set 1 paper 1 sol guide.pdf

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


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


(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


(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


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


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)

set 1 paper 1 sol guide.pdf

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