1

TEMASEK JUNIOR COLLEGE, SINGAPORE JC 2 Mid-Year Examination 2017 Higher 2

MATHEMATICS 9758Paper 1 30 June 2017

Additional Materials: Answer paper 3 hours

List of Formulae (MF26)

READ THESE INSTRUCTIONS FIRST

Write your Civics group and name on all the work that you hand in.

Write in dark blue or black pen on both sides of the paper.

You may use a soft pencil for any diagrams or graphs.

Do not use staples, 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 a question, you are required to present the mathematical steps using mathematical notations and not calculator commands.

You are reminded of the need for clear presentation in your answers.

The number of marks is given in 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 7 printed pages.

[Turn over

2

1 At the beginning of the year, Mr Lui invested a total of $140 000 in three banks. Bank A pays a fixed amount of interest of $x at the end of each year. Bank B pays a compound interest of 1.5% on the amount invested at the end of each year and Bank C pays an interest of 5% on the amount invested at the end of 5 years.

The amount of money Mr Lui invested in Bank B is half of the amount he invested in Bank C. At the end of the first year, Mr Lui has a total of $140800 in the three banks. His investment grew to $141 609 at the end of the second year. Find the total amount of money Mr Lui has in the three banks at the end of 5 years, assuming that he did not make any withdrawal or further investment during the period. [5]

[Solution] Let $a, $b, $c be the amount of money Mr Lui invested in Bank A, B and C respectively.

140000a b c

0.5b c 0.5 0b c

1.015 140800a x b c

22 1.015 141609a x b c

Using GC, a = 20 000, b = 40 000, c = 80 000 and x = 200

Total amount of money Mr Lui has at the end of 5 years

= 5$20000 5($200) (1.015) $40000 (1.05)($80000)

$148091.36

Define the unknowns clearly. Avoid using A, B and C which are the names of the banks. It is more efficient to use GC to solve the simultaneous equations than algebraically. In this context (money), the “non-exact” amount should be rounded off to 2 dec place unless otherwise stated.

Marker’s Comments This question was generally well done. However, some students did not correctly understand the conditions given in the context and thus scored very few marks.

Students who solved the simultaneous equation algebraically instead of using GC wasted precious time (at the expense of other questions) and often make mistakes.

There is also a large number of students who have gotten the simultaneous equations correct but generated wrong answers from GC. Students need to be more careful when entering numbers in GC.

3

2 It is given that 2 3

f ( ) , 02

xx x

x

. Solve the inequality f ( ) 1x using an algebraic

method. [4]

Hence find the range of values of x for which 2f 1x . [2]

2 3

12

x

x

, 0x

2 3

1 02

x

x

2 2 3

02

x x

x

Multiply both sides by 22x ,

2 3 1 0x x x , 0x

Thus, solution is 1x or 0 3x .

Some students sketched this curve

wrongly as

Many students forgot the condition

0x .

For 2f 1x , replace x by 2x .

2 1x or 20 3x

(No real solution 3 0 or 0 3x x 2 0x )

2 1x contributes no solution,

and not “NA” or “reject”.

Solution to 20 3x can be

easily seen from

and not from splitting into

2 0x and 2 3x

Marker’s Comments The first part was generally well done.

Many students gave a poor presentation of the solution to the “Hence” question.

Student’s solution: 1x or 0 3x

Replace x by 2x ,

1x or 0 3x

There should be a clear indication that you are solving the 2nd question by first writing the

question 2f 1x .

0 −1 3

WHY? Is this replacement part of the method for solving the first question ?

3

4

3 The vectors a and b are unit vectors such that × × = 2 +a b b a b .

(i) Find a b b b . [1] 

(ii) Show that the angle between the vectors a and b is 120. [3]

(iii) Find the exact value of a b . [2]

[Solution]

(i) Since a b b

is perpendicular to b

, 0a b b b

Note that cross product is not associative:

a b b a b b

(ii) 2a b b a b

2a b b b a b b

2

0 2a b b

1

2a b

1cos

2a b

, where is the angle between a

and b

1

cos2

120 (Shown)

(iii) 2a b a b a b

= 2 2

2a a b b

1

1 2 1 12

Alternative solution

Using cosine rule, 2 2 2

2 cos60a b a b a b

1

1 1 2 12

Take note that if use cosine rule, must draw diagram to see that the angle

(opposite a b

) is

60o and not 120o.

a

b a+b

120 60

5

Marker’s Comments

(i) Many students leave the answer as 2 1a b

which is not accepted since question did

not say leave answer in terms of a

and b

. Some students went on to wrongly assume

that vector a

and vector b

are perpendicular.

There are also many students who are not aware that cross product is not associative and they did by the WRONG method:

0

0 0

a b b b a b b b

a b

b

(ii) Many did this part (ii) by working backward then they realise that part (i) answer should

be zero. However, they are unable to explain the reason why 0a b b b

. Many

are also stuck at not knowing how to arrive at 1

2a b

.

(iii) Attempt to use cosine rule is common. However, many did not draw diagram and they use 120o instead of 60o.

Mistakes such as b

2 and a b a b

are still common.

In general this is a badly done question. Many students skip part (i) and (iii) with little working for part (ii).

X

6

4 (i) Show that ( 1)! ! !r r r r . [1]

(ii) Find 1! 1 2! 2 2 1 ! 2 1 2 ! 2n n n n in terms of n. [3]

(iii) Hence, find 2

1

2 ! 2n

r

r r

in terms of n. [3]

(iv) Determine with a reason if 1

2 ! 2r

r r

exists. [1]

[Solution]

(i) ( 1)! ! ( 1) ! !r r r r r

1 1 !r r

!r r (Shown)

Alternatively by expansion:

( 1)! ! ( 1) ! !

! ! !

r r r r r

r r r r

!r r

(ii)

1! 1 2! 2 2 1 ! 2 1 2 ! 2n n n n

2

1

!n

r

r r

2

1

( 1)! !n

r

r r

2! 1!

3! 2!

4! 3!

2 1 ! 2 2 !

2 ! 2 1 !

2 1 ! 2 !

n n

n n

n n

(2 1)! 1n

Take note that the last term is given by r = 2n and not r = n.

7

(iii) 2

1

2 ! 2n

r

r r

2 2

2 1

!r n

r

r r

2 2

3

!n

r

r r

2 2

1

! 2 2! 1 1!n

r

r r

(2 2 1)! 1 5n

(2 3)! 6n

We are looking for

2

1

2 ! 2n

r

r r

which is same as

3! 3 4! 4 ..... 2 2 ! 2 2n n Thus any method to change the general term must keep the series the same.

(iv) 2

1

2 ! 2 2 3 ! 6n

r

r r n

Since as ,n 2 3 ! n ,

the sum 1

2 ! 2r

r r

does not exist.

Marker’s Comments

(i) Very well done but there are still some students who do not know the meaning of factorial.

(ii) Very well done except for few cases of ending with r = n. Students must be reminded to show cancellation in their working. There are some students who present their working

in the row form “ 2! 1! 3! 2! ...... 2 ! 2 1 ! 2 1 ! 2 !n n n n ”. Even though

they showed cancellation in the working, this way of arranging the terms should be discouraged as not all students are able to see which terms are being cancelled out.

8

(iii) Very badly done.

Mistake I

Many students replace r by 2r in 2

1

!n

r

r r

. However this leads to mistakes such as:

Did not change upper limit.

Lower limit becomes 1r and students do not know how to handle it.

There are some who successful find 2

1

2 ! 2n

r

r r

by the following correct alternative

method:

2 2 2

1 1

2 2

1

2 2

1

2

1

! 2 ! 2

2 1 ! 1 1! 1 2! 2 2 ! 2

2 ! 2 2 1 ! 6

2 ! 2 2 2 1 ! 6 (Replace 2 by 2 2)

2 3 ! 6

n n

r r

n

r

n

r

n

r

r r r r

n r r

r r n

r r n n n

n

Mistake II Another very common problem is that many students do not know how to do “Hence” after

obtaining2 2

3

!n

r

r r

. The answer 2 3 ! 3!n is written down without any working. A

misconception is due to some students thinking that since lower limit is 3, so can just replace

the 1! in (2 1)! 1!n as 3!.

Other possibility is that the student obtain this answer without using Hence, but using MOD separately instead.

Another common wrong answer is (2 2 2 1)! 1!n

9

(iv) Very badly done. Many students do not understand the meaning of

“ 1

2 ! 2r

r r

exists” . They write down 2 3 ! n and they can wrongly

conclude that 1

2 ! 2r

r r

exists.

Poor presentation such as 2 3 ! 6 6n is common.

There are also a few students who state that there is no common ratio and so

1

2 ! 2r

r r

does not exists, indicating poor understanding of sum to infinity of the

different types of series.

10

5 The parametric equations of a curve C are

4cos 2x a t , 4sin 2y a t

where 1

04

t and a is a positive constant.

(i) Find the equation of the tangent to C at the point P with parameter p. [3]

The tangent at P meets the x-axis and y-axis at point A and B respectively.

(ii) Show that OA + OB depends only on a , where O is the origin. [2] (iii) Find in terms of a, the area of the region bounded by C, the axes and the tangent

of the curve at point P when OA = OB. [4] [Solution]

(i) 3d4 cos 2

d

xa t

t 2sin 2t 3d

4 sin 2d

ya t

t 2cos2t

3

3

d 8 sin 2 cos2

d 8 cos 2 sin 2

y a t t

x a t t

= 2

2

sin 2

cos 2

t

t (or 2tan 2t )

Equation of the tangent at point P (i.e. when t = p) :

2

4 42

sin 2sin 2 cos 2

cos 2

py a p x a p

p

2 4 2 2 2 4cos 2 sin 2 cos 2 sin 2 sin 2 cos 2y p a p p x p a p p 2 2 2 4 4 2cos 2 sin 2 sin 2 cos 2 sin 2 cos 2y p x p a p p a p p

2 2 2 2 2 2cos 2 sin 2 sin 2 cos 2 cos 2 sin 2y p x p a p p p p

2 2 2 2cos 2 sin 2 sin 2 cos 2y p x p a p p

Note: There is no need to

convert 4cos 2t to

2

cos4 1

2

t

using double

angle formula when differentiating. Quite a number of students perform this unnecessary step. A number of students did not sub t = p when finding the equation of the tangent at point P.

11

(ii)  When y = 0, 2 2 2sin 2 sin 2 cos 2x p a p p

2cos 2x a p

Point A = 2cos 2 ,0a p

When x = 0, 2 2 2cos 2 sin 2 cos 2y p a p p

2sin 2y a p

Point B = 20, sin 2a p

2 2cos 2 sin 2OA OB a p a p

2 2cos 2 sin 2a p p

a , depends only on a (Shown)

(iii)

Area = 0

1d

2 2 2

a a ay x

2

0 4 3

4

sin 2 8 cos 2 sin 2 d8

aa t a t t t

2 2

6 8

a a (By GC)

2

24

a units2

Marker’s Comments Overall, this question was very badly done. Although most students knew how to find the gradient, they made mistakes with the differentiation of the trigonometry function. Many also struggled when they tried to simplify the trigonometry terms, especially when trying to find points A and B. Some failed to read the question carefully. They mistook points A and B to be the x and y intercepts of the curve rather than the x and y intercepts of the tangent. Part (iii) was mostly left unattempt. Those who attempt gave the wrong limits (t values) for the integral.

12

6 Mr Ng’s New Year resolution is to learn a new piece of piano music. To achieve that, he comes up with a practice schedule to learn a new piece. He practises for 25 minutes on Day 1, and he increases his practice duration by two minutes each day, until the duration reaches a maximum of one hour. He will then subsequently practise only 1 hour each day.

(i) Show algebraically that Mr Ng first practised for 1 hour on Day 19. [1]

(ii) Show that Mr Ng practised for a total of 60 324N minutes from Day 1 to Day N inclusive, where 19N . [3]

Mr Ng sets a metronome at 100 Beats Per Minute (BPM) throughout his practices. He uses it only during his practices. However, the metronome is faulty. In the first minute, the metronome beats at 100 BPM as set by Mr Ng. For each subsequent minute of use, the number of beats decreases by 0.0005%.

(iii) The number of beats played by the metronome after the last minute of practice on

Day N is 60 324100

Nk

, where 19N . State the value of k . [1]

The metronome is due for calibration when the metronome plays fewer than 99% of Mr Ng’s setting of 100 BPM.

(iv) Determine the day that the metronome is due for calibration. [4]

[Solution] (i) On Day 18, Mr Ng would play for

18 25 (18 1)(2) 59u minutes.

Hence, Mr Ng would play for 1 hour on Day 19.

Students should include a written explanation to answer the question.

(ii)

Method 1 [Direct] From Day 1 to Day 18, Mr Ng would have played a total of

18

1825 59 756

2S minutes.

From Day 19 to Day N, Mr Ng would have played a total of

60 18 60 1080N N minutes

Hence, he would have played a total of 60 1080 756N

60 324N mins(Shown)

Method 2 [Indirect] The number of minutes short of 1 hour played each day follows an AP: 35 (Day 1), 33, 31, … 5, 3, 1 (Day 18).

1835 33 3 1 35 1 324

2 .

If he played for 60 minutes each day from Day 1 to Day N, he

would have played 60N minutes.

He would have played a total of 60 324N minutes. (Shown)

Finding 19S is incorrect

as Day 19 (60 mins) does not follow the AP.

13

(iii) Actual number of beats (accuracy) after Day 19N

60 324100 1 0.000005

N

Hence, 1 0.000005 0.999995k

As 0.999995 is an

exact value, students should not round it off.

(iv)

Method 1 [use result in (iii)] To find the day when the accuracy falls below 1% of 100 BPM, no. of beats in original setting actual no. of beats 1 beat

60 324100 100 1 0.000005 1

N

60 3241 0.999995 0.01

N

60 3240.999995 0.99

N

ln 0.99

60 324ln 0.999995

N

38.9N least integer 39N

Hence, the metronome is due for calibration on Day 39.

Method 2 [use the fact that the number of beats played after each minute follows a GP] Let n be the total number of minutes played.

1

1

1

0.99 100

100 0.999995 0.99 100

0.999995 0.99

ln 0.991

ln 0.9999952011.062

n

n

n

ar

n

n

From (ii), we know that Mr Ng played 60 324N min on Day N.

60 324 2011.062

38.9177

39

N

N

N

Since ln 0.999995 0 ,

the inequality sign must be changed when division is performed. Some students who used this method

mistook n to be the

number of days instead.

We cannot divide n by

60 to find the number of days because Mr Ng did not play for 60 minutes everyday.

Marker’s Comments The question was generally well attempted. Students were able to make use of the results they were required to show to construct their solutions. However, in doing so, students should be very careful of the possible logical lapses in their presentation. In fact, basic algebraic errors caused students to lose their marks even though they were able to answer the question.

14

7 (a) The diagram below shows the sketch of the curve C with equation

3

2 41

xy

x

,

0 1x . Find the exact area of the shaded region R bounded by the curve C, the lines

1

2x , 1y and the y-axis. [3]

(b) The region bounded by the curve 1

1 2y

x

, the line 3x and the x- and y- axes

is rotated through 360o about the y-axis. Find the exact value of the volume of the solid formed. [4]

(c) A wine cask has a radius of 30 cm each at the top and bottom, and a radius of 40 cm at the middle. The height of the cask is 1 m. To find the volume of the cask

by using the formula 2 db

ay x , we lay the cask on its side as shown in the

diagram below. Find a possible equation in the form y = f(x) for the arc AB of the cask. [2]

y = 1

x = x = 1

y

R

x O

15

[Solution] (a) Area of the shaded region R

=

12

30 2 4

11 d

2 1

xx

x

= 1 3

2 2 4

0

1 12 1 d

2 2x x x

1

1 22 4

0

1 14 1

2 2x

141 3

2 12 4

1

43 32

4 2

Use

f '( ) f dn

x x x

= 1f ( )

1

nx

Cn

(b) 1

1 2y

x

1 11

2x

y

When 3x , 1

7y

When 0x , 1y .

Volume of the solid = 1 2 21

7

1d 3

7x y

= 2

1

1

7

1 91 d

4 7y

y

= 1 21

7

2 91 d

4 7y y

y

= 111

7

92ln

4 7y y y

= 1 1 9

1 1 7 2ln4 7 7 7

= 1

3 ln2 7

or 3 ln 72

Common mistakes:

Vol = 2

3

0

1d

1 2x

x

Vol =1 2

0dx y

16

(c) Since points (0, 40) and (−50, 30) and (50, 30) lie on the arc AB, and the arc AB is symmetrical about the y-axis with maximum point at (0, 40),

Method 1:

let equation of the arc AB be: 2y ax b

When x = 0, y = 40 b = 40

When x = 50, y = 30 30 = 502a + 40 1

250a .

Therefore, the equation of the arc AB is

2140

250y x for 50 50x .

Method 2:

let equation of the arc AB be: 2 2

2 21

y x

a b

When x = 0, y = 40 a = 40

When x = 50, y = 30 2 2

2 2

30 501

40 b 2 40000

7b .

Therefore, the equation of the arc AB is 2 2

2

71

40 40000

y x , for 50 50x and 30 40y

Marker’s Comments Part (i) was generally well done with only a minority in each class who failed to recognise

that the integral is of the form f '( ) f dn

x x x and either went to use integration by parts

or wrote their answer as ln 3

2 41 x .

For Part (ii), a few students from each class wrote 2

3

0

1d

1 2x

x

, not knowing that the

question asks for volume of rotation about y axis. Some made mistakes with the limits. For Part (iii), some students did not attempt this part. Some were able to make the correct guess of the type of equation (either parabola or ellipse) but were unable to find the coefficients for the equations.

17

8 (a) Showing your workings clearly, find the complex numbers w and z which satisfy the simultaneous equations

iw z and 2i 2w z . [5]

(b) The complex number v satisfies the equations

3vv and 3arg

2

v

v

.

(i) Find the modulus and argument of v. [3]

(ii) Find the smallest positive integer n such that nv is a real number. [3]

[Solution]

(a) iw z --- (1)

2i 2w z � 2 2iw z --- (2)

Substitute (2) in (1):

2 2i iz z

� 22i 2 i 0z z

� 2 4 4(2i)( i) 2 4 2 2i 1 i

4i 4i 4i 2iz

Thus, 1 i i 11 i

2i i 2z

or 1 i i 1

1 i2i i 2

z

2i 1 i

1 i1+i 1 i

w

2i 1 i

1 i1+i 1 i

w

Alternative solution

iw z --- (1)

2i 2w z � 2

2i

wz

--- (2)

Substitute (2) in (1):

2

i2i

ww

� 2 2 2 0w w

� 2 4 4(2) 2 4 2 2i

1 i2 2 2

w

i.e., 1 iw or 1 iw

� 1 i 2

2iz

1 i 2

2iz

1 i i 11 i

2i i 2

1 i i 1

1 i2i i 2

18

(b)(i) 3vv 2

3v

� 3v since 0v

3

arg2

v

v

� 3arg arg

2v v

� 3arg arg

2v v

� 1 3 3arg

2 2 4v

(b)(ii) nv is a real number

� arg nv k , where k

� argn v k

� 3

4n k

� 4

3

kn , k

Thus smallest positive integer 4(3)

43

n .

Marker’s Comments For (a), some students attempt to solve this problem by first expressing w = a + bi and z = c + di, which results in a system of equations involving 4 unknowns. Many students who use this method did not manage to get the correct answers. This method is not recommended. For (b), students must know what an ‘integer’ means. Some students leave their answer as

4

3n which is clearly not an integer.

19

9 (a) The gradient of a curve C changes with respect to x at a rate of 1

x where 0x .

Given that the line 2 3y x is a tangent to C at the point ex , find the

equation of C where 0x . [5]

(b) A Ricatti equation is a first order differential equation that is quadratic in y . It is of the form

21 2 3 4

df ( ) f ( ) +f ( ) f ( )

d

yx x y x y x

x

where 1 2 3 4f ( ), f ( ), f ( ), f ( )x x x x are functions of x .

It is given that 2 2d 5sin sin + sin cos

d 4

yx x y x x y

x .

By using the substitution sinu x y , show that the given Ricatti equation can

be reduced to 2d 5+

d 4

uu u

x . Hence, find the particular solution of the given

Ricatti equation given that 1y when 6

x

, leaving your answer in terms of .

[7] [Solution]

(a) Rate of change of gradient = 2

2

d 1

d

y

x x

� d

lnd

yx k

x since 0x

When ex , d

2d

y

x �� ln e 2k

�� 1k

� ln 1 dy x x

1

ln dx x x x x cx

��������� lnx x x x c

lnx x c

When x = e, 2e 3y ,

2e 3 eln e c ���� e 3c

Equation of curve C is ln e 3y x x where x > 0.

Common Mistakes

d 1

d

y

x x ,

d 1

d

y

x x ,

d1d

d

yxx x

Did not realise x > 0 and left

answer as d

lnd

yx k

x .

d

ln 2 3d

yx k x

x ( Note that

equation of tangent line d

d

y

x)

Wrong formula when doing integration by part:

1ln dx x x x x c

x

Note: e 2e 3d c

3, 2d c

Many students did not find the value of c.

20

(b) Given: sinu x y

Differentiate wrt x,

d dsin cos

d d

u yx x y

x x

Given DE:

2 2d 5sin sin + sin cos

d 4

yx x y x x y

x

2d 5sin cos sin + sin

d 4

yx x y x y x y

x

Therefore, 2d 5+

d 4

uu u

x ------ (1) (Shown)

2

1 d 1 d

5+

4

u xu u

2

1d 1 d

11

2

u x

u

1 1tan

2u x C

When 6

x

, 1y and thus 1

sin 16 2

u

Hence, 1 1 1tan

2 2 6C

12

C

1

tan12 2

u x

1sin tan

12 2x y x

1cosec tan

12 2y x x

Common Mistakes Did not put brackets at the right

place. E.g. d

cos , sind

yxy x

x

Cannot differentiate

sinu x y     correctly:

E.g. d

cosd

uy x

x  , 

dcos

d

uy x

y

Prove it in long method way.

Wrong separable variable

method: 2

1 5 d d

4u x

u u

Many students cannot do completing square correctly.

  1

tan2

u x C

 

  1 1 1

2 2 4

Did not make y as a subject. Many students solve the DE (1)

wrongly as follow:

When 6

x

,  1y 1

2u

d2

d

u

x  

d4 3

d

y

x

4 3 dy x = …

Marker’s Comments (a) Badly done for this part. Many student do not know that “rate of change of gradient”

is 2

2

d

d

y

x.

(b) Most students can prove the first part result but many of them proved it in a long way. Many students failed to solve the DE.

21

y

x

A(6, 4)

C(3, 0)

2

0

B(0, 4)

10 (a) The graph of f ( )y x is shown below. The curve has a maximum point at

(6,4)A and axial intercepts at (0,4)B and (3,0)C . The lines 2x and

2y are the vertical and horizontal asymptotes respectively.

(i) Sketch the graph of 1

f ( )y

x , giving the equations of any asymptotes, the

coordinates of any turning points and the points of intersection with the axes. [3]

(ii) By sketching the graph of fy x Solve the inequality f 4x . [3]

(b) The graph below shows the curve with equation 2 2

2 2

( )1

x y b

a c

, where a, b and

c are positive constants. Find the values of a, b and c. [3]

Describe a sequence of transformation which would transform the graph of 2

22

13

xy onto the graph of

2 2

2 2

( )1

x y b

a c

. [3]

x

y

O

22

y

x 3 0 2

[Solution]

(a) (i)

(a) (ii)

By sketching out the graph of f ( )y x , range of values of x is

2x or 0x or 2x .

(b) By observation, 2b .

Substitute (1, 2) into 2 2

2 2

( 2)1

x y

a c

,

we have 2 1 1a a since a is positive.

Equations of oblique asymptotes: 2 2c

y x ca

22

2

( 2)curve has equation 1

2

yx

Step 1: Scaling of factor 1

3 parallel to x-axis

Step 2: Scaling of factor 2 parallel to y-axis

Step 3: Translation of 2 units in the positive y-direction

23

Marker’s Comments

(a)(i) The sketching of the 1

f ( )y

x graph is generally well-attempted, with the main

issue being that A and B must be level on the coordinate scale in order to achieve maximum marks.

(a)(ii) Many students sketched the graph of f ( )y x wrongly which resulted in them

having confusion with the inequalities. Most students fail to notice that the

inequality holds when 0x . Students who used “and” instead of “or” for their

solutions were penalised. (b) Descriptions of transformations were allowed if reasonable as there were many

variations. For example, the expected translation was presented as a “move” or “shift”, and a scaling parallel to the y-axis was presented as a scaling up the y-axis. Students lose marks by attaching “units” to scale factors, and by getting the sequence of transformations wrong.

24

11 The function f is given by

22 for 0,

f 12 for 0.

2

x xx

x x

(i) Define 1f in a similar form. [4]

(ii) Sketch the graphs of fy x and 1fy x on the same diagram. Without the

use of graphing calculator, find the exact solution(s) of the equation

1f fx x . [5]

Another function g is given by g( ) ln 5x x , 5x .

(iii) If the domain of f is now restricted to the interval [ 1, 0] , show that the

composite function gf exists. Find an expression for gf x , stating its domain

and range in exact form. [4]

[Solution]

(i) 22 for 0,

f 12 for 0.

2

x xx

x x

Let 22y x 2 2x y

2x y

Since 0,x 2x y .

Also, let 1

2 2 42

y x x y

Hence, 1 2 for 2,f

2 4 for 2.

x xx

x x

25

(ii)

At the intersection, 1f ( ) f ( )x x x .

22 x x

2 2 0x x

1 1 4( 2)

2x

2x   or   1x

Since 0x , 2x .

Also, 1

22

x x        4x

Hence, the exact solutions of 1f fx x are 2, 4.x

26

(iii)   Domain of f restricted to [ 1,0] fR [1,2]

Since f g[1,2] R D ( ,5) , gf exists

gf fD D [ 1,0]

2gf ( ) ln 5 2 , 1 0.x x x

From the graph of gf ( )y x , gfR [ln 3,ln 4]

Marker’s Comments In general, students should improve on their solution (set notations, etc) for function questions.

(i) Most students successfully considered the different domains and were able to express x in terms of y in order to find the inverse function. However, a number of students were unable to determine the domain of inverse function.

(ii) The graphs were not drawn properly in general. Many students did not include details of the graph (part parabolic in shape, axial intercepts, symmetry between

graphs). The part involved the finding the solution of 1f fx x was not well

done. A number of students were not aware that they can solve the question by

equating f x x . Students who ended up with 3 solutions of more, showed

complete ignorance of the graphs they had drawn in the earlier part (2 intersections). A handful of students were penalised for writing out the solution in coordinates, indicating the fact that they did not understand what they were solving.

(iii) Many problems arose when students were unable to write out the domain and range of the functions accordingly and show that the composite function exists. A number of students were not paying attention to the question, resulting in loss of marks, not stating the rule, domain and range of the composite function.