CJC 2019 H2 Mathematics

Prelim Paper 1

1 Differentiate 2

e x with respect to x. Hence find 23e dxx x . [3]

2 Without using a calculator, solve the inequality 2 34 32 1xx xx

. [4]

3 The curve C has the equation 2 2( 5) ( 3) 1

16 9x y .

(i) Sketch C, showing clearly the coordinates of the stationary points and the vertices.

[2]

The region R is bounded by curve C, the line x = 9 and the x axis.

(ii) Find the volume of the solid generated when region R is rotated completely aboutthe x axis. [3]

(iii) Describe fully a sequence of two transformations which would transform the curve

C to the curve 2 2(2 5) 1

16 9x y . [2]

4 (i) Show that 3 3 3 33 ( 1) 1

( 1) ( 1)r r A Br r r r

, where A and B are to be determined. [2]

(ii) Hence find 3 3 3 3 3 37 19 6 (2 1) 1...

(1) (2) (2) (3) (2 ) (2 1)n nn n

. [3]

(iii) Use your answer in part (ii) to find 3 31

3 ( 1) 1 1( 1) 2

r

r

r rr r

. [3]

5 A developer has won the tender to build a stadium. Sunrise Singapore, who would finance the construction, wants to have a capacity of at least 50 000 seats. There is a limited land area to the stadium, and in order for the stadium to have a full sized track and football pitch, the first row of seats can seat 300 people, and every subsequent row has an additional capacity of 20 seats.

(i) What is the least number of rows the stadium must have to meet Sunrise Singapore’s requirement? [3]

Assume now that the stadium has been built with 60 rows, with row 1 nearest to the football pitch.

(ii) For the upcoming international football match between Riverloop FC and Gunners FC, tickets are priced starting with $60 for Category 1, with each subsequent category cheaper by 10%. How much would a seat in the 45th row cost? [1]

Category 1 Rows 1 – 20 Category 2 Rows 21 – 40 Category 3 Rows 41 – 60

There is a total crowd of 51 000 for the football match.

(iii) Assuming that all the tickets from the cheapest category are sold out first before people purchase tickets from the next category, calculate the total revenue collected for the football match. [5]

6 (i) Given that cosxy e x , show that 2 2

22

d d d2d d dy y yy y yx x x

. [4]

(ii) Find the Maclaurin series for cosxy e x , up to and including the term in 3x .

[4]

(iii) Verify the correctness of the expansion found in part (ii) using standard series found in the List of Formulae (MF26). [3]

7 (i) A water trough, shown in the diagram below, in the shape of a triangular prism is used to collect rainwater. The trough consists of two rectangular zinc sheets of negligible thickness, each with fixed dimensions 10 m by 2 m, and two triangular zinc sheets with height h m, as shown in the diagram below. Use differentiation to find the maximum volume of the trough, proving that it is a maximum. [5]

(ii) Water collected in the trough is drained at a rate of 0.001m3 /s into a container consisting of two cylinders, as shown below. The larger cylinder has radius 1 m and height 0.5 m. The smaller cylinder has radius 0.6 m and height 0.5 m. Find the rate at which the depth of water is increasing after 29 minutes. [4]

(iii) Let H be the height of the water level in the container at time t.

Sketch the graph of ddHt

against t. [3]

8 Do not use a calculator in answering this question.

(a) Showing your working clearly, find the complex number z and w which satisfy the simultaneous equations

3 i 12 9i,2 * 3 16 23i.z wz w

[5]

(b) It is given that z1 is a root of the equation 4 3 23 4 8 0z z z , where

1 1 3iz .

(i) Express 4 3 23 4 8 0z z z as a product of two quadratic factors with real coefficients. [4]

(ii) Given that i 51ep q z , determine the exact values of p and q, where

q . [4]

9 Line l1 has equation 1 2

8 13 5

r where is a real parameter, and plane p has equation

170

abr . It is given that l1 lies completely on p and the point Q has coordinates

( 1, 2,3) .

(i) Show that a = – 1 and b = 2. [3]

(ii) Find the foot of perpendicular from point Q to point P. Hence find the shortest distance between point Q and p in exact form. [4]

Given that the shortest distance between l1 and the foot of perpendicular of Q on p is 5 63

.

(iii) Using the result obtained in part (ii) or otherwise, find the shortest distance between Q and l1. [2]

The line l2 is parallel to p, perpendicular to l1 and passes through P and Q.

(iv) Show that the Cartesian equation of the line l2 is 1 2 3

2x y z . [3]

(v) Find the vector equation of line l3, which is the reflection of l2 about p. [3]

10 Tumours develop when cells in the body divide and grow at an excessive rate. If the balance of cell growth and death is disturbed, a tumour may form. A medical scientist investigates the change of the tumour size, L mm at time t days of a particular patient using models A and B. For both of the models, it is given that the initial rate of the tumour size is 1 mm per day when the tumour size is 1 mm.

(i) Under Model A, the scientist observes that the patient’s turmour is growing at a rate proportional to the square root of its size. At the same time, the tumour is reduced by radiation at a rate proportional to its size. It is further observed that the patient’s tumour is decreasing at 2 mm per day when the tumour is 4 mm.

Show that L and t are related by the differential equation

d 3 2dL L Lt

. [2]

(ii) Using the substitution 2L y where y > 0, show that the differential equation in part (i) can be written as

d 3 2d 2y yt

.

Find y in terms of t and hence find L in terms of t only. [7]

(iii) Under Model B, the scientist suggests that L and t are related by the differential equation

2

2 2 2

d 2d (1 )

L tt t

.

Find the particular solution of this differential equation. [4]

(iv) Find tumour sizes predicted by Models A and B in the long run. [2]

Page

1 o

f 29

CATH

OLI

C JU

NIO

R C

OLL

EGE

H

2 M

ATH

EMA

TICS

2

019

JC2

PREL

IMIN

AR

Y EX

AMIN

ATI

ON

SO

LUTI

ON

Q1.

Tec

hniq

ues o

f Int

egra

tion

Ass

essm

ent O

bjec

tives

Solu

tion

Exa

min

er’s

Fee

dbac

kIn

tegr

atio

n by

par

ts

Diff

eren

tiatio

n w

as w

ell

done

exc

ept f

or a

few

st

uden

ts w

ho c

lear

ly d

o no

t kno

w h

ow to

di

ffere

ntia

te e

xpon

entia

l fo

rm.

Man

y di

d no

t obs

erve

th

e w

ord

‘HEN

CE’

in

the

ques

tion.

The

y w

ent

on to

do

by p

arts

, sp

littin

g 2

ex

a

nd

3 x a

nd

was

una

ble

to c

ontin

ue.

22

de

2e

dx

xx

x

2

2

22

22

22

3

2

2 2 2

ed

12

e d

2 1e

e d

2 11

e2

e d

22

11

ee

22

x

x

xx

xx

xx

xx x

xx

xx

x

xx

x

xC

2

2

21

d2

e2

dd

ed

x

x

vu

xx

xu

xv

x

Page

2 o

f 29

Q2.

Ineq

ualit

ies

Ass

essm

ent O

bjec

tives

So

lutio

n E

xam

iner

’s F

eedb

ack

Use

of a

lgeb

raic

met

hod

to so

lve

ineq

ualit

y.

23

43

02

13

31

02

1

xx

xx

xx

xx

212

11

30

21

32

31

10

21

32

30

21

xx

xx

xx

xx

xx

xx

33

2x

o

r

10

2x

Des

pite

rem

inde

rs,

man

y st

uden

ts

still

w

ent

to

cros

s-m

ultip

ly

for

ineq

ualit

y.

Sinc

e th

is q

uest

ions

say

s “w

ithou

t th

e us

e of

a

calc

ulat

or”,

it

is

expe

cted

th

at

stud

ents

fa

ctor

ize

com

plet

ely

and

obta

in t

he r

oots

. In

com

plet

e fa

ctor

izat

ion

will

res

ult

in lo

ss o

f mar

ks.

A

num

ber

of

stude

nts

wro

te

33

2x

o

r

10

2x

a

s

the

final

ans

wer

. Thi

s sh

ould

not

be

as th

e qu

estio

n is

a st

rict i

nequ

ality

to

beg

in w

ith.

Man

y st

uden

ts

also

w

rote

‘a

nd’

inst

ead

of ‘o

r’.

Use

of r

epla

cem

ent

23

32

x

or

2

10

2x

0x

Man

y st

uden

ts c

ould

iden

tify

the

corre

ct re

plac

emen

t but

cou

ld n

ot

go o

n to

obt

ain

the

final

ans

wer

m

ark

for

0x

as

the

answ

er in

the

earli

er p

art w

as w

rong

.

-3

3/2

0

1/

2

+

+

+

Page

3 o

f 29

Q3.

Def

inite

Inte

gral

s + C

onic

s + T

rans

form

atio

nsA

sses

smen

t Obj

ectiv

es

Solu

tion

Exa

min

er’s

Fee

dbac

k St

anda

rd g

raph

- el

lipse

(i)

This

par

t was

gen

eral

ly w

ell a

ttem

pted

with

so

me

stud

ents

losi

ng m

arks

for n

ot

indi

catin

g th

e co

ordi

nate

s of p

oint

s sp

ecifi

cally

requ

este

d by

the

ques

tion.

Th

ere

is a

sign

ifica

nt n

umbe

r of s

tude

nts

who

ske

tche

d a

hype

rbol

a in

stea

d an

d gi

ven

no c

redi

t at a

ll.

Vol

ume

abou

t x-a

xis

(ii)

22

2

2

53

51

39

116

916

53

91

316

xy

xy

xy

y

Vol

ume

(abo

ut x

-axi

s)

22

99

2

55

5π

dπ

39

1 d

16

10.8

4267

894

10.8

to 3

s.f.x

yx

x

This

par

t w

as v

ery

badl

y at

tem

pted

or

not

atte

mpt

ed a

t all

by s

tude

nts.

Mos

t stu

dent

s w

ho a

ttem

pted

the

que

stio

n di

d no

t re

aliz

e w

hich

are

a w

as b

eing

refe

rred

to a

nd d

id n

ot

man

aged

to g

et th

e co

rrect

exp

ress

ion

for y

, bu

t ins

tead

ass

umed

the

nega

tive

squa

re ro

ot

to b

e rej

ecte

d sin

ce th

e y >

0. A

larg

e nu

mbe

r of

stud

ents

are

also

conf

used

abou

t the

lim

its

to b

e us

ed in

the

defin

ite in

tegr

al o

r did

not

re

aliz

e in

the

ir al

gebr

aic

man

ipul

atio

n th

at

22

ab

ab

a

nd th

ey p

roce

ed to

obt

ain

an o

verly

sim

plifi

ed e

xpre

ssio

n. A

smal

l but

si

gnifi

cant

num

ber

of s

tude

nts

set

up t

he

inte

gral

with

res

pect

to

y in

stea

d w

hen

the

axis

of r

otat

ion

was

cle

arly

def

ined

.

Ox

y

(9,3

)

(5,6

) (5,3

)(1

,3)

Ox

y

(9,3

)

x =

9

Vo

VVVVVVVVVVVVVVVVVVVVVVVVVlulululululu

me

me

me

me

me

me

((((((abo

Page

4 o

f 29

Tran

slat

ion

and

Scal

ing

(iii)

Scal

ing

para

llel t

o x

axis

with

scal

e fa

ctor

of ½

follo

wed

by

trans

latio

n of

3 u

nits

in th

e ne

gativ

e y

axis.

OR

Tran

slat

ion

of 3

uni

ts in

the

nega

tive

y ax

is fo

llow

ed b

y a

scal

ing

para

llel t

o x

axis

with

scal

e fa

ctor

of ½

.

This

par

t w

as v

ery

badl

y at

tem

pted

. Man

y st

uden

ts u

sed

thei

r ow

n co

lloqu

ial

term

s su

ch

as

“Stre

tch”

, “M

ultip

le”,

“S

hift”

, “M

ove”

whi

ch w

ere

not

acce

pted

. So

me

used

the

gene

ric te

rm “

Tran

sfor

m”

in p

lace

of

“sc

ale”

and

“tra

nsla

te”

whe

n th

ey w

ere

requ

ired

to

desc

ribe

the

trans

form

atio

n.

Man

y st

uden

ts a

lso

faile

d to

incl

ude

the

key

wor

d “s

cale

fact

or” i

n th

eir d

escr

iptio

n of

the

scal

ing

step

. O

ther

co

mm

on

mist

akes

in

clud

es t

he c

onfu

sion

of ½

vs

2 fo

r sc

ale

fact

or a

nd +

vs

– 3

for t

he m

agni

tude

of t

he

trans

latio

n. Q

uite

a n

umbe

r of s

tude

nts

also

om

itted

the

axis

in w

hich

the

trans

form

atio

n w

as a

pplie

d or

stat

ed th

e w

rong

axi

s.

Page

5 o

f 29

Q4.

Sig

ma

Not

atio

nA

sses

smen

t Obj

ectiv

es

Solu

tion

Exa

min

er’s

Fee

dbac

k Si

mpl

ifyin

g ex

pres

sion

by p

artia

l fr

actio

ns

(i)

Met

hod

1:

2

33

33 3

23

33

33 3

3

33

31

13

31

11

33

1

1

11

11 1

rr

rr

rr

rr

rr

rr

rr

rr

rr

rr

w

here

1

and

1A

B

.

Met

hod

2:

33

33

31

11

1r

rA

Br

rr

r

3

33

11

1r

rA

rBr

By

com

parin

g co

effic

ient

of

0 r:

1A

B

y co

mpa

ring

coef

ficie

nt o

f 3 r:

01

AB

B

33

33

31

11

11

1r

rr

rr

r

This

par

t w

as g

ener

ally

wel

l at

tem

pted

. St

uden

ts

shou

ld c

ompa

re th

e co

effic

ient

s on

both

side

s to

find

the

unkn

owns

.

Sum

mat

ion

of se

ries b

y th

e m

etho

d of

diff

eren

ces

(ii)

3

33

33

3

62

11

719

...1

22

32

21

nn

nn

2

33

1

31

11

n

r

rr

rr

This

par

t was

not

wel

l atte

mpt

ed. T

he re

sult

in (i

) can

be

use

d to

find

the

expr

essio

n fo

r the

ser

ies

with

out

usin

g si

gma

nota

tion.

byyyyyyyyyyytttttttttttttttth

eheheheehehhhhhhehhee

s(((iiii(ii((i

)i)iii)i)i)iiiiiiii)i)i)i)i)i)i)i)i)i)ii)i))))ii)iii))ii)))i

Page

6 o

f 29

2

33

1

11 1

n

rr

r

33

33

33

33

11

12

11

23

...1

12

12

11

22

1

nn

nn

3

11

21

n

Stud

ents

wer

e ge

nera

lly n

ot fa

mili

ar w

ith th

e si

gma

nota

tion

to r

epre

sent

the

serie

s. T

he s

igm

a no

tatio

n sh

ould

be

used

with

inte

gral

lim

its.

A

sign

ifica

nt

perc

enta

ge

of

stud

ents

w

rote

th

e

follo

win

g si

gma

nota

tion:

33

0.5

31

11

n

r

rr

rr

whe

re

the

low

er li

mit

was

0.5

and

the

run

ning

ind

ex w

as

assu

med

to in

crea

se b

y 0.

5 ea

ch ti

me.

Som

e stu

dent

s wer

e al

so c

onfu

sed

by th

e la

st te

rm in

th

e se

ries.

Thou

gh t

hey

reco

gniz

ed t

hat

r ha

s be

en

repl

aced

by

2n, t

hey

did

not t

hink

of 2

n as

the

uppe

r lim

it. In

stea

d th

ey re

pres

ente

d th

e se

ries

wro

ngly

by

the

follo

win

g si

gma

nota

tion

3

31

62

11

22

1

n r

rr

rr

w

hich

onl

y in

clud

ed t

he e

ven

term

s of

the

giv

en

serie

s (i.e

. 2nd

, 4th

, 6th

,…,

th2n

).

Som

e stu

dent

s fo

und

a po

sitiv

e va

lue

of B

in (i

) and

bl

indl

y pe

rfor

med

met

hod

of d

iffer

ence

eve

n th

ough

it

was

not

pos

sible

. C

redi

t w

as n

ot g

iven

in

such

ca

ses.

Lim

its, S

um to

infin

ity a

nd u

se o

f st

anda

rd se

ries e

xpan

sion

(ii

i) A

s 2n

,

31

02

1n

,

31

11

21

n

This

par

t was

not

wel

l atte

mpt

ed.

Stud

ents

wer

e ab

le t

o re

cogn

ize

that

the

que

stio

n as

ked

for t

he lo

ng-te

rm a

vera

ge. H

owev

er, t

hey

wer

e co

nfus

ed b

y th

e ru

nnin

g in

dex

r an

d th

e un

know

n co

nsta

nt n

. A

sig

nific

ant

perc

enta

ge o

f stu

dent

s co

nsid

ered

th

e ca

se

whe

re

r

inst

ead

of

2n

.

As

sioiiioiooo

n nnnn nnnnnnn

Page

7 o

f 29

23

3

3

3

1

31

1

31

11 2

1

31

11 2

1

11

11

...2

22

1 21

11

22

r

r

r

rr

rr

rr

rr

rr

Som

e stu

dent

s di

d no

t kn

ow h

ow t

o pr

esen

t th

eir

anal

ysis

and

sol

utio

ns m

athe

mat

ical

ly f

or t

he f

irst

sum

mat

ion.

Th

ey

wro

ngly

w

rote

3

33

1

31

11

11

21

r

rr

rr

. A

s th

e sy

mbo

l

is no

t a

num

ber,

the

pres

enta

tion

shou

ld

be

33

31

31

11

11

12

1lim n

r

rr

rr

n

.

A

sign

ifica

nt

perc

enta

ge

of

stud

ents

ha

d th

e

misc

once

ptio

n th

at

if

10

2

r

as

r

,

1

10

2

r

r

.

For s

tude

nts w

ho re

cogn

ized

that

they

shou

ld u

se th

e fo

rmul

a fo

r sum

to in

finity

,

som

e w

ere

not

able

to

iden

tify

the

first

ter

m

corr

ectly

whe

re th

ey a

ssum

ed

1a

,

som

e w

ere

not

able

to

iden

tify

the

valu

e of

co

mm

on r

atio

r a

s th

ey w

ere

conf

used

with

the

runn

ing

inde

x r.

Page

8 o

f 29

Q5.

A.P

. & G

.P.

Ass

essm

ent O

bjec

tives

So

lutio

n E

xam

iner

’s F

eedb

ack

Find

ing

sum

of a

fini

te a

rithm

etic

se

ries

(i)

300

a

, 20

d

Su

m o

f sea

ts in

the

first

ro

ws

5000

0n

2

300

120

5000

02n

n

58n

M

inim

um n

o. o

f row

s is 5

8.

n

2

300

120

2nn

57

4902

0 58

50

460

5951

920

Som

e st

uden

ts w

rote

30

01

2050

000

nTn

inst

ead

of

230

01

2050

000

2nn

.

Som

e st

uden

ts pu

t "50

000"

in

stea

d of

"50

000"

.

Find

ing

nth

term

of a

fini

te g

eom

etric

se

ries

(ii)

60a

, 0.

9r

3

1Pr

ice

of se

ats i

n 45

th ro

w60

0.9

$48.

60

Part

(ii) i

s gen

eral

ly w

ell d

one.

Find

ing

nth

term

of a

fini

te a

rithm

etic

se

ries

(iii)

Num

ber o

f sea

ts in

60t

h ro

w30

060

120

14

80

Cre

atin

g ne

w A

P st

artin

g fr

om la

st ro

w,

1480

a

, 20

d

2

1480

120

5100

02n

n

53.2

8n

Num

ber o

f sea

ts in

Cat

egor

y 3

20

20S

214

8020

120

2580

02

Ther

e w

ere

conf

usio

n in

the

num

ber

of se

ats i

n ea

ch c

ateg

ory

as st

uden

ts

tend

to m

ixed

up

the

calc

ulat

ions

for

the

num

ber o

f se

ats

in th

e fir

st a

nd

third

ca

tego

ry.

A

num

ber

of

stud

ents

tri

ed

to

use

Geo

met

ric

Prog

ress

ion

to c

alcu

late

the

num

ber

of s

eats

for e

ach

cate

gory

. NNNN

uNN

Page

9 o

f 29

Num

ber o

f sea

ts in

Cat

egor

y 2

40

2040

SS

214

8040

120

2580

043

600

2580

02

1780

0

Num

ber o

f sea

ts in

Cat

egor

y 1

4051

000

S51

000

4360

074

00

Tota

l rev

enue

274

0060

1780

060

0.9

2580

060

0.9

2659

080

$

Alte

rnat

ive

300

a

, 20

d

Num

ber o

f sea

ts in

Cat

egor

y 3

6040

SS

6040

230

060

120

230

040

120

22

5340

027

600

2580

0

Num

ber o

f sea

ts in

Cat

egor

y 2

40

20S

S20

2760

02

300

201

202

2760

098

0017

800

Num

ber o

f sea

ts in

Cat

egor

y 1

5100

025

800

1780

074

00

Page

10

of 2

9

Tota

l rev

enue

274

0060

1780

060

0.9

2580

060

0.9

2659

080

$

Page

11

of 2

9

Q6.

Mac

laur

in’s

Ser

ies

Ass

essm

ent O

bjec

tives

So

lutio

n E

xam

iner

’s F

eedb

ack

Impl

icit

diff

eren

tiatio

n an

d us

e of

pro

duct

rule

e

cos

xy

x 2

eco

sx

yx

D

iffer

entia

ting

w.r.

t. x

, d

2e

cos

esi

nd

xx

yy

xx

x

2d

2e

sin

dx

yy

yx

x

Diff

eren

tiatin

g w

.r.t.

x ,

22

2

dd

d2

22

esi

ne

cos

dd

dx

xy

yy

yy

xx

xx

x

22

22

2

dd

dd

22

22

dd

dd

yy

yy

yy

yy

yx

xx

x

22

22

dd

d2

24

2d

dd

yy

yy

yy

xx

x

2

22

2d

dd

2d

dd

yy

yy

yy

xx

x

(sho

wn)

A lo

t of c

andi

date

s did

no

t squ

are

the

y an

d th

en

impl

icitl

y di

ffer

entia

te,

but i

nste

ad tr

ied

to

bulld

oze

thei

r way

th

roug

h th

e re

peat

ed

diffe

rent

iatio

n, to

var

ious

ou

tcom

es. M

uch

time

coul

d ha

ve b

een

save

d if

the

corr

ect p

repa

ratio

n fo

r im

plic

it di

ffere

ntia

tion

was

use

d.

Man

y st

uden

ts c

ould

not

ge

t to

the

final

requ

ired

form

of t

he e

quat

ion

beca

use

they

cou

ld n

ot

see

the

appr

opria

te

subs

titut

ion

for

esi

nx

xor

eco

sx

x

Som

e no

tatio

ns w

hich

ar

e no

t cor

rect

are

use

d,

eg:

2d

d(

)(e

cos

)d

dx

yy

yx

xx

Or s

illy

erro

rs li

ke

eco

sx

x be

com

ing

eco

sx

x a

nd

2co

sx

beco

min

g co

s2x

Page

12

of 2

9

Diff

eren

tiatin

g w

.r.t.

x ,

22

23

2

22

32

dd

dd

dd

dd

22

22

dd

dd

dd

dd

yy

yy

yy

yy

yy

yx

xx

xx

xx

x

Whe

n 0

x

,

f0

1

,

1f '

02

,

1f '

'0

4

,

5f '

''0

8

23

15

12

24

3!8

xx

xy

23

51

28

48x

xx

y

Ver

ifica

tion

Usi

ng M

F26,

e

cos

xy

x

1

22

eco

sx

yx

1

23

22

11

28

482

xx

xx

y

2

32

11

12

848

22

xx

xx

y

2

32

3

14

28

848

xx

xx

xy

23

51

28

48x

xx

y

(v

erifi

ed)

Som

e st

uden

ts

misu

nder

stan

d th

e qu

estio

n. T

hey

are

supp

osed

to v

erify

that

th

e ex

pres

sion

in (i

i) is

corr

ect.

They

inte

rpre

ted

it as

che

ckin

g IF

the

expr

essio

n is

cor

rect

and

so

me

conc

lude

d th

at it

is

not.

A w

orry

ing

situ

atio

n is

that

man

y st

uden

ts

com

mit

basic

alg

ebra

ic

mist

akes

, lik

e sa

ying

3

11

32

22

11

13

3x

xx

x

Page

13

of 2

9

OR

Usi

ng M

F26,

e

cos

xy

x

11

23

22

21

12!

3!2

xx

xy

x

12

32

32

12

62

2x

xx

xy

x

3

1 2

13x

yx

33

3

2

3

11

12

21

23

2!3

11

32

22

...3!

3

xx

yx

x

xx

32

31

11

...2

68

16x

xy

xx

23

51

28

48x

xx

y

(v

erifi

ed)

Som

e st

uden

ts g

ave

too

man

y te

rms a

nd h

ad a

ha

rd ti

me

man

ipul

atin

g.

They

hav

e to

real

ize

that

th

ey c

an tr

unca

te th

e se

ries f

or te

rms b

eyon

d

3 x. S

ome

stud

ents

tru

ncat

e to

o m

uch

too

early

, end

ing

up w

ith th

e in

corr

ect a

nsw

er to

o.

A fe

w st

uden

ts sq

uare

th

eir e

xpre

ssio

n in

(ii)

to

com

pare

with

the

stan

dard

ser

ies e

xpan

sion

of

2 yan

d co

nclu

ded

it is

corre

ct. C

redi

t is n

ot

give

n be

caus

e 2

2A

B

does

not

nec

essa

rily

impl

y A

B

Page

14

of 2

9

Q7.

App

licat

ions

of D

iffer

entia

tion

Ass

essm

ent O

bjec

tives

So

lutio

n E

xam

iner

’s F

eedb

ack

Max

ima/

Min

ima

prob

lem

s(i)

V

olum

e of

trou

gh

2

12

104

2V

hh

2

104

Vh

h

1

22

2d

110

410

42

d2

Vh

hh

hh

1

22

22

d10

410

4dV

hh

hh

To m

axim

ize

volu

me

of tr

ough

, d

0dV h

1

22

22

104

104

hh

h

2

210

104

hh

24

2h

2

reje

ct

2 si

nce

0h

h

OR

22

2

22

4

100

4

400

100

Vh

h

Vh

h

3d

280

040

0dV

Vh

hh

To m

axim

ize

volu

me

of tr

ough

, d

0dV h

240

0(2

)0

hh

2 re

ject

2

and

0si

nce

0h

hh

h

Man

y st

uden

ts w

ere

not a

ble

to

form

the

expr

essio

n of

the

volu

me

corr

ectly

.

This

is th

e m

ost c

omm

only

seen

mist

ake:

2

110

42

Vh

h

.

Whe

n it

com

es to

diff

eren

tiatio

n,

som

e stu

dent

s mad

e m

istak

e in

the

chai

n ru

le w

hile

per

form

ing

the

prod

uct r

ule.

Exam

ple:

2h

was

ofte

n m

isse

d ou

t.

ToTTTTT 4444044 h

Page

15

of 2

9

1st d

eriv

ativ

e te

st:

h

22

2

d dV +

0 -

OR

2n

d der

ivat

ive

test

:

32

22

22

d10

42

12d

Vh

hh

h

Whe

n 2

h

, 2

2d

0d

V h

It is

a m

axim

um v

alue

. 310

24

220

mV

Mos

t stu

dent

s wer

e ab

le to

ver

ify

the

max

imum

vol

ume

but t

he

pres

enta

tion

was

qui

te b

ad fo

r so

me

scrip

ts.

Som

e st

uden

ts lo

st on

e m

ark

as

they

forg

ot to

find

the

max

vol

ume.

Rat

e of

cha

nge

(ii)

3d

0.00

1 cm

/sdV t

Afte

r 29

min

,

30.

001

6029

1.74

mV

Vol

ume

of la

rger

cyl

inde

r,

3

22

ππ

10.

51.

5707

96m

LVr

h

Thus

the

wat

er le

vel i

s at t

he sm

alle

r cyl

inde

r afte

r 29

min

utes

.

2

π0.

6SV

h

2

dπ

0.6

dSV h

dd

dd

dd

SS

S

S

VV

ht

ht

Gen

eral

ly q

uite

OK

. Mos

t stu

dent

s at

tem

pted

this

part

usin

g a

varie

ty o

f m

etho

ds to

obt

ain

the

answ

er.

Com

mon

mist

ake:

stud

ents

forg

ot to

ch

eck

the

wat

er le

vel a

t 29

min

utes

an

d ju

st m

ade

thei

r ow

n as

sum

ptio

n w

hile

maj

ority

atte

mpt

ed t

o fo

rm

the

expr

essio

n of

the

vol

ume

by

cons

ider

ing

both

lar

ge a

nd s

mal

l co

ntai

ners

.

SV S dddddd dV d

Page

16

of 2

9

2

d1

0.00

1d

π0.

6sh t

fo

r 0.

6r

1 m

/s36

0π

Sket

ch g

raph

(ii

i) Fo

r the

larg

er c

ylin

der:

2

π1

LVh

2

dπ

1d

LV h

dd

dd

dd

LL

L

L

VV

ht

ht

2

d1

0.00

1d

π1

Lh t

fo

r 1

r

1 m

/s10

00π

For t

he sm

alle

r cyl

inde

r:

πd

1d

360

sh t

Tim

e ta

ken

to fi

ll th

e la

rger

cyl

inde

r

2π

10.

5=

1570

.796

s=26

.179

938

min

=26.

2m

in0.

001

Tim

e ta

ken

to fi

ll up

the

smal

ler c

ylin

der

2

π0.

60.

5=

565.

4867

s9.

4247

8m

in0.

001

Tota

l tim

e ta

ken

to fi

ll up

the

cont

aine

r 26

.179

938+

9.42

478

min

=35

.6m

in

Man

y st

uden

ts a

ttem

pted

thi

s pa

rt an

d al

so a

ble

to o

btai

ned

at le

ast 1

or

2 o

ut o

f 3 m

arks

.

ToTTTTTT ===

Page

17

of 2

9

Com

mon

m

istak

e:

(ext

ra v

ertic

al

line)

Man

y lo

st th

e 1

mar

k as

they

did

n’t

man

age

to fi

nd th

is.

t (m

in)

O

2

6.2

35.

6

Page

18

of 2

9

Q8.

Com

plex

Num

bers

Ass

essm

ent O

bjec

tives

So

lutio

n E

xam

iner

’s F

eedb

ack

Solv

ing

of s

imul

tane

ous e

quat

ions

(a)

3

i12

19i

1z

w

2*

316

23i

2z

w

13:

93i

3657

i3

zw

2

i:2i

*3i

16i

23z

w

4

34

:92i

*73

i59

zz

Let

i,

so *

iz

xy

zx

y

9

i2i

i73

i59

xy

xy

Com

pari

ng r

eal a

nd im

agin

ary

part

s, 9

259

and

92

73x

yy

x

So

lvin

g,

5an

d 7

xy

5

7iz

From

(2),

2*

316

23i

zw

11

1623

i2

*16

23i

25

7i3

3w

z

23i

w

Maj

ority

of

the

cand

idat

es l

ost

thei

r fin

al m

ark

beca

use

they

sim

ply

cann

ot c

opy

the

ques

tion

prop

erly

. M

any

wro

te 9

i in

stea

d of

19i

, th

is i

s un

acce

ptab

le!

Som

e ca

ndid

ates

miss

out

on

the

inst

ruct

ion

“Do

not u

se c

alcu

lato

r in

answ

erin

g th

e qu

estio

n”, t

hey

still

w

rite

“usi

ng

GC

” w

hile

so

lvin

g th

e sim

ulta

neou

s eq

uatio

n af

ter

com

parin

g re

al a

nd

imag

inar

y pa

rts.

Som

e ca

ndid

ates

too

k a

long

er w

ay b

y le

tting

i

za

b

a

nd

iw

cd

and

so

lved

fo

r 4

unkn

owns

. It i

s a

tedi

ous

proc

ess

and

they

end

ed

up u

sing

GC

to so

lve.

Thi

s is n

ot re

com

men

ded.

Som

e ca

ndid

ates

mad

e a

conc

eptu

al e

rror

. Afte

r m

akin

g z t

he su

bjec

t fro

m e

quat

ion

(1):

112

19i

3z

w

, the

y w

ent o

n to

conc

ude

that

*1

1219

i3

zw

w

hich

is in

corr

ect!

They

can

not d

o th

is st

ep h

ere

as w

is a

com

plex

nu

mbe

r.

Com

plex

con

juga

te ro

ots

(b)(

i) Si

nce

all c

oeffi

cien

ts a

re re

al, t

he o

ther

root

is

1

3i

.

21

3i1

3i2

4z

zz

z

4

32

22

34

82

4z

zz

zz

zaz

b

Man

y ca

ndid

ates

are

uns

ure

abou

t fa

ctor

s an

d ro

ots.

Som

e ca

ndid

ates

gav

e th

e co

njug

ate

as 1

3i

(neg

ate

the

real

par

t) w

hich

is w

rong

.

ts s ss

(b((((((((((((b)()()()()()(

i)i)i)i)i)i) Si

n

Page

19

of 2

9

Com

parin

g co

effic

ient

of

3:

z 3

21

aa

Com

parin

g co

effic

ient

of

2:

z

42

42

ba

b

Com

parin

g co

effic

ient

of

:z 0

24

2b

ab

4

32

22

34

82

42

zz

zz

zz

z

Can

dida

tes w

ho d

id n

ot g

et fu

ll cr

edit

stru

ggle

d at

al

gebr

aic

man

ipul

atio

ns

whe

n co

mpa

ring

coef

ficie

nts/

long

div

isio

n.

Ther

e is

no n

eed

to so

lve

for t

he ro

ots.

Com

plex

num

bers

in e

xpon

entia

l fo

rm

(b)(

ii)iq

5 1ep

z

2

i3

11

3i =

2e

z

5

2i

iq3

ee

2e

p

10

iiq

3e

e 3

2ep

e 3

2ln

32p

p

1012

23

33

q

Man

y le

ft th

is bl

ank.

Can

dida

tes

who

exp

ande

d th

is us

ing

Car

tesi

an

form

ofte

n st

rugg

led,

the

y ar

e re

com

men

ded

to

conv

ert t

o m

odul

us-a

rgum

ent f

orm

.

Som

e fo

rgot

abo

ut th

e po

wer

5.

Can

dida

tes

need

to

prac

tice

mor

e on

fin

ding

ar

gum

ents

of c

ompl

ex n

umbe

r. 1

13i

z

li

es

in

the

2nd

quad

rant

an

d he

nce,

11

3ar

gta

n1

z

(whe

n fin

ding

ba

sic

angl

e, a

lway

s 1

tan

the

pos

itive

real

/ im

par

ts)

Som

e di

d no

t kn

ow h

ow t

o co

nver

t q

into

the

pr

inci

pal r

ange

. The

y ne

ed to

subt

ract

mul

tiple

s of

2

Page

20

of 2

9

Q9.

Vec

tors

(Lin

es &

Pla

nes)

Ass

essm

ent O

bjec

tives

So

lutio

n E

xam

iner

’s F

eedb

ack

Line

lies

on

plan

e(i)

1

12

12

:8

18

35

35

l

r :17

0ap

b

r

Sinc

e 1l

lies c

ompl

etel

y on

p,

1 817

30a b

817

1a

b

Sinc

e 1l l

ies i

n p,

2 1an

d5

0a b

a

re p

erpe

ndic

ular

,

2 10

50a b

20

ab

2b

a

Subs

titut

e2

ba

into

(1),

1617

aa

17

17a

1a

and

2

2b

a

(S

how

n)

Mos

t stu

dent

s w

ere

able

to u

se th

e in

form

atio

n pr

ovid

ed to

form

ulat

e 2

equa

tions

in a

and

b, a

nd p

roce

eded

to

solv

e si

mul

tane

ousl

y.

Alte

rnat

ive

met

hod

of o

btai

ning

the

two

equa

tions

w

as

to

subs

titut

e di

ffere

nt v

alue

s of

in

ord

er t

o ob

tain

diff

eren

t poi

nts o

n 1l.

SuSSSSSaa

Page

21

of 2

9

Foot

of p

erpe

ndic

ular

(ii

) 1 2 3

OQ

Let N

be

the

foot

of p

erpe

ndic

ular

from

poi

nt Q

to p

.

Equa

tion

of th

e lin

e pa

ssin

g th

ough

N a

nd Q

is

11

22

,3

0

1 22 3

r

W

hen

the

line

pass

es th

roug

h N

on

p

11

22

217

30

1

44

17

53

17

5

20

4

Com

mon

mist

ake

was

for

stu

dent

s to

erro

neou

sly

assu

me

that

poi

nt Q

lies

on t

he p

lane

. A

qui

ck c

heck

w

ould

hav

e sh

owed

oth

erw

ise.

A h

andf

ul o

f stu

dent

s w

ere

able

to

visu

aliz

e th

e re

latio

nshi

p be

twee

n 1l

, p

and

Q t

o di

scov

er t

hat

the

perp

endi

cula

r dist

ance

can

be

foun

d ef

ficie

ntly

us

ing

the

leng

th

of

proj

ectio

n fo

rmul

a.

11

02

810

33

0AQ

OQ

OA

22

1 20

020

ˆ10

45

51

20

AQn

N

Q

p

A=

Q

Page

22

of 2

9

11

45

22

28

63

33

ON

Shor

test

Dist

ance

=

QN

ON

OQ

22

51

62

33

4 8 0 48

80 45

Shor

test

dis

tanc

e (ii

i)

Shor

test

dist

ance

bet

wee

n Q

and

1l

Ther

e w

ere

som

e stu

dent

s w

ho

iden

tifie

d th

e le

ngth

s to

be fo

und

for

(ii)

and

(iii)

wro

ngly

(i.e

. gav

e (ii

i) an

swer

for (

ii) a

nd v

ice

vers

a). T

hey

wer

e st

ill g

iven

the

cred

it.

Alte

rnat

ive

met

hods

(e

.g.

usin

g le

ngth

of p

roje

ctio

n) w

ere

acce

pted

.

N

Q

pM

Page

23

of 2

9

22

54

56

3

225

806

9

150

809

290 3

o

r 1

870

3

or 9

.83

(3 s.

f)

Equa

tion

of li

ne

(iv)

Dire

ctio

n ve

ctor

of

2l2

11

25

0

02

50

15

41

10 5 5

25

1 1

Vec

tor e

quat

ion

of th

e lin

e 2l is

1

22

1,

31

rt

t

Mos

t st

uden

ts

wer

e un

able

to

vi

sual

ize

the

rela

tions

hip

betw

een

21,an

d l

lp

.

They

trie

d to

wor

k ba

ckw

ards

usin

g th

e gi

ven

Car

tesi

an

equa

tion

to

obta

in th

e ve

ctor

equa

tion

of

2lin

an

atte

mpt

to

fu

lfill

the

ques

tion

requ

irem

ent o

f sho

win

g th

e pr

oces

s to

find

2l.

rrrrrr

Page

24

of 2

9

11

22

22

33x

tx

ty

tt

yz

tt

z

Car

tesia

n eq

uatio

n of

the

line

2l is

1

23

2xy

z

Ref

lect

ion

of li

ne a

bout

a p

lane

(v

)

Usi

ng R

atio

The

orem

Sinc

e N

is th

e m

id-p

oint

of

OQ

and O

Q

,

1 2O

NO

QO

Q

2O

QO

NO

Q

Mos

t st

uden

ts w

ere

unab

le

to

visu

aliz

e th

e sit

uatio

n w

hen

a lin

e pa

ralle

l to

a pl

ane

is re

flect

ed in

the

plan

e. T

hey

did

not r

ealis

e th

at th

e lin

e an

d re

flect

ed l

ines

bot

h sh

are

the

sam

e di

rect

ion

vect

or.

Stud

ents

w

ho

mad

e m

istak

es

in

prev

ious

par

ts (

e.g.

pos

ition

vec

tor

of fo

ot o

f per

pend

icul

ar) w

ere

give

n cr

edit

for

thei

r at

tem

pt t

o ap

ply

Rat

io T

heor

em.

N

Q

O

Q’

N

Q

p

Q’

SiSSSSSS

Page

25

of 2

9

51

92

62

143

33

OQ

Vec

tor e

quat

ion

of th

e lin

e 3l is

9

214

1,

31

rs

s

Page

26

of 2

9

Q10

. Diff

eren

tial E

quat

ions

Ass

essm

ent O

bjec

tives

So

lutio

n E

xam

iner

’s F

eedb

ack

Form

ulat

e of

diff

eren

tial e

quat

ion

(i)

Gro

wth

rate

L

Des

truct

ion

rate

L

d dL

aL

bLt

Whe

n 1

L

, d

1dL t

1(1)

ab

Whe

n 4

L

, d

2dL t

44

22

42

(2)

ab

ab

Solv

ing,

3,

2a

b

d3

2dL

LL

t

(s

how

n)

Som

e stu

dent

s di

d no

t m

anag

e to

fo

rmul

ate

two

equa

tions

with

tw

o un

know

ns a

s the

y se

t a

b

Mos

t of

thos

e w

ho m

anag

ed to

get

th

e tw

o eq

uatio

ns

with

tw

o un

know

ns

proc

eede

d to

ge

t th

e co

rrect

val

ues

of a

and

b.

Man

y st

uden

ts d

id n

ot a

ttem

pt th

is

part.

Solv

ing

diff

eren

tial e

quat

ion

(ii)

Usi

ng su

bstit

utio

n:

2L

y

Diff

eren

tiate

with

resp

ect t

o t:

dd

2d

dL

yy

tt

2d

23

2d

d2

32

dd

32

d2

yy

yy

ty

yt

yy

t

Usi

ng su

bstit

utio

n: y

L

Diff

eren

tiate

with

resp

ect t

o t:

1 2

d1

dd

2d

yL

Lt

t

1 2d

13

2d

2 33

22

32 2

yL

LL

t

Ly

y

Mos

t st

uden

ts u

sed

2L

y

. Th

e m

inor

ity w

ho u

sed

yL

a

re a

lso

able

to g

et th

e co

rrec

t equ

atio

n.

222222 d

Page

27

of 2

9

d3

2d

2y

yt

2

d1

32

dyy

t

2d

d1

d3

2d

2 d

32

ln3

2

32

e

32

ee

32

e, w

here

e

23

e3

e2

tC C

t

tC

t

tyt

ty

t yt

Cy y

tC

y y yA

Ay

A Ay

2

3e

2

3e

2

t

t

Ay

L

AL

W

hen

0t

, 1

L

, 2

03

e1

2A

Sinc

e 0

y

, 3

11

2A

A

23

e 2

t

L

Mos

t st

uden

ts a

re a

ble

to i

dent

ify

this

as a

sepa

ratio

n of

var

iabl

es D

E.

Man

y st

uden

ts d

id n

ot i

nclu

de t

he

mod

ulus

sign

for t

he lo

garit

hm.

Am

ong

thos

e w

ho d

id h

ave

the

mod

ulus

sig

n, m

any

rem

oved

the

m

odul

us a

nd d

id n

ot p

ut

Man

y st

uden

ts st

oppe

d he

re w

ithou

t fin

ding

A .

111111 Sin

Page

28

of 2

9

(iii)

2

22

2

d2

d1

Lt

tt

Inte

grat

e bo

th si

des w

ith re

spec

t to

t,

22

22 1

2

2

d2

dd

1 21

d

11

11

Lt

tt

t tt

t

tC

Ct

Whe

n d

0,

1,

1dL

tL

t

,

2

11

10

0

C

C

2

d1

d1

L tt

In

tegr

ate

agai

n bo

th si

des w

ith re

spec

t to

t,

2

11 d

1 tan

Lt

t tD

Whe

n 0,

1,

tL

11

tan

01

DD

1ta

n1

Lt

Mos

t stu

dent

s co

uld

not i

dent

ify th

e fo

rm

f'f

nx

x a

nd t

hus

coul

d no

t int

egra

te.

Som

e stu

dent

s w

ho

proc

eede

d ei

ther

m

isse

d ou

t or

di

d no

t un

ders

tand

th

e ph

rase

‘p

artic

ular

so

lutio

n’,

thus

th

ey

did

not

subs

titut

e in

the

initi

al v

alue

s.

WWWWWW 1 D

Page

29

of 2

9

Lim

its

(iv)

As

t

,

For M

odel

A:

23

9e

0,2

4t

L

For M

odel

B:

1π

πta

n,

12

2t

L

Com

mon

m

istak

e w

as

usin

g 90

inst

ead

of 2

. Ang

les

shou

ld b

e in

radi

ans

for c

alcu

lus q

uest

ions

.

Page

1 o

f 20

CATH

OLI

C JU

NIO

R C

OLL

EGE

H

2 M

ATH

EMA

TICS

2

019

JC2

PREL

IMIN

AR

Y EX

AMIN

ATI

ON

SO

LUTI

ON

Q1.

Sys

tem

of L

inea

r E

quat

ions

Ass

essm

ent O

bjec

tives

So

lutio

n E

xam

iner

’s F

eedb

ack

Solv

e sy

stem

of l

inea

r equ

atio

ns.

131

11

4a

bc

497

6416

ab

c

864

251

264

ab

c

From

G.C

., 1

1,2,

0.25

ab

c

1,2,

4a

bc

Mos

t ca

ndid

ates

wer

e su

cces

sful

with

th

e qu

estio

n ex

cept

tho

se w

ho d

id n

ot

know

how

to

hand

le t

he t

rans

form

ed

coor

dina

tes

or th

e re

plac

emen

t of

1/c

as

anot

her v

aria

ble.

The

re w

ere

also

a lo

t of

alge

brai

c m

anip

ulat

ion

erro

rs

whi

ch

refle

cted

the

lac

k in

num

erac

y sk

ills

of

the

cand

idat

ure.

Th

ere

is al

so

a si

gnifi

cant

gr

oup

of

cand

idat

es

who

at

tem

pted

to

so

lve

the

equa

tions

m

anua

lly i

nste

ad o

f us

ing

the

GC

, an

d m

ost o

f the

m w

ere

unsu

cces

sful

.

Page

2 o

f 20

Q2.

Vec

tors

(Bas

ic)

Ass

essm

ent O

bjec

tives

So

lutio

n E

xam

iner

’s F

eedb

ack

Con

cept

of p

aral

lel v

ecto

rs(i)

k

m

p q

whe

re

\0

m

=

pq

m

nk

p

=q

q w

here

\0

mn

k

=

Sinc

e n

p

q, p

and

q a

re p

aral

lel v

ecto

rs.

This

ba

sic

ques

tion

prov

ed

to b

e di

ffic

ult

for

mos

t ca

ndid

ates

, w

here

onl

y a

hand

ful

prov

ided

a c

ompl

ete

solu

tion

with

ex

plan

atio

n.

Maj

ority

of

th

e ac

cept

ed

answ

ers

simpl

y co

ntai

n th

e ke

y w

ord

“par

alle

l” w

ith

ambi

guou

s st

atem

ents

and

wer

e gi

ven

the

bene

fit o

f do

ubt.

Use

of s

cala

r pro

duct

(ii

) k

p

=p

qq

Sinc

e p

and

q ar

e pa

ralle

l vec

tors

from

par

t (i),

0

or

180

so

cos

1

, so

pq

pq

.

kp

=p

qq

2k

q

sin

ce

0

p

2k

q

k

q

sin

ce

0k

.

This

que

stio

n w

as e

ither

not

atte

mpt

ed o

r ve

ry b

adly

at

tem

pted

by

the

cand

idat

ure.

Alm

ost

ever

y ca

ndid

ate

perfo

rmed

div

ision

on

the

vect

ors

dire

ctly

, w

hich

is

conc

eptu

ally

w

rong

, in

stea

d of

ap

plyi

ng

mod

ulus

th

roug

hout

to

redu

ce t

he e

quat

ion

to t

hat

cons

istin

g of

on

ly s

cala

rs a

nd th

e us

ual a

lgeb

raic

man

ipul

atio

n ca

n be

ap

plie

d. O

ther

err

ors

incl

ude

assu

min

g th

at t

he v

ecto

rs

are

in

the

sam

e di

rect

ion

(exc

ludi

ng

oppo

site

as

a po

ssib

ility

) and

com

parin

g co

effic

ient

s.

Page

3 o

f 20

Q3.

App

licat

ions

of D

iffer

entia

tion

(f’ g

raph

)A

sses

smen

t Obj

ectiv

es

Solu

tion

Exa

min

er’s

Fee

dbac

k Sk

etch

ing

grap

h of

f'x

(i)

G

ener

ally

stu

dent

s di

d qu

ite o

k fo

r th

is pa

rt. S

ome

did

not p

rese

nt t

he

turn

ing

poin

ts in

coo

rdin

ates

form

.

Gra

phic

al in

terp

reta

tion

of

f'0

x

(ii)(a

) 1

2x

o

r 2

5x

Qui

te b

adly

atte

mpt

ed.

Gra

phic

al in

terp

reta

tion

of

f''

0x

(ii)(

b)2

x

Qui

te b

adly

atte

mpt

ed.

y

1y

2x

1,0

5,

0

Page

4 o

f 20

Q4.

Fun

ctio

nsA

sses

smen

t Obj

ectiv

es

Solu

tion

Exa

min

er’s

Fee

dbac

k C

ondi

tion

for e

xist

ence

of i

nver

se

func

tion

(i)

Any

hor

izon

tal l

ine

cuts

the

grap

h of

f at

mos

t onc

e, f

is a

one-

one

func

tion,

-1

fex

ists

.

Bad

ly a

ttem

pted

.

Com

mon

mist

akes

:

Man

y us

ed o

nly

one

coun

ter

exam

ple

to sh

ow th

at f

is on

e-on

e.

Abl

e to

reso

lve

the

mod

ulus

bas

ed o

n th

e do

mai

n an

d ab

le to

find

the

inve

rse

func

tion.

(ii)

3y

xx

1

2

2

2

1f

3 30

39

02

4

39

24

39

Sinc

e2,

24

39

f(

),

D(0

,2]

24

yx

xx

xy

xy

xy

xx

y

xx

Bad

ly d

one.

Man

y di

d no

t kno

w

how

to re

solv

e th

e m

odul

us b

y lo

okin

g at

the

dom

ain.

Som

e m

ade

alge

brai

c sl

ip a

nd

henc

e no

t abl

e to

per

form

co

mpl

etin

g th

e sq

uare

to m

ake

x th

e su

bjec

t.

How

ever

maj

ority

wer

e ab

le to

find

th

e do

mai

n.

3 2

x

y

Page

5 o

f 20

Sket

ch o

f pie

ce w

ise

func

tions

(ii

i) M

ost

stud

ents

atte

mpt

ed t

his

part

and

wer

e ab

le t

o ge

t ei

ther

1 o

r 2

mar

ks.

Stud

ents

nee

d to

pay

mor

e at

tent

ion

to th

e pr

esen

tatio

n su

ch a

s to

lab

el t

he c

oord

inat

es o

f th

e en

d po

ints

, inc

lusi

ve o

r exc

lusi

ve.

Com

posi

te fu

nctio

n w

ith p

iece

wis

e.

(iv)

2

gf2

32

23

1

xx

x

xx

Bad

ly a

ttem

pted

. Man

y st

uden

ts d

id

not s

eem

to u

nder

stan

d th

e co

nditi

on o

f the

exi

sten

ce o

f co

mpo

site

func

tion.

R

ange

of c

ompo

site

func

tion.

(v

) gf

R(

2,2]

Bad

ly d

one.

Stu

dent

s w

ho u

sed

the

map

ping

met

hod

wer

e ab

le t

o ge

t th

e an

swer

eas

ily.

(4,2

/5)

( 2,2

)

x

y y

=(x

)

( 0,-2

)

Page

6 o

f 20

Q5.

Par

amet

ric

Equ

atio

nsA

sses

smen

t Obj

ectiv

es

Solu

tion

Exa

min

er’s

Fee

dbac

k Sk

etch

par

amet

ric

grap

hs

(i)

At

π

,

1

sin

π1

x

,

πco

sπ

π1

y

1,π

1

At

π 2

, π

1si

n0

2x

,

ππ

πco

s2

22

y

π

0,2

Man

y ca

ndid

ates

fai

led

to g

ive

the

coor

dina

tes

in

the

exac

tfo

rm,

as

requ

ired

by

the

ques

tion.

They

are

als

o un

able

to

sket

ch

the

curv

e in

th

e sp

ecifi

ed

dom

ain.

Endp

oint

s m

ust

be

clea

rly

labe

lled

with

sol

id c

ircle

s si

nce

both

end

poin

ts a

re in

clud

ed.

Gra

dien

t of p

aram

etric

eq

uatio

ns

(ii)

1sin

,co

s,

xy

dd

cos

,1

sind

dx

y

d1

sin

sin

1d

cos

cos

y x

As

π 2

,

πco

s0

2

,

d1

sin

dco

sy x

.

The

tang

ent i

s ve

rtica

l.

OR

The

tang

ent i

s par

alle

l to

y-ax

is.

Find

ing

d dy x w

as g

ener

ally

wel

l-

atte

mpt

ed.

How

ever

so

me

cand

idat

es a

re s

till c

onfu

sed

over

di

ffere

ntia

tion

of

simpl

e tri

go

term

s and

car

eles

snes

s.

The

last

pa

rt w

as

badl

y at

tem

pted

. C

andi

date

s fo

und

the

valu

e of

d dy x

but

did

not

pro

ceed

to e

xpla

in “

wha

t hap

pene

d to

the

tang

ent”

. Th

ose

who

wro

te “

grad

ient

of

tang

ent

tend

s to

inf

inity

” w

ere

give

n th

e cr

edit.

H

owev

er,

“tan

gent

te

nds

to

infin

ity”

is no

t ac

cept

ed

as

tang

ent i

s a li

ne!

Ox

y O

OOOOR

OR

OOOOR

OOOOOOOO

ThThhThThTThTThTTThThThThhThThThThhe e e eee ee e e e eeeee e

tattttttttttttttttttngngngngngng

enenenenenent t t t t t i

s p

Page

7 o

f 20

Cand

idat

es s

houl

d le

arn

how

to

spel

l “ve

rtica

l” p

rope

rly.

Equa

tion

of n

orm

al

(iii)

At P

, p

,

1

sin

,co

sp

pp

Gra

dien

t of n

orm

al

cos

1si

npp

Equa

tion

of n

orm

al a

t P:

cos

cos

1si

n1

sinp

yp

px

pp

At y

-axi

s, 0

x

:

cos

cos

01

sin1

sin

cos

cosp

yp

pp

py

pp

py

p

0,Q

p

Maj

ority

of t

he c

andi

date

s cou

ld

give

the

grad

ient

of n

orm

al b

ased

on

thei

r res

ults

in (i

i), w

hile

so

me

still

stru

ggle

d.

Man

y w

ent o

n to

ass

ume

valu

es

of p

to b

e 0,

4 o

r 2

whi

ch is

not n

eede

d. T

hey

mer

ely

used

th

e re

sults

0,p

to v

erify

inst

ead

of sh

owin

g.

Alg

ebra

ic m

anip

ulat

ion

need

s to

be fu

rther

stre

ngth

ened

.

Are

a in

volv

ing

para

met

ric

equa

tions

(iv

) In

tegr

atin

g ab

t y-a

xis:

Th

is

was

ba

dly

atte

mpt

ed,

show

ing

that

ca

ndid

ates

ha

ve

very

wea

k fo

unda

tion

in f

indi

ng

area

s in

volv

ing

para

met

ric

equa

tions

.

Dia

gram

has

to b

e ai

ded

to s

olve

th

is pa

rt.

Ox

y

OOOOOOOOOOOOOOOO

Page

8 o

f 20

π 2co

s

π 2

π2

2

π2

2 π 2

area 1

cos

1si

n d

2 1d

cos

1si

n

d2

d1

sin

cos

cos

1si

nd

22

1si

n2co

s1

2sin

sin

d2

41

sin2

1co

s2co

s1

2sin

d2

42

1si

n23

cos2

cos

2sin

d2

42

2

pp

p

p

p p p

pp

pp

xy

yp

px

pp

p

pp

pp

pp

π 2π 2

1si

n23

sin2

cos

2cos

24

24

1si

n23

ππ

sin

π3

sin2

cos

2cos

2cos

24

22

24

24

1si

n23π

3si

n2

cos

2cos

24

42

43π

33

cos

42

2

p

pp

pp

pp

p

pp

pp

p

pp

For

thos

e w

ho a

ttem

pted

, th

ey

wro

te

the

limits

w

rong

ly:

π 2 d

px

y

inst

ead

of

π 2co

s d

ppx

y

due

to

conf

usio

n be

twee

n y-

limits

and

-li

mits

.

Som

e di

d no

t inc

lude

the

area

of

the

trian

gle.

Do

not

chan

ge

the

para

met

ric

equa

tion

into

Car

tesia

n eq

uatio

n un

less

you

are

spe

cific

ally

told

to

do so

!

Page

9 o

f 20

Inte

grat

ing

abt x

-axi

s:

1si

n

0 π 2 π 2

2π 2

2π

ππ

22

2

area

1 d

cos

1si

n2

d1

d

2co

s1

sin

d2

11

cos

cos

dco

ssi

nsi

nco

s2

21

sin2

cos

cos

dco

ssi

n2

4

1si

nsi

n d

cos

dco

s2

p

p p

p

pp

p

yx

pp

pp

xy

pp

p

pp

pp

pp

pp

pp

p

p

ππ

22

π 2

sin2

sin

4

ππ

1co

s21

sin

2si

nsi

nco

s d

cos

sin

22

22

4