XJC H2 Math - Set 1 - P1 (ANS)

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 1 1 MATHEMATICS Topic identification and short answers Qn Topic(s) Part Answers 1 Complex numbers (cartesian form) 𝐴=−1 𝐵=1 2 Sequences and series (arithmetic series) (a) 𝑎=𝑚2+(𝑛−1)(𝑚+𝑛)𝑚𝑛 𝑑=−2(𝑚+𝑛)𝑚𝑛 (b) [shown] 3 Complex numbers (modulus, real and imaginary, argument); Graph (sketching, conics section, transformation) (a) [shown] 𝑦=±√3(𝑥−1) (b) (c) −𝜋3<arg(𝑧−1)<𝜋3 Paper 9758/01 Set I – Paper 1

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 1 2 4 Differentiation (implicit functions, equations of tangents) (a) d𝑦d𝑥=2𝑥+3𝑦+42𝑦−3𝑥 (b) [shown] (c) 3𝑥+5𝑦+2=0 11𝑥+28𝑦+46=0 5 Complex numbers (polar form, modulus and argument, Argand diagram) (a) 𝑢−𝑟𝑢+𝑟=itan12𝜃 (b) [shown] Right angle at 𝑊𝑂𝑍 (or 𝑍𝑂𝑊) (c) ∠𝑂𝑊𝑍=12𝜃 |𝑢−𝑟|=2𝑟sin12𝜃 |𝑢+𝑟|=2𝑟cos12𝜃 6 Vectors (ratio theorem, angle between two vectors, scalar products) (a) 𝜆=|𝐛||𝐚|+|𝐛| (b) [shown] 7 Functions (inverse, composite); System of linear equations (a) (i) f−1(𝑥)=𝑥+12𝑥−1, 𝑥∈ℝ, 𝑥≠12 (a) (ii) g(𝑥)=4𝑥−3𝑥−1 (b) (i) h(2)=0.5 h(0.5)=−1 h(−1)=2. (b) (ii) h(𝑥)=1−1𝑥 8 Integration techniques (by substitution, by parts); Definite integrals (area of a region) (a) 𝐼0=𝜋𝑎28 (b) [shown] (c) 5𝜋16 units2 9 Vectors (three dimensions, vector product, magnitude, angle); (a) [shown] 𝑏=9.5 (b) 𝑝=78; 𝑞=78; 𝑟=932 Area≈132.98 units2≈133 units2 (c) Distance ≈14.60407≈14.6 units

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 1 3 10 Differentiation (parametric functions, stationary points, tangents); Graphs (parametric equation, transformations); Definite integrals (volume of revolution) (a) d𝑦d𝑥=𝑡2−12𝑡 (b) 𝑡1=e−1 𝑡2=e (c) Angle=49.6° (d) [shown] 𝑘=𝑎2(e+e−1) (e) Volume=14𝜋𝑎3(8−𝑒2+5𝑒−2) units3 11 Differential equations; Differentiation (local maxima and minima, connected rates of change) (a) (i) d𝑅d𝑡=0.6𝑅−0.4𝑅𝑊 d𝑊d𝑡=−0.8𝑊+0.2𝑅𝑊 (a) (ii) [shown] (b) (i) Rabbit: smallest = 900, largest = 10,700. Wolf: smallest = 300, largest = 4,500. (b) (ii) 𝐵2

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 1 4 Suggested solutions and post-mortem Qn Suggested Solutions Post-mortem 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 This question mainly assesses on complex numbers expressed in cartesian form. Candidates are expected to recall the factor-remainder theorem from O–Level Additional Mathematics, which will lead to a system of two complex equations relating 𝐴 and 𝐵. The remaining steps follow accordingly with algebraic elimination and substitution.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 1 5 2 (a) [4] 𝑆𝑛=𝑚→𝑛2(2𝑎+(𝑛−1)𝑑)=𝑚→2𝑎+(𝑛−1)𝑑=2𝑚𝑛 𝑆𝑚=𝑛→𝑚2(2𝑎+(𝑚−1)𝑑)=𝑛→2𝑎+(𝑚−1)𝑑=2𝑛𝑚 Eliminating 2𝑎, (𝑛−1)𝑑−(𝑚−1)𝑑=2𝑚𝑛−2𝑛𝑚 (𝑛−𝑚)𝑑=2(𝑚2−𝑛2𝑚𝑛)=2(𝑚+𝑛)(𝑚−𝑛)𝑚𝑛 ∴𝑑=−2(𝑚+𝑛)𝑚𝑛 Substituting 𝑑 into 𝑆𝑛, 2𝑎=2𝑚𝑛−(𝑛−1)(−2(𝑚+𝑛)𝑚𝑛)=2𝑚2+2(𝑛−1)(𝑚+𝑛)𝑚𝑛 ∴𝑎=𝑚2+(𝑛−1)(𝑚+𝑛)𝑚𝑛 This question mainly deals with arithmetic series. It should be relatively straightforward to find an expression for 𝑆𝑚 and 𝑆𝑛. Afterwards, elimination and substitution can be done to yield 𝑎 and 𝑑. 2 (b) [2] 𝑆𝑚+𝑛 =𝑚+𝑛2[2(𝑚2+(𝑛−1)(𝑚+𝑛)𝑚𝑛)+(𝑚+𝑛−1)(−2(𝑚+𝑛)𝑚𝑛)] =(𝑚+𝑛)[𝑚2𝑚𝑛+(𝑛−1)(𝑚+𝑛)𝑚𝑛−(𝑚+𝑛−1)(𝑚+𝑛)𝑚𝑛] =(𝑚+𝑛)[𝑚2𝑚𝑛+(𝑚+𝑛)(𝑛−1−𝑚−𝑛+1)𝑚𝑛] =(𝑚+𝑛)[𝑚2𝑚𝑛+(𝑚+𝑛)(−𝑚)𝑚𝑛] =(𝑚+𝑛)[𝑚2−𝑚2−𝑚𝑛𝑚𝑛] =−(𝑚+𝑛) Like the first part, the second part entails making use of the formula for arithmetic series. As a side, a subsequential “show” question such as this one may sometimes prove to be a gainful hindsight, as it helps verify whether previously found results are correct (thereby securing marks in previous parts) so that no errors are carried forward to latter parts (thereby securing marks in latter parts).

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 1 6 3 (a) [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). This question assesses on complex operations, and conic equations, particularly on complex moduli and the equation of a hyperbolic curve respectively. It is perhaps most straightforward to begin substituting 𝑧=𝑥+i𝑦 into the given complex equation. The absence of an imaginary number, coupled with 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 asymptotes of a hyperbola is routine in A–Levels – nothing difficult for well-prepared candidates.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 1 7 3 (b) [2] As hinted by the key phrase “part of the hyperbola”, not all points on in (a) will satisfy the given relationship in 𝑧. The left-hand side of the equation |𝑧+3|=2Re(𝑧) contains a modulus, which is always non-negative, and therefore it follows that Re(𝑧)≥0. Hence, only the part where 𝑥≥0 on the curve is relevant. 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 clearly shown, uncompromised by existing axes. 3 (c) [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 1 8 4 (a) [1] By implicit differentiation, 2𝑥+3𝑦+3𝑥d𝑦d𝑥−2𝑦d𝑦d𝑥+4=0 2𝑥+3𝑦+4=d𝑦d𝑥(2𝑦−3𝑥) ∴d𝑦d𝑥=2𝑥+3𝑦+42𝑦−3𝑥 Implicit differentiation should pose little challenge to careful candidates. 4 (b) [3] At the tangent point (𝑥,𝑦), the tangent equation is 𝑦+4=𝑚(𝑥−6), where 𝑚 is the gradient. Now since d𝑦d𝑥=𝑚, →2𝑥+3𝑦+42𝑦−3𝑥=𝑦+4𝑥−6 →(𝑥−6)(2𝑥+3𝑦+4)=(2𝑦−3𝑥)(𝑦+4) →2𝑥2+3𝑥𝑦+4𝑥−12𝑥−18𝑦−24=2𝑦2−3𝑥𝑦+8𝑦−12𝑥 →2(𝑥2+3𝑥𝑦−𝑦2)+4𝑥−26𝑦−24=0 Since (𝑥,𝑦) lies on 𝐶, we use the equation of 𝐶 to obtain 𝑥2+3𝑥𝑦−𝑦2=1−4𝑥. →2(1−4𝑥)+4𝑥−26𝑦−24=0 →2−4𝑥−26𝑦−24=0 →2𝑥+3𝑦=11 This question assesses candidates on forming tangent line equations and finding relationships between their gradients and their intersection point. Successful candidates will recognise, after equating the gradient function with the gradient of a general line passing through (6,−4), that the equation of 𝑪 itself helps eliminate implicit terms in the intermediate steps. With some algebraic manipulation, the result follows. 4 (c) [3] 2𝑥+3𝑦=11→𝑥=−13𝑦+112 Substituting 𝑥 into the equation of 𝐶, (−13𝑦+112)2+3(−13𝑦+112)𝑦−𝑦2=1−4(−13𝑦+112) Using calculator, 𝑦=−1 or 𝑦=−13. Substituting these 𝑦 values into 2𝑥=3𝑦=11, we have