2007 CJC Prelims Paper 2 Solution

2007 CJC Prelims Paper 2 Solution - Prelim 07 H2 Maths...

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Prelim 07 H2 Maths Paper 2 Marking Scheme 1(i) dx du xe e dx dy u u [M1] x dx dy 4 [A1] (ii) xdx dy 4 C x y 2 2 [M1] C x xe u 2 2 x C x e u 2 C x C x u , 2 ln is an arbitrary constant. [A1] [1 mark for correct shape] Curve cuts x-axis at (-1/2, 0) and (1, 0). Vertical asymptotes are x = 0 and x = 2 1 . [A1] 2(a) (i) i z 3 1 4 = k i i e e 2 3 2 2 [M1] = k i i e e 2 3 2 2 = ) 3 1 ( 2 2 k i e 6 ) 3 1 ( 4 1 2 k i e z , k = 0,-1 , 1, - 2 [A2] (ii) i z z z 3 1 2 16 4 2 0 4 2 4 4 8 [M1] Therefore 6 ) 3 1 ( 4 1 2 k i e z , k = 0, -1, 1, - 2 [A1] and 6 ) 3 1 ( 4 1 2 k i e z , k = 0, -1, 1, -2 [A1] (b) Circle centre ( 2, 2), rad = 2 [B1] Shading the circle [B1] PC = 64 + 36 = 10 [M1] Max dist = PA = 12 [A1] Min. dist = PB = 8 [A1] 3 (i) Let AB = k 1 2 2 . b = k 2 k 2 k + 2 1 0 = k 2 2 k + 1 2 k [M1] ( k 2) + 2(2 k + 1) 2( 2 k ) = 18 [M1] 9 k = 18 k = 2 b = 0 5 4 [A1] (ii) AB = 2 1 + 4 + 4 = 6. [B1] (iii) BA = 2 AC a b = 2( c a ) [M1] 2 c = 3 a b 2c = 3 2 1 0 0 5 4 c = 3 1 2 . [A1]
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(iv) CD = 2 5 3 3 1 2 = 5 6 5 CB = 0 5 4 3 1 2 = 3 6 6 = 3 1 2 2
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This note was uploaded on 08/29/2009 for the course MA 9740 taught by Professor Moe during the Summer '07 term at Singapore Management.

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2007 CJC Prelims Paper 2 Solution - Prelim 07 H2 Maths...

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