Queensway Sec Prelim 2024 Add Math P1 Solutions

2024 4E5N Prelim P1 Solutions No. Solution 1(a) 𝑓(𝑥) = 3𝑥5 − 11𝑥3 + 30𝑥2 + 39 = (𝑥 − 1)(𝑥 + 3)𝑄(𝑥) + 𝑎𝑥 + 𝑏 When 𝑥 = 1, 61 = 𝑎 + 𝑏 ----- eqn (1) When 𝑥 = −3, − 123 = −3𝑎 + 𝑏 ----- eqn (2) (1) – (2) 184 = 4𝑎 𝑎 = 46 𝑏 = 15 1(b) 𝑓(𝑥) = (𝑥 − 1)(𝑥 + 3)𝑄(𝑥) + 46𝑥 + 15 𝑓(𝑥) − 3 = (𝑥 − 1)(𝑥 + 3)𝑄(𝑥) + 46𝑥 + 15 − 3 Remainder = 46𝑥 + 15 − 3 = 46𝑥 + 12 2(a) 𝑉 = 3 ( ℎ2 4 + 8𝜋 ℎ3) 𝑑𝑉 𝑑ℎ = 3 2 ℎ − 72𝜋 ℎ4 When ℎ = 4, 𝑑𝑉 𝑑ℎ = 6 − 9𝜋 32 = 192−9𝜋 32 (𝑜𝑟 5.1164) 𝑑ℎ 𝑑𝑡 = 𝑑ℎ 𝑑𝑉 × 𝑑𝑉 𝑑𝑡 𝑑ℎ 𝑑𝑡 = 32 192−9𝜋 × 35 = 1120 192−9𝜋 𝑐𝑚/𝑠 (𝑜𝑟 6.84 𝑐𝑚/𝑠) 2(bi) 𝑦 = 2−5𝑥 𝑒𝑥 𝑑𝑦 𝑑𝑥 = 𝑒𝑥(−5)−(2−5𝑥)𝑒𝑥 (𝑒𝑥)2 = 5𝑥−7 𝑒𝑥 For decreasing function, 5𝑥−7 𝑒𝑥 < 0 𝑥 < 7 5 (𝑜𝑟 1.4) 2(bii) When 𝑦 = 0, 2 − 5𝑥 = 0 𝑥 = 0.4 𝑑𝑦 𝑑𝑥 = −3.35 (3𝑠. 𝑓) 3(a) 1−3𝑥−3𝑥2 𝑥(𝑥+1)2 = 𝐴 𝑥 + 𝐵 𝑥+1 + 𝐶 (𝑥+1)2 1 − 3𝑥 − 3𝑥2 = 𝐴(𝑥 + 1)2 + 𝐵𝑥(𝑥 + 1) + 𝐶𝑥 When 𝑥 = −1, 𝐶 = −1 When 𝑥 = 0, 𝐴 = 1 When 𝑥 = 1, 𝐵 = −4 1−3𝑥−3𝑥2 𝑥(𝑥+1)2 = 1 𝑥 − 4 𝑥+1 − 1 (𝑥+1)2

2 3(b) ∫ 1−3𝑥−3𝑥2 2𝑥(𝑥+1)2 𝑑𝑥 = 1 2 ∫ 1−3𝑥−3𝑥2 𝑥(𝑥+1)2 𝑑𝑥 = 1 2 ∫ 1 𝑥 − 4 𝑥+1 − 1 (𝑥+1)2 𝑑𝑥 = 1 2 ln 𝑥 − 2 ln(𝑥 + 1) + 1 2(𝑥+1) + 𝑐 [Accept ln √𝑥 − ln(𝑥 + 1)2 + 1 2(𝑥+1) + 𝑐 ] 4(a) 42 + (ℎ − 8)2 = (𝑘 − 4)2 + 82 16 + ℎ2 − 16ℎ + 64 = 𝑘2 − 8𝑘 + 16 + 64 ℎ2 − 16ℎ = 𝑘2 − 8𝑘 ℎ2 − 𝑘2 = 16ℎ − 8𝑘 (shown) 4(bi) When ℎ = 1, 1 − 𝑘2 = 16 − 8𝑘 𝑘2 − 8𝑘 + 15 = 0 (𝑘 − 5)(𝑘 − 3) = 0 𝑘 = 5 𝑜𝑟 𝑘 = 3 (𝑟𝑒𝑗𝑒𝑐𝑡𝑒𝑑 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑑𝑖𝑎𝑔𝑟𝑎𝑚) Let 𝐴(0, 𝑦) (1 − 𝑦)2 = 52 + 𝑦2 1 − 2𝑦 + 𝑦2 = 25 + 𝑦2 𝑦 = −12 ∴ 𝐴(0, −12) 4(bii) Area = 1 2 | 0 −12 5 0 4 8 0 1 0 −12 | = 1 2 |44 − (−60)| = 52 𝑢𝑛𝑖𝑡𝑠2 5(a) Area = 1 2 (7)(7) sin 𝜃 + 1 2 (7)(5.6) sin(90 − 𝜃) + 1 2 (5.6)(8) sin 𝜃 + 1 2 (8)(8) sin(90 − 𝜃) = 49 2 sin 𝜃 + 98 5 cos 𝜃 + 112 5 sin 𝜃 + 32 cos 𝜃 = 51.6 cos 𝜃 + 46.9 sin 𝜃 (shown) 5(b) 𝑄 = 51.6 cos 𝜃 + 46.9 sin 𝜃 = 𝑅 cos(𝜃 − 𝛼) 𝑅 = √51.62 + 46.92 = 69.729 tan 𝛼 = 46.9 51.6 𝛼 = 42.268 ∴ 51.6 cos 𝜃 + 46.9 sin 𝜃 = 69.7 cos(𝜃 − 42.3°) 5(c) Max value of Q = 69.7 cos(𝜃 − 42.268°) = 1 𝜃 = 42.268 Corresponding value= 42.3° 5(d) maximum value of 1 𝑄2+3 = 1 0+3 = 1 3 :

3 6(a) 𝑣 = 4𝑒−𝑡 − 1 2 𝑒2𝑡 𝑎 = −4𝑒−𝑡 − 𝑒2𝑡 When 𝑡 = 0.5, 𝑎 = −5.14 𝑚/𝑠2 6(b) 𝑑𝑎 𝑑𝑡 = 4𝑒−𝑡 − 2𝑒2𝑡 When 𝑑𝑎 𝑑𝑡 = 0, 4𝑒−𝑡 = 2𝑒2𝑡 𝑒3𝑡 = 2 𝑡 = 1 3 ln 2 𝑑2𝑎 𝑑𝑡2 = −4𝑒−𝑡 − 4𝑒2𝑡 Whe