XJC H2 Mathematics 9758 – Set II: Paper 2 (Answers)

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) (i) (ii) (𝜋4+12)𝑘2 units2 (iii) −(𝜋+2)𝑘2 2 Sequences and series (geometric series, method of differences) 𝑎𝑛=2𝑛−1−(𝑛−1)𝑛2 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); (i) [shown] (ii) |𝐛−𝐚|2=|𝐚|2+|𝐛|2−2|𝐚||𝐛|cos𝜃 𝜃=72° (iii) 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 (i) [shown] (ii) 𝑥 𝑃(𝑋=𝑥) 3 1−3𝑝+3𝑝2 4 3𝑝−9𝑝2+12𝑝3−6𝑝4 5 6𝑝2−12𝑝3+6𝑝4 (iii) 0.354≤𝑝≤0.646 8 Normal distribution (i) [shown] (ii) 𝜎≈50.01044≈50.0 (iii) 𝑘=12 𝑚=325 Probability=1.53×10−6 9 Binomial distribution (i) [shown] (ii) [shown] Most probable value of 𝑋=100 (iii) Required probability ≈0.0781 (iv) 𝑘𝑚𝑎𝑥=10 10 Sampling; Hypothesis testing (i) [shown] (ii) E(𝑥)=95.02 Var(𝑥)=2.5796 (iii) 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. (iv) 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 (i) 𝐴 is Quick Sort. 𝐵 is Merge Sort. 𝐶 is Selection Sort. (ii) [shown] (iii) Algorithm 𝐴, Quick Sort; Best-case 𝑇=−0.248+(2.46×10−4)𝑛log2𝑛 𝑛≈1277 bytes. (iv) [shown] (v) 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 (i) [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 (ii) [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 (iii) [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 (i) that the integrated result must be negative, and from (ii) 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 [6] 𝑢𝑛+2=2𝑢𝑛+1−𝑢𝑛+2𝑛−1 (𝑢𝑛+2−𝑢𝑛+1)−(𝑢𝑛+1−𝑢𝑛)=2𝑛−1 𝜈𝑛+1−𝜈𝑛=2𝑛−1 ∑[𝜈𝑟+1−𝜈𝑟]𝑛𝑟=1=∑(2𝑟−1)𝑛𝑟=1 ⎣⎢⎢⎢⎢⎢⎡𝜈2−𝜈1+𝜈3−𝜈2+𝜈4−𝜈3⋮+𝜈𝑛−1−𝜈𝑛−2+𝜈𝑛−𝜈𝑛−1+𝜈𝑛+1−𝜈𝑛 ⎦⎥⎥⎥⎥⎥⎤=2(2𝑛−12−1 )−𝑛 𝜈𝑛+1−𝜈1=2𝑛+1−2−𝑛 𝜈𝑛−𝜈1=2𝑛−2−(𝑛−1) 𝜈𝑛=2𝑛−2−(𝑛−1)+𝜈1 𝑢𝑛+1−𝑢𝑛=2𝑛−2−𝑛+1+(2−1) 𝑢𝑛+1−𝑢𝑛=2𝑛−𝑛 ∑[𝑢𝑟+1−𝑢𝑟]𝑛𝑟=1=∑(2𝑟−𝑟)𝑛𝑟=1 ⎣⎢⎢⎢⎢⎢⎡𝑢2−𝑢1+𝑢3−𝑢2+𝑢4−𝑢3⋮+𝑢𝑛−1−𝑢𝑛−2+𝑢𝑛−𝑢𝑛−1+𝑢𝑛+1−𝑢𝑛 ⎦⎥⎥⎥⎥⎥⎤=2(2𝑛−12−1 )−𝑛(𝑛+1)2 𝑢𝑛+1−𝑢1=2𝑛+1−2−𝑛(𝑛+1)2 𝑢𝑛−1=2𝑛−2−(𝑛−1)𝑛2 𝑢𝑛=2𝑛−1−(𝑛−1)𝑛2 This question mainly deals with method of differences. After obtaining an expression for 𝜈𝑛+1−𝜈𝑛, method of differences can then be employed to yield an expression for 𝑢𝑛+1−𝑢𝑛. At this point, a second method of differences follows to obtain expression for 𝑢𝑛 in terms of 𝑛. Along the way, appropriate base modification is required, i.e., changing 𝑢𝑛+1→𝑢𝑛.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 2 6 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 when to best use it, along with common pitfalls to avoid: Using second derivative for simplified expressions Given a simplified expression for a first-order derivative, differentiating a second time proves more timesaving than sign test. There must be evident engagement with unknown constants, if present, to arrive at the conclusion. This can be achieved by considering the boundary values of these constants, as contextually implied, and thereafter explaining briefly how they affect the value of the second derivative (in this case, the phrase “since 𝑟 is positive”).

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 2 7 [continued] Using sign test 𝜃 𝜃𝑚𝑎𝑥− 𝜃𝑚𝑎𝑥 𝜃𝑚𝑎𝑥+ sin𝜃 – – – 3cos𝜃+1 – 0 + 1−cos𝜃 + + + d𝑉d𝜃=2𝑟33sin𝜃(3cos𝜃+1)(1−cos𝜃) + 0 – Slope / — \ ∴𝜃=109.5° gives maximum volume. Using a graphical method, 𝑉=2𝑟33sin2𝜃(1−cos𝜃)→𝑉𝑟3=23sin2𝜃(1−cos𝜃) 𝑉𝑟3 is maximum at 𝜃≈1.911 ∴𝑉 is maximum at 𝜃≈1.911×(180°𝜋)≈109.5° Maximum volume =2𝑟33(1−cos2𝜃)(1−cos𝜃)=2𝑟33(1−19)(1+13)=6481𝑟3 units3 Using sign test for factorised expressions Given a factorised expression for a first-order derivative, sign test (or first derivative test) will save the hassle of applying tedious chain rules in further differentiation. To show the nature of the stationary value, a sign test table must appreciate the factors that are significant to the sign change. Avoid presenting substandard sign test tables that merely show the signs of the derivative values and the corresponding slopes. There must be some form of appreciation that only specific factors contribute significantly to the sign. Alternati