
ase
Gre
en
Blu
e R
ed
No.
of w
ays
1 4
1 1
53
230
41
1
××
=
2 3
2 1
53
260
32
1
××
=
3 3
1 2
53
230
31
2
××
=
4 2
3 1
53
260
32
1
××
=
5 2
2 2
53
220
22
2
××
=
6 1
3 2
53
25
13
2
××
=
Tota
l 17
5
Ex
clud
ing
the
gree
n ba
lls, w
e on
ly h
ave
1,
1, 2
, 2, 3
. Sin
ce w
e ar
e ig
norin
g th
e co
lour
s of
the
balls
, we
are
form
ing
3-di
git n
umbe
rs fr
om th
e
5 di
gits
1, 1
, 2, 2
, 3.
Cas
e 1:
All
3 di
gits
are
dis
tinct
.
T
he 3
dig
its a
re 1
, 2, 3
and
the
num
ber o
f way
s of
arr
angi
ng th
em a
re 3
!
C
ase
2: 2
dig
its a
re id
entic
al.
Ste
p 1:
Cho
ose
2 di
gits
that
are
iden
tical
(1, 1
or 2
, 2):
2
S
tep
2: C
hoos
e a
digi
t fro
m th
e re
mai
ning
dig
its
(1
, 3 o
r 2, 3
): 2
Ste
p 3:
Arr
ange
the
3 ch
osen
dig
its in
a ro
w:
3!
32!=
N
o. o
f way
s =
()(
)3!
3!2
26
1218
2!
+=
+=