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           [M1] [A1] 6(d) 2 3 3 2 2 2( 2)(2 5) 3, 2(1)(1) 2 zy dz ydy dz dz dy dt dy dt dyy dt dyxx dt when x dz dy dt dy dy dt            [M1] [A1] Qn 6 (a)     2 2 5 8 23xx (b) 23 8x (c) 0.05 units/s2

7(a) 1 1 3 2 3 122 5125 25 55 5 55 3 12 2 5 125 0.3807 0.381 (3 . ) x x x x x x x xx x sf              [M1] [M1] [A1] [A1] 7(b) 2 16 2 2 16 2 2 1 4 22 3 4 2 2 log log loglog log loglog 4 log log log 3 log4 3 lg 3lg 4 lg 2 4lg 2 xx xx x xx xx x x xx         [M1] [M1] [A1] [M1] Qn 7 (a) 0.381

8(a)   1212 1 3 12 2 r r rrT C x x     [B1] 8(b)       1212 1 3 12 12 412 4 4 12 8 2 12 2 2 Term with : 12 4 4 2 Coef. of (2) 16896 r r rr r r r r T C x x Cx xr r xC            0 0 12 6 3 Term with : 12 4 0 3 Coef. of (2) 14080 xr r xC    Term independent of x = 1(16896) + (1)(14080) = 2816 [M2] [M1] [M1] [A1] 8(c) n = 12 – 4r 4r is even => n = 12 – 4r is even. Therefore, there are no terms containing xn whereby n is odd. [M1] [A1] Qn 8 (a)   1212 1 3 12 2 r r rrT C x x     (b) 2816 (c) There are no terms containing xn whereby n is odd.

9(a) 2 2 2 2 sec 1 sec 1 1 1sec 1 cos 1 1 cos 2 1 12 2 cos 2 3 xLHS x x x x RHSx            [M1] [M1] [A1] 9(b) 2 2 sec 2 4 1 sec 2 7 24 cos 4 3 7 14 4cos 4 12 1cos 4 2 3 5 7 114 , , , 3 3 3 3 5 7 11, , ,12 12 12 12 x x x x x ref x x              [M1] [M1] [A1] [A1] 9(c) 2 2 8 4 8 No. of solutions = 16 xx          [A1] Qn 9 (b) 5 7 11, , ,12 12 12 12     (c) 16 solutions

10(a) cos 4cos4 sin 7sin 7 7sin 4cos OB OB ADO OA OA AB OA OB                 [M1] [A1] 10(b) 22 1 sin( ) 7 4 65 4tan 29.7447 65 sin( 29.7 ) AB R R AB              [M1] [M1] [A1] 10(c) Max length of AB = 65 Max occurs when sin( 29.744 ) 1   sin( 29.744 ) 1 90 29.744 90 90 29.744 60.256 60.3 (1 ) ref dp              [A1] [M1] [A1] 10(d) 65 sin( 29.744 ) 55 55sin( 29.744 ) 65 66.906 29.744 66.906 66.906 29.744 37.162 37.2 (1 ) ref dp                 [M1] [M1] [A1] Qn 10 (b) 65 sin( 29.7 )AB   (c) Max AB = 65 ;  = 60.3 (d)  = 37.2