XJC H2 Math - Set 2 - P2 (ANS)

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 2 1 MATHEMATICS Topic identification and short answers Qn Topic(s) Part Answers Section A: Pure Mathematics [40 marks] 1 Graph (sketching, transformation); Sequences and series (geometric series) (a) (b) (𝜋4+12)𝑘2 units2 (c) −(𝜋+2)𝑘2 2 Connected rates of change (a) [shown] (b) [shown] 68 seconds 3 Differentiation (maxima and minima) 𝜃=109.5° gives maximum volume. Maximum volume=6481𝑟3 units3 4 Complex numbers (cartesian form, conjugate) (a) [shown] (b) 𝛾=±25𝑞i 5 Vectors (two dimensions, modulus); (a) [shown] (b) |𝐛−𝐚|2=|𝐚|2+|𝐛|2−2|𝐚||𝐛|cos𝜃 𝜃=72° (c) Regular pentagon −2𝑘 2𝑘 4𝑘 𝑂 𝑦 𝑥 (3𝑘,𝑘2) (−𝑘,2𝑘) (𝑘,𝑘) 𝑦=𝑓(𝑥) Paper 9758/02 Set II – Paper 2

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 2 2 Section B: Probability and Statistics [60 marks] 6 Probability (permutations and combinations) Probability=1.83×10−37 7 Discrete random variables (a) [shown] (b) 𝑥 𝑃(𝑋=𝑥) 3 1−3𝑝+3𝑝2 4 3𝑝−9𝑝2+12𝑝3−6𝑝4 5 6𝑝2−12𝑝3+6𝑝4 (c) 0.354≤𝑝≤0.646 8 Normal distribution (a) [shown] (b) 𝜎≈50.01044≈50.0 (c) 𝑘=12 𝑚=325 Probability=1.53×10−6 9 Binomial distribution (a) [shown] (b) [shown] Most probable value of 𝑋=100 (c) Required probability ≈0.0781 (d) 𝑘𝑚𝑎𝑥=10 10 Sampling; Hypothesis testing (a) [shown] (b) E(𝑥)=95.02 Var(𝑥)=2.5796 (c) Since 𝑛 is a multiple of 50, 𝑛>30, and thus the sample size is large enough so that, by Central Limit Theorem, the sample mean approximately follows a normal distribution. (d) 𝑛𝑚𝑖𝑛=17450

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 2 3 11 Linear regression (a) 𝐴 is Quick Sort. 𝐵 is Merge Sort. 𝐶 is Selection Sort. (b) [shown] (c) Algorithm 𝐴, Quick Sort; Best-case 𝑇=−0.248+(2.46×10−4)𝑛log2𝑛 𝑛≈1277 bytes. (d) [shown] (e) Suggestion: 𝑆=0.368+0.540ln𝑇 (Any suitable suggestion acceptable.) 1.2 7.5 0.3 1.3 𝑂 𝑆 / bytes 𝑇 / ms

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 2 4 Suggested solutions and post-mortem Qn Suggested Solutions Comments Section A: Pure Mathematics [40 marks] 1 (a) [2] To begin, it must first be understood that the given piecewise function gives rise to vertically scaled sketches of the same graph in different periods. Be mindful of the domain of the sketch and indicate inclusion of the point (−2𝑘,0) by means of a solid point, and exclusion of the point (4𝑘,0) by means of a hollow point. 1 (b) [1] Required area = Area of a quarter circle + Area of a right triangle =14𝜋𝑘2+12𝑘2 =(𝜋4+12)𝑘2 units2 There may be a tendency to resort to integration to find the required area due to keywords such as finding “the exact area bounded”. Considering the mark for this part, candidates may want to resort to simpler ways to find the area required. As far as presentation is concerned, take care to write mensuration values (such as length, area and/or volume) with appropriate units. 1 (c) [2] Since 𝑦=f(𝑥) is entirely blow the 𝑥–axis, ∫𝑓(𝑥)∞−2𝑘d𝑥=−(𝜋4+12)𝑘2(2+1+12+14+⋯) =−(𝜋4+12)𝑘2⎝⎜⎜⎛21−12 ⎠⎟⎟⎞ =−(𝜋4+12)𝑘2(4)=−(𝜋+2)𝑘2 Any successful response to this part would recognise from (a) that the integrated result must be negative, and from (b) that the area in one period is subsequently half of that from the previous period, which calls for geometric sum to infinity. −2𝑘 2𝑘 4𝑘 𝑂 𝑦 𝑥 (3𝑘,𝑘2) (−𝑘,2𝑘) (𝑘,𝑘) 𝑦=𝑓(𝑥)

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 2 5 2 (a) [2] The shortest track 𝐴𝐵 can be drawn as a straight line on a sector formed by flattening the curved surface of the conical mountain. Slant height of mountain 𝐴𝐶=√852+7202=725 Using cosine rule, cos∠𝐶𝐴𝐵=𝐴𝐶2+𝐴𝐵2−𝐵𝐶22(𝐴𝐶)(𝐴𝐵) cos∠𝐶𝐴𝐷=𝐴𝐶2+𝐴𝐷2−𝐶𝐷22(𝐴𝐶)(𝐴𝐷) Since ∠𝐶𝐴𝐵=∠𝐶𝐴𝐷, →𝐴𝐶2+𝐴𝐵2−𝐵𝐶22(𝐴𝐶)(𝐴𝐵)=𝐴𝐶2+𝐴𝐷2−𝐶𝐷22(𝐴𝐶)(𝐴𝐷) →𝐴𝐶2+𝐴𝐵2−𝐵𝐶2𝐴𝐵=𝐴𝐶2+𝐴𝐷2−𝐶𝐷2𝐴𝐷 →7252+𝐷2−𝑀2𝐷=7252+𝑠2−𝑧2𝑠 This question may prove challenging. Success in this question highly relies on the acknowledgement that any mensuration on the surface of a non-flat three-dimensional shape is only accurate when done on its net or flattened surface. Once candidates appreciate this, identifying the shortest possible track 𝐴𝐵 becomes intuitive. Candidates may also realise that the result to be shown bears striking resemblance to the less-known form of the cosine formula cos𝜃=𝑎2+𝑏2−𝑐22𝑎𝑏. Observing this form on both sides of the equation to be shown should sufficiently prompt candidates to apply this rule twice on some angle that would summon the desired constants and variables. Should candidates find little progress in this question, candidates are encouraged to skip ahead, especially if the following parts calls for making use of given results. 𝐶 𝐴 𝐵 𝑀 𝑠 𝑧 𝐷 𝐴′

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 2 6 2 (b) [4] Let gondola’s altitude be ℎ. Given that 𝐴𝐶=725, we have: →ℎ720=725−𝑧725=1−𝑧725 →dℎd𝑡=−720725d𝑧d𝑡 ∴Given dℎd𝑡=−4, d𝑧d𝑡=14536 (shown) With 𝐷=900 and 𝑀=300, the result in (i) becomes: →4982512𝑠=7252+𝑠2−𝑧2 →4982512d𝑠d𝑡=2𝑠d𝑠d𝑡−2𝑧d𝑧d𝑡 Given d𝑠d𝑡=1 and d𝑧d𝑡=14536, →4982512=2𝑠−14518𝑧 →𝑧=36145𝑠−2989558 Substituting back, we have: →4982512𝑠=7252+𝑠2−(36145𝑠−2989558)2 →𝑠≈67.829 metres (GC) Since speed is 1 metre per second, required time to nearest second =68 seconds This question assesses on related rates of change with a slight nudge to implicit differentiation. Success on the “show” part will confirm for the candidates the meaning of “altitude” and its “decreasing” rate. Once candidates also identify relevant similar shapes involving the altitude, candidates should be able to show the same rate of change of 𝑧. The remaining steps follow with appropriate differentiation and substitution. Candidates are reminded that, despite the allowance of graphical calculator in this question, it is best to delay all rounding until the final step to avoid marginal error. 𝐴 𝐶 gondola ℎ 𝑧 720

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 2 7 3 [8] 𝑂𝐹=𝑟cos(180°−𝜃)=−𝑟cos𝜃 𝐸𝐹=𝑟sin(180°−𝜃)=𝑟sin𝜃 ∴ Volume 𝑉 of the pyramid =13(12⋅2𝐸𝐹⋅2𝐸𝐹)(𝑂𝐴+𝑂𝐹) =13(2𝑟2sin2𝜃)(𝑟−𝑟cos𝜃) =2𝑟33sin2𝜃(1−cos𝜃) For maximum volume, d𝑉d𝜃=2𝑟33[2sin𝜃cos𝜃(1−cos𝜃)+sin2𝜃(sin𝜃)]=0 2sin𝜃cos𝜃(1−cos𝜃)+sin2𝜃(sin𝜃)=0 sin𝜃(2cos𝜃−2cos2𝜃+1−cos2𝜃)=0 sin𝜃(−3cos2𝜃+2cos𝜃+1)=0 sin𝜃(3cos𝜃+1)(1−cos𝜃)=0 Since 𝜃≠0°,180°, reject sin𝜃=0 and cos𝜃=1 ∴𝜃𝑚𝑎𝑥=cos−1(−13)=90°+cos−1(13)≈109.5° Using second derivative When 𝜃=𝜃𝑚𝑎𝑥, d2𝑉d𝜃2=2𝑟33[cos𝜃(−3cos2𝜃+2cos𝜃+1)+sin𝜃(6cos𝜃sin𝜃−2sin𝜃)] =2𝑟33[cos𝜃(−3cos2𝜃+2cos𝜃+1)+(1−cos2𝜃)(6cos𝜃−2)] =2𝑟33[(−13)(−3(19)+2(−13)+1)+(1−(19))(6(−13)−2)] =2𝑟33[−329]=−6427𝑟3<0,since 𝑟 is positive. ∴𝜃=109.5° gives maximum volume. When demanded an expression of mensuration values – such as perimeter, area, or volume – in terms of length (in this case, 𝑟) and angle (in this case, 𝜃), candidates may expect that trigonometry will be handy. Upon discovering the stationary values of 𝑉, there must be some evidence of appreciation towards edge cases, i.e., making use of sin𝜃≠0 and cos𝜃≠1, or the corollary that 𝜃≠0° and 𝜃≠180° (which may also be apparent from the range 0°<𝜃<180°) to give reasoned rejections and arrive at a feasible value. Be reminded to provide non-exact angles in degrees up to one decimal point as appropriate. Current trend suggests that A–Levels tend to relieve candidates of the need to proof the nature of stationary points. Nonetheless, it is good to be familiar with different methods of proving such points – suggestions provided lists down 3 methods and a