2013 Preliminary Examination H2 Mathematics 9740 Paper 1 (Solutions)

1 If two curves y = ax3 + bx and y = ln (px

2) intersect at the points where x = 1, x = 2 and

x = 4, find the values of a, b and p, correct to 2 decimal places. [4]

[Solution]

At the points of intersection, ax3 + bx = ln (px2)

When x = 1, a + b = ln p a + b ln p = 0 ------- (1)

When x = 2, 8a + 2 b = ln (4p) 8a + 2 b ln p = ln 4 ------- (2)

When x = 4, 64a + 2 b = ln (16p) 64a + 2 b ln p = ln 16 ------- (3)

Using GC (with unknowns a, b and ln p):

a = 0.01

b = 3.557 b = 12.65

ln p = 3.545 p = 34.63

Markers Comments: About 50% of the cohort could not solve the question. They tried to solve it without using GC, and working is very tedious.

2 A police officer discovered the dead body of a murdered man. The body was located in a

room that was kept constant at 20oC. According to Newtons law of cooling, the body will

radiate heat energy into the room at a rate proportional to the difference in temperature

between the body and the room. Let T oC be the body temperature at time t minutes.

(i) Write down a differential equation relating T and t. [1]

(ii) By solving this differential equation, show that 20 e k tT A where A and k are

constants. [3]

The coroner arrived at the scene at 10.40 pm and the body temperature of the victim was

measured to be 34oC. He made another measurement of the temperature after 90 minutes and

temperature was 30oC. It is assumed that the victims temperature was 37oC at the time of

death. Find the time of the victims death. [3]

[Solution]

(i) d

20 , 0d

Tk T k

t

(ii) 1

d d20

T k tT

ln 20 T kt c or ln 20 > 20T kt c T

20 e k t cT

20 e ec k tT

20 e where = ekt cT A A

Let x be the time from death to the point where his body temperature reaches

340C.

When t = 0, 37 = 20 + AA = 17

When t = x, 34 20 17e kx 17e 14 (1)kx

When t = x + 90, 90

30 20 17ek x

90

17e 10 (2)k x

90

(1) e 14Taking , we have

(2) 10e

kx

k x

9014

e10

k

7

90 ln5

k

1 7

ln90 5

k

Sub into (1) and solving for x, 1 14

ln17

xk

= 51.9 52

The victim died at 9.48pm.

Markers Comments: (i) Well done. (ii) Generally well done except for a minority of students who did not put the arbitrary

constant after integrating. Many students (about 70% of the cohort) get the last part correct. Those who did not

get it correct make the following mistakes: Sub in the wrong condition (when t = 90, T = 30) Did not write down the time.

3 The rth term of a sequence is given by 2 2

8

(2 1) (2 1)r

ru

r r

, for r = 1, 2, 3, ..

The sum of the first n terms is denoted by Sn.

(i) Given that 18

9S and 2

24

25S , find the exact values of S3 and S4. [1]

(ii) By considering the values of 1 nS for n = 1, 2, 3 and 4, or otherwise, make a

conjecture for a formula for Sn in terms of n. [1]

(iii) Prove your conjecture by the method of mathematical induction. [4]

(iv) Find the smallest integer n for which the sum Sn differs from 1 by less than 105

.

[2]

[Solution]

(i) 18

9S , 2

24

25S , S3 =

2 224 24 48

25 495 7 and S4 =

2 248 32 80

49 817 9

(ii) 1 21 1

19 3

S

2 21 1

125 5

S

3 21 1

149 7

S

4 21 1

181 9

S

2

11

2 1nS

n

Hence, a conjecture for Sn

2

11

2 1n

.

Markers Comments:

Part (i) and (ii) are generally well done except for some weaker students who mistook nS as nU

and hence they cannot make any conjecture at all. Quite a number of students did not make Sn in

terms of n and leave their conjecture as

2

11

2 1nS

n

.

One common wrong conjecture is

2

11

2nS

n

. These students only check that they can get

1 S1 but not the rest. They either did not recognize the odd numbers or they do not know how to

write down the general form of an odd number.

(iii) Let Pn be the statement: Sn 21

1(2 1)n

, for all positive integers n.

When n = 1, 18

LHS9

S and 2

1 8RHS 1

3 9 . P1 is true.

Assume Pk is true for some positive integer k. i.e. 21

1(2 1)

kSk

When n = k +1, need to prove that Pk+1 is true,

i.e.

1 2 2

1 11 1

2 32 1 1kS

kk

1 1LHS k k kS S u

2 221 8 1

1(2 1) 2 1 2 3

k

k k k

221 8 1

1 1(2 1) 2 3

k

k k

2

22

1 2 3 8 11

(2 1) 2 3

k k

k k

2

22

1 4 12 9 8 81

(2 1) 2 3

k k k

k k

2

22

1 4 4 11

(2 1) 2 3

k k

k k

2

22

2 111

(2 1) 2 3

k

k k

2

11

2 3k

= RHS

Hence, 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 all

n+.

Markers Comments:

Generally well done for (iii) but very few students get full mark for this part because of poor presentations and conclusion.

There is a large number of students who wrote or or n n n . They are very confused with the notation.

There are students who are confused between Pn and Sn and some used Sn to define the statement as well.

There are still a few students who prove P1 by just writing down LHS = 8

9 = RHS, without showing

any substitution, and hence lose the 1 mark here.

These steps of expansion must be

shown! Students must not skip these

steps in proof.

Even though almost all students did assume Pk true for some (which a few students wrote all)

k , quite a number of students made mistakes such as 1k k kS S u and 1 1k kS S u which

shows that they have no conceptual understanding of the essence of MI at all. The mistake in the latter can also be seen in conclusion where some students interpret P1 true and Pk true Pk+1 true to be same as Pk and P1 true Pk+1 true.

Another common mistake is that students forgot to change the sign to negative when they write

221 8 1

1 1(2 1) 2 3

k

k k

. This is most common among those who did not introduce the use of

bracket, this confusion of whether the numerator need to change to negative sign is very common

among students indicating very poor algebraic skills:

2

22

2 3 8 11

(2 1) 2 3

k k

k k

(iv) 1 Sn < 105

52

110

(2 1)n

2 5(2 1) 10n

Method 1:

Using GC, when

n = 157, 2 5(2 157 1) 99225 10

n = 158, 2 5(2 158 1) 100489 10 Least n = 158

Method 2:

By sketching the graph of 2 5(2 1) 10y n

For 2 5(2 1) 10 0n , 157.6n

Least n = 158

Method 3: 2 5(2 1) 10 0n

2 54 4 1 10 0n n (n +158.6) (n 157.6) > 0

n > 157.6 or n < 158.6 (rejected)

Least n = 158

157.6 n

y

158.6 157.6 n

Should be ve

Markers Comments: Very few students score the full 2 marks for this part (iv). Many students are stuck with

5

2

110

(2 1)n

and do not know how to proceed. For students from better classes, many

students proceed to solve 2 5(2 1) 10n by square root both sides without giving detailed

working of how they get to5

2(2 1) 10n . Some students who know how to use GC to get

158 did not show graph or the table of values. The students also did not look carefully at

the value displayed by the GC, When n = 158, 2 5(2 158 1) 1 10 instead of 100489 and hence they thought it has not exceeded 105, ended up getting n = 159 instead of 158 as the answer.

4 (a) The sum, nS , of the first n terms of a sequence is given by 23 4 2nS n n where

n 2. Show that this sequence is an arithmetic progression. [3]

(b) A large volume of water consists of x kg of impurities. In a water purification process,

1

r of the impurities, where , 1r r is removed in the first stage. In each subsequent

stage, the amount of impurities removed is 1

r of that removed in the preceding stage.

Assume no other impurities are added in the process.

(i) Find the amount of impurities removed in the second and third stages of

purification in terms of x and r. [1]

(ii) Find the total amount of impurities removed by the end of the nth

stage. [2]

(iii) Find the range of values of r such that at least half of the impurities will remain no

matter how many stages are used. [2]

[Solution]

(a) 2 2

1 3 4 2 3( 1) 4( 1) 2 6 1n n nT S S n n n n n

1 6 1 6( 1) 1n nT T n n

= 6, which is independent on n this progression is an arithmetic progression.

(b)(i) In the 2nd

stage, amount of impurities removed = 2

x

r

In the 3rd stage, amount of impurities removed = 3

x

r

(ii) Total amount of impurities removed by the end of the nth

stage of purification

2 n

x x x

r r r

11

11

nx r

r

r

1

1

n

n

x r

r r

(iii) Assuming that we can carry out the process of purification indefinitely.

Total amount of impurities remained

2 3

x xxx

r r r

1

1 11

x xx x

r r

r

For 1

1 2

xx x

r

, [or alternatively

1

1 21

x

r x

r

]

1

1 2

3

xx

r

r

Markers Comments: (a) This supposed standard question is very badly answered.

Many students dont know that 1n n nT S S . An alarmingly number of students

used 1n n nT S S or dont even know how to start.

To show that the sequence is an arithmetic progression, there are many students

who used 1 6 1n nS S n and concluded that this is a term that is independent of

n, which clearly is not the case. This shows that they do not understand the concepts and just memorised the formula.

There is also a number of students who used 2 1 3 26, 6,T T T T and then

concluded that 1 6n nT T , which shows that the sequence is an A.P. Students

need to know that they cannot generalise like this.

Some students went to show 1

constantn

n

T

T instead.

(b) Generally ok for (i). The common errors spotted are these:

In the 2nd stage, amount of impurities removed = 2

1

r

In the 3rd stage, amount of impurities removed = 3

1

r

Students misunderstood the question and wrote down the total amount removed instead of the amount removed in the respective stages.

For (ii), it is very surprising that there are still a substantial number of students who do not know the formula for the sum of the first n terms of a geometric progression. Other than that, it is quite well-answered. For (iii), many students (more than half) do not understand the question and thought that the question is asking for at least half of the impurities removed. Many also dont know that they are supposed to use the sum to infinity of a geometric progression. In short, this part is very badly answered.

5 A graphic calculator is not to be used in answering this question.

Two complex numbers p and q are given by 1 ip and 1 3iq , and p

zq

.

(i) Express z in the form x + yi, where x and y are exact real values to be determined. [2]

(ii) By considering the moduli and arguments of p and q, find the exact values of |z| and

arg z, where arg z . [4]

(iii) Hence, show that 11 6 2

sin12 4

. [3]

[Solution]

(i) 1 i 1 3i 1 3i i 3 1 11 3 1 3 i4 4 41 3i 1 3i

pz

q

(ii) 2 21 1 2p , arg p = 1tan 14

.

2

21 3 2q , 12

arg tan 33 3

q

.

2 2 11

, arg arg arg2 4 3 12

pz z p q

q

(iii) From part (i) and (ii),

1 11 3 1 3 i4 4

2 11 11

cos isin2 12 12

Comparing the imaginary parts, 2 11 1sin 1 32 12 4

Since 11 11

sin sin12 12

,

we get 2 3 111 3 1 6 2

sin12 4 42 2

Alternative Method (Using Argand Diagram)

OR

OR

Markers Comments:

For part (i):

Most of the students were awarded full marks.

1 i 3i 1 1 3i i 3 1 11 3 1 3 i4 4 43i 1 3i 1

pz

q

The above is an alternative method which many students used. However, a fair number of

students forgot the negative sign (in red) in the denominator.

Re(z)

Im(z) From diagram,

142

2

3 111sin

12

2 3 1

4

6 2

4

Re(z)

Im(z) 142

2

1 311sin

12

1 311sin

12 2 2

3 111 6 2sin

12 42 2

11sin sin

12 12

14

22

1 3

3 1

2 2

6 2

4

Re(z)

Im(z)

For part (ii):

Many students had no problems finding p and q .

Many students displayed a common misconception in finding arg p and arg q .

It is wrong to write 13

arg tan1

q

. For the function 1tan x to exist, the function

tan x must be one-one. This means that the domain for the function tan x should be

restricted to ,2 2

, which is known as the principal range . Hence, the range for the

function 1tan x is also ,2 2

. This means that the function 1tan x will always lead to

values only within the first and fourth quadrants. Hence, in general, to find the argument

of a complex number z x yi , students should first find the basic angle 1tany

x ,

which is the acute angle between OZ and the horizontal real axis. Next, students should

use one of the four formulae for finding the argument in the four quadrants based on .

Some students used

argarg

arg

pz

q , which is wrong.

A number of students used 2 2

1 11 3 1 3

4 4z

and

1

11 3

4arg tan1

1 34

z

which lead to the correct final answer. However, these

students were not awarded any marks because they did not consider p , q , arg p and

arg q as required by the question.

For part (iii):

Method 1: To find 11

sin12

, students should note that 11

arg12

z

. Hence, they

should consider the imaginary component of z, which would involve 11

sin12

, and use

11 11sin sin

12 12

.

Method 2: Students could observe that *11

arg12

z

and proceed to use the Argand

diagram for both z and z* .

Method 3: For z x yi , students could use sin argy

zz

, regardless of the sign of

the numerator y. A similar formula is cos argx

zz

. Cross-multiplication of these two

formulae would lead to the imaginary and real components of z respectively.

Method 4: Students could observe that 11

sin sin12 12

, where

12

is the basic angle.

Quite a number of students used 11

sin sin12 12

, which is wrong. It should be

11sin sin

12 12

.

Students should note that 1 3 0 and 1 3 0 , so they should not use 1 1 34

and 1 1 34

as the lengths of any right-angled triangle. Instead, they should use

1 1 34

and 1 1 34

which are positive.

Students must know that z is the distance from O to Z. Many students did not realise that

they had already found the length of OZ as z in part (ii), so they wasted their time trying

to find 2 2

1 11 3 1 3

4 4z

.

This is a Show question. Students should note that the skipping of essential steps could

lead to marks being deducted. Hence, for Show questions, students must present their

working coherently without any gaps in understanding.

6 Let f ( ) 1 1r r r r . Show that f ( ) f ( 1) 3 1r r r r . [1]

(i) Find 1

1 .n

r

r r

[2]

(ii) Using the result obtained in part (i), deduce that 2

1

1( 1)(2 1)

6

n

r

r n n n

. [2]

(i) Find the sum of the series

22 2 2 2 2 2 21 3 2 3 3 4 5 3 6 3 1 ,n n where n is odd. Give

your answer in a fully factorised form. [4]

[Solution]

f ( ) f ( 1) 1 1 2 1r r r r r r r r

1 1 2r r r r

3 1r r

(i) 1 1

11 f f 1

3

n n

r r

r r r r

1

f (1) f 03

1

f (2) f 13

1

f ( 1) f 23

n n

1

f ( ) f 13

n n

1

f ( ) f 03

n

1

1 1 03

n n n 1

1 13

n n n

(ii) 1

11 1 1

3

n

r

r r n n n

21

11 1

3

n

r

r r n n n

21 1

11 1

3

n n

r r

r n n n r

21

1 11 1 ( 1)

3 2

n

r

r n n n n n

21

11 2 1 3

6

n

r

r n n n

21

11 2 1

6

n

r

r n n n

(deduced)

(iii) 22 2 2 2 2 2 21 3 2 3 3 4 5 3 6 3 1n n

22 2 2 2 2 2 2 2 21 2 3 4 5 2 2 4 6 1n n

1

222

1 1

2 2

n

n

r r

r r

21 1 1 1 1

1 2 1 2 2 1 2 16 6 2 2 2

n n nn n n

1 4 1 1

1 2 16 3 2 2

n nn n n n

1 1

1 2 1 1 16 3

n n n n n n

1

1 2 1 2 16

n n n n 1

1 4 16

n n n

Markers Comments: Most students were able to answer up to part (ii) and achieve 5 marks out of 9. Only a

small number of students attempted part (iii) successfully. Common mistakes seen were:

(1) Omission of brackets (very common): Eg. 1 1

11 f f 1

3

n n

r r

r r r r

(2) Omission of the steps in reducing 1 1

1 1 ( 1)3 2

n n n n n to 1

1 2 16

n n n .

(3) Wrong limits in the summation notation for

22 2 2 2 2 2 21 3 2 3 3 4 5 3 6 3 1n n .

Eg. 22

1 1

(2 1) 3 2n n

r r

r r

Very few students regrouped into the form

22 2 2 2 2 2 2 2 21 2 3 4 5 2 2 4 6 1n n

7 (a) The diagram below shows the graph of y = f(x). The graph has a minimum point at (1,

3) and intersects the axes at x = 2 and y = 4. The equations of the asymptotes are y =

2x and x = 1.

On separate diagrams, sketch the graphs of

(i) y = f (x), [2]

(ii) 1

f ( )y

x , [2]

(iii) y2 = f (x). [2]

In each case, give if possible, the equations of the asymptotes, the coordinates of the

turning points and the coordinates of the points where the graph crosses the x- and y-

axes.

(b) A curve undergoes the transformations A, B and C in succession:

A: a translation of 2 units in the negative direction of the x-axis.

B: a scaling parallel to the x-axis by a factor 2.

C: a translation of 1 unit in the positive direction of the y-axis.

The equation of the resulting curve is 2

2 94

y xx

.

Find the equation of the curve before the three transformations were effected.[3]

y

x 0

(1, 3) 4

x= 1

2

y = 2x

[Solution]

(a) (i)

(ii)

(iii)

y

x 0 1

x= 1

y = 2

y = f (x)

x= 2

y

x 0 1

(1, )

y

x 0

(1, ) 2

x= 1

2

y2 = f (x)

2 (1, )

(b) 2

2 94

y xx

22 8

4y x

x

2 1

2 2 8 4 82 4 2

y x xx x

1 14( 2) 8 4

( 2) 2y x x

x x

The equation of the curve before the transformations were effected is 1

4y xx

Markers Comments:

7ai) was done well with occasional presentation issues such as omitting the equation of the

asymptotes

7aii) Many students faced difficulties with the oblique asymptotes. The new asymptote y =

0 (or 1

2y

x out of syllabus) was often left out. Students should refer back to the meaning

of the transformation to get ideas on how to deal with the oblique asymptote. Since the

transformation is a reciprocal transformation, students should consider the reciprocal of

large y values as x increases. Doing so, they should realise that the resultant are small y

values. As the y values are small such as 1

4 and

1

3, students drew the graph close to the x

axis, making it difficult to read their graph. Students are advised to stretch this region for

clarity even if meant that the scaling might be off by a little.

7aiii) Many students drew a roughly correct shape for this question. However, most had

problems with the oblique asymptote again. As it is not within syllabus to know that there

will be a non-linear asymptote 2y x , students need not draw this asymptote at all.

Students were not penalised if they drew this asymptote as a straight line but were warned.

Some students forgot that at 2x , they were supposed to draw a rounded looking curve

at the x axis, not sharp.

7b) concepts were mostly in place. The alarming fact was that many students copied from

the question wrongly and some read the transformation wrongly. Students are advised to

read the question slowly and underline the key information before proceeding.

replace y by y + 1

replace x by 2x

replace x by (x2)

8 A curve is defined by the parametric equations

2

2 2

1 4 , , 0 1.

1 1

t tx y t

t t

(i) Show that the equation of the normal to the curve at the point where 1

2t is given by

2 6

3 5y x . [5]

(ii) Sketch the curve and the normal in part (i) on the same diagram. [2]

(iii) Show that the area bounded by the curve, the normal and the axes can be expressed in

the form f ( ) db

at t c , where a, b and c are constants to be determined. Hence evaluate

this area. [3]

[Solution]

(i)

2 2

22

2 1 2 1d

d 1

t t t tx

t t

3 3

22

2 2 2 2

1

t t t t

t

2

2

4

1

t

t

2

22

4 1 2 4d

d 1

t t ty

t t

=

2 2

22

4 4 8

1

t t

t

=

2

22

4 4

1

t

t

2

22 2

22

1d d d 4 4 1

d d d 41

ty y x t t

x t t t tt

At t = 1

2, x =

3

5, y =

8

5

21

1d 32

1d 2

2

y

x

Gradient of the normal = 2

3

Equation of the normal at t = 1

2:

8 2 3

5 3 5y x

2 6

3 5y x (Shown)

(ii)

2

1

y

x

3 85 5,

(iii) Area = 35

11 6 8 3d

2 5 5 5y x

0

22 21

2

21 4 4 d

25 1 1

t tt

t t

= 0.447295 + 0.84

= 1.29 units 2 (Using GC)

1 21

, 0,2 25

a= b= c=

Markers Comments:

Many students did not simplify their working for d

d

y

xand attempted to obtain the form directly.

This lead to many careless mistakes. Students also did not attempt to show full working to getting

d 3

d 2

y

x

and attempted to bulldoze their way through in hope that the marker will not check

through their working. This is an incorrect practice that is undesirable. A minority of the students

did not know that the gradient of the normal is 1

d 3

d 2

y

x

.

Some students fail to label the graph properly. A large portion did not know how to do parametric

integration. Some attempted to do

0

21

2

4d

1

tt

t which is conceptually incorrect. A large majority did

not change limits and around a quarter of the cohort identified the wrong region to integrate.

In sum, students really need to learn to simplify their steps, show clear and full working and revise

their parametric integration.

9 (i) Find 2e cos 4 dx x x . [4]

(ii) Sketch the curves y = ex sin 2x and y = e

x for 0

2x

on a single diagram. Find the

exact x-coordinate of the point of intersection between the two curves. [3]

(iii) Find the exact volume of the solid generated when the region bounded by the two

curves and the y-axis is rotated through 2 radians about the x-axis. [4]

[Solution]

(i) 2 2 21 1

e cos 4 d e cos 4 e 4sin 4 d2 2

x x xx x x x x

2 21 e cos 4 + 2 e sin 4 d

2

x xx x x

2 2 21 1 1

e cos 4 + 2 e sin 4 e 4cos 4 d2 2 2

x x xx x x x

2 2 21

e cos 4 + e sin 4 4 e cos 4 d2

x x xx x x x

2 21

5 e cos 4 d e cos 4 2sin 42

x xx x x x + c

2 21

e cos 4 d e cos 4 2sin 4 + 10

x xx x x x c

Alternatively,

2 2 21 1

e cos 4 d sin 4 e sin 4 2e d4 4

x x xx x x x x

2 21 1e sin 4 e sin 4 d

4 2

x xx x x

2 2 21 1 1 1

e sin 4 cos 4 e cos 4 2e d4 2 4 4

x x xx x x x

2 2 21 1 1

e sin 4 + e cos 4 e cos 4 d4 8 4

x x xx x x x

2 25 1

e cos 4 d e cos 4 2sin 44 8

x xx x x x + c

2 21

e cos 4 d e cos 4 2sin 4 + 10

x xx x x x c

(ii)

y

x

y = ex

y = ex sin 2x

0

1

Students are advised to

simplify before the next

integration.

1. Label graphs and x- and y- axes.

2. Give the values of the intercepts.

3. Sketch for the given domain 02

x

.

4. The curve y = ex sin 2x is not symmetrical

in the interval.

5. The curve y = ex is a strictly increasing

function and touches the curve y = ex sin 2x

at one point only.

.

At the point of intersection, ex sin 2x = e

x

ex (sin 2x 1) = 0

ex = 0 or sin 2x = 1

22

x

4

x

(iii)

Volume of solid generated 4 42 2

0 0e d e sin 2 dx xx x x

4 42 2 2

0 0e d e sin 2 dx xx x x

4 42 2

0 0

1 cos 4e d e d

2

x x xx x

4 2 20

1e e cos 4 d

2

x x x x

4 42 2

0 0

1 1 1e + e cos 4 2sin 4

2 2 10

x x x x

using (i) answer

02 21 1 1

e 1 e e2 2 10

21

2e 310

units3

Markers Comments:

i. This part is generally well done. Many students used integration by parts but made mistakes in the

differentiation/integration of the two functions. A number of students did not simplify before the

next integration and thus made more careless mistakes. Many students did not have the final

constant of integration.

ii. The curves were generally well sketched though many were incomplete with missing intercepts or

unlabelled graphs or axes. A number of students wrongly sketched the curve y = ex sin 2x to be

symmetrical in the interval (like a quadratic curve) or undefined at x = 2

when it is not. Some

students sketched more than the required domain. Some sketches of the curve y = ex intersected the

curve y = ex sin 2x wrongly at more than one point or at the maximum point.

Majority of the students solved the equation ex sin 2x = ex to find the point of intersection of the two

curves but cancelled out ex on both sides of the equation without any explanation. Some students

did not show any working for finding the x-coordinate.; some did not give the exact form.

y

x

y = ex

0

1

y = ex sin 2x

no solution as ex > 0

Use double angle formula:

2 1 cos 4sin 22

xx

Answer need to be simplified

iii. This part is badly attempted; weaker students did not even attempt to answer this part. Two

common wrong answers for the volume of solid were 42

0e e sin 2 d x x x x

and

4 4

0 0e d e sin 2 dx xx x x

, indicating students poor grasp of the concept on volume of

revolution. Some students were able to give the correct definite integral but did not know how to

find the integral 4 2 2

0e sin 2 dx x x

. Less than 10% of the cohort was able to get to the final exact

answer.

10 The functions f, g and h are defined by

f : 2 , , x x x x k ,

2

1g : , x x

x ,

2

1h : , ,

4 4x x x k

x x

.

State the largest value of k such that 1f exists. [1]

Use this value of k for the following parts.

(i) Find 1f in a similar form. Hence find the range of values of x such that

1f fx x . [4]

(ii) Determine the range of values of x for which f gx x , giving your answers in

exact form. [4]

(iii) Show that the composite function gf exists and explain if the functions h and gf are

equal. [2]

[Solution]

Largest 2k

(i) Since 2x , f ( ) 2x x

Let 2y x 2x y

1f : 2 , 0x x x

Df = , 2 and 1fD 0, . For 1f fx x , 0 2x

Markers comments for (i):

1. First part was generally well-done with almost all being able to state the correct largest value of k.

2. Many students encountered difficulties in finding the 1f of a modulus function, reflecting their weak concepts in the modulus function. Many started by squaring both sides which is not the correct approach. The best method is to spilt up the modulus function based on the given domain. Some errors:

2 2y x - thought that by squaring it removes the modulus sign which is

incorrect.

2 2y x x y , Hence 1f : 2 x x

Alternative solution:

Let 2y x

2y x or y = (x 2)

x = y +2 (rejected since 2x ) or x = y +2

3. Surprisingly, not many of them understood what the question requires by finding 1f

in a similar form. Similar form in this case is to defined the function exactly as how it was written in the question, with the rule and domain stated. However, many

failed to do so with some merely giving an expression for 1f ( )x without stating the

domain.

4. The last part was badly done with many candidates either did not attempt or approach the question wrongly.

Many started equating 1f ( ) f ( )x x or f ( )x x which provide only a single answer,

whereas the question specifically asks for range of values of x. They failed to recognise

that the rules of f and 1f are the same and to find 1f ( ) f ( )x x , they just need to

find the common points using their domains.

(ii) Method 1: Algebraic Method

2

12x

x

3 22 1 0x x

21 1 0x x x by long division

1 5 1 5

1 02 2

x x x

1 5

02

x

or 0 1x or 1 5

22

x

[Note that 0x and 2x ( fD ,2 )

Method 2: Graphical Method

From the graph, for f gx x ,

1 5

02

x

or 0 1x or 1 5

22

x

2 1

y = g(x)

y = f(x)

y

x

0

To find the intersection points:

2

12x

x

3 22 1 0x x

21 1 0x x x 1 5

1or 2

x x

1 0

2 x

Markers Comments for (ii): 1. This part was badly done. Firstly, many students started with the wrong approach and

secondly, they are still not competent in solving cubic inequalities.

2. Some wrong approach:

2

2

2

12x

x

2 2

1 12x

x x - students who used this method need to know that the

inequality, 2

12x

x gives no solution

3. For the remaining students population who managed to solve the cubic inequality, most missed out the condition that 0x and 2x .

4. Despite the question asking for answers in exact form, there are still a handful of students who went to use GC to solve and leave their answers in non-exact form.

11 The equations of two planes p1 and p2 are

2 2 0x y z ,

3x + y + z = ,

respectively, where and are constants.

The line l is the line of intersection between p1 and p2 and has equation

1 1

3 4 5

x y z .

(i) Show that = 1 and = 2. [3]

(ii) Find the exact value of the cosine of the acute angle between p1 and p2. [2]

(iii) The point A with position vector 4j +2k lies on p2. Find the exact shortest distance from

A to l. Hence deduce the exact value of the perpendicular distance from A to p1.[5]

(iv) Another plane p3 contains the origin.

(a) When p3 and p2 have no common point, state a vector equation of p3. [1]

(b) When p3 is perpendicular to p2, A lies on p3. Find the position vector of the

common point of intersection between p1, p2 and p3. [3]

[Solution]

p1 : 2 2 0x y z and P2 : 3x + y + z = ,

i.e.

2

1 0

2

r and

3

1

r

(i) Equation of line l is 1 1

3 4 5

x y z . i.e.

1 3

0 4 ,

1 5

t t

r

Since l lies on p2, l is perpendicular to the normal of p2.

3 3

4 1 0

5

9 4 5 0 1 (shown)

Since (1, 0, 1) lies on p2, 3(1) 0 1(1) 2 (shown)

Alternative method 1

Let t = 0 and t = 1 Points (1, 0, 1) and (4, 4, 6) lie on line l and plane p2.

3(1) + 0 + (1) = and 3(4) + (4) + (6) =

= 1 and = 2

Students are advised to let

1 1

3 4 5

x y z = t before

obtain the vector equation of

line l.

Students should not use here

as has been defined as an

unknown coefficient of z in P2

Alternative method 2

Let

2 3 3

1 1 4

2 5

k

for some k

2 3

2 6 4

5 5

k

1, 2 3k

1 (shown)

Since (1, 0, 1) lies on p2, 3(1) 0 1(1) 2 (shown)

(ii) Let be the acute angle between p1 and p2.

2 3

1 1

2 1 7 7 11cos

339 11 3 11

(iii) Let

1

0

1

OB

and N be the foot of from A to line l.

Perpendicular distance from A to l,

AN =

3

4

5

50

AB

1 3

4 4

1 5

50

3

8 1

1

50

8 11

50

4 22

5

Hence, Shortest distance from A to p1 , |AF|

= sinAN

2

4 22 7 111

5 33

8

3

At this step, we only know that

2 3

1 1

2

is parallel to

3

4

5

.

They may not be equal.

Question asking for exact value of

cosine of the acute angle, not .

Need to put module for the dot

product since the angle is acute.

B(1, 0, 1)

A(0, 4, 2)

l

N

A

d

l P1

P2

N F O

AN is on the third side

of AB and d, use cross

product and not dot

product.

Alternatively,

Since N lies on l,

1 3

4 for some

1 5

t

ON t t

t

1 3

4 4

5 1

t

AN ON OA t

t

Since AN is perpendicular to line l,

1 3 37

4 4 4 025

5 1 5

t

t t

t

425

7225

125

AN

Perpendicular distance from A to l = 352 4 22

5 5AN

(iv) (a) p3 is parallel to p2 and contains the origin.

Equation of plane p2:

3

1 0

1

r

(b) p3 contains points O & A OA is parallel to p3 .

p3 is perpendicular to p2 p3 parallel to

3

1

1

3n =

0 3 1

4 1 6 1

2 1 2

Equation of p3 is

1

1 0

2

r

Using GC, solve

1

2

3

: 2 2 0

:3 2

: 2 0

p x y z

p x y z

p x y z

Position vector of the point of intersection =

81

1217

2

or

42

617

1

A

l p1

p2

p3

B(1, 0, 1)

A(0, 4, 2)

l

d

N

Need to give answer in vector equation

form as stated in the question.

RHS of the equation has to be 0 as p2 contains origin.

Do not write .

They are not equal.

Markers Comments:

i. - Many students made careless mistake in forming the vector equation of l. They wrote it as

1 3

0 4

1 5

t

r .

- Those who use Alternative method 2 to show = 1, many of them just let

2 3 3

1 1 4

2 5

.

Some of them obtained

2 3

2 6 4

5 5

and just conclude that 2 = 3 = 1.

- Some students after substitute

1 3

4

1 5

t

t

t

into equation of P2 , get 3 1 3 4 1 5 t t t and

do not know how to continue. Since there are only 2 unkowns and to solve, they just needs to

choose any 2 values of t to obtain 2 equations to solve for and .

- Some students after substitute

1 3

4

1 5

t

t

t

into equation of P1 , get 2 6 4 2 10 0t t t and

wrongly conclude that t = 0. It should be t can be any real number.

ii. This part is generally well done. Many students did not read the question carefully, they started with

1

2 3

1 1

2 1cos

9 11

and just give the value for .

Many of them did not put module and give negative value for cos.

Many students did not simplify the answer for cos, left part of the answer as 9 in steads of 3.

iii. Students make the following mistakes :

AN = dAB , AN = dOA , AN = dAB , or AN = d

d

AB

AB

.

Many students either did not answer the 2nd part of part (iii) or they think the both parts have the same answer.

iv. (a) Many students did not give correct form for equation of plane p2 or RHS of equation 0. (b) This part is badly attempted; many students cannot get correct normal vector for p3.