ANGLO-CHINESE JUNIOR COLLEGE MATHEMATICS DEPARTMENT MATHEMATICS Higher 2 Paper 1 18 August 2015 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
Transcript
ANGLO-CHINESE JUNIOR COLLEGE
MATHEMATICS DEPARTMENT
MATHEMATICS
Higher 2
Paper 1 18 August 2015
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
Anglo-Chinese Junior College
H2 Mathematics 9740: 2015 JC 2 Preliminary Examination Paper 1
Page 2 of 6
ANGLO-CHINESE JUNIOR COLLEGE
MATHEMATICS DEPARTMENT
JC2 Preliminary Examination 2015
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 /3
2 /5
3 /6
4 /6
5 /8
6 /6
7 /9
8 /8
9 /9
10 /9
11 /10
12 /12
13 /9
Summary of Areas for Improvement
Knowledge (K) Careless
Mistakes (C)
Read/Interpret
Qn wrongly (R) Formula (F) Presentation (P)
/ 100
Index No: Form Class: ___________
Name: _________________________
Anglo-Chinese Junior College
H2 Mathematics 9740: 2015 JC 2 Preliminary Examination Paper 1 Page 3 of 6
1 Use the substitution 23u x to find 3x √ 23 x dx. [3]
2 Using an algebraic method, solve the inequality 3
1 2 1
x x
x x
. [3]
Hence solve the inequality 3
1 2 1
x x
x x
[2]
3 ABCD is a rectangular field whose sides, AB and BC, measure 2a m and a m respectively. A
road runs along the side AB. A man, starting from A, wishes to reach the opposite corner C in
the shortest possible time. He can walk along the road at 100 m per minute and across the
field at 60 m per minute. Find an expression for the time, in minutes, he will take if he walks
along the road to P, a point x m from B, and then across the field from P to C. [2]
Use differentiation to find, in terms of a, the value of x for the time taken to be the shortest
possible. Find, also, the shortest possible time taken, and prove that it is the minimum. [4]
4 In triangle ABC, AC = 1, BC = 3 and angle CAB = radians.
(i) Show that cosAB √ 29 sin . [3]
(ii) Given that is a sufficiently small angle, show that 2AB a b c , for constants a,
b and c to be determined. [3]
5 Given that 12 tand
1 ed
xyx
x
, where 1tan x denotes the principal value, and that y = 1 when
x = 0, show that 2
2
2
d d1 1 2
d d
y yx x
x x . [2]
By repeated differentiation of this result, find the Maclaurin series for y, in ascending powers
of x, up to and including the term in x3. [3]
Verify that 1tane xy
is a solution of the differential equation 12 tand
1 ed
xyx
x
. [1]
Show that the series expansion for 1tane x
, up to and including the term in 3x , can be
expressed as 1tan 3e ex x kx
, where the numerical value of k is to be determined. [2]
[Turn Over
Anglo-Chinese Junior College
H2 Mathematics 9740: 2015 JC 2 Preliminary Examination Paper 1 Page 4 of 6
6
The diagram shows the curve with equation cos cos2
xy x for 0 x . The curve crosses
the y-axis at 1y and the x-axis at x and x .
(i) Find 2
3
3
cos cos d2
xx x
,
leaving your answer in the form a b√2 + c√3, where a, b and c are rational numbers to
be determined. [5]
(ii) Explain why 2
3
3
cos cos d2
xx x
is smaller than your answer in (i). You may make
reference to the graph. [1]
7 A sequence 1u , 2u , 3u , … is given by
1 1 2
2 12 and ln 1 2 for 2.
1n n
nu u u n
n
(i) Use the method of mathematical induction to prove that for all positive integers n,
2 lnnu n n . [4]
(ii) Hence find 2
15
e run
r r in terms of n. [3]
(iii) Give a reason why the series 2
1
e ru
r r
converges, and write down its exact value. [2]
8 In order to model a particular predator-prey relationship, a biology student came up with the
following differential equations:
d
1d 100
x x
t (A)
d
100d
yx
t (B),
where the variables x and y denote the number (in thousands) of predator and prey
respectively, t days after the start of the observation. There were 50 000 predators at the start
of the observation.
(i) By solving equation (A), show that 0.01100 e tx k , where k is a constant to be
determined. [4]
(ii) What can you say about the population of the predator after several years? [1]
(iii) In the long run, the model shows that number of prey approaches 5 million. Using your
answer in (i), find y in terms of t. [3]
O x
1
cos cos2
xy x
y
Anglo-Chinese Junior College
H2 Mathematics 9740: 2015 JC 2 Preliminary Examination Paper 1 Page 5 of 6
9 The curve C is given by the equation y =
1
x+
2
x2, x ¹ 0 .
(i) Without using a calculator, find the set of values that y can take. [2]
(ii) Sketch the curve C, stating the equations of any asymptotes and the coordinates of any
turning points and points of intersection with the axes. [3]
Given that the solution of the inequality 2
2
1 2ax bx c
x x is the set
: 1.5 1or 1x x x k ,
find the values of a, b and c. [3]
Hence find the value of k. [1]
10 The points A and B have coordinates ( 1, 3, 3) and (1, , 5)k respectively, where k .
(a) (i) Find the length of projection of OB on OA in terms of k. [2]
(ii) State the value of k that gives the shortest length of projection. State also the
relationship between OA and OB at this value of k. [2]
(b) Let 2k . The point C is the reflection of the origin O in the line AB.
(i) Find the position vector of C. [3]
(ii) Find the exact area of the quadrilateral OACB. [2]
11 The complex numbers p and q are given by 2ik and 3 3i respectively, where k ,
0k .
(a) P( )x is a polynomial of degree n with real coefficients where the coefficient of nx is 1.
Given that p and q are roots of P( ) 0x , state the least possible value of n. For this
value of n, express P( )x as a product of quadratic factors with real coefficients. [3]
(b) (i) The complex number 2i
2
q
p has modulus
9
4 and argument , where .
Without using a calculator, find the exact values of k and . [4]
(ii) Solve the equation 2
4i2
qz
p , expressing your answers in the form
ier where
0r and . [3]
12 The curve C has parametric equations 2
, 1 1
u ux y
u u
, where 1u .
(i) Express d
d
y
x in terms of u. [2]
(ii) Given that u is increasing at a rate of 2 units per second, find the rate at which d
d
y
x is
increasing when u = 1. [2]
(iii) Find the acute angle between the tangent at x = 0.5 and the normal at u = −0.5. [4]
(iv) The distinct points P and Q on the curve have parameters p and q respectively. If the
tangents at P and Q intersect the y-axis at the same point, show that 0p q . [4]
[Turn Over
Anglo-Chinese Junior College
H2 Mathematics 9740: 2015 JC 2 Preliminary Examination Paper 1 Page 6 of 6
13 The owner of a newly opened café decided to rent a painting from an artist as part of the
decoration of the café. They set about drafting up a contract for the terms of the rental.
The artist proposed a rental contract (Version 1) stating that the owner will pay the artist $15
for the 1st day of rental and for each subsequent day, the daily rental cost will increase by
$0.50.
(i) On which day of the rental will the owner first have to pay the artist more than $39 as
the daily rental rate? [2]
The owner proposed an alternative contract (Version 2), where the daily rental rate is such
that on the nth day of the rental, the amount of money, in dollars, the owner has to pay to the
artist is given by the function
2
12000f ( ) .
4 4 3n
n n
(ii) Express f ( )n in the form 2 1 2 3
A B
n n
, where A and B are constants to be
determined. [1]
(iii) Hence show that with Version 2 of the contract, the total amount of money the artist
will receive at the end of m days of rental is
12000 14000 .
2 1 2 3
m
m m
[3]
(iv) The artist accepted Version 2 of the contract, and terminated the contract at the end of k
days. Given that the artist received more money in total from Version 2 than if he had
chosen Version 1, find the largest possible value of k. [3]
- End of Paper -
2015 H2 Mathematics Prelim P1 solutions
1 2 2 d3 , 3 , 2 .
d
uu x x u x
x
1
3 2 2
1 3
2 2
5 3
2 2
5 32 22 2
13 d 3 d
2
13 d
2
1
5
13 3 .
5
x x x u u u
u u u
u u c
x x c
2
2
3
1 2 1
30
2 1 1
30
2 1 1
( 1) ( 3)(2 1)0
(2 1)( 1)
( 1)0
(2 1)( 1)
x x
x x
x x
x x
x x
x x
x x x x
x x
x
x x
12
1, 1x x
Replace x by x
12
12
1 12 2
1, 1
1 , 1
, 1
x x
x x
x x
3
2 2
2 2
2 2
2 2
2 2 2
2 2
2
100 60
1 1 2
100 60 2
10 0
100 60
1
53
25 9
16 9
3 3 or (rejected)
4 4
a x x aT
dT x
dx x a
dT x
dx x a
x
x a
x x a
x a
x a x a
B x A P 2a − x
C D
a
22 2
2 2 2 2 2
2 2 2
3/ 22 2
2
3/ 22 2
21 1
60 2
60
60
x xd Tx a
dx x a x a
x a x
x a
a
x a
2
2
3 3When , 0 is min when
4 4
d Tx a T x a
dx
2
21 3 1 3min 2
100 4 60 4
80 48
minutes30
T a a a a
a a
a
4(i)
22 2
2
In , cos , sin
In , 3 9 sin
cos 9 sin (shown)
CNA AN CN
CNB NB CN
AB AN NB
OR
22 2
2
2
2
2
By cosine rule, 3 1 2 1 cos
2 cos 8 0
2cos 4cos 32
2
cos cos 8 (since 0)
cos 9 sin (shown)
AB AB
AB AB
AB
AB AB
AB
4(ii)
2
12 2 2
11 2 2
2 2
22
2
cos 9 sin
11 9
2
11 9 1
2 9
1 11 3 1 ...
2 2 9
24
3
2 4, 0,
3
AB
a b c
B A
3 C
1 θ
N
5
1
1
2 tan
2 tan2
2 2
22
2
1
1 21
1 1 2 (shown)
x
x
dyx e
dx
d y dy ex x
dx dx x
dy
dx
d y dyx x
dx dx
22
2
3 2 22
3 2 2
3 22
3 2
1 1 2
1 2 1 2 2
i.e. 1 1 4 2
d y dyx x
dx dx
d y d y d y dyx x x
dx dx dx dx
d y d y dyx x
dx dx dx
2 3
2 3When 0, 1, 1, 1, 1
dy d y d yx y
dx dx dx
2 3
2 3
1 ...2! 3!
1 ...2 6
x xy x
x xx
1 1 1tan tan 2 tan
2
1 1 (verified)
1
x x xdy dyy e e x e
dx x dx
1
1
1
2 3tan
2 3
2 3 2 3 3tan
3tan
1 ...2 6
1 ...2! 3!
1 ... 1 ...2 6 2! 3! 3
3
1
3
x
x
x x
x x
x xy e x
x xe x
x x x x xe e x x
xe e
k
6(i) Method I (Factor Formula)
0 or .2
y x x
2
3
3
2
2 3
3 2
2
2 3
3 2
2
2 3
3 2
cos cos d2
cos cos d cos cos d2 2
1 3 3cos cos d cos cos d
2 2 2 2 2
1 2 3 1 2 32sin sin 2sin sin
2 2 3 2 2 2 3 2
1 2 2 1 22 1 3 0 2
2 3 3 2 3
5
xx x
x xx x x x
x x x xx x
x x x x
24 12 3 unit .
6 3 2
Method II (Other trigonometric identities)
0 or .2
y x x
2
3
3
2
2 22 3
3 2
2
3 32 3
3 2
22
3
2
23
2
cos cos d2
cos 2cos 1 d cos 2cos 1 d2 2 2 2
2cos cos d 2cos cos d2 2 2 2
2cos 1 sin cos d2 2 2
2cos 1 sin cos d2 2 2
cos 2cos s2 2
xx x
x x x xx x
x x x xx x
x x xx
x x xx
x x
2
2 22 3
3 2
2
2 33 3
3 2
2
in d cos 2cos sin d2 2 2 2
4 42sin sin 2sin sin
2 3 2 2 3 2
2 1 3 22 1 3 2
3 6 2 3
5 4 12 3 unit .
6 3 2
x x x xx x
x x x x
(ii) 2
3
3
cos cos d2
xx x
is the absolute difference between the areas above and below the x-
axis, while (i) measures the sum of the areas above and below the x-axis.
7(i) Let Pn be the proposition 2 ln for .nu n n n
When n = 1, LHS = 1 2u (given).
RHS = 2 ln1 1 2 .
Since LHS = RHS, P1 is true.
Assume Pk is true for some , i.e. k 2 lnku k k .
We want to show that Pk+1 is also true, i.e.
1 2 ln( 1) ( 1) 2ln( 1) 2( 1)ku k k k k .
LHS = 1 2
2 1ln 1 2k k
ku u
k
2
2 12 ln ln 1 2
kk k
k
2
2
2 12ln 2 ln 2
k kk k
k
2 22ln 2 ln 1 ln 2k k k k
2ln 1 2 1k k
= RHS. Pk is true Pk+1 is true.
Since P1 is true and Pk is true Pk+1 is true, by mathematical induction, Pn is true for
for all n .
(ii) 22ln 2 22
2 2 215 15 15 15
.e e e
eru r r rn n n n
r
r r r r
r
r r r
2
1430 2
2
28 2
30 28
28 2
2
2
15 15
e
e 1 e
1 e
1 e.
e e
e e.
e 1
er
n
n
n
n nr
r r
(iii)
2
2
11
ee r
r
r
u
r r
converges as 2e 1 .
2
2 2
2
1
e 1
1 e e 1e r
r
.
(Alternatively, students may use part (ii) by taking n to infinity, and then adding on the
sum of the first 14 terms.)
8(i) d 100
d 100
x x
t
0.01
0.01
1 d 0.01 d
100
ln 100 0.01
100 e , e
100 e .
t k
t
x tx
x t k
x A A
x A
When 0, 50 50.t x A 0.01100 50e .tx
(ii) As , 100.t x
The population of the predators approaches 100 000 after several years.
(iii) 0.01d50e
d
ty
t
.
0.01
0.01
d d 50e d
d
5000e .
t
t
yx t
x
y c
As , 5000 5000t y c . 0.015000e 5000.ty
9(i) 2
1 2y
x x
Method 1
2 2
2
1 22 2 0y yx x yx x
x x
For range, there must be solutions for x, 18
Discriminant 1 4( )( 2) 0y y
Solution set = 18:y y
Method 2
2 3
1 40 4
dyx
dx x x
2
2 3 4
2 12d y
dx x x . When 4,x
2
2
2 12 10
64 256 64
d y
dx
1( 4, )
8 is minimum point
Thus Solution set = 18:y y
(ii)
Asymptotes: 0, 0x y
At 29
1.5,x y , 1, 1x y , 1, 3x y
Therefore, substituting into the quadratic curve,
18
( 4, )
x
y
( 2,0)
29
2.25 1.5
1
3
a b c
a b c
a b c
From GC, 2029 9, 1,a b c .
Plot the graph 2 202
9 9y x x .
From GC, 6k
10(a)(i) Length of projection of OB on OA
OA
OBOA
.
1 113
1 9 9 5 3k
.
1 3 15
19
k
14 3
19
k
10(a)(ii)
Shortest length of projection is 0, and it occurs when 143
k .
OA and OB would be perpendicular.
10(b)(i) 1 1 22 3 5
5 3 2AB
Equation of AB is
1 23 5 , 3 2
r
Let N be the foot of perpendicular from
O to AB.
Then
1 23 53 2
ON
for some λ
ON AB ⇒
1 2 23 5 5 03 2 2
.
⇒ 2 4 15 25 6 4 0
⇒ 33 11
⇒ 13
OC 2ON
23
5323
1
2 3
3
12
43 11
or
2 / 3
8 / 3
22 / 3
A B
C
O
N
10(b)(ii) Area of quadrilateral
= 12
2 OA OB × OR 12
2 OA AB ×
1 23 53 2
×
6 15( 2 6)5 6
2 2 221 8 1
506 11(a) Since P( )x has only real coefficients and p and q are complex roots of P( ) 0x , then
*p and *q are also complex roots of P( ) 0x .
Hence least n = 4
P( )x * *x p x p x q x q
[( ) 2i][( ) 2i][( 3) 3i][( 3) 3i]x k x k x x
2 2 2 2[( ) (2i) ][( 3) (3i) ]x k x
2 2 2( 2 4)( 6 18)x kx k x x
11(b)(i) 2ip k ⇒
2 4p k
3 3iq ⇒ 2 23 3 3 2q
; 1 3
3 4arg( ) tanq
Method 1: 2i
2
q
p
2
2
q
p
2
2
(3 2)
2 4k
2
9 9
4 4k
Method 2: 2i
2
q
p
2i(3 3i)
2( 2i)k
i(9 18i 9) ( 2i)
.2( 2i) ( 2i)
k
k k
2
18( 2i)
2( 4)
k
k
2
9( 2i)
4k
k
2i
2
q
p
2 2
2
92
4k
k
2
9 9
4 4k
216 4k
16 4k 2 3 or 2 3k k (NA since 0k )
Method 1:
When 2 3k ,
1 262 3
arg( ) tanp
Method 2:
When 2 3k ,
2i
arg2
q
p
2i
arg2
q
p
2arg(i) arg( ) arg(2 )q p
22 4 6
6
1 2tan
2 3
6
11(b)(ii) 24i
2
qz
p
2
4 1 i
2 i i
qz
p
4z
2i
2
q
p
2 i69
e4
k
, 0, 1, 2k
z
1i4 24 29
e4
k
, 0, 1, 2k
13 11 23i i i i
24 24 24 243 3 3 3e , e , e , e
2 2 2 2
12(i)
2 2
22 2
2 2
22
2 2
1 1
1 1 1
1 2 2
1 1 1
2 1 2
1 1
u uu dxx
u du u u
u u uu dy u uy
u du u u
dy u uu u
dx u u
12(ii)
2, 1
2 2 2
8 units per second
duu
dt
d dy d dy du
dt dx du dx dt
u
12(iii) 1When (i.e. 1), 3
2
1Gradient of tangent at is 3
2
1 1 3When , 1
2 4 4
1 4Gradient of normal at is
2 3
dyx u
dx
x
dyu
dx
u
1 1 4 acute angle between them is tan 3 tan
3
71.57 53.13 18.4
12(iv) Equation of tangent at P is
22 2
1 1
p py p p x
p p
At y-axis, when x = 0,
2 22
22 1
1 1 1
p p p p ppy p
p p p
Tangent at P cuts the y-axis at 20, p
Similarly, tangent at Q cuts the y-axis at 20, q
2 2
Since , 0 (shown)
p q p q
p q p q p q
13(i) To find least n such that
15 1 0.5 39nT n .
0.5 24.5
49.
n
n
Therefore, on the 50th day of rental the owner will first have to pay the artist more than
