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

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 Complex numbers (cartesian form) 𝐴=−1 𝐵=1 2 Complex numbers (modulus, real and imaginary, argument); Graph (sketching, conics section, transformation) (i) [shown] 𝑦=±√3(𝑥−1) (ii) (iii) −𝜋3<arg(𝑧−1)<𝜋3 3 Differentiation (maxima and minima) ℎ=√3𝑟 gives minimum volume. Minimum volume=4√3𝑟3 units3 4 Complex numbers (cartesian form, conjugate) (i) [shown] (ii) 𝑦=1+2𝑥+(2−𝑎22)𝑥2+(43−𝑎2)𝑥3+⋯ (iii) e2𝑥sin𝑎𝑥=𝑎𝑥+2𝑎𝑥2+⋯ 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 5 Vectors (three dimensions, vector product, magnitude, angle); (i) [shown] 𝑏=9.5 (ii) 𝑝=78; 𝑞=78; 𝑟=932 Area≈132.98 units2≈133 units2 (iii) Distance ≈14.60407≈14.6 units Section B: Probability and Statistics [60 marks] 6 Normal distribution (i) 𝜇=6.667𝑎; 𝜎=0.667𝑎 (ii) 7 Binomial distribution (i) [shown] (ii) 𝑝≈0.968 or 0.252 (iii) 12<𝑝<1 8 Discrete random variables (i) [shown] (ii) E(𝑋)=4𝑁2+3𝑁−16𝑁 (iii) [shown] 𝑚=71 9 Probability (i) P(𝐵)=25; P(𝐴∩𝐵)=730 (ii) [shown] (iii) P(𝐴∩𝐵∩𝐶)=115 (iv) 720≤P(𝐴′∩𝐵′∩𝐶′)≤1330 10 Sampling; Hypothesis testing (i) • 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). (ii) [shown] Var(𝑥)=98.75 (iii) 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 (i) (ii) [shown] The square of the residual horizontal distances between the relevant data points and the best-fit line is minimised. (iii) Required value =8.2 degree Celsius (allow marginal error) This value is unreliable. (iv) 𝑡=417.40−259𝑇 𝑟 =−0.9561828875≈ 0.956 (v) 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 [3] By remainder theorem, 𝑃(𝑧)=(𝑧2+1)𝑔(𝑧)+𝐴𝑧+𝐵 Divided by (𝑧+i), remainder is 1+i → 𝑃(−i)=−𝐴i+𝐵=1+i Divided by (𝑧−i), remainder is 1−i→𝑃(i)=𝐴i+𝐵=1−i Adding the two results to eliminate 𝐴i, −𝐴i+𝐵+𝐴i+𝐵=1+i+1−i 2𝐵=2→𝐵=1 Substituting 𝐵=1 into previous result, −𝐴i+1=1+i→𝐴=−1 While this question mainly tests on complex numbers expressed in cartesian form, it also demands familiarity with factor-remainder theorem from O–Level Additional Mathematics. After using the remainders to deduce a system of two complex equations relating 𝐴 and 𝐵, their values can be found accordingly with algebraic elimination and substitution. 2 (i) [3] Writing 𝑧 as 𝑥+i𝑦, where 𝑥=Re(𝑧) and 𝑦=Im(𝑧), |𝑥+i𝑦+3|=|(𝑥+3)+i𝑦|=2𝑥 ∴√(𝑥+3)2+𝑦2=2𝑥 Squaring both sides, 𝑥2+6𝑥+9+𝑦2=4𝑥2 3𝑥2−6𝑥−9−𝑦2=0 3𝑥2−6𝑥+3−𝑦2=12 3(𝑥2−2𝑥+1)−𝑦2=12 (𝑥−1)24−𝑦212=1 Finding asymptotes, (𝑥−1)24=𝑦212 𝑦=±√3(𝑥−1) ∴ 𝑥 and 𝑦 are related by the equation of the hyperbola with asymptotes 𝑦=±√3(𝑥−1). It is perhaps most straightforward to begin by writing 𝑧 in its cartesian form to uncover the relationship between 𝑥 and 𝑦 from the given complex equation. Afterwards, the absence of an imaginary number and the presence of a squared term in the cartesian equation of a hyperbola should sufficiently hint the use of modulus operation. The remaining part on finding the asymptotes is routine in A–Levels and can be deduced from this cartesian equation.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 2 5 2 (ii) [2] As hinted by the key phrase “part of the hyperbola”, not all points on in (i) will satisfy the given relationship in 𝑧. The left-hand side of the equation |𝑧+3|=2Re(𝑧) contains a modulus, which can only be positive, and therefore implies for the right-hand side that Re(𝑧)≥0. Since 𝑥=Re(𝑧), it should be apparent that the part of the hyperbola that is relevant in the sketch is the part where 𝑥≥0. A handy technique when sketching a graph with asymptotes is to start with the asymptotes, followed by the graph itself, and finally the two axes. When done in this order, the asymptotical behaviour of the graph is prioritised in the sketch and not compromised by any existing axes. 2 (iii) [2] From the sketch in (ii), it can be seen that arg(𝑧−1) has an asymptotic behaviour. ∴−tan−1(√3)<arg(𝑧−1)<tan−1(√3) ∴−𝜋3<arg(𝑧−1)<𝜋3 It may be useful to consider graph transformation in this question. Note that the value of arg𝑧 is measured with respect to the origin 𝑂, which implies that arg(𝑧−1) takes reference from the point (1,0). The range of values can then be deduced through the appropriate translation. The relevant angles can be obtained from the gradient of the asymptotes. Candidates may find it noteworthy that angle calculations using gradient values are becoming more common in recent A–Levels. 𝑦 𝑥 𝑂 𝑦=√3(𝑥−1) 𝑦=−√3(𝑥−1) 1 √3 −√3

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. As is the trend for most optimisation problems, this question concerns packing geometrical shapes, and hence would call for some level of fluency in geometry and trigonometry that is considered routine in H2 Mathematics. When finding an expression for the volume of the octahedron in the beginning, caution is advised when relying on only the diagram given in the question, as there is risk of misinterpretation – in this case, potentially mistaking 𝑟 to be equal to half of the side of the pyramid’s square base. It is advisable to produce a preliminary cross-sectional diagram to cross-check the contact points of the inscribed sphere with the octahedron. All relevant lengths can be deduced using geometry, and eventually the volume of the inscribing octahedron can be obtained in terms of ℎ and 𝑟, which is to be differentiated accordingly. When differentiating expressions involving many letters, candidates must be attentive in distinguishing the variable of differentiation from the remaining letters which represent 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 conside