Ch.A12 Detailed Solutions
Chapter A12 More about Graphs of Functions
Exercise A12A
1.
Draw the line y = –5 on the given graph.
(p.218)
Refer to the figure.
1M
From the figure, the required solution is
y
0 ≤ x ≤ 120 or 240 ≤ x ≤ 360.
1A
8
y = –x3 + 4x + 2
6
3.
Refer to the figure.
y
4
y=2
2
0
–1
–2
8
y=6
4
x
1
2
–2
y = –2
–6
–3
–4
–5
–2
–1
–4
y = –8
–8
y = a(x – h)2
–x3 + 4x + 2 > –2
–12
Draw the line y = –2 on the given graph.
1M
From the figure, the required solution is
–2.5 ≤ x < 2.4.
(a)
From the figure, the coordinates of the vertex are
(–3, 8).
1A
x – 4x ≥ 0
∴
h = –3
–x + 4x ≤ 0
∵
The graph passes through (0, –10).
–x + 4x + 2 ≤ 2
∴
–10 = a(0 + 3)2 + 8
3
(b)
3
3
Draw the line y = 2 on the given graph.
1M
–2.0 ≤ x ≤ 0.0 or 2.0 ≤ x ≤ 2.5.
1A (for both correct)
1M
1A
2
y = –2(x + 3) + 8
= –2(x2 + 6x + 9) + 8
Refer to the figure.
= –2x2 – 12x – 10
y
2
x2 + 6x = –1
–2x2 – 12x = 2
x
60
120
180 240 300 360
–2x2 – 12x – 10 = –8
y = –1
–2
Draw the line y = –8 on the given graph.
–4
y = –5
–6
–8
k=8
a = –2
1A
(b)
0
and
–10 = 9a + 8
From the figure, the required solution is
2.
x
–4
–x3 + 4x > –4
(a)
0
1M
From the figure, the required solution is
x = –0.2 or x = –5.8.
y = 4 cos x° – 3
(c)
1A
Let z = x – 1. Then,
z2 + 6z > –8
(a)
2
–2z – 12z < 16
2cos x° < 1
2
–2z – 12z – 10 < 6
4cos x° < 2
Draw the line y = 6 on the given graph.
4cos x° – 3 < –1
Draw the line y = –1 on the given graph.
From the figure, the required solution is
1M
From the figure, the required solution is
60 < x < 300.
(b)
1M
z < –4.0
∴
1A
or z > –2.0.
1A
x – 1 < –4.0
or
x – 1 > –2.0
x < –3.0
or
x > –1.0
1A
–2cos x° ≤ 1
4cos x° ≥ –2
4cos x° – 3 ≥ –5
© Educational Publishing House Ltd
1
Conquering HKDSE Exam Mathematics (Compulsory Part) Book A
4.
(a)
∵
Volume of the original cylinder
Exercise A12B
= 4 × volume of the smaller cylinder
1.
π (x2)(x) = 4π (32)(2x – 6)
∴
(p.226)
The two 5-sided polygons show a reflection transformation.
1M
axis of symmetry
x3 = 36(2x – 6)
x3 – 72x + 216 = 0
(b)
1
x3 – 72x + 216 = 0
x3 – 72x + 266 = 50
Draw the line y = 50 on the given graph.
y
y = x3 – 72x + 266
∴
The answer is B.
300
2.
200
100
The plane figure is rotated anticlockwise about O through 90°.
y = 50
1M
x
0
4
2
6
8
90°
From the figure, the required solution is
x = 3.8 or x = 6.0.
O
1A
∴
5.
(a)
2x + 4
×8
Radius of the liquid surface =
16
= (x + 2) cm
1M
3.
The answer is A.
The plane figure is rotated clockwise about O through 135°.
Consider the volume of the liquid.
4 3 4
4
1
π x + π (2)3 + π (2)3 = π (x + 2)2(2x + 4) 1M
3
3
3
3
135°
2x3 + 32 = (x2 + 4x + 4)(x + 2)
O
2x3 + 32 = x3 + 6x2 + 12x + 8
x3 – 6x2 – 12x + 24 = 0
(b)
1
∴
x3 – 6x2 – 12x + 24 = 0
x3 – 6x2 – 12x + 64 = 40
4.
Draw the line y = 40 on the given graph.
1M
y
60
The answer is C.
For option I:
The plane figure does not have rotational symmetry.
For option II:
y = x3 – 6x2 – 12x + 64
The plane figure has rotational symmetry.
y = 40
40
centre of rotation
20
0
x
1
2
3
4
5
6
–20
For option III:
The plane figure has rotational symmetry.
From the figure, the required solution is
x = 1.3.
1A
centre of rotation
∴
2
The answer is C.
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Ch.A12 Detailed Solutions
5.
For option III:
The number of axes of reflectional symmetry of the
octagon is 2.
The parallelogram has reflectional symmetry and
rotational symmetry.
4
6
∴
6.
∴
The answer is A.
The number of folds of rotational symmetry of the
rhombus is 2.
11.
The answer is D.
For option I:
The triangle has reflectional symmetry but does not have
rotational symmetry.
centre of rotation
6
6
∴
The answer is A.
For option II:
7.
The triangle has reflectional symmetry and rotational
symmetry.
The coordinates of the image are (–5, –4).
∴
The answer is D.
6
8.
Coordinates of the image = (–1 – 2[–1 – (–4)], –3)
6
= (–7, –3)
∴
6
The answer is B.
For option III:
9.
Distance between P and Q = 2[1 – (–4)]
The triangle does not have reflectional symmetry and
rotational symmetry.
= 10
∴
10.
The answer is D.
For option IV:
The triangle has reflectional symmetry but does not have
rotational symmetry.
For option I:
The parallelogram has reflectional symmetry and
rotational symmetry.
9
6
axes of symmetry
9
∴
The answer is B.
centre of rotation
4
12.
4
For option II:
The parallelogram has reflectional symmetry and
rotational symmetry.
For option I:
Size of each exterior angle of the regular 10-sided polygon
=
360°
= 36°
10
∴
(sum of ext. ∠s of polygon)
I is true.
For option II:
4
Size of each interior angle of the regular 10-sided polygon
=
4
© Educational Publishing House Ltd
∴
(10 − 2) × 180°
= 144°
10
(∠ sum of polygon)
II is true.
3
Conquering HKDSE Exam Mathematics (Compulsory Part) Book A
For option III:
r2 = ( 3 ) 2 + (−1) 2
The number of axes of reflectional symmetry is 10.
13.
∴
III is not true.
∴
The answer is A.
(Pyth. theorem)
r= 4 = 2
tan θ =
−1
3
θ = 330°
For option I:
(n − 2) ×180° 5 × 360°
=
n
n
∴
The polar coordinates of the image of P are (2, 330°).
∴
The answer is B.
n – 2 = 10
n = 12
∴
17.
I is true.
For option II:
Let (r, θ) be the polar coordinates of the image of A.
Number of diagonals of the polygon =
12 × 11
− 12
2
Note that 180° < θ < 270°.
r2 = (−3) 2 + (−3 3 ) 2
= 54
∴
Coordinates of the image of A = (−3,−3 3 )
r = 36 = 6
II is not true.
For option III:
tan θ =
The number of folds of rotational symmetry of the
polygon is 12.
14.
(Pyth. theorem)
∴
III is not true.
∴
The answer is A.
−3 3
= 3
−3
θ = 240°
∴
The polar coordinates of the image of A are (6, 240°).
∴
The answer is D.
Coordinates of B = (5, –2 + 4)
18.
= (5, 2)
Let (r, θ) be the polar coordinates of the image of Q.
Coordinates of the image of B = (5, 2 – 2[2 – (–1)])
Note that 270° < θ < 360°.
= (5, –4)
∴
Coordinates of the image of Q = (4 3 ,−4)
r2 = (4 3 ) 2 + (−4) 2
The answer is C.
(Pyth. theorem)
r = 64 = 8
15.
Coordinates of the reflection image of P
= (9, –6)
1
3
θ = 330°
Coordinates of Q = (9, –6 – 5)
= (9, –11)
16.
=−
4 3
= (–3 + 2[3 – (–3)], –6)
∴
−4
tan θ =
∴
The polar coordinates of the image of Q are (8, 330°).
∴
The answer is A.
(a)
Coordinates of Q = (8, 2)
1M
 −2 + 8 8 + 2 
Coordinates of M = 
,
 = (3, 5)
2 
 2
1A
The answer is B.
Coordinates of the image of P = (−(− 3 ),−1)
19.
= ( 3 ,−1)
Let (r, θ) be the polar coordinates of the image of P.
Note that 270° < θ < 360°.
(b)
Slope of OM =
5−0 5
=
3−0 3
Slope of PQ =
8−2
3
=−
− 2−8
5
y
θ
x
0
r
image of P
4
∵
Slope of OM × slope of PQ =
∴
OM is perpendicular to PQ.
1M
5  3
×  −  = −1
3  5
1A
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1M
Ch.A12 Detailed Solutions
20.
(a)
(b)
21.
(a)
The coordinates of B are (–6, 8).
1A
Exercise A12C
The coordinates of C are (–6, –8).
1A
1.
(p.239)
For option I:
Note that △ABC is a right-angled triangle with ∠ACB = 90°. f(x − 5) = (x − 5)[(x − 5) – 5]
= (x − 5)(x − 10)
1
Area of △ABC = (BC)(AC)
2
≠ g(x)
1
∴ I is not true.
1M
= [8 – (–8)][6 – (–6)]
2
For option II:
1A
= 96
f(x + 5) = (x + 5)[(x + 5) – 5]
The coordinates of P' are (2, 3).
1A
The coordinates of Q' are (–3, –2).
1A
= x(x + 5)
= g(x)
(b)
∴
Let M be the mid-point of PQ.
For option III:
 −2 + 3 3 + (−2)   1 1 
,
Coordinates of M = 
 = , 
2  2 2
 2
1
2 =5
Slope of P'M =
1 3
2−
2
f(−x) = −x(−x − 5)
= x(x + 5)
= g(x)
3−
Slope of P'Q' =
II is true.
1M
3 − (−2)
=1
2 − (−3)
2.
∴
III is true.
∴
The answer is D.
For option I:
f(2x) = log 2x
∵
Slope of P'M ≠ slope of P'Q'
∴
P'Q' does not pass through M.
∴
∴
P'Q' is not the perpendicular bisector of PQ. 1A
For option II:
1M
= g(x)
I is true.
2f(x) = 2log x
Concept Check
≠ g(x)
(p.238)
∴
Mary’s solution
For option III:
y = f(x)
f(x) + log 2 = log x + log 2
translating rightwards by 5 units
= log 2x
= g(x)
y = f(x − 5)
reflecting with respect to the y-axis
y = f(−x − 5) ≠ f(5 − x)
∴
II is not true.
∴
III is true.
∴
The answer is C.
∵
g(x) = 2f(x)
∴
The graph of y = g(x) is obtained by vertically
Mary’s solution is not correct.
Peter’s solution
3.
y = f(x)
stretching the graph of y = f(x) by a factor 2.
reflecting with respect to the y-axis
y = f(−x)
∴
The answer is C.
∵
g(x) = f(3x)
∴
The graph of y = g(x) is obtained by horizontally
translating leftwards by 5 units
y = f [−(x + 5)] = f(−x − 5) ≠ f(5 − x)
∴
Peter’s solution is not correct.
4.
compressing the graph of y = f(x) by a factor 3.
∴
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The answer is C.
5
Conquering HKDSE Exam Mathematics (Compulsory Part) Book A
5.
f(x) =
∴
The graph of y = f(x) is obtained by vertically
compressing the graph of y = g(x) by a factor 3.
∴
∴
1
g(x)
3
∵
= 720°
∴
The coordinates of the vertex of the graph of y = f(x) are
(0, 0).
∴
Period of the function =
By considering the vertices of the two graphs, we can
easily see that:
∴
The answer is B.
∴
∴
9.
12.
g(x) = f(x + 4) + 3
The answer is A.
The coordinates of the vertex of the graph of y = f(x) are
(2, −4).
The coordinates of the vertex of the graph of y = g(x) are
(0, 0).
8.
13.
6
1
× 360° = 180°
2
The given graph can be obtained by following the steps
below:
(1)
vertically stretch the graph of y = sin x° by a factor 2
(2)
translate the resulting graph of y = 2sin x° leftwards
by (90 − 60) = 30 units
∴
The figure shows the graph of y = 2sin (x° + 30°).
∴
The answer is D.
The given graph can be obtained by following the steps below:
By considering the vertices of the two graphs, we can
easily see that:
(1)
vertically stretch the graph of y = cos x° by a factor 3
(2)
The graph of y = g(x) can be obtained by translating the
graph of y = f(x) leftwards by 2 units, and then upwards by
4 units.
translate the resulting graph of y = 3cos x° rightwards
by (230 − 180) = 50 units
∴
The figure shows the graph of y = 3cos (x° − 50°).
∴
The answer is C.
∴
g(x) = f(x + 2) + 4
∴
The answer is B.
14.
The given graph can be obtained by following the steps below:
(1)
reflect the graph of y = cos x° with respect to the x-axis
The given graph can be obtained by following the steps
below:
(2)
(1)
translate the graph of y = f(x) leftwards by 3 units
(3)
(2)
translate the resulting graph of y = f(x + 3)
downwards by 1 unit
vertically stretch the resulting graph of y = −cos x° by
a factor 4
translate the resulting graph of y = −4cos x° leftwards
by (180 − 120) = 60 units
∴
The figure shows the graph of y = −4cos (x° + 60°).
∴
The resulting graph of y = f(x + 3) – 1 can be
obtained.
∴
The answer is B.
∴
The answer is C.
15.
The given graph can be obtained by following the steps below:
(1)
reflect the graph of y = sin x° with respect to the x-axis
The given graph can be obtained by following the steps
below:
(2)
(1)
reflect the graph of y = f(x) with respect to the y-axis
(3)
(2)
translate the resulting graph of y = f(–x) upwards by
1 unit
(4)
∴
The resulting graph of y = 1 + f(–x) can be
obtained.
∴
horizontally compress the resulting graph of y = −sin x°
by a factor 360 ÷ 120 = 3
vertically stretch the resulting graph of y = −sin 3x° by
a factor (4 − 1) = 3
translate the resulting graph of y = −3sin 3x° upwards
by 4 units
The figure shows the graph of y = 4 − 3sin 3x°.
∴
The answer is A.
∴
The answer is D.
16.
10.
To obtain the graph of the function y = 3cos (2x − 60°),
one of the steps is to horizontally compress the graph of
y = cos x by a factor 2.
The coordinates of the vertex of the graph of y = g(x) are
(−4, 3).
The graph of y = g(x) can be obtained by translating the
graph of y = f(x) leftwards by 4 units, and then upwards by
3 units.
7.
The answer is D.
The answer is A.
11.
6.
Period of the function = 2 × 360°
x
+ 1, one of
2
the steps is to horizontally stretch the graph of y = sin x by
a factor 2.
To obtain the graph of the function y = sin
The given graph can be obtained by following the steps below:
(1) horizontally stretch the graph of y = sin x° by a factor
720 ÷ 360 = 2
x°
(2) vertically stretch the resulting graph of y = sin
by a
2
factor [−1 − (−5)] = 4
© Educational Publishing House Ltd
Ch.A12 Detailed Solutions
(3)
translate the resulting graph of y = 4sin
x°
2
Put (0, −4) into y = h + k tan 2x°.
−4 = h + k tan 2(0)°
downwards by 5 units
∴
The figure shows the graph of y = −5 + 4sin
∴
The answer is D.
x°
.
2
h = −4
<0
∴
17.
The given graph can be obtained by following the steps
below:
(1) vertically stretch the graph of y = cos x° by a factor 3
(2) horizontally stretch the resulting graph of y = 3cos x°
by a factor 720 ÷ 360 = 2
x°
upwards
(3) translate the resulting graph of y = 3cos
2
by (2 + 8) ÷ 2 = 5 units
1M
x°
∴ The resulting graph of y = 5 + 3cos
can be
2
obtained.
∴
a=5
1
k=
2
and
Suppose that tan α ° =
(a)
But from the given graph, we can see that y < −2 when
x=
α
2
III is not true.
∴
The answer is A.
1A + 1A
∵
f(2) = 4
∴
a(2)2 + 12(2) – 8 = 4
a = –3
Suppose the graph of y = tan 3x° is transformed to the
given graph.
∵
From the given graph, we can see that y decreases
as x increases.
The graph of y = tan 3x° is reflected with respect to
the x-axis to the given graph, i.e. k < 0.
∴
∴
1A
II is true.
Put (0, 2) into y = h + k tan 3x°.
g(x) = –f(x + 5)
= –[–3(x + 5)2 + 12(x + 5) – 8]
2 = h + k tan 3(0)°
1M
h=2
= 3x2 + 30x + 75 – 12x – 60 + 8
= 3x2 + 18x + 23
>0
1A
∴
19.
(a)
(i)
f (x) = x2 – 8x
= (x2 – 8x + 42) – 42
1M
= (x – 4) – 16
(ii)
The coordinates of the vertex are (4, –16).
But from the given graph, we can see that y < 2 when
The coordinates of the vertex of the graph of
x=
1A
h(x) = 2f(x – 5)
= 2[(x – 5)2 – 8(x – 5)]
1
β
. When x = ,
k
3
 β °
y = 2 + k tan 3   = 2 + 1 = 3
3
1A
y = g(x) are (4 + 3, –16 ÷ 4), i.e. (7, –4).
(b)
I is true.
Suppose that tan β ° =
2
∴
.
∴
4a + 16 = 4
(b)
2
α
. When x = ,
k
2
 α °
y = −4 + k tan 2   = −4 + 2 = −2
2
21.
18.
I is true.
1M
β
3
.
∴
III is not true.
∴
The answer is B.
(a)
f (x) = –x2 – 6x – 7
= 2x2 – 20x + 50 – 16x + 80
= 2x2 – 36x + 130
1A
22.
= –(x2 + 6x + 32) + 32 – 7
20.
= –(x + 3)2 + 2
Suppose the graph of y = tan 2x° is transformed to the
given graph.
∵
∴
∴
From the given graph, we can see that y increases as
x increases.
The graph of y = tan 2x° must not be reflected with
respect to the x-axis to the given graph, i.e. k > 0.
II is not true.
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1M
∴ Coordinates of V = (–3, 2)
(b)
1A
g (x) = f (x – 5) + 3
= –[(x – 5) + 3]2 + 2 + 3
1M
= –(x – 2)2 + 5
1A
7
Conquering HKDSE Exam Mathematics (Compulsory Part) Book A
(c)
(i)
The axis of symmetry of the graph of y = f (x)
25.
(a)
h(x) = g(2x) + k
is x = –3.
g(–3) = –(–3 – 2)2 + 5 = –20
Coordinates of M = (–3, –20)
∵
M is the image of the point N on the
graph of y = f (x).
∴
Coordinates of N = (–3 – 5, –20 – 3)
= (–8, –23)
(ii)
(b)
∴
1A
= 3(2x)2 + 6(2x) – 2 + k
1M
= 12x2 + 12x – 2 + k
1A
∵
∴
The graph of y = h(x) does not have x-intercepts.
∆<0
i.e.
122 – 4(12)(–2 + k) < 0
5–k<0
k>5
1A
1A
Obviously, if N(–8, –23) is reflected with
respect to the axis of symmetry x = –3 to
obtain the image P, then P lies on the graph of
y = f (x) and VP = VN.
Thus, it is possible that VP = VN, where P lies
on the graph of y = f (x).
26.
(a)
g(x) = –f(x – k)
= –2(x – k + 3)2 + 4
∵
∴
1M
g(0) = –14
2
–2(0 – k + 3) + 4 = –14
Let (p, –23) be the coordinates of P.
p + (−8)
= –3
2
1M
1M
2
(k – 3) = 9
k=6
1M
(b)
or
k = 0 (rejected) 1A
h(x) = g(x + 5) – 8
= –2[(x + 5) – 6 + 3]2 + 4 – 8
p =2
1M
2
∴
Coordinates of P = (2, −23)
= –2(x + 2) – 4
1A
∵
The graph of y = h(x) has the maximum point at (–2, –4),
and the graph of y = f(x) has the minimum point at (–3, –4).
23.
(a)
f (x) = 2x2 – 4x – 5
∴
= 2(x2 – 2x + 1) – 2 – 5
1M
= 2(x – 1)2 – 7
∴ Coordinates of the vertex = (1, –7)
(b)
1A
27.
(a)
g(x) = f(x – 4) + k
(i)
 y = x 2 − 6 x + m ...... (1)

 y = 4 x − 5 ...... (2)
Put (1) into (2).
2
(c)
1A
The graph of y = h(x) does not intersect the graph of y = f(x).
= 2[(x – 4) – 1] – 7 + k
1M
= 2(x – 5)2 – 7 + k
1A
∵
The graph of y = g(x) touches the x-axis.
∴
–7 + k = 0
x2 – 6x + m = 4x – 5
x2 – 10x + m + 5 = 0 ...... (3)
∵
(ii)
(a)
2
(–10) – 4(1)(m + 5) = 0
1A
The y-intercept of the graph of y = g(x) is
50.
1A
m = 20
1A
Coordinates of A = (5, 15)
1A
Put m = 20 into (3).
x2 – 10x + 20 + 5 = 0
f(x) = –3x2 – 12x + 4
= –3(x2 + 4x + 4) + 12 + 4
1M
20 – m = 0
g(0) = 2(0 – 5)2 = 50
∴
24.
∆ =0
∴
i.e.
k= 7
The two graphs touch each other.
(x – 5)2 = 0
1M
x=5
= –3(x + 2)2 + 16
Put x = 5 into (2).
∴ Coordinates of the vertex = (–2, 16)
1A
g(x) = –f(x) – k
1M
y = 4(5) – 5 = 15
(b)
= 3(x + 2)2 – 16 – k
(c)
8
∴
1A
g(x) = 0
3(x + 2)2 – 16 – k = 0
3(x + 2)2 = 16 + k
∵ k>0
1M
∴ The above equation has two distinct real roots.
∴ The graph of y = g(x) must cut the x-axis.
∴ The claim is agreed.
1A
(b)
f(x) = x2 – 6x + 20
= (x2 – 6x + 9) – 9 + 20
1M
2
= (x – 3) + 11
∴
The coordinates of the vertex of the graph of
y = f(x) are (3, 11).
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Ch.A12 Detailed Solutions
30.
From (a), the coordinates of the vertex of the graph
of y = g(x) are (5, 15).
∴
28.
(a)
f(x) = 0
(a)
2
x – 4x – 3 = 0
The graph of y = g(x) is obtained by translating
the graph of y = f(x) rightwards by 2 units, and
then upwards by 4 units.
1A
(x2 – 4x + 4) – 4 – 3 = 0
(x – 2)2 – 7 = 0
1A
f(x) = k1(x + 1)2 + k2, where k1 and k2 are constants.
∵
f(1) = 19
∴
k1(1 + 1)2 + k2 = 19
x=2± 7
∴
∴
PQ = ( 2 + 7 ) − ( 2 − 7 )
=2 7
(b)
4k1 + k2 = 19 ...... (1)
∵
(i)
1A
g(x) = f(x + n) + m
g(x) = 0
f(–2) = 10
2
(x + n – 2) – 7 + m = 0
k1(–2 + 1)2 + k2 = 10
k1 + k2 = 10 ...... (2)
(x + n – 2)2 = 7 – m
1M
(b)
and
f(x) = 3(x + 1)2 + 7
(i)
g(x) = f(–x) – 4
(ii)
∴ P'Q' = ( − n + 2 + 7 − m ) − ( − n + 2 − 7 − m )
k2 = 7
∴
=2 7−m
1A
∵ P'Q' =
= 3(–x + 1)2 + 7 – 4
1M
= 3(x – 1)2 + 3
1A
∴ 2 7−m =
∴
(ii)
1M
The smallest value of k is 6.
From (b)(i), g(x) = (x + n – 2)2 – 1
∵
The axis of symmetry of the curve C2 is x = –3.
∴
2 – n = –3
1A
31.
(a)
(i)
(x – h)2 = −
∵
By comparing the coefficients of x, we have
b+6 =1
(ii)
f(x) = 0
(x + 1)(x – 2)(x – 3) = 0
1M
or
x=2
or
x=3
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(b)
1A
k
2
The roots are 2 – i and 2 + i.
∴
h=2
The equation 2(x – h)2 + k = 0 has two
nonreal roots.
∴
The graph of y = 2(x – h)2 + k does not cut
the x-axis.
1M
∴
The claim is disagreed.
and
k=2
1A (for both correct)
1A
Suppose that the graph of y = 2(x – 2)2 + 2 is translated
leftwards by a units, and then upwards by b units. The
x-intercepts of its image are 3 and 6. Then,
2(3 + a – 2)2 + 2 + b = 0 ...... (1)
The x-intercepts of the graph of y = g(x) are
–1 + 3, 2 + 3 and 3 + 3, i.e. 2, 5 and 6.
1M
∵
1A
1M
k
2
x =h± −
The coefficient of x in the expansion of
(x + 1)(x2 + bx + 6) is b + 6.
(x + 1)(x2 – 5x + 6) = 0
1A
2(x – h)2 + k = 0
1A + 1A
b = −5
1M
n =5
By comparing the coefficients of x3 and the constant
terms, we have
a = 1 and c = 6
(c)
1A
m=6
The equation has two distinct integral
roots.
x = –1
1M
= [x – (2 – n)]2 – 1
k −3
(x – 1) =
3
∵
(2 7 )
7–m=1
It is possible to obtain the graph of
y = g(x) by only translating the graph of
y = f(x).
1A
2
(b)
1
7
3(x – 1)2 + 3 = k
(a)
PQ
g(x) = k
(iii)
29.
1
1M
7
The graph of y = g(x) can be obtained by
translating the graph of y = f(x) rightwards by
2 units, and then downwards by 4 units. 1M
∴
1M
x=−n+2± 7−m
Solving the equations (1) and (2), we have
k1 = 3
1M
1A
2(6 + a – 2)2 + 2 + b = 0 ...... (2)
9
Conquering HKDSE Exam Mathematics (Compulsory Part) Book A
(2) – (1):
2(4 + a)2 – 2(1 + a)2 = 0
1M
(5 + 2a)(3) = 0
a =−
We can also find b by putting a = −
∴
10
5
2
5
into (1) or (2).
2
The graph of y = 2(x – 2)2 + 2 could translate
to a position such that the x-intercepts of its
image are 3 and 6.
1A
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