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2014-1-JOH-SULTAN ISMAIL-A Flipbook PDF
2014-1-JOH-SULTAN ISMAIL-A
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1(a)
1(b)
30
B1
4
x ⇒ y = z
1 2 2 11 1 1 1 3 −5 9 3 b 1 4
3 PQ = k I 1 2 2 −5 a 4 1 0 0 1 1 3 8 −5 c = k 0 1 0 0 0 1 4 c 3 b 1 c a12 : a − 10 + 8 = 0 - - - -(1) a32 : 3a − 5b + 4 = 0 - - - -(2) a =2, b=2 a31 : −15 + 8b + c = 0 c = −1 a11 : k = −5 + 16 − 2 = 9 1 2 2 1 Q −1 = 1 1 3 9 3 b 1 1/ 9 2 / 9 2 / 9 = 1/ 9 1/ 9 1/ 3 1/ 3 2 / 9 1/ 9 −5 a 4 x 11 8 − 5 c y = −5 , 4 c z 4 c x 1 y = 2 z 3 x = 1, y = 2, z = 3
2
z =8
4(a) zz = z = 64 M1
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D1 (End-point dotted, symmetry in y = x ) 3
D1 ( graph of f −1 )
D1 (graph of f )
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2014-1-JOH-SULTAN ISMAIL-MARKING SCHEME f ( f −1 ( x)) = x [ f −1 ( x)]2 + 2 = x
y
20
x
Since f is defined for x ∈ [−5, −1] , f −1 ( x) < 0 f −1 ( x ) = − x − 2 R f = [3, 27]
20 10
10
f −1 : x a − x − 2, x ∈ [3, 27]
−10 −10
2(a) T2 = ar = 24 T5 = ar 4 = −3 1 r = − , a = −48 2 S∞ − Sn < 0.005
−48 −48[1 − (−0.5)n ] − < 0.005 1 − (−0.5) 1 − (−0.5) 32(0.5)n < 0.005 log(0.00015625) n> log(0.5) n > 12.64 the smallest value of n = 13
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4 3 2π arg(−4 + 4 3i ) = π − tan −1 = 4 3
A1
−4 + 4 3i = 42 + (4 3) 2 = 8
2π 2π z = 8 cos + i sin 3 3 10π z 5 = 85 cis 3 2π = 32768cis − 3 2π 2π = 32768 cos − i sin 3 3
4(b)
2π w3 = 8cis 3
4
6
3
2π / 3 + 2kπ w = 2cis , k = 0,1, 2 3 2π 8π 14π w1 = 2cis , w2 = 2cis , w3 = 2cis 9 9 9 2π 8π 4π w1 = 2cis , w2 = 2cis , w3 = 2cis − 9 9 9 w = 1.5321 + 1.2856i, −1.8794 + 0.6840i, 0.3473 − 1.9696i M1 M1
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Vertices = (0, −5), (0,5) A1
x
−5 −4 −3 −2 −1 −1 1 2 3 4 5 6 −2 −3 −4 −5 Let (a, b) be a point on the ellipse
4 3 2 1
c 2 = 25 − 9, ⇒ c = ±4 Foci = (0, −4), (0, 4) y
5(a) Centre = (0, 0)
(b) a 2 b2 25 + = 1, ⇒ b 2 = 25 − a 2 - - - (1) 9 25 9 (d1 + d 2 ) 2 = ( a 2 + (b + 4) 2 + a 2 + (b − 4) 2 ) 2 = a 2 + (b + 4) 2 + a 2 + (b − 4) 2 + 2 a 2 + (b − 4) 2 a 2 + (b − 4) 2
2
−1 2 2 2 2 2 2 2 = 2 + 3 + 4 1 + 2 + 2 cos θ −2
1
= 2a 2 + 2b 2 + 32 + 2 (a 2 + b 2 + 16) 2 − (8b) 2 - - - - (2) Substitute (1) into (2), … … … 50 16 = 2a 2 + 50 − a 2 + 32 + 2 (9 + a 2 ) 2 9 9 50 32 = 2a 2 + 50 − a 2 + 32 + 18 + a 2 9 9 = 100 ∴ (d + d ) 2 is constant.
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2 3 4
5
−2 + 6 − 8 = 3 29 cos θ
4 cos θ = − 29 87 θ = 104.34° The acute angle is θ = 75.66°
2
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B1 (or long division)
Section B [15 marks] ( MARK ONLY ONE question in this section – the FIRST ONE AFTER QUESTION 6 )
1
or x = 3 9 = 1.277 x = 9 9 = 1.277 cos( x − 17°13 ') < −0.5 .
3
(a − b − c)2 = (2 b c ) 2 = 4bc log 3 x 2 + log 3 x = log 27 9 log 3 9 log 3 x 3 = log 3 27 1 2 3 log 3 x = log 3 9 or 3 log 3 x = 3 3 1 2 or log x = log 3 9 log 3 x = 9 9
a −b−c = 2 b c
a =b+c+2 b c
x3 + 2 x 2 + 2 x + 3 2 3 ≡ x+ + x2 + 2 x − 3 x −1 x + 3 a= b+ c ( a )2 = ( b + c )2
x3 + 2 x 2 + 2 x + 3 x3 + 2 x 2 − 3x + 5 x + 3 7(a) ≡ x2 + 2 x − 3 x2 + 2 x − 3 5x + 3 ≡ x+ x2 + 2 x − 3 5x + 3 5x + 3 A B ≡ ≡ + x 2 + 2 x − 3 ( x − 1)( x + 3) x − 1 x + 3 5 x + 3 ≡ A( x + 3) + B ( x − 1) x = 1, 8 = 4 A ⇒ A=2 x = −3, − 12 = −4 B ⇒ B=3
7(b)
7(c)
7(d)
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−3
−2
−1
2 1
−1 −2
y
1
2
x 3
−3 cos( x − 17°13 ') = −0.5 = − cos 60° x − 17°13 ' = −120°,120° x = −102°47 ',137°13 '
32 + 122 + 62
AB = 4i − 9 j − k − (i + 3 j + 5k ) = 3i − 12 j − 6k 3i − 12 j − 6k
5 2 1
AD = BC 1 2 4 OD − 3 = −1 − −9 5 3 −1 −1 OD = 11 = −i + 11j + 9k 9
DABC is a parallelogram,
p=2
21 4 21 2 21 = i− j− k 21 21 21 AC = kAB p 1 3 −1 − 3 = k −12 3 5 −6 −4 = −12k or −2 = −6k 1 k= 3 1 p − 1 = (3) 3
The unit vector =
Solution set = { x : −180° ≤ x < −102 °47 ',137 °13 ' < x ≤ 180°}
(iii)
(ii)
8(a) (i)
(b)
4 2 −3 1 1 −2 2 −1 3
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4 2 −3 → 0 2 −5 0 −4 9 4 R2 − R1 → R2 2 R3 − R1 → R3
4 2 −3 R3 + 2 R2 → R3 → 0 2 −5 0 0 −1 −z = 3 2 y − 5z = 3 4 x + 2 y − 3z = 5 x = 2, y = −6, z = −3
5 3 −3 5 3 3
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