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       25 19 23, , ,12 12 12 12    5 19 23, , ,24 24 24 24     [4] 9 (a) 7( )(2 )a x x  = 6 5 27 77( ) ( ) ...1 2( ) 2 2 2x xa x                        = 2( ) 128 448 672 ...a x x x       Since coefficient of x2 = 616, 672a + 448 = 616 672a = 168 14a [4]  a 1

(b) 2412nxx     55426152nnT xx       Since 6T is a constant, then power of x = 0,  2(n – 5) + 5(–4) = 0 2n = 30 n = 15  10245615152T xx        =  202013 00332xx    = 3 00332 [4] 10 150000ptA e (i) When t = 0, A = 150 000 Amount paid for the car = $150 000 [1] (ii) When t = 24, A = 120 000,  24150 000e 120 000p 24120 0004e150 000 5p  424 ln5p  1 4ln24 5p = 0.009 297 After 40 months, A = 0.009 297 40150 000e  = 103 415.545 = 103 416 (nearest dollar)  The value of the car = $ 103 416. [3] (iii) When the value drops to $ 60 000, 0.009 297150 000e 60 000t 0.009 29760 0002e150 000 5t  20.009 297 ln5t  1 2ln0.009 297 5t  = 98.55  The car is 99 months. [2] (iv) Graph of A against t [[2][ 11 (a) 21lnxyx   for 0 < x < 1 ln(1 ) 2 lny x x   d1 2d 1yx x x   ddyx = 2(1 )(1 )x xx x   = 2(1 )xx x For 0 < x < 1, x – 2 < 0 1 – x > 0 x > 0  ddyx= ( )( )( )   ddyx < 0  y is a decreasing function. [5] (b) Surface area of cylindrical ice block, A = 2π r h + 2π r2 = 2π r (2r) + 2π r2 = 6π r2 d12dArr Using chain rule, d d dd d dA A rt r t   d24 12drrt   h = 8  r = 4 d24d 12 (4)rt  = 12 = –0.159 (3 s.f.) Thus, r is decreasing at 0.159 cm/s. [4] A t –150 000 – [2]

12 L2 : 112 9yx  (a) When x = 0, y = 9. When y = 0, x = 12.  C = (0, 9) and D = (12, 0) Midpoint of CD, A = 0 12 9 0,2 2  =  96,2 Gradient of CD = 9 00 12= 34 gradient of L1 = 43  equation of L1, is  9462 3y x   When x = 0, y = 9 782 2     B =  70,2 [3] (b) Since gradient of L1 = 43, 4tan3   = 53.1 [2] (c) A=  96,2, B =  70,2 and D = (12, 0) Area of triangle ABD = 12 6 0 1219 720 02 2 =    154 21 0 0 0 422       = 37.5 units2 [2] (d) E lies on the x-axis, y = 0. By similar triangles, 129272AEAEEB EE = 97  AE : EB = 9 : 7 [2] 13 (a) AB = CD = 2CM OCM : sin2CM  CM = 2 sin  AB = 4 sin  AD = 2AN OAN :  2sin2 2AN   2sin2AN  = 2 cos  AD = 4 cos  P = AB + 2BC + 2OC = 4 sin 2(4 cos ) 2(2)   = 4(2 cos sin 1)   (shown) [3] (b) Let 2 cos sin cos( )R       2 22 1 5R   1tan 26.5652     2cos sin 5 cos( 26.6 )      [2} (c)  4 5 cos( 26.565 ) 1P        P = 12.5 4 5 cos( 26.565 ) 4 12.5    12.5 4cos( 26.565 )4 5   basic angle = 18.134 26.565 18.134      = 44.699 or 8.431   = 44.7 or 8.4 [3] (d) max P =  4 5 1 = 12.944 < 13  Nicholas’ claim is correct. [2]  A B  E(x, 0)  E1 E2 A B C D O 2 2 M N