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### Edexcel_WS_C4_PaperL

Course: MATH C4, Spring 2010
School: Cambridge College
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Word Count: 398

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Solutions Worked Edexcel C4 Paper L 1. 1 dy = 2x dx y ln y = x 2 + c y = e, x = 1: ln e = 1 + c c=0 ln y = x 2 y=e x2 3. (a) (, 0) (b) dy = x cos x + sin x dx dy 0.02 when x = 2.02, dx dy -0.03 when x = 2.04, dx (c) Area = x sin x dx = 0 0 (1) (4) d x (- cos x)dx = -x cos x - 0 dx 0 (- cos x) dx 0 y (4) = -x cos x + sin x 1 = (4) 1 x 4. (2) 0 1 172 - 1 2 17 (a) 17 (17 - 1)(17 + 1) 2 = 16...

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Solutions Worked Edexcel C4 Paper L 1. 1 dy = 2x dx y ln y = x 2 + c y = e, x = 1: ln e = 1 + c c=0 ln y = x 2 y=e x2 3. (a) (, 0) (b) dy = x cos x + sin x dx dy 0.02 when x = 2.02, dx dy -0.03 when x = 2.04, dx (c) Area = x sin x dx = 0 0 (1) (4) d x (- cos x)dx = -x cos x - 0 dx 0 (- cos x) dx 0 y (4) = -x cos x + sin x 1 = (4) 1 x 4. (2) 0 1 172 - 1 2 17 (a) 17 (17 - 1)(17 + 1) 2 = 16 18 = 2 17 17 = 4 18 = 4 9 2 = 12 2 (3) 1 dx (a) V = y 2 dx = (t 2 + 1)2 dt dt -2 0 0 2. dx =1 dt 1 1 2 (b) (1 - x) 2 = 1 + (-x) + - 2 1 2 (5) 2 1 1 (-x)2 = 1 - x - x 2 - . . . 2 8 1 1 577 2 172 578 (2) (b) -2 t2 + 1 2 1 1 (c) put x = 2 , 1 - 2 17 17 t 4 + 2t 2 + 1 dt 0 -2 1 2 1- dt = -2 17 577 578 12 2 = 2 t5 + t3 + t 5 3 17 577 2 12 578 = 206 15 (2) 2 577 408 34 (5) = 0- 2 1 (-32) + (-8) + (-2) 5 3 5. (a) d x dx 1 x 1 sin kx dx = sin kx - sin kx dx k k k = 1 x sin kx + 2 cos kx + c k k = 1 ( - 2) 8 1 - (d) AP = 2 4 (4) - - - - As |OQ| = |OA| we have AP = PQ 1 2 position vector of Q is 1 + 2 4 -1 3 = 3 3 7. (a) dy dy - 2x + x +y =0 dx dx dy (1 + x) = 2x - y dx 2x - y dy = dx 1+x 9 1, 2 9 2 = -5 = 1+1 4 2- (2, 1, 1) P Q (b) 4 1 1 x sin 2x + cos 2x = sin + 0 - 0 + 2 4 8 2 4 0 (4) A (1, 5) 1, O (c) cos 2x = 2 cos2 x - 1 4 2 cos2 x = cos 2x + 1 4 (4) 2x cos2 x dx = 0 0 (x cos 2x + x)dx 4 0 x2 1 = ( - 2) + 8 2 2 1 = ( - 2) + 8 32 (4) (6) 6. 3 1 3 - (a) AB = 6 , line through A and B is r = -1 + 6 12 -5 12 3 1 2 (b) If P lies on AB 1 = -1 + 6 12 -5 -1 The equation is satisfied for x, y, z with = 3 2 - - (c) OP AB = 1 6 = 6 + 6 - 12 = 0 12 -1 OP is perpendicular to AB 1 3 (2) (b) Gradient at (1) (c) dy = 0 where y = 2x (from (a)) dx substitute y = 2x into y - x 2 + xy = 8 (2) 2x - x 2 + x(2x) = 8 x 2 + 2x - 8 = 0 (x + 4)(x - 2) = 0 x = -4, 2 (2) stationary points are (-4, -8) and (2, 4) (4) Ed C4 paper L 8. (a) dy (x + 1) 1 - x 1 1 = = 2 dx (x + 1) (x + 1)2 at x = 1, 1 dy = dx 4 let u = x + 1 du =1 dx when x = 1, x = 0, (3) area (1) = 1 u=2 u=1 2 2 gradient of normal is -4 1 = -4(x - 1) 2 y = -4x + 4 1 2 equation of normal is y - u-1 du = u 2 1 1- 1 1 du u (b) normal cuts x-axis where y = 0 i.e. 9 4x = 2 x= 9 8 1 2 (1, 0) y P 1, 1 2 Q 9 ,0 8 x = u - ln u = 1 - ln 2 = 2 - ln 2 - (1 - ln 1) 1 area (2) = 2 base height of = 1 1 1 1 = 2 8 2 32 1 32 (8) shaded area = area (1) + area (2) 1 shaded area = 1 - ln 2 + area (1) = 0 x dx x+1 = 0.338 units2 Ed C4 paper L
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