XJC H2 Math - Set 3 - P2 (ANS)

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 2 1 MATHEMATICS Topic identification and short answers Qn Topic(s) Part Answers Section A: Pure Mathematics [40 marks] 1 Complex numbers (cartesian form, conjugate); System of linear equations 𝑢=3+2i 𝑤=5−i 𝑧=−4+3i 2 Differentiation (connected rates of change) 𝑠=14√𝑥2+42 (or equivalent) 𝑇=256 3 Integration techniques (by substitution); Definite integrals (volume of revolution) (a) [shown] (b) 23𝜋2−√34𝜋 units3 4 Vectors (two dimensions, collinearity) (a) [shown] (b) 𝑂𝑌⃖⃖⃖⃖⃑=(𝑎𝑎−𝑐)𝐫−(𝑐𝑎−𝑐)𝐩 𝑂𝑍⃖⃖⃖⃖⃖⃑=(𝑏𝑏−𝑐)𝐫−(𝑐𝑏−𝑐)𝐪 [shown] (c) 𝑐=𝑎𝑏2𝑎−𝑏 5 Maclaurin series; Differential equations (a) [shown] (b) 𝑦=1−2𝑘2𝑥2+23𝑘2(𝑘2−1)𝑥4+⋯ (c) 𝑦=cos(2𝑚sin−1𝑥) (d) 𝑘=𝑚 Paper 9758/02 Set III – Paper 2

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 2 2 Section B: Probability and Statistics [60 marks] 6 Probability (permutations and combinations) Number of ways =1016 7 Hypothesis testing; Sampling (a) The probability of obtaining a sample which gives a test statistic that is as extreme or more extreme than the one being observed, assuming that the null hypothesis is true. (b) 𝐻0∶𝜇=30 and 𝐻1∶𝜇≠30 𝑛=100 𝛼=5 Reject the null hypothesis and conclude that there is sufficient evidence at 5% significance level that the average duration of uploaded videos differs from 30.0 seconds. (c) The second sample is large and yields 𝑝-value >0.05 8 Discrete random variables (expectations); Probability (a) [shown] (b) 𝑎=7 (c) Least integer 𝑦=2 9 Linear regression; Normal distribution (normal curve) (a) (b) In the sketch, f(𝑥) increases from 𝑥=59 to 𝑥=61, then decreases from 𝑥=61 to 𝑥=64. This implies that f(𝑥) is maximum at either 𝑥=61 or somewhere else in the range 59<𝑥<64. (c) 𝜇=62 𝑎≈0.0399 𝑏≈−0.00499 (d) 𝑥≈78.6 or 45.4 Both estimates are equally reliable (e) (71.8,0.0247) or (52.2,0.0247)

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 2 3 10 Binomial distribution (a) The event that the stock price rises from one period to the next is independent of any other event where the price rises from any other period to the next. (b) [shown] (c) P(sell at period 2𝑘)=(12)2𝑘[(2𝑘𝑘+1)+(2𝑘𝑘+2)] P(sell at period 2𝑘+1)=(12)2𝑘+1[(2𝑘+1𝑘+2)+(2𝑘+1𝑘+3)] (d) 𝑡odd=13 or 15 𝑡even= 6 or 8 Sell at 𝑡=13 or 15 to get 𝑟3𝑆0 or 𝑟5𝑆0 with probability 0.2444 Sell at 𝑡=6 or 8 to get 𝑟2𝑆0 or 𝑟4𝑆0 with probability 0.3281 (Any suggestion acceptable if justified using values above.) 11 Normal distribution (a) (b) Required probability ≈0.158 (c) [shown] Required least integer 𝑚=94 A fruit of either kind (apple or pear) will almost absolutely exceed 94 grams. To have a higher chance of distinguishing the fruit correctly, the vendor may use a larger value of 𝑚. (d) 𝑘=346

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 2 4 Suggested solutions and post-mortem Qn Suggested Solutions Comments Section A: Pure Mathematics [40 marks] 1 [5] Let 𝑢=𝐴+𝐵i, 𝑤=𝐶+𝐷i and 𝑧=𝐸+𝐹i (where 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, 𝐹∈ℝ). 𝑢−𝑤+𝑧=−6+6i (𝐴+𝐵i)−(𝐶+𝐷i)+(𝐸+𝐹i)=−6+6i (𝐴−𝐶+𝐸)+(𝐵−𝐷+𝐹)i=−6+6i →𝐴−𝐶+𝐸=−6 →𝐵−𝐷+𝐹=6 3𝑢+i𝑤−2i𝑧=16+19i 3(𝐴+𝐵i)+i(𝐶+𝐷i)−2i(𝐸+𝐹i)=16+19i (3𝐴−𝐷+2𝐹)+(3𝐵+𝐶−2𝐸)i=16+19i →3𝐴−𝐷+2𝐹=16 →3𝐵+𝐶−2𝐸=19 i𝑢∗−2𝑤∗−3𝑧=4−8i i(𝐴−𝐵i)−2(𝐶−𝐷i)−3(𝐸+𝐹i)=4−8i (𝐵−2𝐶−3𝐸)+(𝐴+2𝐷−3𝐹)i=4−8i →𝐵−2𝐶−3𝐸=4 →𝐴+2𝐷−3𝐹=−8 Solving simultaneously using GC, 𝐴=3,𝐵=2→𝑢=3+2i 𝐶=5,𝐷=−1→𝑤=5−i 𝐸=−4,𝐹=3→𝑧=−4+3i As apparent, this question assesses candidates’ ability to form a correct system of linear equation in terms of cartesian complex numbers. Great attentiveness to sign changes is required, especially those that arise due to multiplication by the imaginary unit and conjugation. The following parts on comparing complex number parts and solving the resulting system of six linear equations are the less arduous parts of the question. This question is only made possible owing to the limitations of graphing calculators (that are approved within the H2 Mathematics syllabus), particularly the inability to accept complex numbers as direct inputs to the simultaneous equation solver.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 2 5 2 [6] Consider the three-dimensional diagram. (All lengths in metres.) By similar triangles, →𝑠𝑠+√𝑥2+42=1.89=15 →5𝑠=𝑠+√𝑥2+42 ∴𝑠=14√𝑥2+42 Differentiating with respect to time, 𝑡 →d𝑠d𝑡=14(12)(𝑥2+16)−12(2𝑥)d𝑥d𝑡 →d𝑠d𝑡=𝑥4√𝑥2+16 (2)=𝑥2√𝑥2+16 →(d𝑠d𝑡)2=𝑥24𝑥2+64 →4(d𝑠d𝑡)2𝑥2+64(d𝑠d𝑡)2=𝑥2 →𝑥2=64(d𝑠d𝑡)2 1−4(d𝑠d𝑡)2 Substituting rates of changes of 𝑠 to obtain 𝑥, d𝑠d𝑡=−0.3→𝑥2=64(−0.3)21−4(−0.3)2=9→𝑥=−3 (∵d𝑠d𝑡<0) d𝑠d𝑡=0.4→𝑥2=64(0.4)21−4(0.4)2=2569→𝑥=163 (∵d𝑠d𝑡>0) ∴𝑇=(163+3)2=256 The relationship between 𝑠 and 𝑥 is most obvious in a three-dimensional diagram, and hence candidates are not to rely only on the two-dimensional diagram provided by the question. Candidates should then more easily notice the required relationship through Pythagorean Theorem and similar triangles. Upon differentiating with respect to time and finding the connected rates of change, candidates can deduce an expression for 𝑥 in terms of d𝑠d𝑡 and hence use it to find the exact value for 𝑥 given the value of d𝑠d𝑡. Candidates who are mindful of the sign of d𝑠d𝑡 will notice that this will affect the sign of 𝑥 and in extension where it lies on the path with respect to 𝑋. In turn, they should be able to deduce the correct signs, following which the exact value of 𝑇 can be deduced. LAMP 9 MAN 4 1.8 𝑠 𝑥 𝑋

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 2 6 3 (a) [4] 𝑥=sin2𝜃 →√𝑥=sin𝜃→𝜃=sin−1√𝑥 →𝑥=1−cos2𝜃→cos𝜃=√(1−𝑥) →d𝑥=2sin𝜃cos𝜃d𝜃 ∫√(1−𝑥𝑥)d𝑥 =∫√(1−sin2𝜃sin2𝜃)(2sin𝜃cos𝜃)d𝜃 =∫√(cos2𝜃sin2𝜃)(2sin𝜃cos𝜃)d𝜃 =∫(cos𝜃sin𝜃)(2sin𝜃cos𝜃)d𝜃 =∫2cos2𝜃d𝜃 =∫cos2𝜃+1d𝜃 =12sin2𝜃+𝜃+𝐶 =sin𝜃cos𝜃+𝜃+𝐶 =(√𝑥)(√(1−𝑥))+sin−1√𝑥+𝐶 =√(𝑥−𝑥2)+sin−1√𝑥+𝐶 Integration by trigonometric substitution to yield an integrand in the form of a double-angle trigonometric function is routine within the scope of H2 Mathematics. Upon correctly deriving intermediary results from the substitution, candidates should successfully obtain the integration result 12sin2𝜃+𝜃+𝐶. At this juncture, candidates may encounter difficulty in returning the result in terms of 𝑥. Successful attempts to this question will indicate some association between the substitution and right triangles, from which the relationship between 𝑥 and 𝜃 comes most naturally. It is highly likely for unaware candidates to conclude their working prematurely and leave their final answer in terms of 𝜃. To curb this tendency, candidates may wish to immediately make a mental note to back-substitute upon seeing a substitution and an indefinite integral. 𝜃 √𝑥 1 √(1−𝑥)

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 2 7 3 (b) [3] To find the limits of integration, consider sketching the curve 𝑥𝑦4+𝑥=1 using the parametric equations 𝑥=1𝑡4+1,𝑦=𝑡 for 𝑡∈ℝ Hence, volume obtained when revolving the shaded region =𝜋∫√(1−𝑥𝑥)134 d𝑥 =𝜋[√(𝑥−𝑥2)+sin−1√𝑥]341 =𝜋[0+𝜋−(√34+13𝜋)] =23𝜋2−√34𝜋 units3 Although the bulk of this question concerns evaluating definite integrals, candidates may instead find it more difficult to define the region to be revolved, having been given a curve equation that cannot be directly expressed in the form 𝑦=f(𝑥) expected by the graphing calculator. To sketch curves with equations of the form 𝑥=f(𝑦) using a graphing calculator, candidates may consider sketching the following instead: (1) a parameterised version of the equation in the form 𝑦=𝑡