Edexcel_WS_C4_PaperK

# Edexcel_WS_C4_PaperK - Worked Solutions Edexcel C4 Paper K...

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Worked Solutions Edexcel C4 Paper K 1. ( a ) V = 4 3 πr 3 , d V d r = 4 πr 2 (1) ( b ) d V d t = d V d r × d r d t given r = 10 and d r d t = 0 . 1 , d V d t = 4 π × 10 2 × 0 . 1 = 40 π cm 3 s 1 (4) 2. ( a ) 2 x + x · 1 y d y d x + ln y + d y d x = 0 d y d x x y + 1 = − ( 2 x + ln y) d y d x = − ( 2 x + ln y) x y + 1 at (3 , 1) gradient = − 6 + 0 3 + 1 = − 3 2 (3) ( b ) 6 x + 2 x d y d x + y · 2 10 y d y d x + 16 d y d x = 0 d y d x ( 2 x 10 y + 16 ) = − ( 6 x + 2 y) d y d x = 0 when 6 x + 2 y = 0 or y = − 3 x substitute y = − 3 x into equation of curve, 3 x 2 + 2 x( 3 x) 5 ( 9 x 2 ) + 16 ( 3 x) = 0 48 x 2 48 x = 0 48 x(x + 1 ) = 0 x = 0 or 1 (5) 3. ( a ) at P y = 0. cos t = 0 t = π 2 at t = π 2 , x = π 2 4 coordinates of P are π 2 4 , 0 (2) ( b ) (i) area = y d x d t d t = π 2 0 cos t · 2 t d t (2) (ii) A = π 2 0 2 t d d t ( sin t) d t (By parts) = 2 t sin t π 2 0 2 π 2 0 sin t d t = 2 t sin t + 2 cos t π 2 0 = π + 0 ( 0 + 2 ) = π 2 (5)

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4. ( a ) f (x) = ( 1 9 x 2 ) 1 2 = 1 + 1 2 9 x 2 + 1 2 3 2 2 9 x 2 2 = 1 + 9 2 x 2 + 243 8 x 4 (3) ( b ) valid for | 9 x 2 | < 1 | x 2 | < 1 9 , | x | < 1 3 (1) ( c ) (i) ( 1 + 3 x) 1 2 ( 1 3 x) 1 2 × ( 1 + 3 x) 1 2 ( 1 + 3 x) 1 2 = 1 + 3 x ( 1 9 x 2 ) (2) (ii) ( 1 + 3 x) 1
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