Page 1 of 32 CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS JC1 PROMOTIONAL EXAMINATION SOLUTIONS 2014 1 System of Linear Equations No . Assessment Objectives Solution Feedback Able to formulate a system of linear equations from a problem involving three unknowns based on the conditions given, so as to solve the unknowns. [H.O.T.] Given that 1 n n u au bn c + = + + for any n + ∈ . At 1 n = , since 2 8 u = and 1 2 u = , hence 8 (2) (1) a b c = + + , i.e. 2 8 a b c + + = . At 2 n = , since 3 17 u = and 2 8 u = , hence 17 (8) (2) a b c = + + , i.e. 8 2 17 a b c + + = . At 3 n = , since 4 32 u = and 3 17 u = , hence 32 (17) (3) a b c = + + , i.e. 17 3 32 a b c + + = . Consolidating these equations yields the system of linear equations 2 8 8 2 17 17 3 32 a b c a b c a b c + + = + + = + + = . Solving this system (via the G.C.) yields 2, 3, 7 a b c = =- = . More than half of the students were able to solve this question correctly. The most common mistake was letting the coefficient of b be 1 n + instead of n . Another common error was overlooking the fact that n + ∈ , and wrongly defining 0 u to be 0. These students incorrectly concluded that c =2. Other students did not realise this was a question on system of linear equations, and did not formulate any equations. Less common mistakes include treating the coefficient of b , n , as an unknown constant instead of substituting n =1, 2 & 3 accordingly. Few students formulated a correct system of linear equations, yet did not solve the system accurately, possibly due to error while handling the G.C. Some students, having formed the correct system of equations, proceeded to solve the system without the use of G.C. Where no careless mistakes were made, students were not penalised. However, solving manually left them less time for other questions.
Transcript
Page 1 of 32
CATHOLIC JUNIOR COLLEGE
H2 MATHEMATICS
JC1 PROMOTIONAL EXAMINATION SOLUTIONS 2014
1 System of Linear Equations
No
.
Assessment Objectives Solution Feedback
Able to formulate a system of
linear equations from a problem
involving three unknowns based
on the conditions given, so as to
solve the unknowns.
[H.O.T.]
Given that 1n nu au bn c+ = + + for any n+∈� .
At 1n = , since 2 8u = and 1 2u = ,
hence 8 (2) (1)a b c= + + , i.e. 2 8a b c+ + = .
At 2n = , since 3 17u = and
2 8u = ,
hence 17 (8) (2)a b c= + + , i.e. 8 2 17a b c+ + = .
At 3n = , since 4 32u = and
3 17u = ,
hence 32 (17) (3)a b c= + + , i.e. 17 3 32a b c+ + = .
Consolidating these equations yields the system of
linear equations
2 8
8 2 17
17 3 32
a b c
a b c
a b c
+ + =
+ + = + + =
.
Solving this system (via the G.C.) yields
2, 3, 7a b c= = − = .
More than half of the students were able
to solve this question correctly. The most
common mistake was letting the
coefficient of b be 1n + instead of n .
Another common error was overlooking
the fact that n +∈� , and wrongly
defining 0u to be 0. These students
incorrectly concluded that c =2. Other
students did not realise this was a
question on system of linear equations,
and did not formulate any equations.
Less common mistakes include treating
the coefficient ofb , n , as an unknown
constant instead of substituting n =1, 2 &
3 accordingly. Few students formulated a
correct system of linear equations, yet
did not solve the system accurately,
possibly due to error while handling the
G.C.
Some students, having formed the
correct system of equations, proceeded to
solve the system without the use of G.C.
Where no careless mistakes were made,
students were not penalised. However,
solving manually left them less time for
other questions.
Page 2 of 32
2 Differentiation
No
.
Assessment Objectives Solution Feedback
Able to differentiate simple
functions defined implicitly 3 sec
d d3 ln 3 sec tan
d d
y
y xy x
y yy x x x
x x
= −
= + −
( ) d3 ln 3 sec tan
d
y yx y x x
x− = −
( )d sec tan
d 3 ln 3y
y y x x
x x
−=
−
About one quarter could do this question successfully.
The main problem was that students cannot remember
the formula to differentiate 3y.
A handful of students by-passed the need for the
formula by doing:
d
dx 3
y = ln 3d
ed
y
x= ln 3d
ed
y
x= ln 3 d
e (ln 3)d
y y
x
= d
3 (ln 3)d
y y
x.
Some students re-wrote sec x as 1
cos x and used
quotient rule to differentiate although the formula is
given in MF15.
Many students have problems with the rules of
logarithm. Common errors include:
- ln (xy - sec x ) = ln xy – ln sec x
- ln (xy - sec x ) = ln ( xy
sec x ) or
ln xy
ln sec x
- d
dx y ln 3 =
dy
dx ln 3 + y (
1
3 )
- d
dx y ln 3 =
dy
dx ln 3 + y
- d
dx 3
y = y (3
y – 1)
dy
dx
- d
dx y ln 3 =
dy
dx
- d
dx 3
y = 3
y dy
dx
- d
dx y ln 3 =
1
y dy
dx
- d
dx sec x = ln (sec x + tan x)
Able to make d
d
y
x the subject of
formula
Page 3 of 32
3 Graphing Techniques I
No
.
Assessment Objectives Solution Feedback
Able to sketch ellipse and
exponential graphs. (i)
( )
( )
2 221
4 9
Ellipse, centre 2,0
x y++ =
−
Most students are able to score 2 to 3
marks out of 4. At least half of the
students failed to indicate the y-intercept
of the graph of32 exy += − .
Common errors:
- For the graph of 32 exy
+= − :
� Omission of y-intercept.
� Omission of horizontal asymptote
y = 2, or labelling the HA as x = 2
� Drawing an additional vertical
asymptote x = 0 (or x = 3) when
there should be no vertical
asymptote.
� Identifying the x-intercept as
2.30− instead of 2.31− .
� Axial intercepts given in 2 s.f.
instead of 3 s.f.
� Drawing 32 exy
−= − instead of 32 ex
y+= − .
- For the graph of ( )2 22
14 9
x y++ = :
� Identifying the centre as (2,0)
instead of ( 2,0)− .
� Failure to indicate the coordinates
of the stationary points.
� Drawing a hyperbola instead of
an ellipse.
x
y
O
( )222
14 9
x y++ =
32 exy += −
Page 4 of 32
Able to use G.C. to solve
intersections of graphs.
(ii) ( )
( ) ( )
( ) ( )
( )
( )
3
3
3
3
22
22
22
1
2
3
2
2
21 (1)
4 9
e 2 (2)
Substitute (2) into (1):
e 221
4 9
9 2 4 2 e 36
Sketch in G.C.:
e 3
99 2
4
99 2
4
Solving, 1.42 or 3.73
x
x
x
x
x y
y
x
x
y
y x
y x
x
+
+
+
+
++ = − − −
= − + − − −
− +++ =
+ + − =
= − +
= + − +
= − − +
= − −
Alternative:
( ) ( )22
1
2
( 3)
Sketch in G.C.:
9 2 4 2 e
36
Solving, 1.42 or 3.73
xy x
y
x
+= + + −
=
= − −
About a quarter of students are able to
score the full 2 marks. Most students
were able to show how
( ) ( )22 39 2 4 2 e 36xx
++ + − = is obtained
from the 2 equations in (i), but they were
unable to continue to solve correctly for
the values of x.
Instead of finding the x-coordinate of the
points of intersection, some students
solved for the x-intercepts of
( )222
14 9
x y++ = or 3e 2x
y+= − + .
Common errors:
� Rounding off error, giving 3.72−
instead of 3.73− , and 1.41−
instead of 1.42− .
� Solutions given in 2 s.f. instead
of 3 s.f.
� Considered
( )3 29e 3 9 2
4
xx
+− + = + − + only.
� Final answer given in coordinates
form instead of x-values only.
� Indicating the number of
solutions, but did not continue to
find the solutions.
Page 5 of 32
4 Arithmetic Progression & Geometric Progression
N
o.
Assessment Objectives Solution Feedback
Able to formulate an equation in
terms of r, using common
difference between consecutive
terms in an A.P.
(i)
( ) ( )
( ) ( ) ( )
4 3 3
3 2
2
2
2
1 1
Since 0, 0,
1 1 1
Since is non-zero, 1,
1
1 0 (shown)
ar ar ar ar d
ar r ar r
a r
r r r r
d r
r r
r r
− = − =
− = −
≠ ≠
− = − +
≠
= +
− − =
Alternative:
( )
( ) ( )
3 2
3 2
2
2
4 3 3
4 3 3 0
2 1 0
Since 0, 0,
2 1 0
1 1 0
Since is non-zero, 1,
1 0 (shown)
ar ar ar ar d
ar ar ar ar
ar r r
a r
r r
r r r
d r
r r
− = − =
− − + =
− + =
≠ ≠
− + =
− − − =
≠
− − =
This question was poorly done as most
students could not recognize that the
common difference of AP with the terms
in GP i.e. 5 4 4 2T T T T− = − .
Many thought that the first term of AP is a
but it was not!
Next, many could not simplify the
expression further and left the answer as 3 22 1 0r r− + = and therefore, could not
factorize ( )1r − out. Some left the answer
ignored the term ( )1r − while only a
handful of students could explain that
1 0r − = was rejected because 1r ≠ as d is
non-zero, i.e. if r = 1, the terms in GP will
be constant and therefore, the 3
corresponding terms in AP will be the
same which means that d is zero.
Alternative method (as produced by most
students):
Let the first term and common difference
of AP be b and d respectively.
( )
3 3 3
4
3 4
(1)
2 2
Sub (1) into (2) :
2
b ar
b d ar d ar b ar ar
b d ar d ar
ar ar ar ar
=
+ = ⇒ = − = −
+ = + =
+ − =
�
Page 6 of 32
Able to deduce the meaning of
convergence of a geometric series,
where 1r < and reject the
solution where r > 1.
(ii)
( ) ( ) ( )( )
2
2
1 0
1 1 4 1 1
2
1 5 1 5or
2 2
(rejected since 1)
r r
r
r r
r
− − =
− − ± − − −=
+ −= =
<
Most students who did not manage to
show (i) did not continue further. Some
mange to continue this part and obtain the
correct answer. However, most did not
explain clearly why 1 5
2r
+= was
rejected as they did not state 1r < . Many
students wrote reasons e.g. r > 0 as
rejected 1 5
2r
−= instead.
Some students could not apply the
quadratic formula correctly whereas
others gave the answer in decimal when
the question asked for “exact form”.
Able to apply the formulae of sum
of first n terms of an arithmetic
series with the respective first term
and common difference.
(iii)
( )
( ) ( )
3
3
2
2
First term of AP:
1
1
Common Difference:
1 1
1
1 51
2
1 5
2
ar rr
ar ar
r rr r
r
=
=
−
= −
= −
−= −
−=
Some students did not attempt this part of
the question as they did not know how to
start the question.
Most students could apply the formula of
sum of first n terms of AP, however,
majority of them got the first term and
common difference incorrect. The first
term of AP is not a, but ar (second term of
GP) which is equals to 1, hence ending up
with the answer 85 105 5− .
A handful of the students applied the
wrong formula and stated the formula for
sum of first n terms of AP instead i.e.
( )1
1
na r
r
−
− .
Majority of students are weak in their
algebraic manipulation i.e.
Page 7 of 32
( ) ( )20
20 1 52 1 20 1
2 2
1 510 2 19
2
115 95 5
S −
= + −
−= +
= −
( )2
21 51 5
2 2
− −=
failing to square the
denominator 2.
Page 8 of 32
5 Applications of Differentiation
No
.
Assessment Objectives Solution Feedback
Able to find equation of tangent of
curve where the curve is defined
parametrically.
(a) ( )d
6cos 3d
xθ
θ= ( ) ( )d
6sin 3 cos 3d
yθ θ
θ=
At π
12θ = ,
d π 1sin
d 4 2
y
x
= =
π
2sin 24
x
= =
2 π 94 sin
4 2y
= + =
Equation of tangent : ( )9 12
2 2y x
− = −
1 7
22y x= +
Most students were able to differentiate
to get d
d
y
x . Common mistakes: Not
multiplying by 3, not applying chain
rule for ( )2sin 3θ
Most students able to evaluate d
d
y
x , x ,
y at 12
πθ = .
Some students used y mx c= + instead
of the usual tangent eqn, but were still
able to obtain the answer.
Some students used an alternative
method, changing the parametric eqns
to Cartesian before proceeding. Most of
did so were able to get the solution.
Able to interpret derivative as a
physical rate of change of a
physical quantity.
(b) 3d2 0.5 2.5cm /s
d
V
t= − − = −
So 7
2 7
15 30r h h= =
2
2
31 1 7 49
3 3 30 2700V r h h h h
ππ π = = =
Many variations of
d
d
V
t , e.g. 2.5, 1.5,
-1.5. Mostly put 2.5.
Many students differentiated
21
3V r hπ= w.r.t. h to get 21
3V rπ=
Many students found d
d
r
t instead of
d
d
h
t
Page 9 of 32
Able to use concept to solve
problems involving connected
rates of change.
( )2 2d 49 493
d 2700 900
Vh h
h
π π= =
( )2
d d d 9002.5
d d d 49
h h V
t V t hπ= ⋅ = ⋅ −
When 7h = , ( )
( )2
d 9002.5
d 49 7
h
t π= ⋅ −
2250
or 0.298cm/s2401π
= − −
Drink level is dropping at rate of 2250
or 0.298cm/s2401π
Many students did not answer the
question and left their answer as
d0.298
d
h
t= −
Page 10 of 32
6 Maclaurin’s Series
No
.
Assessment Objectives Solution Feedback
Able to perform implicit
differentiation. 2
e cosx
y x=
2 2
2
de sin 2e cos
d
2 e sin
x x
x
yx x
x
y x
= − +
= −
( )
( )
( )
22 2
2
2
2
d d2 e cos 2e sin
d d
d2 2e sin
d
d=2 2e sin 4 4
d
d4 5 shown
d
x x
x
x
y yx x
x x
yy x
x
yy x y y
x
yy
x
= − +
= − −
− + − + −
= −
Most students are able to get the answer to
d
d
y
x. A handful of students who take ln to
both sides of the equation and perform
implicit differentiation ended up getting
stuck when finding 2
2
d
d
y
x. A significant
number of students are not able to
manipulate and show the result given in
the question. Nevertheless, they move on
to attempt the rest of the question with
only a small number giving up totally.
Able to find the first few terms of
a Maclaurin’s series by repeated
implicit differentiation.
3 2
3 2
d d d4 5
d d d
y y y
x x x= −
2
2
3
3
When 0,
1
d2
d
d3
d
d2
d
x
y
y
x
y
x
y
x
=
=
=
=
=
By Maclaurin’s Expansion,
2 3
2 3
3 21 2
2! 3!
3 11 2
2 3
y x x x
x x x
= + + + +
= + + + +
�
�
A small number of students could not
obtain 3
3
d
d
y
x even when the expression for
2
2
d
d
y
x is given. Common mistakes include:
(a)
23
3
d d d4 5
d d d
y y y
x x x
= −
(b) 3 2
3 2
d d d d4 5
d d d d
y y y y
x x x x
= −
.
When finding the value of y when 0x = ,
some students begin with 2ey = instead
of 1y = .
When apply the Maclaurin Expansion,
some students have problems with the
factorials.
Page 11 of 32
Able to find the first few terms in
a binomial expansion.
Able to show a function is strictly
increasing.
(i) ( ) ( )3
22
2 2
3 1
3 2 21 1
2 2!
3 31
2 8
kx kx kx
kx k x
+ = + + +
= + + +
�
�
Since the third terms are both equal,
( )
2
2
3 3
2 8
4
2
2 since 0
k
k
k
k k
=
=
= ±
∴ = >
( ) ( )3
2f 1 2x x= +
( ) ( )( )1
23
f 2 1 22
3 1 2
10 for all
2
x x
x
x
′ = +
= +
> > −
Since ( ) 1f 0 for all
2x x′ > > − ,
f is strictly increasing for all 1
2x > − .
Most students are able to apply the
Binomial Expansion, with a handful of
them writing
( )3
22
3 1
3 2 21 1
2 2!kx kx kx
+ = + + + +�,
thus leading to a wrong conclusion.
A significant number of students made the
mistake of comparing the fourth term (i.e.
the term in 3x ) instead of the third term.
Majority of the students are unable to
show that f is strictly increasing. Common
mistakes include:
(a) Assuming that ( )f 0x′ > and show
that 1
2x > − , instead of assuming
1
2x > −
and show that ( )f 0x′ > ,
(b) Substituting a value of x (i.e. 1x = )
and show that f(1) > 0 and therefore
strictly increasing,
(c) Sketch out a portion of the graph of f
and conclude strictly increasing,
(d) Showing that ( ) ( )3
2f 1 2 0x x= + >
when 1
2x > − instead of showing
Page 12 of 32
( ) 1f 0 for all
2x x′ > > − ,
(e) Finding range of validity to be
1 1
2 2x− < < and conclude that it is strictly
increasing since 1
2x > − lies in the range
of validity.
Page 13 of 32
7 Vectors (Lines)
No
.
Assessment Objectives Solution Feedback
Able to find equation of a straight
line given a point and a direction
vector.
(i) Equation of line l:
1 1
r 1 1 ,
3 1
λ λ = − +
∈�
�
This part of the problem is generally
well-attempted.
Errors committed in the specification of
a vector equation for the required line l
were commonly notational :
• Erroneously equating the line l with
the vector expression for the
position vector of a general point on
that line, as in
1 1
1 1 ,
3 1
l λ λ − +
=
∈�
• Erroneous omission of “ λ ∈� ”.
Able to find the foot of the
perpendicular from a point to a
line.
(ii) Let the foot of the perpendicular from B to the line l be F.
Since F lies on the line l,
1 1
1 1 for a
3 1
valueOF λ λ = − +
����
1
1
3
λ
λ
λ
+ − +
=+
BF is perpendicular to line l
The large majority of candidates
(70~80% of all candidates??) managed
to correctly write down a vector
expression for
1 1
1 1
3 1
OF λ = − +
����
using the fact that F lies on line l,
where it is then crucial to set up some
condition that would enable one to
determine the particular value of λ that
corresponds to point F on the line.
Common errors in erroneously
formulating scalar-product equations
Page 14 of 32
5 3 0
14
1
1 0
1
1 7 1
1 4 1 0
3 6 1
6 1
5
3
1 0
3 1
6
0
14
3
BF
λ
λ
λ
λ
λ
λ
λ λ λ
λ
λ
⋅ =
+ − + − ⋅ = +
− + − + ⋅ =
− + − + =
− +
− +
=
−
=
+
����
1 14 / 3 171
1 14 / 3 113
3 14 / 3 23
OF
+ = − + = +
����
The coordinates of F is 17 11 23
, ,3 3 3
.
Alternative Method :
Note that AF����
is the projection of vector AB����
on line l.
6
5
3
AB OB OA
= − =
���� ���� ����.
such as :
• 0OB OF⋅ =���� ����
• 0AB OF⋅ =���� ����
• 0OF m⋅ =����
� (where
1
1
1
m
=
�,
a direction vector for line l)
points to difficulty with which a
number of affected candidates
experienced in identifying a relevant
pair of perpendicular vectors, i.e. BF����
and m�
.
Such difficulty is also evident amongst
candidates whose scripts reflect serious
conceptual inaccuracy with erroneous
descriptions such as “the line is
perpendicular to point B” prior to the
formulation of one of the erroneous
scalar-product equations.
While some others were able to
correctly point out that vector BF����
is ⊥
to the line l, this relationship was mis-
represented algebraically, as in :
•
actually
6 1 1
5 1 1 0
3 3 1
OF
λ
λ λ
λ
− + − + ⋅ − + =
− + ����
��
Page 15 of 32
As such
1
ˆ ˆ( ) , where 1
1
6 1 11 1
5 1 13 3
3 1 1
114
13
1
AF AB m m m
= ⋅ =
= ⋅
=
���� ����
� � �
Then,
1 1 1714 1
1 1 11 .3 3
3 1 23
OF OA AF= +
= − + =
���� ���� ����
The coordinates of F is 17 11 23
, ,3 3 3
.
indicating difficulty that such
candidates faced in eliciting a relevant
property (direction vector) of the line
that would befit the characteristic of
being ⊥ to BF����
.
Given that the relevant scalar-product
equation involving λ is correctly
formulated however, occasional
numerical calculation errors would be
the main cause for not being able to
accurately determine the position
vector OF����
and thereby the coordinates
of F.
While some candidates used an
alternative method in the approach to
finding the foot of perpendicular (from
B to l), a chief source of error is in
using
OF OB BF= +���� ���� ����
, but erroneously taking
BF����
or FB����
to be ( )AB m����
�.
Able to find shortest distance from
a point to a line using its foot. (iii) “Hence” method :
BF is the shortest distance from B to the line l BF=����
Given that the coordinates of F were
accurately found, most candidates were
able to correctly approach this part.
Common error(s) :
• Mistaken shortest distance from B
to l as OF����
.
Page 16 of 32
6 14 / 3
5 14 / 3
3 14 / 3
41 1
1 423 3
5
or 2.16units
− + = − + − +
− = − =
“Otherwise” method :
Shortest dist. from B to line l
= Perp. dist. from B to l
1
ˆ , where 1 is a direction vector of line
1
6 11
5 1 3
3 1
21 1 14
3 1433 3
1
AB m m l
= × =
= ×
= − = =
����
� �
• Some correctly pointed out that the
shortest required is BF����
, but
simply proceeded with calculating
its length without determining the
vector BF����
first, and erroneously
ended up computing the length of
OF����
instead.
Even among candidates who failed to
correctly determine the coordinates of
the foot-of- ⊥ F from B to line l in the
earlier part, the majority knew the
appropriate approach (i.e. in
determining BF����
first, then its length.),
but ended up arriving at a wrong
answer due to the adoption of
incorrectly calculated coordinates of F
for this part.
Able to find the image of a line
reflected in a line.
[H.O.T.]
(iv) Let 'B be the image.
F is the midpoint of B and 'B
This part of the entire question proved
to be a problem for a significant
proportion of candidates (30-40% ??).
Common error(s) include :
In visualizing the problem, as evident
Page 17 of 32
( )1'
2
' 2
17 72
11 43
23 6
131
103
28
OF OB OB
OB OF OB
= +
= −
= −
=
���� ���� �����
����� ���� ����
The image 'B is 13 10 28
, ,3 3 3
.
Equation of line AB reflected in line l
1 1
1 1 ,
3
',
131
103
28
101
133
19
10
1
3
1
1 ,
3
11
1 ' , ' , where '3
3
3
19
s s
s
r OA sAB s
r
r s
s s sr s
∈
= − + − ∈
= − + ∈
= +
−
= − + ∈ =
���� ����
�
�
�
�
�
�
��
Cartesian equation: 1 1 3
10 13 19
x y z− + −= =
in some of the diagrams drawn (though
no marks were awarded/deducted for
diagrams).
• Not recalling that pt A is actually on
line l as mentioned in the earlier
part of the qn (and thereby taking
an extra, unrequired step of finding
its image in line l).
In the application of the Mid-point
Theorem to find image B′ of B in line l.
• Erroneously using 2
OB OBOA
′+=
���� ��������
where the mid-point of pts. B and B′
should be point F instead.
• Erroneously using
1
1
12
AB AB
′=
+���� ����
where it should have been
114
, which is 13
1
AF
����.
• Erroneously using 2
A ABFO
B ′+=
���� ��������
where the three points should have
been the same.
In formulating a vector equation
representing line l.
Page 18 of 32
• Erroneously choosing OB′����
, AB����
or
6
5
3
,
1
1
1
etc. as a direction vector
of the reflected line.
Not converting into Cartisian eqn. the
vector eqn found for the reflected line.
Page 19 of 32
8 Vectors (Lines & Planes)
N
o.
Assessment Objectives Solution Feedback
Able to find angle between a line
and a plane. (i) Let α be the angle between 1p normal and 1l direction
vector.
5 11 1
1 3
6 2
20cos
62
0.8246088483 o
cos62 14
14
r 47.25
α
α
α
= ⋅
=
= °
The angle between the 1p and 1l 0.746rad or 42.82
πα= − = °
Many students were not able to
obtain full marks for this part.
They found the angle between the
normal and the directional vector
of line and stopped there.
Able to find the line of intersection
between two planes. (ii)
The line of intersection, 2l , between 1p and 2p is
1 2
r 12 4 ,
0 1
µ µ
− − = +
∈
�
� .
Majority of the students were able
to get this part correct.
Common mistakes 1) Pressed the GC wrongly.
Quite a number of student
pressed 2,1,0,0 for the last
row.
2) A significant number of
students reuse the parameter
λ . Marks were awarded for
this part but no credit would
be given for (iii).
3) A significant number of
students did not present the
equation of the line correctly.
Example of mistakes are
missing the r , did not define
the domain of their parameter,
i.e µ ∈� .
Page 20 of 32
Able to find intersections between
two lines without the use of G.C. (iii) Assume that
1l and 2l intersect.
4 1 1 2
5 3 12 4
7 2 0 1
λ µ
− − + = +
1 2 2 54 λ µ λ µ= − − ⇒ + = −+ (1)
12 4 3 45 73λ µ λ µ= + ⇒ − =+ (2)
27 2 7λ µ λ µ+ = ⇒ − = − (3)
(1) × 3 – (2): 11
10 225
µ µ= − ⇒ = −
Substitute 11
5µ = − into (1):
11 32 5
5 5λ λ + − = − ⇒ = −
Substitute 3
5λ = − and
11
5µ = − into (3):
R.H.S3 11
L.H.S. 2 15 5
.
= − − − =
≠
Hence 1l and 2l do not intersect and they do not have a
common point.
This part of badly done by
students. Common mistakes:
1) Poor algebraic
manipulation
2) Did not show working
clearly.
3) Used the GC and did not
show any working despite
the question stating that all
working must be clearly
shown.
4) Wrongly claiming that
since the lines are not
parallel hence they do not
intersect.
Able to find equation of plane.
[H.O.T.]
(iv) Since 1p , 2p and 3p do not have any common points of
intersection.
2l is parallel and not intersecting 3p .
Equation of 3p :
4 1 2
r 5 3 4 ,
7 2 1
,α β α β
− = + +
∈�
3
1 2 5 1
3 4 5 5 1
2 1 10 2
n
− − − = = − = −
×
�
This part of the question is badly
done. Only a minority of the
students got this correct. Many
students left this part empty.
Page 21 of 32
Equation of 3p :
1 4 1
1 5 1 5
2 7 2
r
− − ⋅ − = ⋅ − = �
Caretsian equation of 3p : 2 5x y z− − + =
Page 22 of 32
9 Functions
No
.
Assessment Objectives Solution Feedback
Able to graph a rational function
involving quadratic polynomials.
Able to find range of a function by
graph.
Able to test for the existence of a
composite function.
(i)
The shape of the curve with symmetry in y-axis and
approaching a horizontal asymptote at the two ends
has to be shown properly.
The labelling of intercepts and the asymptote is not
required.
fR [ 1,1)= −
gD ( 2, )= − ∞
f gSince R D , gf exists.⊆
Most students were able to give a close
sketch of the graph of the function
though some graphs are of the wrong
curvature because the students did not
adjust the Window of GC to gain a better
look.
A number of students didn’t indicate the
horizontal asymptote, which is rather
crucial if one wants to identify from the
graph the range of the function. This
made many fail to find the upper bound.
Most students knew how to carry out the
test for the existence of the composite
function. However, many were not
familiar with the interval notations, the
use of which is indeed a basic math skill
at this level.
Able to solve inequalities by
algebraic method or graphical
method.
Method 1: Algebraic
(ii)
2
2
1gf ( ) ln 2 1
1
xx
x
−= + > +
2 2
2
1 2( 1)e
1
x x
x
− + +>
+
2 23 e e 1x x> + − 2
(3 e) e 1x− > −
2 e 16.0993
3 ex
−> ≈
−
2.47 or 2.47x x> < − .
Most could at least give a correct
expression of the composite function,
which means they could write down the
correct inequality explicitly.
But this part is rather poorly done despite
the fact that either an algebraic or a
graphical approach would lead to a
solution.
Many students had exhibited very poor
algebraic skills in the attempt to use an
algebraic approach. Some didn’t know
Page 23 of 32
that they could get rid of the positive
denominator on the left. Some wrongly
changed the direction of the inequality
when they exponentiated both sides.
Some reached 2 e 1
3 ex
−>
− but didn’t know
how to solve this simple inequality.
Method 2: Graphical
Alternative: 2
2
1gf ( ) ln 2 1
1
xx
x
−= + > +
Sketch 2
2
1ln 2 1
1
xy
x
−= + − +
By G.C./graph, the solution is
2.47 or 2.47x x> < − .
As a rule of thumb in A Levels, if the
question didn’t forbid the use of GC, a
graphical approach is usually a better
choice for solving inequality, simply
because the complexity of solving with
GC doesn’t quite depend on the algebraic
complexity of the inequality.
Yet, among those students who opted for
the graphical method, a number of them
were not adept at adjusting the window
settings appropriately so as to gain a
better view of the graph. As a result,
these students became victims of the
initial small pixelated graph and wrongly
concluded that there was no solution.
Method 3: Graphical Alternative: 2
2
1gf ( ) ln 2 1
1
xx
x
−= + > +
Sketch 2
1 2
1ln 2
1
xy
x
−= + +
, 2 1y =
Page 24 of 32
By G.C./graph, the solution is
2.47 or 2.47x x> < − .
Able to restrict the domain of a
function.
Able to test for the existence of the
inverse of a function.
Able to find the inverse of a
function, its domain and range.
(iii) The horizontal line 0y = will cut the graph twice so
the function is not one-one. Thus the inverse
function doesn’t exist.
The greatest value of k is 0.
Let 2
2
1
1
xy
x
−=
+,
2 21yx y x+ = −
2(1 ) 1y x y− = +
2 1
1
yx
y
+=
−
1
1
yx
y
+= ±
−
Since f ,0D ( ]= −∞ ,
1
1
yx
y
+= −
−
1 1f ( )
1
xx
x
− +∴ = −
−
1 ffD R [ 1,1)− = = −
This part tests a few very standard skills
required of the topic on Functions. All
students are expected to answer these
routine questions perfectly.
Most candidates provided satisfactory
answers to the first two questions. One
common misconception observed is the
following. The fact that the graph of the
function fails the horizontal line test does
not lead to the non-existence of its
inverse directly. The test only indicates
whether the function itself is one-one or
not.
Many students still don’t know that
solving 2 1
1
yx
y
+=
− gives two opposite
roots. This is unacceptable.
Page 25 of 32
1 ffR D ( ,0]− = = −∞
Able to determine the relationship
of two curves on Cartesian
coordinates.
Able to formulate and solve an
equation
[H.O.T.]
(iv)
2
2
11
1
xmx
x
−= −
+
2 3 21 1x mx mx x− = + − − 3 22 0mx x mx− + =
2( 2 ) 0x mx x m− + =
0x = or 2 2 0mx x m− + =
Since there are three points of intersection, 2 2 0mx x m− + = has two distinct roots, i.e.,
22 4 4
2
mx
m
± −= .
On one hand, 24 4 0m− >
2 1m <
1 1m− < < .
On the other hand,
Since 22 4 4
02
mx
m
± −= ≤ ,
22 4 4 0m± − > ,
so 0m < .
Hence, 1 0m− < < .
This part is intended to be challenging. It
is an outgrowth of a few problems the
students encountered in the topic
Graphing Techniques I, where they were
required to relate the solutions to an
algebraic equation to the points of
intersection of two curves. Here the
students need to proceed in an opposite
direction, which means that given the
number of points of intersection of two
curves, the students should solve the
equation to obtain the conditions on the
unknown m.
While only a few students managed to
make significant progresses in their
solutions, many did have the correct idea
to get started by equating the two curves
to examine the equation. However, once
again, poor algebraic skills in handling
the equation or later the inequality
created by the discriminant hampered the
progress of many partial solutions.
Page 26 of 32
10 Graphing Techniques II
No
.
Assessment Objectives Solution Feedback
Able to combine transformations
involving translation along x−axis
and scaling parallel to the x–axis.
(a)(i) ( )f 1y x= − −
Correct transformations:
( )
( )
( )
f
translate 1 unit along the positive -axis
f 1
reflect in the -axis
f 1
y x
x
y x
y
y x
=
↓
= −
↓
= − −
Common mistake:
( )
( )
( )
f
reflect in the -axis
f
translate 1 unit along the positive -axis
f 1
y x
y
y x
x
y x
=
↓
= −
↓
= − −
It should be translate 1 unit along the
negative x-axis, i.e. ( )( )f 1x− +
Able to sketch ( )fy x= − when
given ( )fy x= .
(ii) ( )fy x= −
A number of students drew the graph of
( )2 fy x= .
Others drew the correct graph but added in
the asymptote of 2y = , which was
unnecessary.
Common mistakes:
• Taking square root of the x-coordinates
instead of the y-coordinates
• Coordinates are marked out wrongly,
often neglecting the positive or negative
signs.
x
y
O
x
y
O
Page 27 of 32
Able to sketch ( )f 'y x= when
given ( )fy x= .
(iii) ( )f 'y x=
Generally well attempted but a number of
students drew the graph ( )1
fy
x= instead. A
handful also drew the graph of ( )1fy x−= .
Students should be aware that this is a
derivative graph.
Able to reverse a series of
transformation to obtain the
original equation.
(b) 2 2
2
2
2
2
2
4
III' : scaling parallel to -axis by a scale factor of 2
42
II' : translation of 3 units in the negative -direction
34
2
I' : translation of 2 units in the positive -direction
3
2
x y
x
xy
x
xy
y
x
+ =
↓
+ =
↓
+ + =
↓
+
( )22 4y+ − =
Students must read question carefully. 2 2 4x y+ = is the resulting curve and
students were supposed to find the original
curve. Many students assume that 2 2 4x y+ = is the original curve.
The right way to solve would be to undo step
III, followed by step II then step, i.e. in a
systematic manner.
x
y
Page 28 of 32
11 Mathematical Induction/Sigma Notation
No
.
Assessment Objectives Solution Feedback
Able to find the exact terms of a
sequence defined by a recurrence
relation.
(i) 1
2 1
3 2
4 3
5 4
1
6
1 1
(3)(4) 4
1 3
(4)(5) 10
1 1
(5)(6) 3
1 5
(6)(7) 14
u
u u
u u
u u
u u
=
= + =
= + =
= + =
= + =
There were many students who did not
understand the idea of a recurrence
relation. They substituted n 2= to
obtain 1u instead of substituting n 1= .
There were some students who did not
write down the values exactly.
Able to observe the number
pattern generated by the terms in
the sequence and make a
mathematical conjecture.
[H.O.T.]
(ii) 1
2
3
4
5
12
3
1 22
2 4
32
5
2 42
3 6
52
7
22
, 1,2( 2)
n
n
u
u
u
u
u
nu
n
nu n n
n
+
=
= =
=
= =
=
=+
= ≥ ∈+
�
Many students were credited for
considering the values of n
2u in order to
make the conjecture for nu .Students
need to realize that the conjecture for n
u
has to be expressed in terms of n only,
not in terms of n 1u + or n 1u − .
Page 29 of 32
Able to prove conjecture by
mathematical induction.
[H.O.T.]
(iii)
Let Pn be the statement
2( 2)n
nu
n=
+for all
, 1n n+∈ ≥�
When n =1, L.H.S. of 1
P =1
1
6u =
R.H.S. of 1
P =1 1
2(1 2) 6=
+
Since L.H.S.=R.H.S., 1
P is true.
Assume P k is true for some k
+∈� , i.e.
2( 2)k
ku
k=
+
We are required to prove that 1Pk + is also true, i.e.
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