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

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 1 1 MATHEMATICS Topic identification and short answers Qn Topic(s) Part Answers 1 Graphs and transformations Required sequence of transformations: 1st: Translation of 12𝜋 units to the positive 𝑥–direction 2nd: Scaling of scale factor 12 parallel to the 𝑥–axis 3rd: Translation of 1 unit to the positive 𝑦–direction 4th: Scaling of scale factor 12 parallel to the 𝑦–axis (Other possible sequences acceptable.) 2 Maclaurin series (i) cos(𝑥+sin𝑥)≈1−2𝑥2+𝑥4 (ii) 𝜋3−𝜋3324+𝜋538880 or 2sin12 3 Sequences and series (arithmetic series) 𝑢𝑛=5+(𝑛−1)3 4 Differentiation (concept); Maclaurin series (small angle approximation) (i) [shown] (ii) [shown] 5 Inequalities (i) [shown] (ii) 𝜋4<𝑥≤𝜋3 or 2𝜋3≤𝑥<3𝜋4 6 Sequences and series (geometric series); Complex numbers (polar form, modulus); Differentiation (i) [shown] (ii) [shown] (iii) [shown] 7 Integration techniques (by parts); Definite integrals (volume of revolution) (i) 𝐼1=tan−1𝑐 (ii) [shown] (iii) Volume=3748𝜋−564𝜋2 units3 Paper 9758/01 Set II – Paper 1

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 1 2 8 Functions (inverse, composite, domain restriction); Inequalities (a) (i) [shown] (a) (ii) Rh=(−∞,𝑞2−6] [shown] (a) (iii) gh(𝑥)=𝑥2−2𝑞𝑥+13|𝑥2−2𝑞𝑥+16|=𝑥2−2𝑞𝑥+13𝑥2−2𝑞𝑥+16 Rgh=[13−𝑞2|16−𝑞2|,1)=[13−𝑞216−𝑞2,1) (Other equivalent forms acceptable.) (b) (i) XY(2)=8 YX−1(5)=7 X−1X−1Y(4)=2 (b) (ii) Both Y−1 and YX does not exist. 9 Graph (parametric); Differentiation; Definite integrals (i) (ii) 3𝛼𝑦2=𝑥(𝑥−𝛼)2 (Other equivalent forms acceptable.) (iii) Surface area=𝜋3𝛼2 units2 𝑦 𝛼 0 𝑥

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 1 3 10 Differential equations; Differentiation (connected rates of change); Graph (i) [shown] (ii) 𝑘=[𝑠𝐴𝜆+(𝑘01−𝑎−𝑠𝐴𝜆)𝑒−𝜆(1−𝑎)𝑡]11−𝑎 (iii) (iv) (Any of) increase 𝑠 / 𝐴 / 𝑎 or decrease 𝜆 Assumption: other constants remain unchanged 11 Vectors (tree dimensions, angle between two lines, cartesian equations, distances) (i) [shown] (ii) [shown] 𝛿=𝜋6 (iii) The direction vector of 𝑚 is not parallel to (10−1). (iv) 𝑃∶(2cos𝜃+2√3,2√2sin𝜃,2cos𝜃−2√3) (v) 𝐶∶(2√3,0,−2√3) |𝐶𝑃⃖⃖⃖⃖⃑|=2√2 units As 𝜃 varies, 𝑃 is a circle with radius 2√2 units centred at 𝐶. (vi) 2√6 units 𝑘 425 𝑂 𝑡 𝑘=649 𝑘=(83−3415𝑒−34𝑡)2

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 1 4 Suggested solutions and post-mortem Qn Suggested Solutions Post-mortem 1 [4] Using trigonometric identities in MF26, sin2𝑥=12(1−cos2𝑥)=12(1+sin(2𝑥−𝜋2)) Transforming from 𝑦=sin𝑥, 1 ←←←→𝑦=sin(𝑥−𝜋2) 2 ←←←→𝑦=sin(2𝑥−𝜋2) 3 ←←←→𝑦=1+sin(2𝑥−𝜋2) 4 ←←←→𝑦=12(1+sin(2𝑥−𝜋2))=sin2𝑥 Required sequence of transformations: 1st: Translation of 12𝜋 units to the positive 𝑥–direction (with 𝑦–axis invariant) 2nd: Scaling of scale factor 12 parallel to the 𝑥–axis 3rd: Translation of 1 unit to the positive 𝑦–direction (with 𝑥–axis invariant) 4th: Scaling of scale factor 12 parallel to the 𝑦–axis (Invariant axes for scaling are optional.) While this question mainly tests on graph transformation, it also requires fluency in trigonometric manipulation. Valid sequences of transformations may vary up to of 5 or 6 steps, but a 4-step sequence is feasible, as hinted by the marks. Great care must be taken to avoid flipping signs or using an inappropriate negative scale factors and/or translation units. As a side, it is a practice in some JCs to include, on top of the scale factor and the parallel axis, the invariant axis as a required scaling property. The invariant axis is the reference line to which the distance from all points on the graph is being scaled. Although invariant axes are an optional detail for graph transformations in A–Levels, it must be noted that scaling can happen with respect to any line normal to the parallel axis.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 1 5 2 (i) [3] cos(𝑥+sin𝑥) =cos(2𝑥−𝑥33!+⋯) =1−(2𝑥−𝑥33!+⋯)2 2!+(2𝑥−𝑥33!+⋯)4 4!−⋯ =1−(2𝑥−𝑥33!+⋯)(2𝑥−𝑥33!+⋯)2+(2𝑥−⋯)424−⋯ =1−12(4𝑥2−23𝑥4+⋯)+124(16𝑥4−⋯)−⋯ =1−2𝑥2+13𝑥4+23𝑥4+⋯≈1−2𝑥2+𝑥4 Maclaurin expansions for nested standard functions such as this one is most amiably done by expanding the inner function first (in this case, the sine function), followed by the outer function (in this case, the cosine function.) Where expansions are abridged, be reminded to use ellipses or approximation signs as appropriate. 2 (ii) [3] Considering expansion ∫[cos(𝑥+sin𝑥)+cos(𝑥−sin𝑥)]𝜋60 d𝑥 =∫1−2𝑥2+𝑥4+cos(𝑥−(𝑥−𝑥36+⋯))𝜋60 d𝑥 =∫1−2𝑥2+𝑥4+cos(𝑥36−⋯)𝜋60 d𝑥 =∫1−2𝑥2+𝑥4+1−(𝑥36−⋯)2 2!+⋯𝜋60 d𝑥 ≈∫2−2𝑥2+𝑥4𝜋60 d𝑥=[2𝑥−23𝑥3+15𝑥5]0𝜋6=𝜋3−𝜋3324+𝜋538880 Otherwise (direct integration) ∫[cos(𝑥+sin𝑥)+cos(𝑥−sin𝑥)]𝜋60 d𝑥 =2∫cos𝑥cos(sin𝑥)𝜋60 d𝑥=2[sin(sin𝑥)]0𝜋6=2sin12 There are two ways to obtain the value required: by further expansion, or by direct integration. However, be wary of the fact that values obtained from abridged expansions, although expressed in exact form, are still approximations. Unless the exact form is obtained from direct integration, approximations require the use of the appropriate sign. As a side, integrals that evaluate to self-nested functions has appeared in past A–Levels and preliminary examinations, one example being ∫1𝑥ln𝑥 d𝑥, from which this question was inspired.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 1 6 3 [6] Sum of the arithmetic progression 𝑆𝑛 is given by: 𝑆𝑛=𝑛2(2𝑎+(𝑛−1)𝑑)=𝑎𝑛+𝑑𝑛2(𝑛−1)=𝑎𝑛+12𝑑𝑛2−12𝑑𝑛 Using 𝑛=𝑝,2𝑝 and 3𝑝, we have: 𝑆𝑝=𝑎𝑝+12𝑑𝑝2−12𝑑𝑝=185 𝑆2𝑝=2𝑎𝑝+2𝑑𝑝2−𝑑𝑝=670 𝑆3𝑝=3𝑎𝑝+92𝑑𝑝2−32𝑑𝑝=1455 Solving the three equations simultaneously using GC, 𝑎𝑝=35+12𝑑𝑝,𝑑𝑝2=300 Considering integer values of 𝑑 and 𝑝 such that 𝑑𝑝2=300, 𝑑 300 75 12 3 𝑝 1 2 5 10 𝑑>𝑝 𝑑>𝑝 𝑑>𝑝 𝒅<𝒑 ∴𝑑=3,𝑝=10 ∴𝑎=3510+32=5 General term 𝑢𝑛 of the arithmetic sequence is: 𝑢𝑛=5+(𝑛−1)3 This question concerns the topic of arithmetic progression and systems of linear equations. With the formula for arithmetic series a system of linear equations in terms of (the products of) 𝑎, 𝑑 and 𝑝, can be obtained. The remaining follows using GC, or through manual derivation by hand. Since there are limited pairs of integers 𝑑 and 𝑝 such that 𝑑𝑝2=300, it is possible to proceed by simply listing out all possible cases to arrive at the final general formula. As a side, this question serves as an example in which the system of linear equation formed does not directly yield the final values of the variables of interest. Such questions may expect candidates to use knowledge from other topics to find the required values.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set II – Paper 1 7 4 [2] From the diagram,f(𝑥+ℎ)−f(𝑥)ℎ is the gradient of the line passing (𝑥,f(𝑥)) and (𝑥+ℎ,f(𝑥+ℎ)). As ℎ→0, the point (𝑥+ℎ,f(𝑥+ℎ)) approaches (𝑥,f(𝑥)) such that the line becomes tangent to the curve at (𝑥,f(𝑥)). The gradient of this tangent corresponds to the derivative of f at that point 𝑥, namely f′(𝑥). This question concerns the concept of differentiation as a slope of the tangent, which is a prerequisite knowledge under O–Level Additional Mathematics. Candidates must show using the sketch that the expression will form a line which increasingly appears tangential as ℎ→0. [4] When f(𝑥)=cos𝑏𝑥, f′(𝑥)=limℎ→0cos𝑏(𝑥+ℎ)−cos𝑏𝑥ℎ =limℎ→0cos(𝑏𝑥)cos(𝑏ℎ)−sin(𝑏𝑥)sin(𝑏ℎ) −cos𝑏𝑥ℎ =limℎ→0cos(𝑏𝑥)[cos(𝑏ℎ)−1]−sin(𝑏𝑥)sin(𝑏ℎ)ℎ Since 𝑏ℎ is small as ℎ→0, using small angle approximation, f′(𝑥) =limℎ→0cos(𝑏𝑥)[1−𝑏2ℎ22−1]−𝑏ℎ sin(𝑏𝑥)ℎ =limℎ→0−𝑏2ℎ2cos(𝑏𝑥)−𝑏si