Page 1 of 11
S.5 Revision on Selected Topics
Name:________________ Class :_____________ (
)
A. Transformations
Transformations of the graph of y  f (x)
Translation
1
2
Reflection
along x-axis
along y-axis
Enlargement / Reduction
Enlarge to 2 times along y-axis
Enlarge to 2 times along x-axis
Reduce to ½ of the original along y-axis
Reduce to ½ of the original along x-axis
1. [09-10 F.4 Standardized test 2]
2
If the graph of y  ( x  2)  1 is translated 1
unit to the right and then reflected about yaxis, which of the following is the function
after the transformation?
2
A. y  ( x  1)  1
2
B. y  ( x  3)  1
2
C. y  ( x  1)  1
2
D. y  ( x  3)  1
2. [09-10 F.4 Final]
When the graph of y  9 x  5 is translated 4 units
to the left and then reflected about the y-axis, its
equation becomes
A. y  9 x  31 .
B. y  9 x  31.
C. y  9 x  41.
D. y  9 x  41 .
3. [09-10 F.4 Final]
The figure shows two parabolas having the same
axis of symmetry. Which of the following is / are
true?
y
A.
B.
C.
D.
4.
b
2a
I.
h
II.
III.
a2
ck 0
I only
II only
I and II only
I, II and III
[09-10 F.4 Final]
The figure shows the graphs of y  f (x) and
y  p . The solution of the equation f ( x  1)  p
is
y
y = f(x)
A.
B.
C.
D.
p 1.
p 1 .
q  1.
q 1.
y =p
q
O
5. [08-09 F.4 Final]
If the graph of y  3(1  x)  5 is reflected in the
y-axis and then translated 2 units upward, the
equation of the new graph is
2
y = 2(x – h)2 + k
y = ax2 + bx + c
O
x
x
A.
B.
C.
D.
y  3(3  x) 2  5 .
y  3(1  x) 2  3 .
y  3(3  x) 2  5 .
y  3(1  x) 2  7 .
Page 2 of 11
6. [08-09 F.4 Mid-year]
If the graph of y = g (x) is obtained by translating
the graph of y = f (x) 1 unit to the left, then g (x) =
A. f (x) – 1.
B. f (x – 1).
C. f (x + 1).
D. f (x) + 1.
C.
7. [07-08 F.4 Mid-year]
Given a point (2, –4) on the graph y = f (x). Find
the new position of the given point when
f(x) is translated 4 units to the right first and then
reflected along the y-axis.
A. (–6, –4)
B. (–2, 4)
C. (2, –4)
D. (6, 4)
D.
8. [07-08 F.4 Final]
The point A(2, 6) lies on the curve y = g(x). Find
the new position of point A after the
transformation y = 2g(x) + 1.
A. (4, 6)
B. (4, 6)
C. (2, 11)
D. (2, 11)
9. [08-09 F.4 Standardized test 1]
The figure shows the quadratic graph of y=f (x).
10. [06-07 F.4 Mid-year]
If the graph of y  f (x) is reflected in the y axis,
and then translated 4 units upward, it becomes the
graph of y  g (x) . g (x) 
A. f (x)  4 .
B. f (x)  4 .
C.  f ( x)  4 .
D.  f ( x)  4 .
11. [07-08 F.4 Final]
In the figure, the graph of y = f(x) with vertex at
(0, 0) is translated to the position with vertex at
(3, 4). The new equation of the curve y = g(x) is
yy
y = g(x)
y = f(x)
Which of the following may represent the graph of
y = f (x)?
4
A.
x
0
B.
A.
B.
C.
D.
y
y
y
y
f ( x  3)  4 .
f ( x  3)  4 .
f ( x  3)  4 .
f ( x  3)  4 .
3
x
Page 3 of 11
12. [06-07 F.4 Final]
16. [04-05 F.4 Mid-year]
The graph of the function y  3x 2  2 is first
translated 2 units to the right, and then reflected in
the y-axis. The function represented by the final
graph is
A. y  3x 2  12x  10
B.
y  3x 2  12x  10
C.
y  3x 2  12x  10
D.
y  3x 2  12x  10
The graph of y  x  2 x  1 is translated 2 units
to the left. What is the equation of the graph after
translation?
2
A. y  x  2 x  1
2
B. y  x  4 x  9
2
C. y  x  6 x  7
2
D. y  x  2 x  5
2
17. [04-05 F.4 Final]
The figure shows the graph of y = f (x). Which of
the following may be the graph of y = f (–x)?
13. [05-06 F.4 Mid-year]
The graph of f ( x)  3x  2 is translated 5 units
downwards and 3 units to the left to become
g (x) . Find g (x) .
A.
B.
C.
D.
y
y = f(x)
3 x 3  3
3 x 3  3
3 x 3  3
3 x 5  5
x
O
A.
y
14. [05-06 F.4 Final]
The figure shows two quadratic curves. Curve P is
represented by y  a( x  h) 2  k . Which of the
following is the possible presentation of curve Q?
y = f(-x)
x
y
A.
1
y  a ( x  h) 2
2
B.
y  2a( x  h) 2
C.
y  2a( x  h) 2
y  2a( x  h) 2  k
D.
O
Q
B.
y
P
y = f(-x)
x
0
x
O
15. [05-06 F.4 Final]
The figure shows the graphs of y = f (x) and y = h
(x). The function h (x) is a transformation of the
function f (x) = x 2 . Which of the following is a
possible equation of h (x)?
y
C.
y
y = f(-x)
O
y = f (x)
y = h(x)
D.
0
(– 5, –2)
A.
h (x) =  3 ( x  5)  2
B.
h (x) = 3 ( x  5)  2
C.
h (x) = 2 ( x  5)  2
D.
h (x) = 2 ( x  5)  2
y
x
y = f(-x)
2
2
2
2
x
O
x
Page 4 of 11
18. [04-05 F.4 Final]
If the graph y = f (x) undergoes transformations to
the graph y = 2f (x) – 1, find the coordinates of the
point (2, –2) after such transformations.
A. (2, –1)
B. (2, –3)
C. (2, –5)
D. (4, 6)
21. The figure shows the graph of y  3 sin 2 x .
The point P is
Trigonometric functions
19. The figure shows the graph of
A.
B.
C.
D.
E.
240 ,3
135 ,3
240 ,1
135 ,1
270 ,1
o
o
o
o
o
22. The figure shows the graph of the function
A.
B.
C.
D.
E.
y  3 cos x , 0  x  360
y  3 sin x , 0  x  360
y  2  sin x , 0  x  360
y  2  cos x , 0  x  360
y  3  sin x , 0  x  360
360o
180o
20. Let k be a constant and  90    90 . If the
figure shows the graph of y  k sin x o   ,
then
o
o


A. y  cos x
B. y  cos x 


y  cos90  x 
y  cos180  x 
C. y  cos 90 o  x
D.
E.
o
o
23. The figure shows the graph of the function
A.
B.
C.
D.
k = -2 and  = -30o
k = -2 and  = 30o
k = 2 and  = -30o
k = 2 and  = 30o
A. y  cos
B. y 
xo
2
1
cos x o
2
C. y  cos x o
D. y  2 cos x o
E. y  cos 2 x o
Page 5 of 11
24. The figure shows the graph of y 
1
cos 2 x
2
26. Which of the following may represent the
graph of y  cos x o for 0  x  90 ?
The point P is
A.
90 , 2
o
1

B. 180 o , 
2



C. 180 o , 1
1

D.  360 o , 
2

E.
360 , 1
o
25. In the figure, f x  
27. Which of the following may represent the
graph of y  tan x o for 0  x  90 ?
x 1

2 2
1
sin 2 x 
2
1
x 1
sin 
2
2 2
1
1
sin x 
2
2
1
1
sin 2 x 
2
2
A. sin
B.
C.
D.
E.
Page 6 of 11
28. The figure shows the graph of the function
B.
C.
D.
E.
180o
x
2
y  2 sin x
y  1  sin x
y  1  cos x
y  1  cos x
A. y  sin
360o
29.
30.
A(6, 6) is a point on a rectangular coordinate plane. B is the reflection of A with respect to the line x  2 and
C is the reflection of A with respect to the line y = 2.
(a) (i) Write down the coordinates of B and C.
(ii) O is the origin of the coordinate plane. Are O, B and C collinear? Explain your answer.
(4 marks)
(b) It is known that D is a point lying on the x-axis such that DB BC.
(i) Find the area of CBD.
(ii) E is a point lying on BD such that the area of CED is 24 square units. Find the coordinates of E.
(5 marks)
31. The graph of y  x 2  4 x  12 is reduced along the y-axis to k times the original, where k is positive number. It
1  12k
is then translated upwards by
units and the graph of y  g (x) is obtained.
4k
(a) Find g (x) .
(1 mark)
(b) Find the range of the values of k, such that the graph of y  g (x) is always above the x-axis. (3 marks)
32. [HKDSE SP – 8]
In Figure 3, the coordinates of the point A are (-2, 5) . A is rotated clockwise about the origin O
y
through 90° to A’. A” is the reflection image of A with respect to the y-axis.
(a) Write down the coordinates of A’ and A”.
A(2, 5)
(b) Is OA” perpendicular to AA’? Explain your answer.
(5 marks)
O
Figure 3
x
Page 7 of 11
33. [HKDSE 2014 – 8]
B. Polar Coordinates
1. [HKDSE 2013 – 6]
2. [HKDSE PP – 6]
In a polar coordinate system, the polar coordinates of the points A, B and C are (13, 157°) , (14, 247°) and
(15, 337°) respectively.
(a) Let O be the pole. Are A, O and C collinear? Explain your answer.
(b) Find the area of DABC .
(4 marks)
3. In a polar coordinate system, O is the pole. The polar coordinates of the point A and B are (6, 30º) and
( 6 3 , 300º) respectively.
(a) Find the area of AOB .
(b) If C is a point on AB such that OC  AB , find the polar coordinates of C.
4. In a polar coordinate system, O is the pole. The polar coordinates of the point A and B are (2, 90º) and (4,
150º) respectively.
(a) Find the area of AOB .
(b) Find the polar coordinates of the circumcentre of AOB .
C. Ratio of areas
1. [HKDSE 2013 – 18]
In the figure, ABCD is a trapezium with AD // BC
and AD : BC = 2 : 3. Let E be the mid-point of BC.
AC and DE intersect at F. If the area of ∆CEF is
36 cm2, then the area of the trapezium ABCD is
A
A.
B.
C.
D.
216 cm2.
264 cm2.
280 cm2.
320 cm2.
D
2. [HKDSE 2012 – 17]
In the figure, ABCD is a parallelogram. E and F
are points lying on AB and CD respectively. AD
produced and EF produced meet at G. It is given
that DF : FC = 3 : 4 and AD : DG = 1 : 1. If the
area of ∆DFG is 3 cm2, then the area of the
G
parallelogram ABCD is
F
B
E
C
A.
B.
C.
D.
12 cm2.
14 cm2.
18 cm2.
21 cm2.
D
A
F
E
C
B
3.
[HKDSE PP – 17]
In the figure, ABCD is a rectangle. E is the
mid-point of BC. F is a point lying on CD such
that DF 2CF. If the area of CEF is 1 cm2, then
the area of AEF is
F
D
A.
B.
C.
D.
C
2
2 cm .
3 cm2.
4 cm2.
6 cm2.
Page 8 of 11
7. [HKCEE 2008 – 21]
In the figure, ABCD is a parallelogram. M is a
point lying on BC such that BM : MC = 1 : 2. If
BD and AM intersect at G and the area of ∆BGM is
1 cm2, then the area of the parallelogram ABCD is
A.
B.
C.
D.
E
A
B
9 cm2.
11 cm2.
12 cm2.
24 cm2.
B
M
G
A
4. [HKDSE SP – 23]
In the figure, ABCD is a parallelogram. F is a point
lying on AD. BF produced and CD produced meet
at E. If CD : DE = 2 : 1, then AF : BC =
D
8. [HKCEE 2008 – 51]
In the figure, AB is the tangent to the circle at B
and ADC is a straight line. AB : AD = 2 : 1, then
area ∆ABD : area of ∆BCD =
E
A.
B.
C.
D.
1 : 2.
2 : 3.
3 : 4.
8 : 9.
B
F
A
A.
B.
C.
D.
D
C
B
5. [HKCEE 2011 – 19]
In the figure, ABCD is a parallelogram. E is the
mid-point of AB. F and G are points lying on CD
such that DF = FG = GC. BG and CE intersect at
H. If the area of ∆BCH is 6 cm2, then the area of
the quadrilateral EFGH is
F
D
A.
B.
C.
D.
G
C
H
A
B
E
6. [HKCEE 2010 – 26]
In the figure, ABCD is a rectangle. If M is a point
lying on AC such that DM is perpendicular to AC,
then AM : MC =
D
A.
B.
C.
D.
3 : 4.
4 : 3.
9 :16.
16 : 9.
1 : 2.
1 : 3.
1 : 4.
2 : 3.
A
C
D
9. [HKCEE 2007 – 19]
In the figure, ABCD is a parallelogram. E is a point
lying on AB. If BD and EC intersect at F, then the
ratio of the area of ∆DEF to the area of ∆CBF is
A
A.
B.
C.
D.
E
1 : 1.
1 : 2.
2 : 1.
2 : 3.
C
6 cm
M
A
8 cm
B
B
F
C
D
10 cm2.
12 cm2.
15 cm2.
16 cm2.
C
10. [HKCEE 2004 – 17]
In the figure, ABCD is a parallelogram. E is a point
lying on AD such that AE : ED = 1 : 3. If the area
of ∆ABE is 3 cm2, then the area of the shaded
region is
A
A.
B.
C.
D.
9 cm2.
15 cm2.
21 cm2.
24 cm2.
E
B
D
C
11. [HKCEE 2004 – 18]
In the figure, AD and BC meet at E. If CE : EB =
3 : 1, then area ∆ABD : area of ∆CDE =
B
A
A.
B.
C.
D.
1 : 1.
1 : 3.
2 : 3.
4 : 9.
E
C
D
12. [HKCEE 2003 – 18]
In the figure, AEDC is a parallelogram. If AB : BC
= 1 : 2 and AF : FE = 2 : 1, then area ∆ABF : area
of ∆BCD =
B
A
C
A. 1 : 2.
B. 1 : 3.
F
C. 1 : 4.
E
D. 2 : 9.
D
Page 9 of 11
14. [HKCEE 2001 – 50]
In the figure, ADB, BEC and CFA are straight lines.
If the area of ∆ABC is 225 cm2, then the area of the
parallelogram DECF is
A
D
13. [HKCEE 2002 – 44]
In the figure, ABCD is a parallelogram. E and F
are points on AD and BC respectively such that
AB // EF. EF meets AC at G. If and AG : GC =
1 : 2, then area of ABFG : area of EGCD =
A
A.
B.
C.
D.
1 : 2.
1 : 4.
3 : 4.
5 : 8.
E
B
A.
B.
C.
D.
E.
D
F
E
20 cm
30 cm
C
81 cm2.
108 cm2.
126 cm2.
136 cm2.
162 cm2.
G
B
F
C
D. Centres of Triangles
1. [HKDSE 2013 – 43]
Let O be the origin. If the coordinates of the points
A and B are (0, 12) and (30, 12) respectively, then
the y-coordinate of the cicumcentre of ∆OAB is
A. 6.
B. 8.
C. 12.
D. 15.
2. [HKDSE PP – 42]
Let O be the origin. If the coordinates of the points
A and B are (18, –24) and (18, 24) respectively,
then the x-coordinate of the orthocentre of OAB
is
A. –14.
B. 10.
C. 12.
D. 25.
3. [HKCEE 2010 – 51]
Let O be the origin. If A and B are points lying on
the x-axis and the y-axis respectively such that the
equation of the circumcircle of OAB is
x 2  y 2  16x  12 y  0 , then the equation of the
straight line passing through A and B is
A. 3x  4 y  48  0 .
B. 3x  4 y  48  0 .
C. 4 x  3 y  48  0 .
D. 4 x  3 y  48  0 .
4.
[HKCEE 2009 – 51]
In the figure, G is the centroid of ∆ABC. AG, BG
and CG are produced to meet BC, AC and AB at L,
M and N respectively. If BL = 13 cm, BN = 5 cm
and CM = 12 cm, find the area of ∆ABC.
A
N
B
A.
B.
C.
D.
M
G
L
C
60 cm2.
120 cm2.
180 cm2.
240 cm2.
5. [HKCEE 2009 – 52]
The coordinates of two vertices of a triangle are
(–4, –8) and (6, 2). If the coordinates of the
circumcentre of the triangle are (k, –4), then k =
A. –1.
B. 0.
C. 1.
D. 2.
6. [HKCEE 2008 – 52]
Let O be the origin. If the coordinates of the points
A and B are (48, 0) and (24, 18) respectively, then
the y-coordinate of the orthocentre of OAB is
A. –7.
B. 6.
C. 8.
D. 32.
7. [HKCEE 2007 – 50]
If ABC is an obtuse-angled triangle, which of the
following points must lie outsideABC?
I. The centroid of ABC
II. The circumcentre of ABC
III. The orthocentre of ABC
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
9.
Page 10 of 11
8. [HKCEE 2006 – 49]
In the figure, ABC is an acute-angled triangle,
AB = AC and D is a point lying on BC such that
AD is perpendicular to BC. Which of the following
must be true?
A
B
A.
B.
C.
D.
D
C
I. The circumcentre of ABC lies on AD.
II. The orthocentre of ABC lies on AD.
III. The centroid of ABC lies on AD.
I and II only
I and III only
II and III only
I, II and III
Figure 1
Figure 1
(b) It is given that OGH  OCD . Penny claims that G is also the orthocentre of OAC . Is
she correct? Explain your answer.
(2 marks)
Page 11 of 11
E. Travel-graphs
1.
[HKDSE SP – 12]
Figure 5 shows the graph for John driving from town A to town D (via town B and town C) in
a morning. The journey is divided into three parts: Part I (from A to B), Part II (from B to C)
and Part III (from C to D).
(a) For which part of the journey is the average speed the lowest? Explain your answer.
(2 marks)
(b) If the average speed for Part II of the journey is 56 km/h, when is John at C?
(2 marks)
(c) Find the average speed for John driving from A to D in m/s.
(3 marks)
2.
[HKDSE 2014 – 10]