XJC H2 Math - Set 1 - P2 (ANS)

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 2 1 MATHEMATICS Topic identification and short answers Qn Topic(s) Part Answers Section A: Pure Mathematics [40 marks] 1 Vectors (two dimensions); Graphs and transformations Required sequence of transformation: 1st:Scaling by a scale factor of 𝑘2 parallel to the 𝑥–axis 2nd:Translation of 𝑘2−1−1𝑘+𝑘 units in the positive 𝑦–direction 2 Graphs (sketching using GC); Inequalities • If 0<𝑎<1, then −1<𝑥≤−𝑎 or 𝑎≤𝑥<2. • If 𝑎=1, then 1≤𝑥<2. • If 1<𝑎<2, then −𝑎≤𝑥<−1 or 𝑎≤𝑥<2. • If 𝑎=2, then −2≤𝑥<−1. • If 𝑎>2, then −𝑎≤𝑥<−1 or 2<𝑥≤𝑎. 3 Differentiation (maxima and minima) ℎ=√3𝑟 gives minimum volume. Minimum volume=4√3𝑟3 units3 4 Complex numbers (cartesian form, conjugate) (a) [shown] (b) 𝑦=1+2𝑥+(2−𝑎22)𝑥2+(43−𝑎2)𝑥3+⋯ (c) e2𝑥sin𝑎𝑥=𝑎𝑥+2𝑎𝑥2+⋯ 5 Sequences and series (geometric series) (a) End of July 2026 (b) [shown] Amount=$9,477.56 (c) August 2029 Paper 9758/02 Set I – Paper 2

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 2 2 Section B: Probability and Statistics [60 marks] 6 Normal distribution (a) 𝜇=6.667𝑎; 𝜎=0.667𝑎 (b) 7 Binomial distribution (a) [shown] (b) 𝑝≈0.968 or 0.252 (c) 12<𝑝<1 8 Discrete random variables (a) [shown] (b) E(𝑋)=4𝑁2+3𝑁−16𝑁 (c) [shown] 𝑚=71 9 Probability (a) P(𝐵)=25; P(𝐴∩𝐵)=730 (b) [shown] (c) P(𝐴∩𝐵∩𝐶)=115 (d) 720≤P(𝐴′∩𝐵′∩𝐶′)≤1330 10 Sampling; Hypothesis testing (a) • The sample is taken from the population of 1 000 chips of the newest model. • The sample size (or number of chips) 𝑛 is greater than 30. • The 𝑛 chips are obtained randomly (or independently of one another with equal probability). (b) [shown] Var(𝑥)=98.75 (c) 𝐻1:𝜇<40; 6.66≤𝛼<13.3

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 2 3 11 Linear regression (a) (b) [shown] The square of the residual horizontal distances between the relevant data points and the best-fit line is minimised. (c) Required value =8.2 degree Celsius (allow marginal error) This value is unreliable. (d) 𝑡=417.40−259𝑇 𝑟 =−0.9561828875≈ 0.956 (e) 𝑘=3 (allow marginal error) The data is selected in the range of 𝑡 where 𝑇 is strictly increasing and such that the product moment correlation coefficient is as close to 1 as possible.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 2 4 Suggested solutions and post-mortem Qn Suggested Solutions Comments Section A: Pure Mathematics [40 marks] 1 [4] Finding the cartesian equation of 𝐿1, 𝐫=(𝑘1)+𝜆(1𝑘) →𝑥−𝑘=𝑦−1𝑘 →𝑦=𝑘𝑥−𝑘2+1 Finding the cartesian equation of 𝐿2, 𝐫=(1𝑘)+𝜇(𝑘1) →𝑥−1𝑘=𝑦−𝑘 →𝑦=1𝑘𝑥−1𝑘+𝑘 Required sequence of transformation: 1st:Scaling by a scale factor of 𝑘2 parallel to the 𝑥–axis 2nd:Translation of 𝑘2−1−1𝑘+𝑘 units in the positive 𝑦–direction This question mainly assesses on converting vector equations into cartesian equations. Admittedly, despite being within the expectations of the syllabus, two-dimensional vector equations rarely appear in A–Levels. The latter part concerning graph transformations should be relatively more routine. Note that when proposing translations involving unknown constants, ensure that it is done (1) with a positive unit of translation, and (2) in the correct direction. (In this case, it is graphically verifiable that 𝑘2−1−1𝑘+𝑘>0 for 𝑘>1.)

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 2 5 2 [5] 𝑎2−𝑥−2𝑥2−𝑥−2≥1 →𝑎2−𝑥−2−(𝑥2−𝑥−2)𝑥2−𝑥−2≥0 →𝑎2−𝑥2𝑥2−𝑥−2≥0 →(𝑎+𝑥)(𝑎−𝑥)(𝑥−2)(𝑥+1)≥0 Principal values are 𝑥=−1,2 and ±𝑎. Given 𝑎>0, there are 5 possible solution intervals: Case Value of 𝑎 Number line (deduced from GC) Solution interval 1 0<𝑎<1 – + – + – -1 -𝒂 𝒂 2 −1<𝑥≤−𝑎 or 𝑎≤𝑥<2 2 𝑎=1 – – + – -1 1 2 1≤𝑥<2 3 1<𝑎<2 – + – + – -𝒂 -1 𝒂 2 −𝑎≤𝑥<1 or 𝑎≤𝑥<2 4 𝑎=2 – + – – -2 -1 2 −2≤𝑥<−1 5 𝑎>2 – + – + – -𝒂 -1 2 𝒂 −𝑎≤𝑥<−1 or 2<𝑥≤𝑎 This question concerns graph sketching and solving inequalities. As this question does not explicitly prohibit the use of graphing calculators (GC), candidates may simply substitute different 𝑎 values to observe the different cases of graphs and deduce the required solution intervals, all without sketching these graphs on paper as it is not required by the question. Having said so, a manual attempt by hand is perfectly welcomed and encouraged. In fact, answers that rely on GC may also benefit from some manual prework which could reveal strategic choices of 𝑎 values.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 2 6 3 [8] Consider the cross-section of the octahedron where it is tangent to the inscribed sphere. By similar triangles, 𝑥𝑟=ℎ√ℎ2−𝑟2→𝑥=𝑟ℎ√ℎ2−𝑟2 ∴ The volume of the inscribing tetrahedron, 𝑉 =2(13)(2𝑥)2(ℎ) =23(4𝑟2ℎ2ℎ2−𝑟2)(ℎ) =8𝑟23(ℎ3ℎ2−𝑟2) Differentiating 𝑉 with respect to ℎ, d𝑉dℎ=8𝑟23(3ℎ2(ℎ2−𝑟2)−ℎ3(2ℎ)(ℎ2−𝑟2)2) d𝑉dℎ=8𝑟23(ℎ2(3ℎ2−3𝑟2−2ℎ2)(ℎ2−𝑟2)2) d𝑉dℎ=83𝑟2(ℎ2(ℎ2−3𝑟2)(ℎ2−𝑟2)2) Questions on maxima and minima and justification of the nature of stationary values are the bread and butter of A–Levels and school examinations. This question concerns packing geometrical shapes, as is the trend for most optimisation problems, and hence would call for fluency in geometry and trigonometry. When finding an expression for the volume of the octahedron in the beginning, caution is advised when relying only on the given diagram in the question, as there is risk of misinterpretation – in this case, potentially mistaking 𝑟 as half of the side of the pyramid’s square base. This could be prevented should candidates produce their own cross-sectional diagram to confirm where the sphere touches the octahedron. All lengths that are relevant to calculate the volume of the octahedron can be found in terms of ℎ and 𝑟 through geometry. When differentiating the volume, given that the letters involved in the differentiation are aplenty, candidates should carefully distinguish the variable of differentiation from the constants – in this case, ℎ is the variable and 𝑟 is a constant. 𝑥 𝑟 ℎ √ℎ2−𝑟2

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 2 7 [Continued] For stationary value of 𝑉,d𝑉dℎ=0 →ℎ2(ℎ2−3𝑟2)=0 →ℎ2=0 or ℎ2=3𝑟2 →ℎ=0 or ℎ=±√3𝑟 Since ℎ>0, ℎ=√3𝑟 Considering sign test, notice that d𝑉dℎ=83𝑟2(ℎ2(ℎ2−3𝑟2)(ℎ2−𝑟2)2)=83(𝑟ℎℎ2−𝑟2)2(ℎ2−3𝑟2) ℎ (√3𝑟)− √3𝑟 (√3𝑟)+ ℎ2−3𝑟2 – 0 + d𝑉d𝜃=83(𝑟ℎℎ2−𝑟2)2(ℎ2−3𝑟2) – 0 + Slope \ — / ∴ℎ=√3𝑟 gives the minimum volume for the inscribing tetrahedron. Minimum volume =8𝑟23⎝⎜⎜⎜⎛(√3𝑟)3 (√3𝑟)2−𝑟2⎠⎟⎟⎟⎞=8𝑟23(3√3𝑟33𝑟2−𝑟2)=8𝑟23(3√3𝑟2)=4√3𝑟3 units3 As per routine, stationary values are found through the roots of the first derivative. In that process, it is considered good practice to justify any root rejections or acceptance in context, which would otherwise be a common cause of penalisation. In this context, it is obvious that non-positive solutions for the height ℎ of the pyramid should be expressly rejected. Proving that the volume found is indeed minimum can be done in a couple of ways: second differentiation or sign test. In this case, the expression for the first derivative lends itself to a factored form that works best with sign test. Be advised that a proper sign test must show how all significant factors contribute to the sign change. If all constants values are known, justifications using derivative values around the stationary point as quoted from a calculator would also be ac