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 Min pt is at x = 4  (3 – m) = 4 m = 7 M1 M1 A1 OR M1(complete the sq) M1 A1 (b) When m = 8, y = x2 – 10x + 13 + p Lies above x-axis, discriminant < 0  (– 10)2 – 4(13 + p) < 0 48 – 4p < 0 p > 12 B1 B1 8(a) ∠ EDC = ∠ DAC (alt seg thm) ∠ DAC = ∠ DCA (AD = CD, isos triangle) ∴ ∠ EDC = ∠ DCA Hence AC is parallel to DE (alt angles) B1 B1 B1 (b) ∠BCD = ∠BAD (angles in semicircle) Let ∠EDC = x ∠EDC = ∠CBD = x (alt seg thm) ∠DCA = ∠DAC = x (shown in part (a)) ∠DAC = ∠ADM = x (alt angle, AD parallel DE) ∠ADM = ∠ABD = x (alt seg thm) ∴ ∠ABD = ∠CBD By AA, triangle ABD is similar to triangle CBD. B1 A1 A1 9(a) Period = 2 p π 212 (shown)6 p p π π = = A1 (b) a = 6 b = 2 B2

4 No. Answers Marks (c) G1: Shape G1: period G1: max & min points (d) 03 00 to 09 00 or 3 a.m. to 9 a.m. 15 00 to 21 00 or 3 p.m. to 9 p.m. B1 B1 10(a) 2 65s t t dt= −+∫ 32 6 532 tt tc=− ++ When t = 0, s = 3 ∴ 3 23 533 ts tt=− ++ M1 A1 (b) When v = 0, t 2 – 6t + 5 = 0 t = 5 or t = 1 When t = 0, s = 3 When t = 1, s = 16 3 When t = 5, s = 16 3− Total dist = 16 163233  −+× = 13 m M1 A1 M1: for t = 1 or 5 M1 A1 (c) a = 2t – 6 At a = 0, t = 3, s = 0 Hence it is nearer to its initial starting position which is 3 m away compared to point B which is 16 3 m away. M1 M1 A1 11(a) Grad = –3 Eqn: 0 = –3(2) + c  c = 6 Equation of AD is y = –3x + 6 B1 B1

5 No. Answers Marks (b) D(0, 6) Midpoint of AD is (1, 3) Grad of bisector = 1 3 Eqn: 13 (1)3 d= +  d = 8 3 Equation of perpendicular bisector is 18 33yx= + M1 B1 A1 (c) 1811 3 33xx−=+ 8111 333 5 2 xx x −= + = y = 7 2 ∴ 57,22C  M1 A1 (d) BC = 1 2 AD B 57 1, 322  +− = B 71,22   (shown) A1 (e) Area = 752 021 22 172 0 6022 = 7.5 sq units M1 A1 12(a) 2 12 (4 5) dy dx x= − For curve to have stationary point, 0dy dx = . In this case, 2 12 0(4 5) dy dx x= ≠ − since 12 ≠ 0. Hence this curve does not have a stationary point. M1 A1 B1 (b) 2 2 12 4 (4 5) 3 (4 5) 9 152 (rej since )24 dy dx x x x or x = = − −= = > y = 1 ∴ P(2, 1) M1 M1 A1

6 No. Answers Marks (c) Eq of normal: 31 (2) 4 c= −+ 5 2c=  35 42yx= −+ When y = 0, x = 10 3 Area = 1 10 5 12 34 × −× = 25 24 or 1. 04 sq units M1 M1 A1 7(a) t 2 4 6 8 10 lg m 2.96 3.04 3.12 3.21 3.29 (b) lglg lg 3 bma t= + lg a = 2.87  a = 741 [accept 724, 733, 750] grad = lg 3.04 2.96 3 42 b −= − b = 1.32 M1 A1 A1 (c) lg m40 = t + 120 lg m = 0.025t + 3 ∴t = 7.5 weeks G1 (correct line drawn on grid) A1

7 lg m t 3.2 2.9 3.1 3.0 0 2 4 6 8 10 × × × × ×