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Set 1 Paper 1
1 © Pearson Education Asia Limited 2012
Set 1 Paper 1
Section A(1)
1. 43)3(2
43
32−−−−
−
−
= baba
ba 1M
75 −= ba 1M
7
5
b
a= 1A
(3)
2.
xbbxb
bxxb
−+=
−+=
23
)2(3
1M
bbxbx 23 −=− 1M
3
)3(
−=
−=−
b
bx
bbx
1A
(3)
3. (a) 22 )5(2510 −=+− aaa 1A (b) 32 250202 bbabba −+− ])5[(2
)2510(2
22
22
bab
baab
−−=
−+−= 1M
)5)(5(2 −+−−= babab 1A (3)
4. The original cost of the computer game
%2040$ ÷= 1M
200$= 1A
∴ The amount that Mary paid
%)201(%)201(200$ −×+×= 1M
192$= 1A
(4)
5. Let x and y be the numbers of $5-coins and $2-coins that Stanley has respectively.
=+
=+
14425
42
yx
yx 1A + 1A
Therefore, we have
144)42(25 =−+ xx 1M
20
144843
=
=+
x
x ∴ The number of $5-coins is 20. 1A
(4)
6. (a) 731 ≤− x or 73
1−−<
−x
x
2
2
63
−≥
−≥
≤−
x
x
x
or
or
or
5
204
2131
−<
−<
−−<−
x
x
xx
1A + 1A
(b) If 4−<x ,
we have 4−<x and ( 2−≥x or x < −5) ∴ The solutions are x < −5. 1A
(4)
7. θ−°=∠ 180BAD (opp. ∠s, cyclic quad.) 1M
°=∠ 90ABD (∠ in semi-circle) 1M
∴
°−=
°−−°−°=∠
∠°=∠+∠+∠
90
90)180(180
) of sum (180
θ
θADB
ABDBADADB △
∵ ))ABBC = (given)
∴
°−=
∠=∠
90θ
ADBBDC
(arcs prop. to ∠s at ⊙ce) 1M
∴
θ
θθ
2270
)90(180
) of sum (180
−°=
−°−−°=∠
∠°=∠+∠+∠
CBD
BCDBDCCBD △
1A
(4)
8. (a) Mean cm 161= , mode cm 154= 1A + 1A
(b) ∵ There are two modes after removing two data.
∴ One of the removed data must be 154 cm. 1A
Let x cm be another removed datum.
2.116120
15422161+=
−−× x 1M
144=x
∴ Another removed datum is 144 cm. 1A
(5)
9. (a) By substituting y = 0 into 01634 =++ yx , we have
4
0164
−=
=+
x
x
∴ The coordinates of A )0 ,4(−= 1A
Slope of L3
4−=
Let (0, b) be the coordinates of B.
3
13
4
04
0
=
−=
−
−−
−
b
b
∴ The coordinates of B )3 ,0(= 1A
(b) ∵ The perpendicular distance from P to L is a constant.
∴ The locus of P is parallel to L.
Let (x, y) be the coordinates of P. Since the locus of P passes through B,
3
4
0
3−=
−
−
x
y 1M
0934 =−+ yx
By substituting (−3, 6) into the equation of the locus of P,
1M
Solution Guide and Marking Scheme
2 © Pearson Education Asia Limited 2012
R.H.S.
3
9)6(3)3(4L.H.S.
≠
−=
−+−=
∴ The locus of P does not pass through (−3, 6). 1A
(5)
Section A(2)
10. (a)
64
1605.2
360901105.1
=
=
=+++
x
x
xx
1A
(1)
(b) (i) Number of students in the school
°
°×−°÷=
360
645.111042 1M
1080= 1A
(ii) Number of students living in Kowloon
288
360
645.11080
=
°
°××=
1A
Let m be the number of students living in Kowloon that leave the school.
2
1
360
645.1
1080
288×
°
°×=
−
−
m
m 1M
13
2160
)1080(2)288(15
=
−=−
m
mm
∵ m is not an integer.
∴ The angle of the sector representing the group of
students living in Kowloon cannot be halved. 1A
(5)
11. (a) Let 221 nknkP += , where k1, k2≠0. 1A
∴
+=
+=
(2)22515954
(1)10010380
21
21
KK
KK
kk
kk 1M
By solving, we have 481 =k and 12 −=k . 1A
∴ The required profit
) thousands560$(or 000 560$
1000])20()20(48$[ 2
=
×−=
1A
(4)
(b) If the company makes a profit of $600 000, we have
......(*)060048
60048
2
2
=+−
=−
nn
nn
0
96
)600)(1(4)48((*) of 2
<
−=
−−=∆
1M
∴ The company could not make a profit of $600 000. 1A
(2)
12. (a)
)12()32)(34()( 2−−+−++= xxxxxf 1M
10852
129123682
23
223
−−+=
−−−−−++=
xxx
xxxxxx
1A
(2)
(b) (i)
0
110)1(8)1(5)1(2
)1()1()1(
23
=
−−−−−+−=
−+−=− gfh
1A
∴ By the factor theorem, x + 1 is a factor of h(x).
0
510)3(8)3(5)3(2
)3()3()3(
23
=
−−−−−+−=
−+−=− gfh
1A
∴ By the factor theorem, x + 3 is a factor of h(x).
(ii)
)(12)32)(3)(1(
)(12)32)(34(
)()()(
2
xgxxxx
xgxxxx
xgxfxh
+−−−++=
+−−−++=
+=
∵ x + 1 and x + 3 are factors of h(x), and g(x) is a
linear polynomial.
∴ g(x) – 2x – 1 = 0 1M
Alternative solution
Let g(x) = ax + b, where a and b are constants.
−=−
−=−
5)3(
1)1(
g
g
−=+−
−=+−
53
1
ba
ba
)2......(
)1......(
(1) – (2):
2
42
=
=
a
a
∴ b = 1
∴ g(x) = 2x + 1 1M
∴ h(x) = (x + 1)(x + 3)(2x – 3) 1A
For h(x) = 0, 2
3or 3or 1 −−=x 1A
(5)
13. (a)
Speed of Kanice km/h 5.2= 1A
∴ Their distance apart
fig.) sig. 3 to(cor. km 729.0
km 60
35
2
15.2
60
355.2
=
××−×=
1A
(2)
(b) Let v km/h be Simon’s speed from 10:35 am to 11:00 am.
25.4
60
25
60
35
2
15.25.2
=
×+××=
v
v 1M
∴ Simon’s speed from 10:35 am to 11:00 am is
4.25 km/h. 1A (2)
1M
Set 1 Paper 1
3 © Pearson Education Asia Limited 2012
(c) New speed of Simon
km/h 675.4km/h %)101(25.4 =+×= 1A
∴ Time spent for Simon to walk from A to B
minutes 57.7272
hour 96212.0
hour 675.460
35
2
15.25.2
60
35
≈
≈
÷
××−+=
1M
∵ Simon reaches B earlier than Kanice by less than 3
minutes. ∴ Simon’s claim is not correct. 1A
(3)
14. (a) (i) )6 ,2(=B and )2 ,6( −−=C 1A + 1A
(ii) Slope of OB 302
06=
−
−=
Slope of BC 1)6(2
)2(6=
−−
−−=
∵ Slope of OB≠slope of BC
∴ O, B and C are not collinear. 1A
(4)
(b) (i) Let (d, 0) be the coordinates of D.
∵ DB ⊥ BC
∴ 8
1102
6
=
−=×−
−
d
d
∴ The coordinates of D are (8, 0).
∴ Area of △CBD
14.
sq.units 2
6)82()]2(6[)]6(2[
2
2222+−×−−+−−
=
×=
BDBC1M
sq.units 48= 1A
(ii) ∵ Area of △CED2
1= area of △CBD
and the heights of △CED and △CBD are the
same.
∴ E is the mid-point of BD. 1M
∴ Coordinates of E
)3 ,5(
2
06 ,
2
82
=
++=
1A
(5)
Section B
15. (a)
2loglog
2log
2727
27
+=
=
E
EN
2loglog3
2
2log
9log
log
2log27log
log
279
27
2
3
27
+=
+=
+=
E
E
E
1M
2log3
227+= M 1M
∴ M and N have a linear relationship. 1A
(3)
(b)
fig.) sig. 3 to(cor. 01.4
2log)7.5(3
227
=
+=N
∴ The magnitude of the typhoon on Scale B is 4.01. 1A
(1)
16. (a)
k
kkkxkx
k
kxxkxg
4
481214
4
121)124()(
22
2
−−+−=
−+−−=
1A
(1)
(b) ∵ The graph of y = g(x) is always above the x-axis.
∴ ∆ of g(x) = 0 is smaller than 0, i.e.
04
48121)(4)4(
22
<
−−−−
k
kkkk 1M
0)14)(116(
011264 2
<+−
<−+
kk
kk
16
1
4
1<<− k 1A
∵ k is a positive number.
∴ 16
10 << k 1A
(3)
17. (a) The required probability
125
60
65
61
64
C
CCCC ×+×= 1M
33
4= 1A
Alternative solution The required probability
8
2
9
3
10
4
11
5
12
65
8
6
9
3
10
4
11
5
12
6××××+×××××= 1M
33
4= 1A
(2)
1M
1M
Solution Guide and Marking Scheme
4 © Pearson Education Asia Limited 2012
(b) The required probability
125
21225
C
C−
−=
125
103
C
C= 1M
33
5= 1A
(2)
(c)
The required probability
125
353
61 2
C
CC ××= 1M
33
20= 1A
(2)
18. (a) (i) °=°−°−°=∠ 854550180PTQ 1A
Consider △PTQ, by the cosine formula,
PTQTQTPTQTPPQ ∠−+= cos))((2222 1M
°−+= 85cos)24)(7(2247 22
fig.) sig. 3 to(cor. 4.24
40728724.24
=
≈PQ
∴ The distance between P and Q is 24.4 m. 1A
(ii) Consider △PTQ, by the sine formula,
PQT
PT
PTQ
PQ
∠=
∠ sinsin 1M
fig.) sig. 3 to(cor. 6.16
sin
7
85sin
40728724.24
°=∠
∠=
°
PQT
PQT
1A
∴ The bearing of P from Q
)323(or W8.42N
W)6.1645(N
°°=
°−°≈
1A
(6)
(b) Let x m be the perpendicular distance from T to PQ.
Consider the area of △PTQ, we have
x××≈°××× 40728724.242
185sin247
2
1 1M
6
856997569.6
>
≈x
∴ Philip’s claim is not correct. 1A
(2)
19. (a) Join OA.
θ=∠OAC (∠in alt. segment) 1M
Consider △ABC,
°=∠
°=°+∠
°=−°+++∠
90
18090
180)290()(
OAB
OAB
OAB θθθ
(∠sum of △)
1M
∴ OB is a diameter of the circle OAB. (converse of ∠ in semi-circle) 1A
(3)
(b) (i) 3230cos
3=
°=OB 1A
∴ The radius and the centre of the circle are 3
and 0) ,3( respectively. 1M
∴ The equation of the circle is 3)3( 22=+− yx . 1A
(or 03222=−+ xyx )
(ii) The coordinates of A
)30sin3 ,30cos332( °°−= 1M
=
2
3 ,
2
3 1A
∵ °=°×−°=∠ 3030290OCA
∴ The equation of the tangent AC
°=
−
−
30tan
2
3
2
3
x
y
1M
13
1
2
3
3
1
2
3
+=
−=−
xy
xy
1A
(or 033 =+− yx )
(7) (c) Let A’ be the point on the circle such that CA’ is another tangent of the circle from C.
∵ OCAAOC ∠=′∠ (tangent’s properties) 1M
∴ The graph of CA’ can be obtained by reflecting the
graph of y = f (x) along the x-axis, i.e. the equation
of CA’ is y = −f (x). 1M
∴ Alvin’s claim is correct. 1A
(3)