2007 TPJC Prelim P2 solution

# 2007 TPJC Prelim P2 solution - Solutions for H2 Mathematics...

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1 Solutions for H2 Mathematics 9740/02 Preliminary Examination 2007 Q1. (i) x k dt dy (ii) hy dt dx xy A xhy k dx dy dx x A dy y B x A y ln 2 2 B x A y 2 2 2 = ) ( 2 2 B x A = ln C x n , where C = B 2 , n = 2 A Q2. 2 1 ) tan 1 ( y Differentiating wrt x , 2 1 1 1 ) tan 1 ( 2 x x dx dy ) tan 1 ( 2 ) 1 ( 1 2 dx dy x Differentiating wrt x , 2 2 2 2 1 1 2 2 ) 1 ( x dx dy x dx y d x 2 ) 1 ( 2 ) 1 ( 2 2 2 2 2 dx dy x x dx y d x (Proved) Differentiating wrt x , 0 ) 3 1 ( 2 ) 1 ( 2 ) 2 )( 1 ( 2 ) 1 ( 2 2 2 2 2 2 2 3 3 2 2 dx dy x dx y d x x dx y d x x dx y d x 0 ) 3 1 ( 2 ) 1 ( 6 ) 1 ( 2 2 2 2 3 3 2 2 dx dy x dx y d x x dx y d x When x = 0, y = 1, 4 , 2 , 2 3 3 2 2 dx y d dx y d dx dy By Maclaurin’s expansion, 2 1 ) tan 1 ( y = ... ) 4 ( ! 3 ) 2 ( ! 2 ) 2 ( 1 3 2 x x x = ... 3 2 2 1 3 2 x x x

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2 Q4. (i) AC = OC – OA = 4 j + 3 k – (2 i + 2 k ) = 2 i + 4 j + k Length of side AC = | AC | = 21 1 16 4 (ii) AB = OB OA = 2 i + k – (2 i + 2 k ) = 4 i i AB.AC = 7 1 8 1 4 2 1 0 4  Let BAC = . cos = 21 17 7 1 16 4 1 16 7 | AC || AB | . 0 3 68 . (iii) Vector perpendicular to AB and AC = AB x AC = 8 3 2 2 16 6 4 1 4 2 1 0 4 x
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2007 TPJC Prelim P2 solution - Solutions for H2 Mathematics...

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