2023 SASS AMath Prelims P2 Ans

1 y = 2x + 5 … (1) y2 = px … (2) Sub (1) into (2): (2x + 5)2 = px 4x2 + 20x + 25 = px 4x2 + (20 – p)x + 25 = 0 Since the line and curve do not meet, D < 0  (20 – p)2 – 4425 < 0 400 – 40p + p2 – 400 < 0 p2 – 40p < 0 p (p – 40) < 0 0 < p < 40  greatest integer value of p = 39 [4] 2 23 4.5 10y x x    (a) When x = 0, y = 10.  Gabriel is 10 m above the water when he first leave the diving board. [1] (b) 23 4.5 10y x x    =  23 1.5 10x x   =   2 21.5 1.53 102 2x        = 23 ( 0.75) 0.562 5 10x       = 23( 0.75) 1.687 5 10x     y = 23( 0.75) 11.687 5x   [2] (c) Since 23( 0.75) 0x  , 11.687 5y Thus, Gabriel will not reach 12 m when executing the dive. [2] 3 35 4xyx (a) ddyx1213 5 4 3 (5 4 ) ( 4)25 4x x xx     = 63 5 45 45 4xxxx  = 3(5 4 ) 6(5 4 ) 5 4x xx x   = 315 6(5 4 )xx [3] (b) When x = 1, y = 3 and d9dyx.  gradient of the normal = 19  Equation of the normal is 13 ( 1)9y x    2819 9y x   [2] 4  22dsin 62dyxx   dsin 6 dd 2yx xx  =  1cos 66 2x c   When 4x, d1d 2yx,  1 1cos2 6c    c = 13  d1 1cos 6d 6 2 3yxx     1 1cos 6 d6 2 3y x x     11 1sin 636 2 3y x x c     When 4x, 1312y, 1131sin12 36 12c     1c  Equation of the curve is  1 1sin 636 2 3y x x     [6] 5 sin(5 ) 100y k t  (a) Since max blood pressure is 120 mmHg, k + 100 = 120  k = 20 and min blood pressure = –20 + 100 = 80 mmHg [2] (b) Period of y = 25= 0.4 min In 0 ≤ t ≤ 0.6, there are 1.5 cycle. [2] p 0 40 o o 2023 Additional Mathematics Preliminary Examination Paper 2 Solutions

(c) 20 sin(5 ) 100 100t  sin(5 ) 0t 5 0, , 2t   t = 0, 0.2, 0.4 Thus for y to be at least 100 mmHg, 0 ≤ t ≤ 0.2  Duration = 0.2 min [2] 6 (a) LHS = 23cos cos 2 1cos 2cosx xx x  = 23cos (2cos 1) 1cos (cos 2)x xx x   = 22cos 3cos 2cos (cos 2)x xx x  = (2 cos 1)(cos 2)cos (cos 2)x xx x  = 2cos 1cosxx = 2cos1cos cosxx x = 2 – sec x [3] (b) 223cos cos 2 1tan seccos 2cosx xx xx x   22 sec tan secx x x   22 sec (sec 1) secx x x    2sec 2sec 3 0x x   (sec x +3)(sec x – 1) = 0 sec x = –3 or sec x = 1 1cos or cos 13x x   basic angle = 70.528 or basic angle = 0 For –180 ≤ x ≤ 180, x = ± (180 – 70.528) or x = 0  x = –109.5, 0 or 109.5 [3] 7 (a) 1253 325nn nn  25 53 3 325nn nn     253 3 1 525nn   53 85nn  53 58n n  5158n [3] (b) 2 4322log (3 2) log 1log 3x x      222232log 12log (3 2)log 4log 3xx   222log ( 1)2log (3 2)2 2xx   22 22log (3 2) log ( 1) 2x x    222(3 2)log 21xx 222(3 2)21xx 2 29 12 4 4( 1)x x x    25 12 0x x  (5 12) 0x x  120 or5x x  But 23x,  125x [4] 8 (a) TUQ = SUR (vertically opposite s) RSU = SPQ (al