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 (alternate segment thm) = TQP (base s of isosceles ) = QTU (alternate angles) TQU and SRU are similar s (AA) [3] (b) Since T is the midpoint of PS and PQ//TU, then by the converse of Midpoint Theorem, U is the midpoint of QS and PQ = 2TU and U is the midpoint of TX  TX = 2TU = PQ i.e. TX // PQ and TX = PQ PQXT is a parallelogram. [2] (c) From (a), RSU = QTU, this indicates the property of angles in the same segment. Thus, there is a circle passing through the points Q, R, S and T. [1] X

9 (a) Surface area of test tube: 22 2 12rh r   26rh r  26rhr [2] (b) V olume of test tube, 3 223V r r h   232236rV r rr     3 2)23(6V r r r    336V r r  [2] (c) 2d6dVrr  = 2(6 )r For stationary value of V, d0dVr.  2(6 ) 0r  ( 6 )( 6 ) 0r r   Since r > 0,  6rcm Stationary value of V = 36 6 63 = 6 6 6 63  = 4 6cm3 22d2dVrr  < 0 since r > 0 Thus, V is a maximum. [4] 10 (a) nx y k n lg x + lg y = lg k  lg y = –n lg x + lg k lg x 0.30 0.60 0.78 0.90 lg y 0.93 0.78 0.69 0.63 (b) –n = gradient of line = 1.08 0.620 0.9 = –0.511  n = 0.51 lg k = lg y – intercept = 1.08  k = 1.0810 = 12.02 [3] (c) When x = 5, lg x = 0.70 lg y = 0.72  y = 0.7210 = 5.25 [2] 11 (a) Let the centre of the circles be (a, a) since it lies on the line y = x. Distance of centre from (0, –3) = 5 2 2( 3) 5a a   2 26 9 5a a a    22 6 4 0a a   23 2 0a a   ( 1)( 2) 0a a   1 or 2a a     The centres of C1 and C2 are (–1, –1) and (–2, –2) respectively. [3] (b) C1 : 2 2( 1) ( 1) 5x y    gradient of line joining centre and (1, 0) = 0 111 1 2  gradient of tangent at (1, 0) = –2  equation of tangent at (1, 0) is 0 2( 1)y x    2 2y x   [2] (c) C1 : 2 2( 1) ( 1) 5x y    C2 : 2 2( 2) ( 2) 5x y    Since the point of intersection lies on the x-axis, y = 0.  2 2( 1) 1 5 ( 1) 4x x       2 2( 2) 4 5 ( 2) 1x x       1 2 1 or 3x x       2 1 1 or 3x x       the point of intersection is (–3, 0). [3] h + r r 0.2 0.4 0.6 0.8 0.2 0.4 0.6 0.8 lg y lg x 0     1.0

(d) Greatest distance between P and Q = distance between the 2 centres +2 5 = 2 2( 1 2) ( 1 2) 2 5      = 2 2 5 units. [2] 12 2369( 3)vt  (a) a = 3d36( 2)( 3)dvtt   = 372( 3)t When t = 0, a = 72 8( 27) 3  Initial acceleration = 83m/s2 [2] (b) When the particle is at rest, v = 0. 2369 0( 3)t   236 9( 3) 0t   2( 3) 4t  3 2t   5 or 1t t   The particle is at instantaneous rest when t = 1 or 5 s. [2] (c) 2369 d( 3)s tt  = 136( 3) 9t t c    = 3693t ct   When t = 0, s = 0.  0 = 12 + c  c = –12  369 123s tt    s = 0 369 12 03tt     (3 4)( 3) 12t t     23 5 12 12t t     23 5 0t t   (3 5) 0t t   t = 0 or 53t Thus the particle returns to O only once at 53t. [3] (d) t = 0: s = 0 v = 0, t = 1: s = 369 122  = –3 m t = 2: s = 3618 121  = 6 m Distance travelled in the first 2 seconds = 32 + 6 = 12 m [3] 13 At C(k, k – 3), k – 3 = (9 – k)(k – 3) (k – 3)(1 – 9 + k) = 0 (k – 3)(k – 8) = 0 Since k > 0, k = 8. A = (6, 0) B = (9, 0) Area of shaded region = Area of rectangle + Area under curve = (8 – 6)(8 – 3) + 98(9 )( 3) dx x x  = 2  5 + 92812 27 dx x x   = 10 + 93286 273xx x       = 10 +  243 6 81 27 9     5126 64 27 83         = 10 + 0 –  83 = 383units2 [8] 0 6 –3 o o   