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SPM TRIAL EXAM 2012 Marking Scheme

Additional Mathematics Paper 2 SECTION A

Question Important Steps Marks 1 y = 3x – 4 1

5x2 – 4x(3x – 4) + (3x – 4)2 = 9 or 2x2 – 8x + 7 = 0

1

)2(2)7)(2(4)8()8( 2 −−±−−

=x 1

x = 2.707, 1.293 1 y = 3(2.707) – 4 , y = 3(1.293) – 4 = 4.121 = – 0.121

1

TOTAL 5

2 (a) Change base of logarithm : 4

22

log (1 2 )log 4log (1 2 ) xx −

− = or

equivalent 1

Use n log x = log x n : 2 log 2 ( 1 – 2x ) = log2 ( 1 – 2x )2 1 Solve : (2x + 5 ) = ( 1 – 2x ) 2 1 x = 22

1 ,− 1

x = 21− 1

(b) 31 1

TOTAL 6

3 (a) – 8 1

(b)

Use Tn = a + (n – 1 ) d : – 8 + ( 22 – 1 ) ( 3 ) 1 Use Sn = ])([ dnan 122 −+ : 2 [2( 8) ( 1)(3)]n n− + − 1 Solve : ])()()([ 31822 −+− nn = 55 1 n = 10 1

(c) T13 1 28 1

TOTAL 7

4 (a)

x = 3 1

(b) 5

59105.9 ⎟⎠

⎞⎜⎝

⎛ −+=median

1

Median = 10.5 1

(c) All midpoints are correct. 1

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1120220

20)22(2)17(4)12(5)7(6)2(3

==++++

=x 1

∑ =++++= 3150)22(2)17(4)12(5)7(6)2(3 222222fx or Σf (x- x )2 = 730

1

22

20220

203150

⎟⎠

⎞⎜⎝

⎛−=σ or 207302 =σ 1

= 36.5 1 TOTAL 8

5

(a) y

xyπ2

=

x

Shape 1

Max/min 1

One period

1

Complete from 0 to

2π 1

(b) Equation xy

π2

= 1

Straight line xyπ2

= 1

2 solutions 1 TOTAL 7

6

(a)

32QSm = 1

)6(231 −=− xy 1

823

−= xy 1

(b) Q(0, − 8) 1

√[(x − 6)2 + (y − 1)2 ] or √[(6 − 0)2 + (1+8)2] 1

x2 − 12x + 36 + y2 − 2y + 1 = 117 1 x2 + y2 − 12x − 2y − 80 = 0 1

TOTAL 7

π 2π π 2

3π 2

3

-3

O

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SECTION B

7 (a)

x 1 1.5 2 2.5 3 3.5 4

log10 y 0.04 0.18 0.28 0.40 0.52 0.62 0.76

1

(b) Plot log10 y against x [ correct axes and uniform scales ] 1 All 7 points plotted correctly 1 Line of best fit 1

(c) (i) 3.72 1

(ii) 22101010 hk xy loglog )(log −= 1

Use :log210 hc −= 202

10 .log −=− h 1 h = 2.5 1

(iii) Use :log210 km = 2402

10 .log =k 1 k = 3.0 1

TOTAL 10

8 (a)

∫−= dxxy 2 1

cxy +−

=22 2

1

92 +−= xy 1

(b)

3 2

0( 9)x dx− +∫

1

33

0

93x x

⎡ ⎤−+⎢ ⎥

⎣ ⎦

1

−)10)(10(21

3 2

0( 9)x dx− +∫ or 50 –

3 2

0( 9)x dx− +∫

or −)10)(10(21

33

0

93x x

⎡ ⎤−+⎢ ⎥

⎣ ⎦ or 50 –

33

0

93x x

⎡ ⎤−+⎢ ⎥

⎣ ⎦

1

= 32

1

(c) Volume = ∫ − dyy)9(π 1

92

0

92yyπ

⎡ ⎤−⎢ ⎥

⎣ ⎦

1

812

orπ equivalent 1

TOTAL 10

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9 (a)

Angle AOB = 6.5/5 1 = 1.3 rad. 1 Angle POQ = 0.8667 rad. 1

(b) MN = 5 × sin(0.8667 rad.) = 3.811 cm or ON = 5 × cos( 0.8667 rad.) = 3.2367 cm

1

Length of arc PQ = 6 × 0.8667 = 5.2002 1 Perimeter = 3.811 + 5.2002 + 1 + (6−3.2367) 1 = 12.77 cm 1

(c) Area of sector OPQ = ½ × 62 × 0.8667 1 Area of shaded region = 15.60 − ½ (3.811)(3.2367) 1 = 9.433 1

TOTAL 10

10 (a) (i)

RTPRPT +=

)148(21 xyRT +−=

1

)148(218 xyyPT +−+=

= yx 47 +

1

(ii)

RS RP PS= +uuur uuur uuur

)14(318 xy +−=

1

yx 8314

−= 1

(b) (i) yhxhPM 47 += 1

(ii) RMPRPM +=

)8314(8 yxky −+=

1

= ykxk )88(314

−+

1

(c) 147

3h k= or hk 488 =−

1

21

=h 1

43

=k 1

TOTAL 10

11 (a)(i)

377

10 )55.0()45.0()7( CXP == 1

= 0.07460 1

(ii)

P(X = 0, 1, 3) = 10C0(0.45)0(0.55)10 + 10C1(0.45)1(0.55)9 + 10C2(0.45)2(0.55)8 1

= 0.09956 1

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(b) (i)

6.0 7.2 8.1 7.2( )1.2 1.2

P z− −< < OR ( 1 0.75)P z− < < 1

= 2266.01587.01 −− 1

= 0.6147 1

(ii) 8.0

6048

2.12.7

==⎟⎠

⎞⎜⎝

⎛ −>tzp 1

842.02.12.7

−=−t 1

190.61896.6 ort = 1 TOTAL 10

SECTION C

12 (a)

a = 1.4 0.6dv tdt= −

1

= 1.4 − 0.6(2) = 0.2 1

(b)

1.4t −0.3t2 + 0.5 = 0 1 (3t +1)(t − 5) = 0 or using quadratic formula 1 t = 5 1

(c)

s = 2(1.4 0.3 0.5)t t dt− +∫ = 0.7t2 − 0.1t 3 + 0.5t + c integrate 1 At t = 0, s = 0 ⇒ c = 0 finding c

or ∫ ∫ +−++−5

0

10

5

22 5.03.04.15.03.04.1 dtttdttt limits √ 1

When t = 5, s = 7.5 m, when t = 10, s = −25 m or substitute t=0, 5, 10 in [0.7t2 − 0.1t 3 + 0.5t] 1

Total distance = 7.5 x 2 + 25 or 7.5 + |−25 − 7.5| 1 = 40 m 1

TOTAL 10

13 (a) 158 100

130x = × 1

= 121.54 1

(b)

150 100 107.14140

A⇒ × = 1

121.54 100 90.03135

B⇒ × = 1

120 100 109.09110

C⇒ × = 1

123 100 102.5120

D⇒ × = 1

(c) 12

)25.102()209.109()503.90()314.107( ×+×+×+×=I 1

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= 99.56 1

(d) 56.99

100120

2013 ×=I 1

= 119.47 1 TOTAL 10

14 (a) (i)

sin sin3012 7ACB∠ °

= 1

∠ACB = 59° 1

(ii) 2 2 24 11.47 12cos2(4)(11.47)

AKB + −∠ = 1

cos ∠AKB = 0.0388 ∠AKB = 87.78° or 87°47’ 1

(iii) ∠ABC = 91° 1

Area rABC = ½ (7)(12) sin 91° or Area of rAKB = ½ (4)(11.47) sin 87.78°

1

Area of quadrilateral = Area rABC + Area of rAKB = 41.99 + 22.92

1

= 64.91 cm2 1

(b)(i)

1

(ii) ∠A’C’B’ = 121° 1 TOTAL 10

15 (a)(i)

x + y ≤ 10 or equivalent 1 y – x ≤ 4 or equivalent 1 x ≤ 2y or equivalent 1

(b) Draw correctly one straight line from the inequalities 1 Draw correctly two more straight line from the inequalities 1 Region R correctly shaded 1

(c)(i) Maximum point ( 3 , 7 ) 1 RM [ 10(3) + 25(7) ] = RM 205 1

(ii) Minimum point (2 , 6 ) 1 RM [ 10(2) + 25(6) ] = RM 170 1

TOTAL 10

B’

A’ C’

12 cm 7 cm

30o

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GRAPH FOR QUESTION 7

0 0.5 1 1.5 2 2.5 3 3.5 4 x

log10 y

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

– 0.1

– 0.2

×

×

×

×

×

×

×

( 0 , – 0.2 )

( 4 , 0.76 )

3.2

0.57

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GRAPH FOR QUESTION 15

y

0 1

1

x 2 3 4 5 6 7 8 9

2

3

4

5

6

7

8

9

10 y – x = 4 = 2y

x + y = 10

x = 2y

R

( 3 , 7 )

10x + 25y = 150 ( 2 , 6 )

y = 6

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