CJC 2019 H2 Mathematics - The Learning Space · 2020. 5. 28. · CJC 2019 H2 Mathematics Prelim...
Transcript of CJC 2019 H2 Mathematics - The Learning Space · 2020. 5. 28. · CJC 2019 H2 Mathematics Prelim...
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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]
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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]
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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]
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(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]
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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]
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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
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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