2023 SASS AMath Prelims P1 Ans

1 222 3 4(2 )(2 )x xx x  = 222Bx CAxx Multiplying by 2(2 )(2 )x x , 2 22 3 4 (2 ) ( )(2 )x x A x Bx C x       Sub x = 2: 24 + 32 + 4 = A(2 + 4) 6A = 18  A = 3 Sub x = 0: 4 = 2A + 2C 2C = 4 – 6  C = –1 Comparing coefficients of x2, A – B = 2  B = 1 222 3 4(2 )(2 )x xx x  = 23 122xxx [4] 2 2242( 1)dxxx =22422 1dx xxx  =2422 11 dxxx    = 4212lnx xx     = 1 14 2ln 4 2 2ln 24 2              = 15 32(2ln 2) 2ln 24 2            = 92ln 24 [4] 3 y = a + 2 tan bx (a) When x = 0, y = –4 a = –4 Since period of y = 480, 180480b  180 3480 8b   34,8a b   [3] (b) Graph of 34 2 tan8y x   4 (a) Equation of the straight line is 9 55 ( 6)8 6Y X   Y = 2X – 7 ln y = 2 ln x – 7 ln y – 2 ln x = –7 27eyx 7 2ey x [4] (b) When Y = –1, X = m, 2m – 7 = –1 m = 3 [1] 5 (a)  23 5 8 5 60y x x     d2 3 5 8 5dyxx   For stationary points, d0dyx.  2 3 5 8 5 0x    8 52 3 5x  4 5 3 53 53 5x   4 5 3 59 5x 3 5 5x  [3] (b)    223 5 3 5 3 5 5x    =   3 5 45 30 5 25   =   3 5 70 30 5  =  210 150 5 70 90   = 60 20 5 y =   60 20 5 8 5 3 5 5 60    =   60 20 5 120 40 5 60    = 20 5 [2] 2ln 7yx  y x – 4 480 240 [2] 2023 Additional Mathematics Preliminary Examination Paper 1 Solutions

6 sin 4y x x (a)  dsin 4 4 cos 4dyx x xx  = sin 4 4 cos 4x x x [2] (b) Integrating,  3 3344 4sin 4 d 4 cos 4 dy x x x x x      333444cos 4sin 4 4 cos 4 d4xx x x x x      3344cos 44 cos 4 d sin 44xx x x x x      3431 1 14 cos 4 d ( 1)3 2 4 2 4x x x                 = 31 16 8 4   = 318 6 [4] 7 Since x – 2 is a factor, P(2) = 0  8a – 36 + 2b – 6 = 0  4a + b = 21 … (1) Since P(x)/x – 3, R = 66, P(3) = 66  27a – 81 + 3b – 6 = 66  9a + b = 51 … (2) (2) – (1): 5a = 30  a = 6 (1): b = 21 – 24  b = –3  a = 6, b = –3 [4] (b) 3 2P( ) 6 9 3 6x x x x    = (x – 2)(6x2 + kx + 3) Comparing coefficients of x: 3 – 2k = –3 k = 3 P(x) = (x – 2)(6x2 + 3x + 3) = 3(x – 2)(2x2 + x + 1) [2] (c) 3 29 6 0ax x bx    3 29 6 0ax x bx     3 2( ) 9( ) ( ) 6 0a x x b x        P(–x) = 0 Since P(x) = 0  x = 2, Then P(–x) = 0  –x = 2  x = –2 [2] 8 (a) sina where 0 <  < 180 (i) cos21a  [2] (ii) cot  = 21aa  [1] (iii) sin (90 – ) = cos  = 21a  [1] (b)  1 1sec 22 43  for –π < ϕ < π  2sec 243   3cos 24 2  basic angle = 6 2 2 24 4 4          24 , , 2 , 26 6 6 6      