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 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 =− ( 2 x + ln y) d y d x ( 2 x + ln ± x y + 1 ² at (3 , 1) gradient ± 6 + 0 3 + 1 ² 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 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 = 0 48 x 2 48 x = 0 48 x(x + 1 ) = 0 x = 0o r 1 (5) 3. ( a )a t Py = 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) = h 2 t sin t i π 2 0 2 # π 2 0 sin t d t = h 2 t sin t + 2 cos t i 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 ² ³ 9 x 2 ´ + ± 1 2 ²± 3 2 ² 2 ³ 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 1 2 × ( 1 + 3 1 2 ( 1 + 3 1 2 = 1 + 3 x p ( 1
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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_PaperK - Worked Solutions Edexcel C4 Paper K...

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