ACSI 2018 Promo Paper 1 ans

1 Y5 HL Mathematics Final Examination 2018 1 [Maximum mark: 5] Well done except fpr some careless mistakes ( 1 √𝑥 − 1) 4 (1 − 𝑥)2 = [( 1 √𝑥) 4 + 4 ( 1 √𝑥) 3 (−1) + (4 2) ( 1 √𝑥) 2 (−1)2 + (4 3) ( 1 √𝑥) 1 (−1)3 + (−1)4] × [1 − 2𝑥 + 𝑥2] = ⋯ 1 + (4 2) ( 1 √𝑥) 2 (−2𝑥) + ( 1 √𝑥) 4 (𝑥2) + ⋯ = ⋯ 1 − 2 × 4! 2! 2! + 1 + ⋯ = 1 − 12 + 1 = −10 2 [Maximum mark: 5] (i) Very well done 1 √4𝑟 − 3 + √4𝑟 + 1 = √4𝑟 − 3 − √4𝑟 + 1 (√4𝑟 − 3 + √4𝑟 + 1)(√4𝑟 − 3 − √4𝑟 + 1) = √4𝑟 − 3 − √4𝑟 + 1 4𝑟 − 3 − (4𝑟 + 1) = √4𝑟 + 1 − √4𝑟 − 3 4 (ii) Most tried to force an AP out of the series which is wrong = 1 4 ∑ (√4𝑟 + 1 − √4𝑟 − 3) 100 1 = 1 4 [(√5 − √1) + (√9 − √5) + (√13 − √9) + (√17 − √13) … . +(√401 − √397)] = 1 4 [−1 + √401] 3 [Maximum mark: 5] Many did not realise that 1 − 𝑡𝑎𝑛𝜃 > 0 and can be multiplied for simple solving. Most did not remember − 𝜋 4 < 𝜃 < 𝜋 4 as the necessary condition − 𝜋 4 < 𝜃 < 𝜋 4 ⇒ −1 < 𝑡𝑎𝑛𝜃 < 1 𝑆∞ = 1 1−𝑡𝑎𝑛𝜃 𝑆∞ < 3 + √3 2 1 1 − 𝑡𝑎𝑛𝜃 < 3 + √3 2 2 < (3 + √3)(1 − 𝑡𝑎𝑛𝜃) [since 𝑡𝑎𝑛𝜃 < 1] 2 − 3 + 3𝑡𝑎𝑛𝜃 − √3 + √3𝑡𝑎𝑛𝜃 < 0 −1 + 3𝑡𝑎𝑛𝜃 − √3 + √3𝑡𝑎𝑛𝜃 < 0 (3 + √3)𝑡𝑎𝑛𝜃 < 1 + √3

2 𝑡𝑎𝑛𝜃 < 1 + √3 3 + √3 𝑡𝑎𝑛𝜃 < (1 + √3)(3 − √3) (3 + √3)(3 − √3) 𝑡𝑎𝑛𝜃 < 3 + 2√3 − 3 6 𝑡𝑎𝑛𝜃 < √3 3 𝜃 < 𝜋 6 Hence − 𝜋 4 < 𝜃 < 𝜋 6 4 [Maximum mark: 6] Implicit differentiation is ok but some had misread question as 𝑑𝑦 𝑑𝑥 = 1 4𝑥3 + 𝑥𝑦2 = 5𝑥𝑦 12𝑥2 + 𝑥 (2𝑦 𝑑𝑦 𝑑𝑥) + 𝑦2 = 5𝑥 𝑑𝑦 𝑑𝑥 + 5𝑦 𝑑𝑦 𝑑𝑥 (2𝑥𝑦 − 5𝑥) = 5𝑦 − 𝑦2 − 12𝑥2 𝑑𝑦 𝑑𝑥 = 5𝑦 − 𝑦2 − 12𝑥2 2𝑥𝑦 − 5𝑥 At P, 𝑦 = 𝑥 Subst 𝑦 = 𝑥 into curve : 4𝑥3 + 𝑥3 = 5𝑥2 𝑥2(𝑥 − 1) = 0 𝑥 = 0 (𝑁. 𝐴. ) 𝑜𝑟 𝑥 = 1 Subst 𝑥 = 1 𝑎𝑛𝑑 𝑦 = 1 into 𝑑𝑦 𝑑𝑥: 𝑑𝑦 𝑑𝑥 = 8 3 Equation of tangent to the curve at P is 𝑦 − 1 = 8 3 (𝑥 − 1) 𝑦 = 8 3 𝑥 − 5 3 or 3𝑦 = 8𝑥 − 5 5 [Maximum mark: 9] (a) About half got this correct. Some tried to fit singles in between couples. Qn did not say singles cannot be together 𝐶1,𝐶2,𝐶3,𝐶4,𝑀, 𝑊 No of ways = 6! × 24 = 720 × 16 = 11520

3 (b) Fairly well done. Some wrongly took last term as 𝑎𝑟𝑘+1 Let Pn be the statement: 𝑆𝑛 = 𝑎(𝑟𝑛−1) 𝑟−1 P1: 𝑆1 = 𝑎(𝑟1−1) 𝑟−1 = 𝑎 Hence P1 is true. Assume Pk is true. 𝑆𝑘 = 𝑎(𝑟𝑘 − 1) 𝑟 − 1 Pk+1: 𝑆𝑘+1 = 𝑆𝑘 + 𝑢𝑘+1 𝑆𝑘+1 = 𝑎(𝑟𝑘 − 1) 𝑟 − 1 + 𝑎𝑟𝑘 = 𝑎𝑟𝑘 − 𝑎 + 𝑎𝑟𝑘(𝑟 − 1) 𝑟 − 1 = 𝑎(𝑟𝑘+1 − 1) 𝑟 − 1 ∴ 𝑃𝑘+1 is true. P1 and Pk are true ⇒ 𝑃𝑘+1 is true, ∴ By Mathematical Induction, Pn is true for all 𝑛 ∈ 𝑍+ 6 [Maximum mark: 8] (i) Mostly well done except a handful who went with memory work and got the wrong formula (𝛼 + 𝛽 + 𝛾)2 2𝑥3 + 𝑥2 − 5𝑥 + 3 = 0 𝛼 + 𝛽 + 𝛾 = − 1 2 𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾 = − 5 2 𝛼𝛽𝛾 = − 3 2 (𝛼 + 𝛽 + 𝛾)2 = (𝛼 + 𝛽)2 + 2(𝛼 + 𝛽)𝛾 + 𝛾2 (𝛼 + 𝛽 + 𝛾)2 = 𝛼2 + 𝛽2 + 𝛾2 + 2(𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾) (− 1 2) 2 = 𝛼2 + 𝛽2 + 𝛾2 + 2 (− 5 2) 𝛼2 + 𝛽2 + 𝛾2 = 21 4 (ii) Both methods were equally popular with the usual careless mistakes {𝑦 − (𝛼 − 1)}{𝑦 − (𝛽 − 1)}{𝑦 − (𝛾 − 1)} = {(𝑦 + 1) − 𝛼