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 When 𝑡 = 1 3 ln 2, 𝑑2𝑎 𝑑𝑡2 < 0 (max) 6(c) When 𝑣 = 0, 4𝑒−𝑡 = 1 2 𝑒2𝑡 𝑒3𝑡 = 8 𝑡 = 1 3 ln 8 = ln 8 1 3 = ln 2 (shown) 6(d) 𝑠 = −4𝑒−𝑡 − 1 4 𝑒2𝑡 + 𝑐 When 𝑡 = 0, 𝑠 = 0, ∴ 𝑐 = 17 4 𝑠 = −4𝑒−𝑡 − 1 4 𝑒2𝑡 + 17 4 When 𝑡 = 0, 𝑠 = 0 When 𝑡 = ln 2 , 𝑠 = 1.25 When 𝑡 = 3, 𝑠 = −96.806 Total distance travelled = 1.25 + (96.806 + 1.25) = 99.3𝑚 7(a) cot 𝐴−tan 𝐴 cot 𝐴+tan 𝐴 = 2𝑐𝑜𝑠2𝐴 − 1 𝐿𝐻𝑆 = cot 𝐴−tan 𝐴 cot 𝐴+tan 𝐴 = cos 𝐴 sin 𝐴 − sin 𝐴 cos 𝐴 cos 𝐴 sin 𝐴 + sin 𝐴 cos 𝐴 = 𝑐𝑜𝑠2𝐴−𝑠𝑖𝑛2𝐴 𝑐𝑜𝑠2𝐴+𝑠𝑖𝑛2𝐴 = 𝑐𝑜𝑠2𝐴 − 𝑠𝑖𝑛2𝐴 = 𝑐𝑜𝑠2𝐴 + 𝑐𝑜𝑠2𝐴 − 1 = 2𝑐𝑜𝑠2𝐴 − 1 = 𝑅𝐻𝑆 Alternative 𝐿𝐻𝑆 = cot 𝐴−tan 𝐴 cot 𝐴+tan 𝐴 = 1 tan 𝐴 −tan 𝐴 1 tan 𝐴+tan 𝐴 = 1−𝑡𝑎𝑛2𝐴 1+𝑡𝑎𝑛2𝐴 = 1−𝑡𝑎𝑛2𝐴 𝑠𝑒𝑐2𝐴 = 𝑐𝑜𝑠2𝐴 − 𝑠𝑖𝑛2𝐴 = 2𝑐𝑜𝑠2𝐴 − 1 = 𝑅𝐻𝑆

4 7(b) cot 𝐴−tan 𝐴 cot 𝐴+tan 𝐴 = cos 𝐴 −𝜋 < 𝐴 < 𝜋 2𝑐𝑜𝑠2𝐴 − 1 = cos 𝐴 (2 cos 𝐴 + 1)(cos 𝐴 − 1) = 0 cos 𝐴 = − 1 2 𝑜𝑟 cos 𝐴 = 1 𝑟𝑒𝑓 𝑎𝑛𝑔𝑙𝑒 = 𝜋 3 𝑜𝑟 𝑟𝑒𝑓 𝑎𝑛𝑔𝑙𝑒 = 0 𝐴 = 2𝜋 3 , −2𝜋 3 𝑜𝑟 𝐴 = 0 𝐴 = −2𝜋 3 , 0, 2𝜋 3 8(a) 𝑦 = tan 𝑥 𝑑𝑦 𝑑𝑥 = 𝑠𝑒𝑐2𝑥 8(b) When 𝑥 = 𝜋 6 , 𝑦 = √3 3 , 𝑑𝑦 𝑑𝑥 = 4 3 𝑦 = 𝑚𝑥 + 𝑐 √3 3 = 4 3 ( 𝜋 6) + 𝑐 𝑐 = √3 3 − 2𝜋 9 𝑦 = 4 3 𝑥 + √3 3 − 2𝜋 9 Accept (𝑦 = 4 3 𝑥 + 3√3−2𝜋 9 ) 𝑜𝑟 (9𝑦 = 12𝑥 + 3√3 − 2𝜋) 8(c) 𝑑2𝑦 𝑑𝑥2 = −2𝑐𝑜𝑠−3𝑥(− sin 𝑥) = 2 sin 𝑥 𝑐𝑜𝑠3𝑥 At 𝑥 = 𝜋 6 , 𝑑2𝑦 𝑑𝑥2 = 1.5396 = 1.54 (3𝑠. 𝑓) 8(d) 𝑑𝑦 𝑑𝑥 = 𝑠𝑒𝑐2𝑥 When 𝑑𝑦 𝑑𝑥 = 0, 1 𝑐𝑜𝑠2𝑥 = 0 Since 𝑠𝑒𝑐2𝑥 = 0 is not defined, ∴ the above conclusion is wrong. 9 ∫ (3𝑥2 − 16𝑥 + 16) 4 3 0 𝑑𝑥 + ∫ (3𝑥2 − 16𝑥 + 16)4 4 3 𝑑𝑥 = [𝑥3 − 8𝑥2 + 16𝑥] 4 3 0 + [𝑥3 − 8𝑥2 + 16𝑥] 4 4 3 = 256 27 + (0 − 256 27 ) = 0 The area above the x-axis, bounded from 𝑥 = 0 to 𝑥 = 4 3 is the same as the area below the 𝒙-axis, bounded from 𝑥 = 4 3 to 𝑥 = 4.

5 10(a) When 𝑦 = 𝑒, 𝑒 = 4 𝑥 𝑥 = 4 𝑒 ∴ 𝐴 ( 4 𝑒 , 𝑒) When x= 2𝑒, 𝑦 = 4 2𝑒 𝑦 = 2 𝑒 ∴ 𝐵 (2𝑒, 2 𝑒) 10(b) 𝐴𝑟𝑒𝑎 = ∫ 4 𝑥 2𝑒 4 𝑒 𝑑𝑥 + ( 4 𝑒) (𝑒) = 4[ln 𝑥]4 𝑒 2𝑒 + 4 = 4 [ln 2𝑒 − ln 4 𝑒] + 4 = 4[ln 2 + ln 𝑒 − ln 4 + ln 𝑒] + 4 = 4[ln 2 − ln 4] + 12 = 4[−ln 2] + 12 = 12 − ln 16 10(c) Area of rect from from y-axis to A = 𝑒 × 4 𝑒 = 4 𝑢𝑛𝑖𝑡𝑠2 Area of whole rect= 2𝑒 × 𝑒 = 2𝑒2 𝑢𝑛𝑖𝑡𝑠2 ∴ 4 < 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑎𝑑𝑒𝑑 𝑟𝑒𝑔𝑖𝑜𝑛 < 2𝑒2 (explained) 11(a) Refer to graph. 𝑦 = 𝐴(1 + 𝑥)𝑛 lg 𝑦 = lg 𝐴 + 𝑛 lg(1 + 𝑥) lg(1+x) 0.301 0.477 0.602 0.699 0.778 0.845 lg y 1.15 1.41 1.60 1.75 1.87 1.97 Correct points plotted Straight line Plot table 11(b) From the graph, lg 𝐴 = 0.7 [Accept 0.68 ≤ lg 𝐴 ≤ 0.72] 𝐴 = 5.01 𝑛 = 𝑔𝑟𝑎𝑑 = 1.51 [Accept 1.49 ≤ 𝑛 ≤ 1.51] 11(c) 𝑦 = 16 1+𝑥 lg 𝑦 = lg 16 − lg(1 + 𝑥) Draw the line lg 𝑦 = lg 16 − lg(1 + 𝑥) 11(d) 𝐴(1 + 𝑥)𝑛+1 = 16 𝐴(1 + 𝑥)𝑛 = 16 1+𝑥 ∴ lg(𝑥 + 1) = 0.2 𝑥 = 0.585

6 END OF PAPER 𝐥𝐠 𝒚 = 𝐥𝐠 𝑨 + 𝒏 𝐥𝐠(𝟏 + 𝒙) 𝐥𝐠 𝒚 = 𝐥𝐠 𝟏𝟔 − 𝐥𝐠(𝟏 + 𝒙)