SPS AM Prelim Ans

Sec 4 Exp AM Prelim Exam Paper 1 Marking Scheme 1(a) (x + 3)(x – 1) f(1) = 0 1 + 4 + 2 + a – b = 0 a – b = –7 ---------(1) f(–3) = 0 (–3)4 + 4(–3)3 + 2(–3)2 + a(–3) – b = 0 81 – 108 + 18 – 3a – b = 0 –3a – b = 9 -------(2) (1) – (2) : 4a = –16 a = –4 b = 3 [M1] [M1] [A1] [A1] 1(b) Long division x2 + 2x + 1 x2 + 2x –3 x4 + 4x3 + 2x2 – 4x  3 (x4 + 2x3 – 3x2) 2x3 + 5x2 – 4x  3 (2x3 + 4x2 – 6x) x2 + 2x  3 ( x2 + 2x  3) 0 f(x) = (x2 + 2x – 3)(x2 + 2x + 1) Other factor is x2 + 2x + 1. [M1] [A1] Qn 1 (a) a = –4, b= 3 (b) the other factor: x2 + 2x + 1   

2(a) Let the perpendicular distance be x.       1 4 2 8 182 4 2 8 36 36 8 4 2 36 8 4 2 8 4 2 8 4 2 288 144 2 64 32 992 2 x x x cm          [M1] [M1] [A1] 2(b) sin 45 9921 2 2 92 9 2 2 99 2 (2)2 9 9 2 x PS PS PS         [M1] [A1] Qn 2 (a)   992 2 cm (b) PS =  9 9 2 cm

3(a)       23 2 3 23 3 5 3 11 2 3 23 1 1 1 (shown) xx x x xx x x x dy eedx e e ee e e e        [M1] [M1] 3(b)   5ln 2 3ln 2 at , 1 33 8 8 gradient of normal 33 equation of normal is 55 8 ln 224 33 8 8 55 ln 233 33 24 P dy e dx e yx yx          [M1] [M1] [A1] 3(c) 8 550, ln 233 24Q  [A1] Qn 3 (b) 8 8 55 ln 233 33 24yx   (c) 8 550, ln 233 24Q 

4(a)       2 2 6 18 let 0, 6 18 0 6 3 0 0 or 3 13 14 0,13 and 3, 14 dy xxdx dy dx xx xx xx yy         [M1] [M1] [A2] 4(b)     2 2 2 2 2 2 12 18 at 0, 18 0 0,13 is maximum point. at 3, 18 0 3, 14 is minimum point. dy xdx dyx dx dyx dx          [M1] [A1] [A1] Qn 4 (a) (0, 13) and (3, 14) (b) (0, 13) max point; (3, 14) min point.

5(a)     2 cos 2 2 sin 2 2 2cos 2 2cos 2 4 sin 2 d x x x x xdx x x x     [M1] [A1] 5(b)      2 2 00 2cos 2 4 sin 2 2 cos 2 2cos 2 4 sin 2 2 cos 2 4 sin 2 2cos 2 2 cos 2 sin 22 2 cos 22 sin 2 2 cos 2 so, 1 4 sin 2 sin 2 2 cos 2 2 2 x x x dx x x c x dx x x dx x x c x x dx x dx x x c x x x c x x x c x x dx x x x x                               [M1] [M1] [M1] [M1] [A1]

6(a)                 23 2 2 2 3 2 5 2 2 5 2 5 6 12 2 5 2 5 8 17 dy x x xdx x x x xx               [M1] [A1] 6(b)         2 2 0, so 2 5 8 17 0 since 2 5 0 for all values of , then 8 17 0. 8 17 0 17 8 dy xxdx xx x x x         [M1] [A1] 6(c) 2 2 0.35 (2 5) (8 17) when 3, 0.35 7 0.35 7 0.05 units/s dy dy dx dt dx dt dxxx dt x dx dt dx dt           [M