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Page 1 of 20
CATHOLIC JUNIOR COLLEGE
H2 MATHEMATICS
JC2 PRELIMINARY EXAMINATION PAPER 1 SOLUTIONS 2015
1 Complex Numbers
No.
Assessment Objectives Solution Feedback
Solve complex roots of quadratic
equations [A02] 2
2 2
4 3 22
w z z
z z z
4 3 22 2 3 10 0z z z z
4 3 2 22 3 3 10 0z z z z z
4 3 2 22 3 10 0z z z z z
2 3 10 0w w
5 2 0w w
5w or 2w 2 5z z 2 2z z 2 5 0z z 2 2 0z z
51
2
1 4z
1 41 2
2z
11
2
2
1
2
7
i1
2
7
Many students did not see the trick of finding 2w to simplify the equation. A large number
of students used long division or rearranging
the equations to get to 2 3 10 0w w .
Many students did not realise that the set of
real numbers is a subset of the set of complex
number and wrongly rejected the solutions
211
2z
.
Some students wrongly rejected the solutions
7i1
2z
not realising that the z is an
element of the set of complex numbers.
Page 2 of 20
2 Vectors
No.
Assessment Objectives Solution Feedback
Ratio theorem, properties of dot
product [A02] Method :
3 2
5OP
b a
Since F lies on line passing through points O and A , OF a
for a lambda value.
3 23
55 5
2FP
a b a
b a
Since FP is perpendicular to OA
2
2
3 20
5 5
3 20
5 5
3 2cos
5 5
3 π 24 1 cos 1 0
5
0
0
3 5
8
5
FP OA
b a a
b a a a
b a a
Therefore 8
5OF a
This question was not well attempted by
students. Only a minority of the students
managed to get the full credits.
Common mistakes:
1) Did not use ratio theorem to find p .
2) Wrongly used the projection vector
formula. Instead of ˆ ˆp a a , a number of
students used ˆ ˆp a a , credit was given to
the use of wrong formula since the angle
be between a and b is acute.
3) A large number of students cannot
differentiate between vector and its
length. Mistakes like
3 4 2 13 2
5 5
b ap was
frequently seen.
4) Many students claimed that length of
projection of p on a is 8
5 hence
8
5p a , full credit was not given as
essential steps are not explained or shown.
5) Many students attempted to use sine and
cosine rule to find the length and were not
successful in doing so.
2
3
Page 3 of 20
Method :
Triangle ABE is similar to triangle APF
3
5
AP AF PF
AB AE BE
cosπ3
1
4 2
2
OE
OB
OE
OE
Then 2 1 1AE OE OA .
Then 3 3 31
5 5 5AF AE
Hence 3 8
15 5
OF OA AF .
5
8OF a
E
2
3
Page 4 of 20
Method (3):
OF is the projection vector of OP on OA .
2
2
3 2
5 1 1
13 2
5
13 cos 2
5 3
13
5
π
14 1 2 1
5
2
8
OA OAOF OP
OA OA
b a a a
b a a a a
b a a a
a
a
cos b aa b
2 a a a
E
2
3
Page 5 of 20
3 Arithemtic & Geometric Progression
No.
Assessment Objectives Solution Feedback
Formula of nth term of arithmetic
series
Solving practical problems using
arithmetic series [A01]
(i) Odd day duration forms AP with 20a , 9d , 38n
38 20 38 1 9
353
U
Most students are able to recall the nU
formula. Those who got it wrong got the
wrong value of n
Formula for sum of a finite arithmetic
series
Solving practical problems using
arithmetic series [A02]
(ii) Odd day duration forms AP with 20a , 9d , 38n
38
382 20 38 1 9 7087
2S
Even day duration form AP with 20a , 10d ,
37n
37
372 20 37 1 10 7400
2S
Total no. = 7087+7400 = 14487
Most students are able to recall the nS
formula. Those who got it wrong got the
wrong value of n for either even or odd day,
or the wrong value of d for the even day
duration.
Few students took a direct sum of 75 days
instead of considering odd and even days,
ending with the wrong answer.
Some students wrongly rounded the exact
answer to 3s.f.
Formula of nth term of geometric
series
Solving practical problems using
geometric series [A02]
(iii) 75th day onwards forms GP with 353a , 0.8r ,
26n
26 1
26 353 0.8
1.33
U
Most students are able to recall the nU
formula. Those who got it wrong got the
wrong value of n
Solving practical problems using
geometric series [A03] (iv) Duration of exercise is too low to be effective towards
the end portion of the 100 days.
OR
Duration of exercise will not be whole number, difficult
to gauge duration accurately
Most students able to answer this part.
Some students gave the wrong answer that the
duration of exercise will go below zero.
Page 6 of 20
4 Mathematical Induction + Sigma Notation + Method of Difference + Conjecture
No.
Assessment Objectives Solution Feedback
[A01] (i) R.H.S. 1
1 2 1 13
r r r r r r
11 2 1
3r r r r
13 1
3r r
1r r =L.H.S.
In general, the whole question was well-done.
Common problems are mainly with the
notations.
(i) Many students thought they could only do
from LHS to RHS and so wasted time working
backwards.
One long-winded method is to expand the
whole expression completely until r3 and then
cancel. Most were correct but time wasted.
Apply the method of difference
[A01]
(ii) 1 2 2 3 3 4 1n n
1
( 1)n
r
r r
1
11 2 1 1
3
n
r
r r r r r r
11 2 3 0 1 2
3
2 3 4 1 2 3
3 4 5 2 3 4
( 1)( 2) ( 1) ( 1)n n n n n n
11 2 0 1 2
3n n n
11 2
3n n n
Most were correct with a minority with
notational errors, especially with n and r.
A small minority could not observe the pattern
and cancel correctly.
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(iii) Let Pn be the statement
1
1( 1)( 2) 1 2 3
4
n
r
r r r n n n n
,where
n
When 1n , L.H.S. 1 2 3 6
1
R.H.S. 1 2 3 4 6 L.H.S.4
1P is true.
Suppose Pk is true for some k , i.e.
1
11 2 1 2 3
4
k
r
r r r k k k k
R.T.P. 1Pk is true, i.e.
1
1
11 2 1 2 3 4
4
k
r
r r r k k k k
L.H.S. 1
1
1 2k
r
r r r
1
1 2 1 2 3k
r
r r r k k k
11 2 3 1 2 3
4k k k k k k k
11 2 3 4
4k k k k R.H.S.
1Pk is true.
Since 1P is true, Pk is true 1Pk is true. By
mathematical induction, Pn is true for all n .
Common error is with the presentation of the
statements.
A small minority could not get the induction
statement correct and confused it with part (ii).
Quite a number did not use n Z+ but used R
or Z.
A few expanded out the whole expression to
compare.
Many could not express the final statement
correctly.
Formulate a conjecture
[A03]
(iv)
1
1 2 3
11 2 3 4
5
n
r
r r r r
n n n n n
Many could not get the conjecture correct.
Some tried to solve it using summation
properties.
Some left it blank, indicating that they could
not understand what is a conjecture.
Page 8 of 20
5 Recurrence Relations
No.
2
Assessment Objectives Solution Feedback
Use a recurrence relation
Identify a geometric progression
given by recurrence relation
[A02]
(i) 1
11
22 2n nv v
1
11
22 1n nv v
1
1
2n nv v
Since 1 1constant
2
n
n
v
v
, sequence V is a GP.
-Many students were able to find the
recurrence relation between the terms in
sequence V. Some didn’t use direct substitutions so they took a detour to reach the
answer. Those who failed to find the relation
mostly made mistakes in calculation.
-The last part was however poorly done.
Though it was apparent that most knew that it
suffices to show that the ratio between
adjacent terms is a constant, many failed to
present the argument in a logical manner.
-Some used the term “common ratio” before they showed that the sequence is GP, which
has to be true first before “common ratio” makes sense. Some stopped at showing the
ratio is half but didn’t say anything about the sequence. These incomplete answers won’t warrant the mark.
-Another few students attempted to deduce
the general formulae for the sequences. One
problem with this approach is that it is not
rigorous because it is done by “observation” or “pattern recognition”. Another problem is that the fact that the derived formulae
resemble the form of the nth term formula for
a GP doesn’t directly lead to the conclusion that the sequence is GP, which by definition
has to be proven to be a sequence with a
constant ratio between adjacent terms.
Understanding of the convergence of
a sequence
[A02]
(ii) Method :
1 1 2 1v u
1
11
2
n
nv
Many unsuccessful attempts in this part
revealed that the students lacked conceptual
understanding of the convergence of a
sequence. They tried instead to find the sum
to infinity of the sequences, which is a totally
different concept.
Page 9 of 20
when n , 1
11
20
n
nv
.
So the limit of sequence V is 0.
Since 2n nu v , when n , 2nu .
So the limit of sequence U is 2.
Method :
Let the limit of U be L . According to the given
recurrence relation, when n ,
11
2L L
So 2L , i.e. the limit of sequence U is 2.
Since 2n nv u , when n , 2 2 0nv .
So the limit of sequence V is 0.
Understanding of the convergence of
a series [A02]
(iii) For V , sum to infinity1
21
12
.
This part was well done in general. Some
insisted that sum to infinity must be a positive
number due to unknown reasons.
Understanding of the convergence of
a series
[A02]
(iv)
1
1( 1) 1
2 1 12 1 2 2
1 2 21
2
n
n nn
r
r
v
1 1 1
12 2 2 2 2
2
nn n n
r r r
r r r
u v v n n
When n , 1
22
0
n
,
2n ,
thus 1
n
r
r
u
, i.e., sum to infinity doesn’t exist.
Many didn’t attempt this part. Those who attempted seriously had some idea about the
right direction to go. However the question
was phrased in such a way that the sum to nth
term expression of sequence U must first be
found accurately, which could be the reason
many could not score any mark.
Page 10 of 20
6 Vectors
No.
Assessment Objectives Solution Feedback
Finding acute angle between 2 planes
[A01]
(i) Let the angle between 1p and 2p be .
1 2 1 2
2 5 2 5 cos
4 7 4 7
40 21
c
cos
0.153 rad or 8
os
78
.8
1 2 1 2
n n n n
Some careless mistakes were seen.
Some students provided the obtuse angle
instead.
Finding the line of intersection
between 2 planes [A01] (ii)
A vector equation of line l is
19 6
6 1 ,
0 1
r .
Some students keyed into the GC wrongly and
achieved the wrong answer.
Intersections between 3 planes [A02] (iii) Since the three planes does not have common point of
intersection, the line l is parallel to 3p .
6 1
1 0
1 3
6 3 0
3
a
a
a
Since 3p contains the point 1,1,1 ,
1 1
1 3
1 3
1 3 3
7
b
b
b
Most students are able to solve this part.
Except some careless mistakes were seen.
Page 11 of 20
Distance between a line and a plane
[A02]
(iv)
18 11
7 319
1 3
18 21 3
19
6
19
619 units
19
h AB
3
n
Only some students are able to solve this part.
A variety of wrong approaches were seen.
Page 12 of 20
7 Maclaurin’s Series + Small Angle Approximations + Binomial Theorem
No.
Assessment Objectives Solution Feedback
Interpret f '( )x and perform simple
differentiation. [A01]
(i) 2
1f( )
1x
x x
2
2
1 2f '( )
1
xx
x x
Most of the students got the answer.
Some of the students cant interpret f '( )x ;
they find the inverse function instead.
Binomial expansion [A01] (ii)
12
f( )
1
x
x x
2 32 2 2
1 2 1 2 31 ( 1)( ) ..........
2! 3!x x x x x x
2 2 3 3
3
1 2 ..........
1
x x x x x
x x
Common mistake:
Did not expand till 3x .
Complete the square
21 3
2 4x
instead.
Some applied differentiation to get the
expansion, leading to much longer working.
Able to link Binomial and maclaurin
series and apply small angle
approximation [A02]
(iii)
22
2
2
(1 2 )sin (3 )
1
f '( ) sin (3 )
1 3 3 ........
1 3 3 ......
xx
x x
x x
x x
x x
Majority can’t see the link to part (i); hence either solved it by using applying binomial
expansion formula or differentiation.
A number of students have no knowledge of
small angle approximation by writting
sin(3 ) xx
Validity of binomial expansion and
understand the use of approximation.
[A03]
(iv) For part (ii), the validity of the expansion: 2 1x x
From G.C., 1.618 0.618x .
Hence, it is not valid to use the answer in part (ii) to find
the approximation value of 1
0f( ) dx x
Badly done.
However some were able to score 1 mark by
writting 2 1x x ; but unable to solve it .
Page 13 of 20
8 Applications of Differentiation
No. Assessment Objectives Solution Feedback
Able to use differentiation to find
minimum area.
(i) 2310
4x y
2
40
3y
x
232
2A x xy
2 2
2
3 40 3 80 32
2 2 33A x x x
xx
2
d 80 33
d 3
Ax
x x
2
2
80 33 0
3
80
3
xx
xx
1/3
3 80 80
3 3x x
OR 2
2 3
d 1603
d 3
A
x x
When
1/380
3x
, 2
2
d0
d
A
x
least amount of material used when
1/380
3x
x 1/3
80
3
1/380
3
1/380
3
d
d
A
x -ve 0 +ve
Slightly more than half of them were able to
solve part (i).
Common mistakes:
1 310
3 2x x y
233
2A x xy (failed to interpret
“open tank”) Weak algebraic manipulation; hence
failed to solve for x.
1/3
3 80 80
3 3x x
Some of them forgot to verify that A is min.
Page 14 of 20
Able to form the equation of volume
for prism and apply connected rate of
change/ implicit differentiation.
(ii) cos6
h
k
2
3k h
21 2 55
2 3 3
hV h h
d 10 d
d d3
V hh
t t
d 3
d 30
h
t h
or
d d d
d d d
h V h
t t V
d 1 1
d 3 10
3
h
t h
d 3
d 30
h
t h
When 1,h d 3
m /d 30
hs
t or 0.0577 m/s
Average. Very few students were able to
score full marks.
Common mistake:
Find V in terms of x (should use
another letter to denote); hence ended
up in longer working.
Some students perform d
d
V
hfrom the
expression of V in terms of 2
variables.
Page 15 of 20
9 Applications of Differentiation + Parametric Equations
No.
Assessment Objectives Solution Feedback
Able to find dy/dx for parametric
functions and use dy/dx to interpret
the tangents to curve.
(i) d2 1
d
xt
t
d2 1
d
yt
t
d 2 1
d 2 1
y t
x t
As 1
2t , tangent to C tends to vertical line.
As 1
2t ,
d0
d
y
x , tangent to C tends to horizontal line.
In general, only 2 students did not attempt this
question. In fact, all who attempted obtained the
correct d
d
y
x.
Many students did not understand the meaning of
tangent and assumed that ‘tangent’ is ‘gradient’. Students exhibited ‘poor’ use of mathematical language in explaining tangent.
Able to apply part (i) answers to show
the features of the curve. (ii)
Students did not understand fully the features of the
curve. The coordinates of the end points of the graph
must be labelled. The tangents at 1
2t and
1
2t
must be clearly shown.
Able to solve the roots between
parametric and Cartesian equations. (iii)
At 1t , 2, 0x y and d 1
d 3
y
x
0 3 2y x
2 23 2t t t t
22 3 0t t
( 1)(2 3) 0t t
31 rej. point or
2t P t
Q3 15
4, 4
Students performed much algebraic error due to
sheer carelessness, which will otherwise allow them
to score full marks.
Convert parametric to Cartesian form. (iv) 22 ,x y t 2
2
x yx y t t
22
2 ,2 2
x yx yx y x y
This simply involves removing the third parameter t
. However, a number of students went on to solve
using differential equations. Other students who tried
to remove the third parameter often forget to put
when taking .
3 1,
4 4
6, 2
2,6
1 3,
4 4
Page 16 of 20
10 Techniques of Integration + Applications of Integration
No.
Assessment Objectives Solution Feedback
Find area bounded by a curve
(i)
1
2
0
1
2
0
12
0
2
8Area d
1
24 d
1
4 ln( 1)
4ln 2 units
xx
x
xx
x
x
The question was generally well done, with
many students being able to use a sketch to
identify the area to be found and write down
Area
1
2
0
8
1
xdx
x
. However, there were
some students who were unable to recognize
the use of '( )
ln ( )( )
f xf x C
f x to integrate
the above, resorting to the use of Integration
by parts, which is erroneous.
(0, 0) x
y
(1, 4)
x=1
y=4
Page 17 of 20
Find volume of revolution about the
x-axis
Perform integration by substitution
(ii) Change of limits: 0, 0x
π
1,4
x
2dtan sec
d
xx
212
2
0
242
2
0
π24
2
0
π4
2
0
8π 4 1 π d1
8tan16π π sec d
tan 1
64 tan16π d
sec
16π 64π sin d
xV x
x
p=16, q = 64, 0,4
a b
There were a number of students who did not
remember to change the limits after
substituting tanx .
Most of the students were able to gain credit
for differentiating tanx w.r.t. .
Some students did not recognize that a
cylinder is obtained when we rotate a
horizontal line about the x-axis. Some forgot
that 2V r h while others confused the
height with the radius of the cylinder. There
were some students who did not know that the
volume of revolution about the x-axis is 2V y dx instead of
2V x dy . In the
simplification of the expression after
substituting tanx , there were some
students who did not recognize the use of the
trigonometric identity: 2 2tan 1 sec .
Students who did not explicitly state the
values of p, q, a and b will lose a mark.
Perform integration using double
angle formulae
(iii) 4
0
π4
0
2 3
1 cos216π 64 d
2
sin 216π 32π
2
116π 32π
4 2
32π 8π units
V
Most students were able to gain credit once
they recognize the use of double angle
formulae to convert 2sin into
1 cos 2
2
.
Most were successful with the integration.
However, many lost a mark due to forgetting
to change limits.
Page 18 of 20
11 Graphing Techniques + Functions
No.
Assessment Objectives Solution Feedback
Determine the asymptotes and turning
point of a graph. [A02] (i)
2
2
2
2
f ( ) Asymptotes: and 0
f '( ) 1
For f '( ) 0, 1 0
(rej since 0)
x a ax x y x x
x x
ax
x
ax
x
x a
x a a x
ay
2 ,2a
a a aa
Most students were able to identify the 2
asymptotes. Some students failed to notice
that they are only required to draw for 0x .
Less than half of the students were able to
find the minimum point.
It will be easier to find f '( )x if students had
simplified f ( )x to 1x ax first.
Common mistakes:
Substituting value for a to find
minimum point. Substitution of value
should only be used to find the shape
of graph.
Use of ruler to draw the curve.
Failure to find the equation of the
oblique asymptotes.
Condition for existence of composite
function, rule and domain of
composite function [A01]
(ii) For fg to exist, g fR D .
fD (0, )
gR (1, )
Since g fR (1, ) (0, ) D , fg exists.
2(e +1)fg : ,
e +1
x
x
ax x
since fg gD D
Common mistakes:
Incorrect condition for existence of fg
Incorrect gR such as (0, ), [2, )
Careless in finding fg(x), either
missing out the power 2 or giving the
denominator as x
Incorrect fgD
Students are required to state fD clearly even
though it was given in the question. Answer
must be in similar form as required by
question.
x
y y = x
Page 19 of 20
(iii) Sketch
2(e +1)
e +1
x
x
ay
2
fg
1As , e 0, hence .
1
R (1 , )
x ax y
a
Alternative
Consider the graph of f ( )y x with restricted domain
of (1, ) [which is gR ]. Since 0 1, then 0 1,a a
so the minimum point in the graph in (i) will not be
considered under the restricted domain.
21
f (1) 11
aa
, hence fgR (1 , )a .
A number of students skipped this part. For
those who attempted, those who did not give
any working were not awarded any marks as
question is to find the range of fg, not state.
For those who attempted the question, most
used the method of sketching fg( )y x .
Common mistakes:
Substituting 1
2a to get the
incorrect range as 3
,2
Failure to find equation of horizontal
asymptote correctly
Checking the endpoints only, without
use of graph
For the students who used the method of
restricting the domain of f. Clear presentation
is especially important. Common mistakes:
Incorrect to say fg fR R because for
function f, fR is [2 , )a .
Finding fg(1) instead of f(1)
Failure to make reference to the graph
of f in (i)
Unclear explanation on how the
method is to be used
x
y
y = 1+a
fg( )y x
x
y
1
Page 20 of 20
Restriction of domain of a function so
that the inverse is a function [A01] (iv) Greatest value of k is a . This is a stating question, no explanation is
required.
Finding the expression and domain of
an inverse function.
[A01]
2
2
2
2
2 2
Let
0
( ) 4(1)( )
2
4 4 or
2 2
rej
x ay
x
xy x a
x yx a
y y ax
y y a y y ax x
x a
(Method 1)
2
2 2
2 2
2 2
0
0 2 2
2 2
or 2 4 2 4
rej
x yx a
y yx a
y yx a
y y y yx a x a
x a
(Method 2)
1
21
ff
4f ( )
2
D R [2 , )
x x ax
a
Some students managed to attempt this part
fully, either by quadratic formula or
completing-the-square approach.
Most students did not manage to reject
correctly as they considered 0x instead of
x a . If we use the minimum point as a
guide (where x a and 2y a ), we have
2
yx . Since the restricted domain is
0 x a , we need 2 4
2
y y ax
in
order to have x a .
Common mistakes:
Poor algebraic manipulation in an
attempt to make x the subject
Giving 1fD as fD instead