Peirce Prelim 2021 S4E Add Math P1 MS

Peirce Secondary School 07 Sept 2021 Preliminary Exam UPDATED 13 Sept Sec 4E/5N Add Math 4049/I Question Answer Marks Partial Marks Guidance 1 Solve the equation . 1 102𝑥. 10 − 80(10𝑥) = −70 6 M1 For splitting Let u = 10𝑥 M1 For substitution 𝑢2 − 8𝑢 + 7 = 0 M1 For quadratic equation. Accept 10𝑢2 − 80𝑢 + 70 = 0 [M2 for (10𝑥)2 − 8(10𝑥) + 7 = 0 𝑖𝑓 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑠𝑢𝑏𝑠𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝑢𝑠𝑒𝑑] (u – 7) (u – 1) = 0 u = 7 or u = 1 M1 For u or 10𝑥 values x = lg 7 or x = 0 x = 0.845 (3 sf) or x = 0 A2 Accept 0.84509804. Reject lg 7 Question Answer Marks Partial Marks Guidance 2 𝑓(𝑥) = 𝑥2 − 5𝑥 + 7 2(i) 𝑏2 − 5𝑏 + 7 = 𝑐2 − 5𝑐 + 7 3 M1 f(b) = f(c) or f(b)– f(c) = 0 or f(c) – f(b)=0 (𝑏 − 𝑐)(𝑏 + 𝑐) = 5(𝑏 − 𝑐) M1 For factorisation Since b≠c, b-c ≠0 Therefore 𝑏 + 𝑐 = 5 (shown) A1 For conclusion 2(ii) 𝑏 = 5 − 𝑐 4(5 − 𝑐)(𝑐) = 21 4 M1 For substitution (2𝑐 − 3)(2𝑐 − 7) = 0 M1 c = 1.5 or c = 3.5 b = 3.5 or b = 1.5 M1 Reject pre-mature rejection of values

Since b > c, b = 3.5 and c = 1.5 A1 Accept fractions Question Answer Marks Partial Marks Guidance 3(i) = 38 + (8 1 )37 (− 𝑥 2) 1 + (8 2 )36 (− 𝑥 2) 2 + (8 3 )35 (− 𝑥 2) 3 + ⋯ 3 M1 For expansion = 6561 – 8748𝑥 + 5103𝑥2 − 1701𝑥3 + ⋯ A2 For 4 terms [A1 for 2-3 terms; A0 if 0-1 term correct.] 3(ii) (𝑥 − 5 𝑥) 2 (3 − 𝑥 2) 8 = (𝑥2 − 10 + 25 𝑥2) (6561 – 8748𝑥 + 5103𝑥2 − 1701𝑥3 + ⋯ ) 3 M1 Expansion of (𝑥 − 5 𝑥) 2 = 𝑥2 − 10 + 25 𝑥2 = … + 87480x – 42525x + … = … + 44955x + … M1√ √ For overall expansion if expansion of (𝑥 − 5 𝑥) 2 Coefficient of x = 44955 A1 Reject 44955x Question Answer Marks Partial Marks Guidance 4(i) When 𝑡 = 0, 𝑀 = 24 𝑔 When 𝑡 = 35, 𝑀 = 12 𝑔 3 M1 For M = 12 g 𝑒−𝑘(35) = 1 2 or 𝑒𝑘(35) = 2 −35 𝑘 = 𝑙𝑛 1 2 or 35 𝑘 = 𝑙𝑛 2 M1 For isolating e or taking ln k = 0.0198042 = 0.0198 (3 s f) A1 Reject 𝑘 = − 1 35 𝑙𝑛 1 2 or 1 35 𝑙𝑛 𝑙𝑛 2 or0.020 Accept 0.01980

4(ii) 𝑀 = 24𝑒−0.0198042(365) 2 M1√ Accept t = 366 days , √ for k = 0.017414028 = 0.0174 g (3 s f) A1 Question Answer Marks Partial Marks Guidance 5 5(i) 𝑑𝑦 𝑑𝑥 = (2𝑥 − 3)(3) − 3𝑥(2) (2𝑥 − 3)2 5 M1 For Quotient or Product Rule applied correctly = −9 (2𝑥−3)2 = −9 4 M1√ √ for equating 𝑑𝑦 𝑑𝑥 to gradient of tangent x = 2.5 or x = 0.5 A1√ √ for both x values y = 3.75 or y = -0.75 A1√ √ for both y values Points are (2.5, 3.75) and (0.5, -0.75) A1 For both points 5(ii) 𝑑𝑦 𝑑𝑥 = −9 (2𝑥−3)2 2 Since (2𝑥 − 3)2 > 0, 𝑥 ≠ 1.5 −9 (2𝑥−3)2 < 0 M1 Reject using one/two values of x to show 𝑑𝑦 𝑑𝑥 < 0. Curve is a decreasing function as 𝑑𝑦 𝑑𝑥 < 0. A1√ √ For correct conclusion 𝑑𝑦 𝑑𝑥 < 0 𝑜𝑟 𝑖𝑚𝑝𝑙𝑖𝑒𝑑 𝑘𝑛𝑜𝑤𝑙𝑒𝑑𝑔𝑒 𝑑𝑦 𝑑𝑥 < 0 .

Question Answer Marks Partial Marks Guidance 6(i) a = 2 2 B1 b = 2 B1 6(ii) 𝑚 = −𝜋 + 𝑝 1 B1