Edexcel_WS_C4_PaperG

Edexcel_WS_C4_PaperG - Worked Solutions Edexcel C4 Paper G...

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Worked Solutions Edexcel C4 Paper G 1. ( a ) d y d x = 2 cos t sin t when t = π 2 , gradient = 0 (2) ( b ) using cos 2 t + sin 2 t = 1 , x 2 + ± y 2 ² 2 = 1 x 2 + y 2 4 = 1 (3) 2. ( a ) 1 x 2 + 3 2 x + 1 + 1 x + 2 (using ‘cover up’ rule) (4) ( b ) # 4 3 ³ 1 x 2 + 3 2 x + 1 + 1 x + 2 ´ d x = µ ln (x 2 ) + 3 2 ln ( 2 x + 1 ) + ln (x + 2 ) 4 3 = ln 2 + 3 2 ln 9 + ln 6 ³ ln 1 + 3 2 ln 7 + ln 5 ´ = ln ³ 2 × 6 5 ´ + 3 2 ln ³ 9 7 ´ = ln ³ 12 5 ´ + 3 2 ln 9 7 (5) 3. ( a ) ( 1 + ax) 6 = 1 + 6 ax + 15 a 2 x 2 (3) ( b ) ( 1 + bx) ± 1 + 6 ax + 15 a 2 x 2 ² = 1 + 6 ax + 15 a 2 x 2 + bx + 6 abx 2 we have 6 a + b =− 9 …[A] 15 a 2 + 6 ab = 24 …[B] from equation [A] b 9 6 a substitute in [B] 15 a 2 + 6 a( 9 6 a) = 24 Hence a 2 ,b = 3 (7) 4. ( a ) = # x d d x ( tan x) d x [By parts] = x tan x # tan x d x = x tan x + ln cos x + c (4) ( b ) # y 1 2 d y = # x sec 2 x d x 2 y 1 2 = x tan x + ln cos x + c y = 4 ,x = 0 : 2 4 = 0 + ln 1 + c c = 4 2 y = x tan x + ln cos x + 4 when x = π 4 ,2 y = π 4 · tan π 4 + ln 1 2 + 4 2 y = π 4 + ln 2 1 2 + 4 2 y = π 4 1 2 ln 2 + 4 y = π 8 1 4 ln 2 + 2 y = ³ π 8 1 4 ln 2 + 2 ´ 2 (6)
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5. ( a ) x 1 1 1 2 2 2 1 2 3 12 x 12 8 6 4.8 4
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This note was uploaded on 06/04/2010 for the course MATH C4 taught by Professor N/a during the Spring '10 term at Cambridge College.

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Edexcel_WS_C4_PaperG - Worked Solutions Edexcel C4 Paper G...

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