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}{(𝑦 + 1) − 𝛽}{(𝑦 + 1) − 𝛾} Subst. 𝑥 = 𝑦 + 1 into 2𝑥3 + 𝑥2 − 5𝑥 + 3 = 0 2(𝑦 + 1)3 + (𝑦 + 1)2 − 5(𝑦 + 1) + 3 = 0 2(𝑦3 + 3𝑦2 + 3𝑦 + 1) + 𝑦2 + 2𝑦 + 1 − 5𝑦 − 5 + 3 = 0 2𝑦3 + 7𝑦2 + 3𝑦 + 1 = 0 Or 2𝑥3 + 7𝑥2 + 3𝑥 + 1 = 0 Alternatively, 𝛼 − 1 + 𝛽 − 1 + 𝛾 − 1 = − 1 2 − 3 = − 7 2 (𝛼 − 1)(𝛽 − 1) + (𝛼 − 1)(𝛾 − 1) + (𝛽 − 1)(𝛾 − 1) = (𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾) − 2(𝛼 + 𝛽 + 𝛾) + 3 = − 5 2 − 2 (− 1 2) + 3

4 = 3 2 (𝛼 − 1)(𝛽 − 1)(𝛾 − 1) = 𝛼𝛽𝛾 − (𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾) + (𝛼 + 𝛽 + 𝛾) − 1 = − 3 2 − (− 5 2) + (− 1 2) − 1 = − 1 2 Equation is 𝑥3 − (− 7 2) 𝑥2 + 3 2 𝑥 − (− 1 2) = 0 2𝑥3 + 7𝑥2 + 3𝑥 + 1 = 0 7 [Maximum mark: 6] (i) 𝑠𝑖𝑛 𝜋 12 = 𝑠𝑖𝑛 (𝜋 3 − 𝜋 4) = sin 𝜋 3 cos 𝜋 4 − cos 𝜋 3 sin 𝜋 4 = √3 2 × √2 2 − 1 2 × √2 2 = √6 − √2 4 Hence arcsin (√6−√2 4 ) = 𝜋 12 (ii) Most not done using complementary angles x = cos 5𝜋 12 = sin 𝜋 12 = √6 − √2 4 8 [Maximum mark: 6] Fairly well done A, B, C, D A B C D f’(x) x

5 Asymptote at y = 0 Shape of graph 9 [Maximum mark: 18] (a) (i) Many did not present answer as a single translation g(𝑥) = 𝑥(𝑥 − 2) = 𝑥2 − 2𝑥 = (𝑥 − 1)2 − 1 𝑥2 is translated (1 0) to (𝑥 − 1)2 and then translated ( 0 −1) to (𝑥 − 1)2 − 1 Transformation is translation ( 1 −1) (ii) A few did not know how to do a reciprocal graph 𝐴′(−1.24, 0.25), 𝐵′(3.24, 0.25), 𝐶′(−0.414, 1), 𝐷′(2.41, 1), 𝐸′(1, −1) (b) (i) Not many got the range of k correct h: 𝑥 → (𝑥 − 1)2 + 2, 𝑥 ∈ 𝑅, 𝑥 ≥ 1 𝑘: 𝑥 → −2𝑥 − 3, 𝑥 ≥ −2 𝑅𝑘 = ]−∞, 1] which is not a subset of the domain of h(x) for hk(x) to exist. (b) (ii) Most could solve this ℎ𝑘(𝑥) = ℎ(𝑎𝑥 + 𝑏) 4𝑥2 + 16𝑥 + 18 = (𝑎𝑥 + 𝑏 − 1)2 + 2 4𝑥2 + 16𝑥 + 18 = (𝑎𝑥 + 𝑏)2 − 2(𝑎𝑥 + 𝑏) + 1 + 2 4𝑥2 + 16𝑥 + 18 = (𝑎𝑥)2 + 2𝑎𝑏𝑥 + 𝑏2 − 2𝑎𝑥 − 2𝑏 + 3 Comparing the constant terms: 𝑏2 − 2𝑏 + 3 = 18 𝑏2 − 2𝑏 − 15 = 0 (𝑏 − 5)(𝑏 + 3) = 0 𝑏 = 5, −3 (N.A.)

6 Also, comparing coefficients of x: 2𝑎𝑏 − 2𝑎 = 16 2𝑎(𝑏 − 1) = 16 𝑎 = 2 (c) Many did not reject the positive answer and most did not get the domain correct Let 𝑦 = 4 (2𝑥−3)2+1 𝑦(2𝑥 − 3)2 = 4 − 𝑦 2𝑥 − 3 = ±√4 − 𝑦 𝑦 𝑥 = 1 2 [3 ± √4 − 𝑦 𝑦 ] 𝑝−1(𝑥) = 3 2 − 1 2 √ 4−𝑥 𝑥 since 𝑅𝑝−1 = 𝐷𝑝 𝑝−1(𝑥) = 3 2 − 1 2 √4 − 𝑥 𝑥 , 0 < 𝑥 ≤ 4 10 [Maximum mark: 16] (a) (i) careless Vector perpendicular to both lines,𝐧 = ( 𝟑 𝟐 −𝟐 ) × ( −𝟑 𝟒 −𝟏 ) = 𝟑 ( 𝟐 𝟑 𝟔 ) (ii) Some took AB as the shortest distance Let A be the point (-3, -4, 6) and B be the point (4, -7, -3). 𝐴𝐵→ = ( 4 −7 −3 ) − ( −3 −4 6 ) = ( 7 −3 −9 ) Shortest distance between the skew lines = |( 7 −3 −9 ) . ( 2 3 6 )| √22 + 32 + 62 = |7 × 2 + (−3) × 3 + (−9) × 6| 7 = 7 Or Plane containing 𝜋1 𝒓𝟏. ( 2 3 6 ) = ( −3 −4 6 ) . ( 2 3 6 ) = −18 Plane containing 𝜋1 𝒓𝟏. ( 2 3 6 ) = ( 4 −7 −3 ) . ( 2 3 6 ) = −31 Shortest distance between the skew lines = perpendicular distance between parallel planes

7 = 2 2 2 18 ( 31) 49 49 7 2 492 3 6 3 6 −− = = =  ++  units (b) (i) Good 𝒑 = ( 1 2 −1 ) + 𝑡 ( 1 −2 3 ) At 𝑡 = 0, 𝒑 = ( 1 2 −1 ) ∴ 𝑃(1, 2, −1) (b) (ii)some switched cartesian to vector form instead of working the other way round 𝑥 = 1 + 𝑡 ⇒ 𝑡 = 𝑥 − 1 𝑦 = 2 − 2𝑡 ⇒ 𝑡 = 2 − 𝑦 2 𝑧 = 3𝑡 − 1 ⇒ 𝑡 = 𝑧 + 1 3 ∴ 𝑥 − 1 = 2 − 𝑦 2 = 𝑧 + 1 3 (b) (iii)(a) Good 2(1 + 𝑡) + (2 − 2𝑡) + 3𝑡 − 1 = 6 𝑡 = 1 (iii) (b) Mostly well done At 𝑡 = 1, 𝒑 = ( 1 2 −1 ) + ( 1 −2 3 ) = ( 2 0 2 ) Coordinates of P are (2, 0, 2) (iii) (c) Mostly well done Distance travelled = ⌈( 2 0 2 ) − ( 1 2 −1 )⌉ = ⌈( 1 −2 3 )⌉ = √12 + (−2)2 + 32 = √14 11 [Maximum mark: 16] (a) Poorly done. Correct ones did not remove 1 1+𝑤3 (1 − 𝑧)6 − 𝑧6 = 0 (1 − 𝑧 𝑧 ) 6 = 1 1, 𝑤, 𝑤2, 𝑤3, 𝑤4 and 𝑤5 are solutions of the complex equation 𝑧6 = 1 Let 𝑤 = 1−𝑧 𝑧 , 𝑧 = 1 1 + 𝑤 𝑧 = 1 1 + 1 , 1 1 + 𝑤 , 1 1 + 𝑤2 , 1 1 + 𝑤3 (𝑁. 𝐴. ), 1 1 + 𝑤4 , 1 1 + 𝑤5 ∴ 𝑧 = 1 2 , 1 1 + 𝑤 , 1 1 + 𝑤2 , 1 1 + 𝑤4 , 1 1 + 𝑤5

8 (b) (i) some had problem with the argument ⌈√3 − 𝑖⌉ = √3 + 1 = 2 𝑎𝑟𝑔(√3 − 𝑖) = − 𝜋 6 Hence √3 − 𝑖 = 2𝑐𝑖𝑠 (− 𝜋 6) By de Moivre’s Theorem, (√3 − 𝑖) 𝑛 = 2𝑛𝑐𝑖𝑠 (− 𝑛𝜋 6 ) = 2𝑛 (𝑐𝑜𝑠 1 6 𝑛𝜋 − 𝑖𝑠𝑖𝑛 1 6 𝑛𝜋) (ii) most could not give final answers (√3 − 𝑖) 𝑛 = 2𝑛 (𝑐𝑜𝑠 1 6 𝑛𝜋 − 𝑖𝑠𝑖𝑛 1 6 𝑛𝜋) is real and negative ⇒ 𝑠𝑖𝑛 𝑛𝜋 6 = 0 and 𝑐𝑜𝑠 𝑛𝜋 6 < 0 𝑛 = 0, 6, 12, 18, 24, . . . .. and 𝑛 = 6, 18, 30, 42, . . . . .. Hence 𝑛 = 6, 18, 30, 42 , . . . . .. Or 𝑛 = 6 × 1, 6 × 3, 6 × 5, 6 × 7, . . .. (iii) numerous different errors Subst. √3 − 𝐼 into 𝑧9 + 16(1 + 𝑖)𝑧3 + 𝑎 + 𝑖𝑏 = 0, 29 (𝑐𝑜𝑠 3 2 𝜋 − 𝑖𝑠𝑖𝑛 3 2 𝜋) + 16(1 + 𝑖) [23 (𝑐𝑜𝑠 1 2 𝜋 − 𝑖𝑠𝑖𝑛 1 2 𝜋)] + 𝑎 + 𝑖𝑏 = 0 512(𝑖) + 16(1 + 𝑖) × 8(−𝑖) + 𝑎 + 𝑖𝑏 = 0 512(𝑖) + 128(−𝑖 + 1) + 𝑎 + 𝑖𝑏 = 0 𝑎 + 𝑖𝑏 = −128 − 384𝑖 ∴ 𝑎 = −128 𝑎𝑛𝑑 𝑏 = −384