XJC H2 Mathematics 9758 – Set I: Paper 1 (Answers)

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 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 Sequences and series (arithmetic series) (i) 𝑎=𝑚2+(𝑛−1)(𝑚+𝑛)𝑚𝑛 𝑑=−2(𝑚+𝑛)𝑚𝑛 (ii) [shown] 4 Differentiation (implicit functions, equations of tangents) (i) d𝑦d𝑥=2𝑥+3𝑦+42𝑦−3𝑥 (ii) [shown] (iii) 3𝑥+5𝑦+2=0 11𝑥+28𝑦+46=0 5 Complex numbers (polar form, modulus and argument, Argand diagram) (i) 𝑢−𝑣𝑢+𝑣=itan(𝛼−𝛽2) (ii) [shown] Right angle at 𝑊𝑂𝑍 (or 𝑍𝑂𝑊) (iii) 𝜃=𝛼−𝛽2 |𝑢−𝑣|=2𝑟sin𝜃 |𝑢+𝑣|=2𝑟cos𝜃 6 Vectors (ratio theorem, angle between two vectors, scalar products) (i) 𝜆=|𝐛||𝐚|+|𝐛| (ii) [shown] 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 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) (i) 𝐼0=𝜋𝑎28 (ii) [shown] (iii) 5𝜋16 units2 9 Differentiation (parametric functions, stationary points, tangents); Graphs (parametric equation, transformations); Definite integrals (volume of revolution) (i) d𝑦d𝑥=𝑡2−12𝑡 (ii) 𝑡1=e−1 𝑡2=e (iii) Angle=49.6° (iv) [shown] 𝑘=𝑎2(e+e−1) (v) Volume=𝜋𝑎34(8−𝑒2+5𝑒−2) units3 10 Differential equations; Differentiation (local maxima and minima, connected rates of change) (i) (a) d𝑅d𝑡=0.6𝑅−0.4𝑅𝑊 d𝑊d𝑡=−0.8𝑊+0.2𝑅𝑊 (i) (b) [shown] (ii) (a) Rabbit: smallest = 900, largest = 10,700. Wolf: smallest = 300, largest = 4,500. (ii) (b) 𝐵2 11 Sequences and series (geometric series) (i) End of July 2026 (ii) [shown] Amount=$9,477.56 (iii) August 2029

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 1 3 Suggested solutions and post-mortem Qn Suggested Solutions Post-mortem 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 1 4 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 1 5 3 (i) [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. 3 (ii) [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 4 (i) [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 (ii) [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 (iii) [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 the following: • When 𝑦=−1, 𝑥=1 and 𝑚=−35 → equation of tangent: 3𝑥+5𝑦+2=0 • When 𝑦=−13, 𝑥=−103 and 𝑚=−1128 → equation of tangent: 11𝑥+28𝑦+46=0 After substituting into 𝐶 accordingly, the resulting quadratic equation can be solved with ease using calculator, though proceeding manually is welcomed albeit not required. The remaining steps follow.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set I – Paper 1 7 5 (i) [4] Write 𝑢=𝑟(cos𝛼+isin𝛼) and 𝑣=𝑟(cos𝛽+isin𝛽). 𝑢−𝑣𝑢+𝑣 =𝑟(cos𝛼+isin𝛼)−𝑟(cos𝛽+isin𝛽)𝑟(cos𝛼+isin𝛼)+𝑟(cos𝛽+isin𝛽) =𝑟[cos𝛼−cos𝛽+i(sin𝛼−isin𝛽)]𝑟[cos𝛼+cos𝛽+i(sin𝛼+isin𝛽)] =−2sin(𝛼+𝛽2)sin(𝛼−𝛽2)+i[2cos(𝛼+𝛽2)sin(𝛼−𝛽2)]2cos(𝛼+𝛽2)cos(𝛼−𝛽2)+i[2sin(𝛼+𝛽2)cos(𝛼−𝛽2)] =2sin(𝛼−𝛽2)[−sin(𝛼+𝛽2)+icos(𝛼+𝛽2)]2cos(𝛼−𝛽2)[cos(𝛼+𝛽2)+isin(𝛼+𝛽2)] =it