![7
10
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
y = 2x − 6 N1
3 x + 24 3 y + 24 K1
(c) = 2 or = −2
7 7
10 38 N1
U − ,−
3 3
K1
( x − 2) + ( y + 2) = 2 ( x − 4) + ( y − 2)
2 2 2 2
(d)
N1
3 x 2 + 3 y 2 − 28 x − 20 y + 72 = 0
10
11 (a) (i) P ( X = 0 ) = C0 (0.6) (0.4) or P ( X = 1) = C1 (0.6) (0.4) K1
10 0 10 10 1 9
P ( X ≥ 2) = 1 − [ P ( X = 0) + P ( X = 1) ]
10 0 10 10 1 9
= 1 ─ C0 (0.6) (0.4) ─ C1 (0.6) (0.4) 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](https://image.slidesharecdn.com/5marksschemeforaddmathspaper2trialspm-100825060715-phpapp02/85/5-marks-scheme-for-add-maths-paper-2-trial-spm-7-320.jpg)
5 marks scheme for add maths paper 2 trial spm
- 1. 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= K1 2( 7 ) 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 2 (a) f ( x ) = −( x 2 − 4 x − 21) K1 2 −4 −4 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) 2 − 25 N1 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. 3 1 n −1 1 K1 = 4 262144 n −1 9 1 1 = 4 4 n −1 = 9 K1 n = 10 12800 N1 S∞ = (c) 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 K1 b) ii) y = 1 − (equation of straight line) π Number of solution = 4 (without any mistake done) N1 7
- 4. 4 5 a) Score 0–9 10 – 19 20 – 29 30 – 39 40 – 49 Number 3 4 9 9 10 N1 1 ( 35) − 7 4 b) Q1 = 19.5 + 10 P1 9 K1 = 21.44 3 ( 35) − 25 Q3 = 39.5 + 4 10 10 K1 = 40.75 Interquatile range = 40.75 − 21.44 K1 = 19.31 N1 6 6 (a) OQ = OA + AQ K1 OQ = (1 − m ) a + m b N1 ~ ~ (b) ( PO + OQ = n PO + OR ) K1 4 OQ = (1 − n ) a + 3n b 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 N1 (ii) OQ = a+ b 11 ~ 11 ~ 8
- 5. 5 7 2 (a)(i) Area = ∫ ( 2 y − y ) dy 2 K1 0 2 y3 = y2 − 3 0 4 2 N1 = unit 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 2 N1 = unit 6 4 7 1 2 (b) Area region Q = − = unit 3 6 6 K1 7 1 = : 6 6 N1 =7:1 1 (c) Volume = π ∫ ( 2 y − y ) dy 2 2 0 K1 1 4 y3 y5 =π − y4 + K1 3 5 0 8 = π unit 3 15 N1 10 8 x 0.000 0.707 1.000 1.414 1.732 N1 1 log10 y 1.000 1.330 1.477 1.672 1.826 N1 P1 (a) P1 P1 P1 Using the correct, uniform scale and axes All points plotted correctly
- 6. 6 Line of best fit K1 N1 1 (b) log10 y = x log10 p + log10 k K1 3 (i) use ∗ c = log10 k N1 k = 10.0 1.83 − 1.0 1 (ii) use * m = = 0.47977 = log10 p 1.73 − 0 3 p = 27.5 10 9 1 K1 (a) ∠COD = 2 π 6 1 N1 = π = 1.047 rad 3 1 20 K1 (b) (i) Arc ABC = 10 π − π or = π 3 3 1 K1 Length AC = 202 − 102 or 20 cos π rad 6 20 1 N1 Perimeter = π + 20 cos π = 38.267cm 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
- 7. 7 10 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 y = 2x − 6 N1 3 x + 24 3 y + 24 K1 (c) = 2 or = −2 7 7 10 38 N1 U − ,− 3 3 K1 ( x − 2) + ( y + 2) = 2 ( x − 4) + ( y − 2) 2 2 2 2 (d) N1 3 x 2 + 3 y 2 − 28 x − 20 y + 72 = 0 10 11 (a) (i) P ( X = 0 ) = C0 (0.6) (0.4) or P ( X = 1) = C1 (0.6) (0.4) K1 10 0 10 10 1 9 P ( X ≥ 2) = 1 − [ P ( X = 0) + P ( X = 1) ] 10 0 10 10 1 9 = 1 ─ C0 (0.6) (0.4) ─ C1 (0.6) (0.4) 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
- 8. 8 t − 4.5 K1 −0.833 = 1.2 t = 3.5004 N1 10 Sub Total No Mark Scheme Marks Mark 12a i) 1 K1 3 (14) (5) sin θ = 21 2 θ = 36.87° or 36° 52 ' ∠ BAC = 180° − 36.87° K1 = 143.13° or 143° 8' 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 = 9.46° or 9° 28' 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° = 152.59° or 152° 35' N1 10
- 9. 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 K1 2 × 100 = 140 25 x = RM 35 N1 10
- 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 = − 6(5) + 21 K1 a 5 = − 9 ms − 2 N1 d) − 3t 3 21t 2 K1 4 S = + − 30t 3 2 21t 2 S = − t3 + − 30t 2 21(3) 2 K1 S 3 = − (3) + 3 − 30(3) = − 22.5 or 2 21(5) 2 S 5 = − (5)3 + − 30(5) = −12.5 2 Total distance = − 22.5 + (− 22.5) − ( −12.5) K1 = 32.5 m N1 10
- 11. 11 Answer for question 14 (a) I I. N1 y I II. N1 I III. N1 (b) Refer to the graph, 1 graph correct K1 3 graphs correct N1 90 Correct area N1 ( (c) i) N1 80 ii) k = 10x + 20y max point ( 20,50 ) N1 70 Max fees = 10(20) + 20(50) K1 = RM 1,200 (20,50) N1 10 60 50 40 30 20 100 10 20 30 40 50 60 70 80 x
- 12. 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 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 0 x