2024 TPSS AMATH P1 PRELIM MS

1 Tampines Secondary School Sec 4&5 Express Additional Math Paper 1 2024 Marking Scheme Total Marks: 90 √ = follow through No. Answers Marks 1 2 2 2 2 2 2 1 ( 1)(4 ) (2 )(1) ( 1) 24 ( 1) xy x dy x x x dx x xx x = + +−= + += + Decreasing function  0dy dx < Since (x + 1)2 > 0, 2x 2 + 4x < 0 2x(x + 2) < 0 ∴−2 < x < 0 M1 A1 M1 A1 2(a) –20x2 + 120x + 3 = –20(x2 – 6x) + 3 = –20[(x – 3) 2 – 32] + 3 = –20 (x – 3) 2 + 183 M1 A1 (b) h1 (0) = 3 h2 (0) = 6.6 TP-1 was fired from a height of 3 metres above ground while TP-2 was fired from a height of 6.6 metres above ground. B1 B1 (c) From TP-1’s max pt (3, 183) and TP-2’s max pt (6, 183), they both reach the same height. TP-1: h = 0  x = 6.02 m TP-2: h = 0  x = 12.1 m > 6.02 m Since TP-2 could reach a further distance from the launched position, compared to TP-1, TP-2 should be acquired. B1 M1 B1 3 Height = 2 2 (2 1) (2 1) x xx + − = 2 22 4 41 ( 21 ) 21 x x AB C x x xx x ++ = ++−− 4x2 + 4x + 1 = Ax(2x – 1) + B(2x – 1) + Cx2 Let x = 0, B = –1 Let x = 1 2 , 4 = 1 4 C  C = 16 Compare coeff of x2, 4 = 2A + 16  A = –6 ∴Height = 2 16 6 1 21x xx −−− M1 A1 M1: either sub mtd or compare coeff A3

2 No. Answers Marks 4a(i) (2 + qx)6 = 64 + 192qx + 240q2x2 + … B3 (ii) (2 + px)(2 + qx)6 = (2 + px)( 64 + 192qx + 240q2x2 + …) Term in x: 384q + 64p = 0  p = –6q Term in x2: 480q 2x2 + 192pqx2 = −168 480q 2 + 192(–6q)q = −168 –480 q2 = −168 11 (rej since 0)22q or q q= = −> p= −3 M1 M1 A1 A1 (b) Tr + 1 = 12Cr (x3)12 − r(−2)r(x−1)r 36 – 3r – r = 0 r = 9 Term = 12C9 (−2)9 = −112 640 M1 M1 A1 5(a) 318'( ) 3 xefx c − = +− = −6e−3x + c When x = 0, f ′(x) = 2, −6e−3(0) + c = 2 c = 8 Stationary point, f ′(x) = −6e−3x + 8 = 0 6e −3x = 8 3 4 3 xe− = x = 14ln33− When x = 14ln33− , f ′′(x) = 18e−3x = 24 > 0  point is minimum M1 M1: subt x = 0 & f’(x) = 2 M1 A1 M1 A1 (b) f ′(x) = −6e−3x + 8 f(x) = 36 3 xe−− − + 8x + d 328 xe xd−= ++ Subt 3 21,e   , 3 3 2 28ede −= ++  d = –8 Hence eqn of curve is 3() 2 8 8 xfx e x −= +− M1 A1

3 No. Answers Marks 6(a) 2 62dy xmdx = +− At x = 4, 0dy dx =  14 – 2m = 0 m = 7 OR y = (x + 3 – m)2 – (3 – m)2 + m + 5