Marking Scheme of Term Exam of F5 Mathematics Compulsory Part 2014 – 15 Section A1 Solution Marks Remarks 1.

15

4

96

610

96

235

yx

yxyx

yxyx

1M

1M

1A

For ppp baab )(

For qp

q

p

aaa

2.

nnppq

nqqnppnqnpqp

nqp

qp

3524

35243254

3254

1M

1M

1A

For moving terms

3. (a)

)76)(76(4936 22

nmnmnm

(b)

)476)(76()67(4)76)(76(

2428)76)(76(24284936 22

nmnmmnnmnmmnnmnm

mnnm

1A

1M

1A

For using the result of (a)

   

4. Let x be the number of $100-notes and y be the number of $500-notes

255

120002400130001005005000500100

xy

yyxyx

∴The actual number of $100-notes is 25.

1A

1A

1M

1A

pp – 1 for any undefined symbol

5.

(a)

1337

37132493104

24)31(3)52(2

4231

352

x

xxx

xx

xx

(b) 1, 2

1M + 1A

1A

1A

6. 0425 2 xpx

20400

0)25)(4(42

2

pp

p

1M + 1A

1A + 1A

1M for using Δ

   

7. (a) Let x kg be Ron’s original weight.

60

7.56)1.01)(05.01(

xx

∴ Ron’s weight was 60 kg.

(b) weight loss

kg3.6

7.5663

1M

1A

1M

1A

1M for 1.05 or 0.9

u – 1 for missing unit

1M for 63

u – 1 for missing unit

8. (a) 5

23

2

xxf

Put x = 2, 52

)2(322

f

81 f

Put x = 4, 52

)4(324

f

112 f

(b) 52

)2(322

xxf

53)( xxf

1M

1A

1A

1M

1A

   

9. (a) A’ = (3, 9) B’ = (5, –2)

(b) Let P be (x, y)

061224442510811896

)2()5()9()3(2222

2222

yxyyxxyyxx

yxyx

2A

1M + 1A

1A

1M for either side correct

 Section A2

Solution Marks Remarks 

10

 

   

11 (a) : = 3 : 5

and ACB : BAC = 3 : 5 (arc prop. to s)

Let ACB = 3x. Then BAC = 5x.

In ABC,

3x + 5x + 60 = 180 ( sum of )

8x = 120

x = 15

BAC = 5(15)

= 75

(b) BOC= 2BAC ( at centre twice at ce)

= 2(75)

= 150

In BOC,

OB = OC (radii of the same circle)

OBC = BCO (base s, isos. )

OCB =

2150180

( sum of )

= 15

1M

1M + 1A

1A

1A

1M

1A

   

12 (a) Since the graph passes through the origin, the y-intercept is 0,

i.e., c = 0

The axis of symmetry is x = 5. 1025 k

(b) Since the graphs passes through the point (2, –16) and (10, 0), we have

)2()2(16)10()10(0

2

2

baba

)2.........(28)1(..........100

baba

(1) – (2) gives 1 a and 10b

(c) The x-coordinate of the vertex = 5

The y-coordinate of the vertex

)5(10)5( 2

25

The coordinates of the vertex are (5, –25).

1A

1A

1M

1A + 1A

1M

1A

For either one correct

   

13

   

14 (a) Slope of BC

41

)4(4)2(0

BCAD

Slope of AD 4

411

The equations of AD:

)0(46 xy

064 yx

Slope of AC 23

0460

ACBE

Slope of BE 32

231

The equation of BE:

8263

)]4([32)2(

xy

xy

0232 yx

1M

1A

1A

(b)

)2..(..........0232)1....(..........064

yxyx

From (1),

)3....(..........46 xy

Substituting (3) into (2),

1614021218202)46(32

xxx

xx

78

x

Substituting 78

x into (3),

7846y

710

The coordinates of P are .7

10 ,78

1M

1A

(c) Slope of CQ

Slope of CP 21

478

07

10

Slope of AB 2)4(0)2(6

Slope of CQ Slope of AB 1221

CQ is perpendicular to AB.

(d) Orthocentre

1M

1A

1A

    

15 (a) )4)(32( ii

232128 iii i1011

(b) )5)(2()4)(32( iibiia

)5)(2()1011( ibiia

bibiaia 1021011 2 ibaba )1010()211(

)5)(2()4)(32( iibiia

ii 4)4(6

ibaba )1010()211(

i1024

By comparing the real and imaginary parts,

10101024211

baba

)2.(....................1)1....(..........24211

baba

:)2(2)1(

2613

22422211

ababa

2a

Substituting 2a into (2),

12 b 1b

1A

1M

1A

1A

For either one correct

   

16. 0922 xx Sum of roots

12

2

Product of roots

9

(a)

22

2)( 2

)9(

)9(2)2( 2

922

(b) Sum of roots

922

Product of roots

1

The required equation is

019222

xx

09229 2 xx

1A

1M

1A

1A

1A

For both sum and product

   

17

18 (a)(i) 854550180PTQ

PTQTQTPTQTPPQ cos))((2222

85cos)24)(7(2247 22

4.24PQ (cor. to 3 sig. fig.)

The distance between P and Q is 24.4 m

(ii) PQT

PTPTQ

PQ

sinsin

PQT

sin

785sin

40728.24

6.16PQT

The compass bearing of P from Q is N 4.28 W.

1M

1A

1M

1A

1A

   

(b) Let x m be the perpendicular distance from T to PQ.

x 4073.242185sin247

21

86.6x which is greater than 6.

Calvin’s claim is not correct.

1M

1A

19 (a) TQR = 100 ( in alt. seg.)

PTQ + 50 = TQR (ext. of )

PTQ + 50 = 100

PTQ = 50

PTQ = TPQ = 50

PQ = QT (sides opp. equal s)

PQT is an isosceles triangle. Marking Scheme

Case 1 Any correct proof with correct reasons. 3

Case 2 Any correct proof without correct reasons. 2

Case 3 Any relevant correct argument with correct reasons. 1

   

19 (b) (i) ATR = TPR + PRT (ext. of )

100 = 50 + PRT

PRT = 50

PRT = TPR = 50

RT = PT (sides opp. equal s) Marking Scheme

Case 1 Any correct proof with correct reasons. 2

Case 2 Any correct proof without correct reasons. 1

(ii) PTQ = RPT = 50 (proved)

TPQ = PRT = 50 (proved)

PQT = 180 PTQ TPQ

= 180 RPT PRT

= RTP

PRT TPQ (A.A.A.)

PQRT

=

TPPR

(corr. sides, ~s)

PQ6 =

6QRPQ

36 = PQ(PQ + 3)

PQ 2 + 3PQ 36 = 0

PQ =

2)36)(1(433 2 cm

= 4.68 cm (correct to 2 d.p.) 

1M

1A

   

19 (c) Join OR and OQ.

QTR = 180 ATR PTQ (adj. s on st. line)

= 180 100 50

= 30

QOP = 2QTR ( at centre twice at ce)

= 2(30)

= 60

OR = OQ (radii of the same circle)

RQO = QRO (base s, isos. )

RQO + QRO + 60= 180 ( sum of )

2RQO = 120

RQO = 60

ROQ is an equilateral triangle. The length of QR equals the radius of the circle.

Marking Scheme

Case 1 Any correct proof with correct reasons. 3

Case 2 Any correct proof without correct reasons. 2

Case 3 Any relevant correct argument with correct reasons. 1