# My Possible Answer - Add Maths Paper 1 - SPM 2011 by nklye

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```									  My Possible Answer & Marking
(Q13 to Q25)
(It has nothing to do with any official answer and marking scheme by the concern authority)

Paper 1
SPM 2011
13. (a) Gradient = − 3 √ 1
√M1
(b) m┴ = ⅓
Eqn: y – 3 = ⅓ (x – 1)
3y = x + 8 √ 2
14. sin 2θ = cos θ
2 sinθ cosθ = cosθ    √M1     √M2 Each basic angle
cosθ (2 sinθ – 1) = 0         √M3 Both basic angles
cosθ = 0 √M1  θ = 90o, 270o
sinθ = ½       θ = 30o, 150o
√M1
θ = 30o, 90o, 150o, 270o √ 4
- 24
15. (a) cot θ = 4              5
┌
B
3    √1                     3   -7
A       ┌                  25
4

(b) sin (A + B) = sinA cosB + cos AsinB
√M1 = (3/5)(- 24/25) + (4/5)(-7/25)
= - 4/5 √ 2

E
16. DF = − OD + DE √M1                      F                   D
= −(3i + 2j) + (−5i + 3j) √M2                      O
= −8i + j √ 3                                 DE = OF
8       h
17. (a) r + s =   ( -2 ) + ( 7 )
=        +
( 8 5 h) √ 1

(b) | r + s | = 13
√ (8 + h)2 + 52 = 13 √M1
(8 + h)2 + 25 = 169
8 + h = √144
h = 4 √ 2 (h = positive value)
18. (a) PR = 6 √M1
tan θ = 6/8 = 0.75
θ = 0.6436 rad. √ 2
(b) PQ = 10(0.6436) = 6.436 √M1
= 6 + 2 + 6.436
= 14.44 √ 2
5x ,     dy
19.   y= 2              = g(x)  ∫ g(x)dx = y
x +1      dx
3                     5x            3
∫ 2g(x)dx = 2y = 2[             ]       √M1
0                     x2   +1       0

5(3)
= 2[      − 0 ] √M2
9+1

= 3 √3
12
20. y = 10 –
x
dy       12                 10x – xy = 12
=        √M1           x(10 – y) = 12
dx       x2                 y = 4 then x = 2
12
=        =   3   √M1
22
∂y = (4 + p ) – 4 = p √M1
∂x   dx            dx ( ∂y )
=       ∂x =
∂y dy              dy
p
√M2 = 1 ( p ) =      √3
3         3
a
21.       ∫ (x + 1) dx
4

x2      a
=   [    + x]        √M1
2       4

a2        42
=   [    + a]–[    + 4 ] √M2
2         2

a2
=    2
+ a − 12    √3
22. ∑ x = 240 , ∑x2 = 9654

240
(a) x = 6 = 40 √ 1

∑x2
(b) σ = √         −   x2
N

9654
= √ 6 − 402 √M1
=3     √2
23. (a)   7P    = 2520     √2
5
√M1

√M1
(b) 2 x 5 x 5 P3 = 600           √2

OR       2   5    4    3        5   √M1
= 600
√2
24. (a) n(S) = 20, n(M) = 6

6    3
P(M) =    =    √1
20   10

(b) P(M and N) = P(M ∩ N)
n(M ∩ N) 2
=         = √M1
n(S)    20
1
= 10 √ 2
25. (a) P(X ≥ 1) = P(X=1) + P(X=2) + P(X=3)
= (12/27) + (6/27) + (1/27) √M1
= 19/27 √ 2

(b) m = P(X = 0) = 1 – P(X ≥ 1)
= 1 – 19/27
= 8/27 √ 1
the end

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