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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 2 2 w z z z z z 4 3 2 2 2 3 10 0 z z z z 4 3 2 2 2 3 3 10 0 z z z z z 4 3 2 2 2 3 10 0 z z z z z 2 3 10 0 w w 5 2 0 w w 5 w or 2 w 2 5 z z 2 2 z z 2 5 0 z z 2 2 0 z z 5 1 2 1 4 z 1 4 1 2 2 z 1 1 2 2 1 2 7 i 1 2 7 Many students did not see the trick of finding 2 w to simplify the equation. A large number of students used long division or rearranging the equations to get to 2 3 10 0 w 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 21 1 2 z . Some students wrongly rejected the solutions 7i 1 2 z not realising that the z is an element of the set of complex numbers.

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Page 1: CATHOLIC JUNIOR COLLEGE - Supergradessupergrades.com.sg/wp-content/uploads/2016/07/CJC_Prelim2015_P1Sol.pdfCATHOLIC JUNIOR COLLEGE H2 MATHEMATICS JC2 PRELIMINARY EXAMINATION PAPER

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.

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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

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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

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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

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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.

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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.

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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.

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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.

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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.

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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.

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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 .

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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.

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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.

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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

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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

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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.

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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

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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: CATHOLIC JUNIOR COLLEGE - Supergradessupergrades.com.sg/wp-content/uploads/2016/07/CJC_Prelim2015_P1Sol.pdfCATHOLIC JUNIOR COLLEGE H2 MATHEMATICS JC2 PRELIMINARY EXAMINATION PAPER

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