![10 (a) T ( 4, 2 ) P1
6+ x 6+ y
= 4, =2 K1
2 2
S ( 2, −2 ) N1
(b) y − 2 2 ( x − 4 )
= K1 K1
= 2x − 6
y N1
3 x + 24 3 y + 24 K1
(c) = 2 or = −2
7 7
10 38 N1
U − ,−
3 3
K1
(d) ( x − 2) + ( y + 2) = 2
2 2
( x − 4) + ( y − 2)
2 2
N1
3 x 2 + 3 y 2 − 28 x − 20 y + 72 =
0
10
11 (a) (i) P ( X 0=
= ) C0 (0.6)0 (0.4)10 or P ( X 1=
10
= ) 10
C1 (0.6)1 (0.4)9 K1
P ( X ≥ 2) = − [ P ( X = + P( X = ]
1 0) 1)
= 1 ─ 10C0 (0.6)0 (0.4)10 ─ 10C1 (0.6)1 (0.4)9 K1
= 0.9983 N1
2
(ii) 800 × K1
5
N1
= 320
(b)(i) P ( −0.417 ≤ z ≤ 1.25 ) K1
= 1 − 0.3383 − 0.1057
= 0.556 N1
(ii) P ( X > t ) =0.7977
Z = −0.833 P1
t − 4.5
−0.833 = K1
1.2
t = 3.5004
N1
10](https://image.slidesharecdn.com/5marksschemeforaddmathspaper2trialspm-100825060520-phpapp01/85/5-marks-scheme-for-add-maths-paper-2-trial-spm-7-320.jpg)
5 marks scheme for add maths paper 2 trial spm
- 1. SPM TRIAL EXAM 2010 MARK SCHEME ADDITIONAL MATHEMATICS PAPER 2 SECTION A (40 MARKS) No. Mark Scheme Total Marks 1 x = 1− 2y P1 2(1 − 2 y ) + y + (1 − 2 y )( y ) = 5 2 2 K1 7y2 − 7y − 3 = 0 − (− 7 ) ± (− 7 )2 − 4(7 )(− 3) y= 2(7 ) K1 y = 1.324 , − 0.324 N1 x = −1.648 , 1.648 N1 OR 1− x y= 2 P1 1− x 1− x 2x 2 + + x =5 2 2 K1 7 x 2 − 19 = 0 − (0 ) ± (0)2 − 4(7 )(− 19) x= 2(7 ) K1 x = −1.648 , 1.648 N1 y = 1.324 , − 0.324 N1 5
- 2. 2 (a) ( f ( x ) = − x 2 − 4 x − 21 ) K1 −4 −4 2 2 = − x 2 − 4 x + − − 21 2 2 N1 = −( x − 2 ) + 25 2 (b) Max Value = 25 N1 (c) f (x ) 25 (2,25) 21 x -3 2 7 Shape graph N1 Max point N1 f ( x ) intercept or point (0,21) N1 d) f ( x ) = ( x − 2 ) − 25 N1 2 7 3 1 1 a) List of Areas ; xy, xy, xy K1 4 16 1 T2 ÷ T1 = T3 ÷ T2 = 4 1 This is Geometric Progression and r = N1 4 n −1 1 25 b) 12800 × = 4 512
- 3. n −1 K1 1 1 = 4 262144 n −1 9 1 1 = 4 4 n −1 = 9 K1 n = 10 12800 N1 (c) S∞ = 1 1− 4 K1 2 N1 = 17066 cm 2 3 7 4 a) 4 cos 2 − 1 − 1 K1 4 cos 2 − 2 ( 2 2 cos 2 − 1 ) N1 2 cos 2θ b) i) 2 1 π 2π -1 -2 P1 - shape of cos graph P1 - amplitude (max = 2 and min = -2) P1 - 2 periodic/cycle in 0 ≤ θ ≤ 2π θ b) ii) y = 1 − (equation of straight line) K1 π Number of solution = 4 (without any mistake done) N1 7
- 4. 5 a) Score 0–9 10 – 19 20 – 29 30 – 39 40 – 49 Number 3 4 9 9 10 N1 1 (35) − 7 P1 b) Q1 = 19.5 + 4 10 9 K1 = 21.44 3 (35) − 25 Q3 = 39.5 + 4 10 K1 10 = 40.75 Interquatile range K1 = 40.75 − 21.44 = 19.31 N1 6 6 (a) OQ = OA + AQ K1 OQ = (1 − m ) a + m b N1 ~ ~ (b) ( PO + OQ = n PO + OR ) K1 OQ = (1 − n ) a + 3n b 4 N1 5 ~ ~ (c) 4 4 K1 (i) − n = 1 − m or 3n = m 5 5 3 1 m= ,n= N1 11 11 N1 8 3 (ii) OQ = a+ b N1 11 ~ 11 ~ 8
- 5. 2 7 = ∫ ( 2 y − y )dy 2 (a)(i) Area K1 0 2 y3 = y2 − 3 0 4 N1 = unit 2 3 1 2 (ii) Area region P = ∫ y dy + ∫ ( 2 y − y )dy 2 K1 0 1 2 1 y3 = × 1× 1 + y 2 − K1 2 3 1 7 N1 = unit 2 6 4 7 1 (b) Area region Q = − = unit 2 3 6 6 K1 7 1 = : 6 6 =7:1 N1 1 (c) Volume π ∫ ( 2 y − y 2 ) dy 2 = 0 K1 1 4 y3 y5 = π − y4 + K1 3 5 0 8 = π unit 3 N1 15 10 8 (a) x 0.000 0.7071 1.000 1.414 1.732 N1 log10 y 1.000 1.330 1.477 1.672 1.826 N1 Using the correct, uniform scale and axes P1 All points plotted correctly P1 Line of best fit P1 1 P1 = (b) log10 y x log10 p + log10 k 3
- 6. (i) use ∗ c = 10 k log K1 k = 10.0 N1 1.83 − 1.0 1 (ii) = use * m = 0.47977= log10 p K1 1.73 − 0 3 N1 p = 27.5 10 9 1 K1 2 π (a) ∠COD = 6 1 N1 = π = 1.047 rad 3 1 20 K1 (b) (i) Arc ABC = π − π or = π 10 3 3 1 Length= AC 202 − 102 or 20 cos π rad K1 6 20 1 Perimeter = + 20 cos π = cm π 38.267 N1 3 6 = (ii) Area of shaded region 1 2 ( ) 2 3 2 102 π − sin π 3 K1 = 61.432cm2 N1 1 (c) ∠CDE = = ∠CAD π rad ( alternate segments ) K1 6 Area = 1 2 ( ) 1 102 π 6 K1 N1 = 26.183cm2 10
- 7. 10 (a) T ( 4, 2 ) P1 6+ x 6+ y = 4, =2 K1 2 2 S ( 2, −2 ) N1 (b) y − 2 2 ( x − 4 ) = K1 K1 = 2x − 6 y N1 3 x + 24 3 y + 24 K1 (c) = 2 or = −2 7 7 10 38 N1 U − ,− 3 3 K1 (d) ( x − 2) + ( y + 2) = 2 2 2 ( x − 4) + ( y − 2) 2 2 N1 3 x 2 + 3 y 2 − 28 x − 20 y + 72 = 0 10 11 (a) (i) P ( X 0= = ) C0 (0.6)0 (0.4)10 or P ( X 1= 10 = ) 10 C1 (0.6)1 (0.4)9 K1 P ( X ≥ 2) = − [ P ( X = + P( X = ] 1 0) 1) = 1 ─ 10C0 (0.6)0 (0.4)10 ─ 10C1 (0.6)1 (0.4)9 K1 = 0.9983 N1 2 (ii) 800 × K1 5 N1 = 320 (b)(i) P ( −0.417 ≤ z ≤ 1.25 ) K1 = 1 − 0.3383 − 0.1057 = 0.556 N1 (ii) P ( X > t ) =0.7977 Z = −0.833 P1 t − 4.5 −0.833 = K1 1.2 t = 3.5004 N1 10
- 8. Sub Total No Mark Scheme Marks Mark 1 12a i) (14) (5) sin θ = 21 K1 3 2 θ = ° or 36° 52 ' 36.87 ∠ BAC 180° − 36.87° = K1 = ° or 143° 8' 143.13 N1 ii) BC 2 = 142 + 52 − 2(14)(5) cos 143.13° K1 2 BC 2 = 333 BC = 18.25 cm N1 iii) sin θ sin 143.13° = K1 2 5 18.25 θ = ° or 9° 28' 9.46 N1 b i) A' 14 cm N1 1 5 cm B' C' ii) ∠ ACB 180° − 143.13° − 9.46° = K1 2 = 27.41° ∠ A ' C ' B ' 180° − 27.41° = = ° or 152° 35' 152.59 N1 10
- 9. Sub Total No Mark Scheme Marks Mark 13 a) 4.55 n 3 = m × 100 or × 100 = 112 K1 3.50 4 m = 130 n = RM 4.48 N1 N1 b) 110(70) + * 130( x) + 120( x + 1) + 112(2) K1 2 = 116.5 7 + x + x +1+ 2 x=3 N1 c i) See 140 P1 3 x (116.5) = 140 K1 100 x = 120.17 / 120.2 N1 ii) x × 100 = 140 K1 2 25 x = RM 35 N1 10
- 10. Sub Total No Mark Scheme Marks Mark 15 a) v 0 = − 30 ms −1 N1 1 b) − 3t 2 + 21t − 30 > 0 K1 2 ( t − 5)( t − 2 ) < 0 2<t<5 N1 c) a = 6t + 21 − K1 3 a 5 = + 21 − 6(5) K1 a 5 = − 9 ms − 2 N1 d) − 3t 3 21t 2 K1 4 S = + − 30t 3 2 21t 2 S =t + − 3 − 30t 2 21(3) 2 S3 = 3 + − (3) − 30(3) = − 22.5 or K1 2 21(5) 2 S5 = + − (5) 3 − 30(5) = −12.5 2 Total distance = − 22.5 + (− 22.5) − (−12.5) K1 = 32.5 m N1 10
- 11. Answer for question 14 (a) I. x + y ≤ 70 N1 y II. x ≤ 2y N1 III. y − 2 x ≤ 10 N1 (b) Refer to the graph, K1 1 graph correct 3 graphs correct N1 90 Correct area N1 (c) i) 15 ≤ y ≤ 40 N1 80 ii) k = 10x + 20y max point ( 20,50 ) N1 70 Max fees = 10(20) + 20(50) K1 (20,50) = RM 1,200 N1 10 60 50 40 30 20 10 x 0 10 20 30 40 50 60 70 80
- 12. log10 y Answer for question 8 2.0 1.9 X 1.8 1.7 X 1.6 1.5 X 1.4 X 1.3 1.2 1.1 1.0 X x 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8