1

SPM TRIAL EXAM 2012 Marking Scheme

Additional mathematics Paper I

Number Solution and marking scheme Sub Marks Full

Marks 1

(a) {p, r, s} (b) {a , b, c, d} (c) Many to One

1 1 1

3

2

(a) 27

B1: 2x – 5 = 2 atau f(x) = 25+x

2

2

3

(a) 5 (b) 42 +x

B2: ( ) 7653 −−=+− xxf B1 : ( ) 53 +− xf

1 3

4

4

p = − 7 , q = − 10 ( both ) B2: p = − 7 or q = −10

B1: 33

7 p−= or

3310 q

=− or 01073 2 =−− xx

or 3(−1)2 + p(1)+q = 0, 3(10/3)2 + p(10/3) + q = 0

3

3

5

h =2 and k = 3 B2 : h = 2 atau k = 3

B1 : 13=

h or 922 =++ kh

3

3

6

45 ≤≤− x B2: ( ) 0)5(4 ≤+− xx or B1 : x2 + x – 20 ≤ 0

3

3

4 -5

2

7

pr12

+ or prrp +2

B3: mm 33

2

2

log3log

2log

2log+

B2: log m2 + 2logm3or mm 3

23

2

2

log3log

log2log+

B1 : log 2 + log32 or 2logm3 or mlog2log

ormlog3log

4

4

8 158 += px B2: )32(43 +=− px B1: 32 or )32(42 +p

3 3

9 a) 3

b) 360

B1: [ ]18 2(3) (18 1)22

+ −

1 2

3

10 13−=a , 4=d (both) B2: 13−=a or 4=d B1: 52 −=+ da or 157 =+ da

3 3

11 (a) r = x2

(b) 31

B1: 2

2

181

xx−

=

1 2

3

12 a) qpx

xy

+= 2

b) 2−=p , 8=q B2 : 2−=p or 8=q

B1: 4204

−

−=p or q+−= )4)(2(0 or 4 = p(2) + q

1 3

4

3

13

2,12

B2: 110)1(3

473

−=⎟⎠

⎞⎜⎝

⎛−

−−×⎟⎠

⎞⎜⎝

⎛−

−

hh or form equation using Pythagoras theorem

B1: 473

−

−

h or

10)1(3

−

−−

h of find the lengths of AB, BC and AC using distance formula

3

3

14 (a) w

1

(b) −2w 1−w2 B1: 21 w− or θcos2w

1 2

3

15

26.57 ° ,116.57 ° , 206.57 ° , 296.57 ° B3: 26.57 ° and 116.57 ° B2: )2)(tan1tan2( +− xx B1: 02tan3tan2 2 =−+ xx

4

4

16

(a) 13 (b) k = −13

B1: ⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

++

25112 k

or jik )25()112( +−+++

1 2

3

17 (a) 4a + 4b (b) ba 42 +−

B1: )44(6 baa ++−

1 2

3

18 8 B2: 9.42)3()3()3(3.13.1 =−+−+−+++ rrrrrr B1: r3.1 or )3(3.1 −r

3 3

19 512

B2: ⎟⎠

⎞⎜⎝

⎛ −+

012)2(32 2

B1:2

02 132 ⎥⎦

⎤⎢⎣

⎡+xx

3 3

4

20

122 −+ aa

B2: [ ]))4()4(()( 22 −+−−+ aa B1: [ ]xx +2

3

3

21 k= 8 B3: (k+1)(k−8) = 0

B2: 22 2 22 5 2 5 6

3 3k k+ + + +⎛ ⎞− =⎜ ⎟

⎝ ⎠

B1: 352 kx ++= or 2222 52 kx ++=∑

4 4

22 1 2( , )3 3−

B2 : x = 13

, y = 2(⅓)(3×⅓−2)

B1 : 12x – 4 = 0

3 3

23 (a) 1 (b) i) 5040

ii) 288 B1: 2 × 3! × 4!

1 1 2

4

24 (a)

154

(b) 35

B1 : 3 2 2 1 3 1 2 21 or 5 3 5 3 5 3 5 3⎛ ⎞⎛ ⎞− × + × + ×⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

1 2

3

25 (a)

53

(b) 625144

B1: 32

25

52

53

⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛C

1 2

3

5