2007 AJC Prelims Paper 1 solutions

2007 AJC Prelims Paper 1 solutions - 2007 H2 Mathematics...

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2007 H2 Mathematics Prelim Paper 1 solutions Solution 1. 1, () 3 2 4 x = 4 (2 -3 ) ( 1 - 2 x ) -3 = 2 34 1 1 3 .... 22 2 ! 2 xx ⎛⎞ −− −−+ + ⎜⎟ ⎝⎠ = 2 13 3 ... 24 4 ++ + for x < 2 Coefficient of x n = 2 1 ( ) [ ] ( ) 3 4 ... 2 1 !2 n n n ⎡⎤ + ⎢⎥ ⎣⎦ = 2 1 (-1) 2n ( ) ( ) n n n + + 2 1 2 1 2 1 = ( 2 1 2 1 2 + + + n n n ) 2. y = (cos –1 x ) 2 dx dy = 2 cos -1 x 2 1 1 x x dx dy x 1 2 cos 2 1 = ( ) y x dx dy x 4 cos 4 1 2 1 2 2 = = (proved) (i) Differentiating wrt x , dx dy dx dy x dx y d dx dy x 4 2 2 1 2 2 2 2 = 2 1 2 2 2 = dx dy x dx y d x When x = 0 , y = (cos –1 0 ) 2 = 4 2 π ; dx dy = - π ; 2 2 dx y d = 2 ; By Maclaurin’s Theorem , y = 4 2 - π x + x 2 + …. (ii) At x = 0, equation of tangent to the curve is y = 4 2 - π x 3a) a) 0 0 4 4 sin sin sin ( sin ) sin (sin ) x xdx x x dx x ππ =−+ ∫∫ 1
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2 Solution = 0 2 0 4 1c o s 2 o s 2 22 x x dx dx π −− −+ ∫∫ = 0 2 0 4 11 sin 2 sin 2 x xxx ⎡⎤ +− ⎢⎥ ⎣⎦ = 1 84 = + b) 2 1 dx x ud u =⇒ = 1 u 2 2 2 d( 1 xx 2 2 u ) x du u u =− 2 1 12 du u = = 1 sin 2 sin ( ) uc c 1 2 x = Alternately : 1 cos ( ) 2 c x + 4. (i) (ii) Direction vector of is 1 l 0 2 5 ⎛⎞ ⎜⎟ ⎝⎠ Equation of is , 1 l 20 35 r ⎛⎞ ⎛ ⎜⎟ ⎜ =− + λ− ⎝⎠ ⎝ % λ ∈ ± 4 13 3 OB = uuur 0 5 & 2 ON =−−λ Then 2 14 2 65 BN ON OB =−= λ 02 1 4 2 2 56 5 BN •− = ⇒− −λ•− 28 4 30 25 0 29 58 2 ⇒+ λ ++ λ = ⇒λ = ⇒λ=− 2 10 4 BN Equation of line BN : 41 13 5 32 r =+ λ %
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3 Solution 5(i) (ii) cos 1 yx =− 1 1 cos 1 1c o s x y Since 11 x π −≤≤ , 1 o s x y :1 c o s , , 1 1 f xx x
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2007 AJC Prelims Paper 1 solutions - 2007 H2 Mathematics...

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