Edexcel_WS_C4_PaperC

# Edexcel_WS_C4_PaperC - Worked Solutions Edexcel C4 Paper C...

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Worked Solutions Edexcel C4 Paper C 1. ( a ) 5 x + 7 (x + 1 )(x + 2 ) = 2 x + 1 + 3 x + 2 (3) y = 2 (x + 1 ) 1 + 3 (x + 2 ) 1 d y d x =− 2 (x + 1 ) 2 3 (x + 2 ) 2 d 2 y d x 2 = 4 (x + 1 ) 3 + 6 (x + 2 ) 3 when x = 1, d 2 y d x 2 = 4 2 3 + 6 3 3 = 13 18 (3) 2. V = 36 h 2 d V d h = 72 h we are given d V d t = 24 d V d t = d V d h × d h d t when h = 2, 24 = 72 × 2 × d h d t d h d t = 24 144 = 1 6 cm s 1 (5) 3. ( a ) ( 1 + 2 x) 1 2 = 1 + ± 1 2 ² ( 2 + ± 1 2 ²± 3 2 ² 2 ( 2 2 + ± 1 2 3 2 5 2 ² 3 . 2 ( 2 3 + ... = 1 x + 3 2 x 2 5 2 x 3 (4) ( b ) valid for 1 2 <x< 1 2 (1) ( c ) ( 1 + ax)( 1 x + 3 2 x 2 5 2 x 3 ) = 1 x + 3 2 x 2 5 2 x 3 + ax ax 2 + 3 a 2 x 3 + 1 + a = 3 , a = 4 coef. of x 3 5 2 + 3 . 4 2 = 7 2 (4) 4. ( a ) when θ = π 2 , x = π + 1 and y = 0 d y d θ sin θ , d x d θ = 2 + cos θ d y d x = sin θ 2 + cos θ when θ = π 2 , gradient of tangent = 1 2 equation of tangent is y 0 1 2 (x π 1 ) 2 y x + π + 1 2 y + x = π + 1 (4) ( b ) At stationary points d y d x = 0 , sin θ = 0 θ = 0 ,π, 2 π θ = 0 ,x = 0 ,y = 1 θ = π, x = 2 y 1 θ = 2 x = 4 y = 1 There are stationary points at ( 0 , 1 ), ( 2 1 ), ( 4 1 ) (4)

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5. (i) x 0 123 ln ( 1 + sin x) 0 0.61056
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## Edexcel_WS_C4_PaperC - Worked Solutions Edexcel C4 Paper C...

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