XJC H2 Math - Set 3 - P1 (ANS)

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 1 1 MATHEMATICS Topic identification and short answers Qn Topic(s) Part Answers 1 Differentiation; Sequences and series (Geometric sequence) 𝑟=−e𝜋 2 Sequences and series (Arithmetic sequence) (a) [shown] (b) 𝑑=2 3 Graphs (sketching); Inequalities (a) (b) 𝑥=0 or 3−√52≤𝑥≤1 or 𝑥≥3+√52 4 Differentiation (maxima and minima) (a) [shown] (b) 𝑥=−1 and 𝑥=2 5 Integration techniques (by parts); Definite integrals (a) ∫{f(𝑥)+g(𝑥)}d𝑥=𝑥ln(ln𝑥)+𝐶 (b) e2ln2−e22+e Paper 9758/01 Set III – Paper 1

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 1 2 6 Differentiation (implicit); Integrals; Graphs (sketching, conics) (a) 𝑘=1 f(𝑥)=𝑥+2𝑥−2 (b) 7 Recurrence relations; Complex numbers (four operations) (a) [shown] Base cases: 𝑢1=1, 𝑢2=5 (b) 𝑟2−2𝑟+5=0 (c) [shown] 𝑚=4 𝛼4+𝛽4=−6 8 Complex numbers (conjugate, argument, modulus, Argand diagram, geometrical effects) (a) [shown] (b) 𝑧3=1 [shown] (c) |𝑧4−𝑧3|=√(5−√52) (d) Required area =58√(10+2√5) units2 (e) Im Re 3 3 3 3 3

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 1 3 9 Differentiation (parametric equations, tangents); Graphs (parametric, conics) (a) d𝑦d𝑥=cos𝜃−1sin𝜃 (b) [shown] (c) 𝑞=𝑝+𝜋 [shown] 𝑦=sin2𝜃 (d) Circle with radius 1 unit and centre (0,0) Clockwise direction 10 Differential equation; Differentiation; Graphs (sketching) (a) d2𝑥d𝑡2+7d𝑥d𝑡+6𝑥=0 (b) 𝑥=4e−𝑡+e−6𝑡 𝑦=2e−𝑡−2e−6𝑡 (c) 11 Functions (inverse, composite) (a) (i) [shown] (a) (ii) Rh−1=[0,1] Df=[0,1] [shown] (a) (iii) Dh−1=[−2,2] Rfh−1=[0,1] [shown] (b) [shown] (c) Q(𝑥)=1−|1−|4𝑥−2||, 0≤𝑥≤1

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 1 4 12 Vectors (three dimensions, normal, angle between two vectors, distances, relationships); Inequalities; Differentiation (implicit, maxima-minima) (a) (√30−1) (b) (001) The orbits are not coplanar. (c) [shown] (d) 14𝜋≤𝑡≤12𝜋 or 54𝜋≤𝑡≤32𝜋 (e) 𝑡=38𝜋,118𝜋 Distance≈2.31 units (f) The perfect condition is not feasible. The statement claims that the distance between the satellite and the starspot is the shortest possible when they are collinear with the centre of the star. However, mathematically the three points cannot be collinear.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 1 5 Suggested solutions and post-mortem Qn Suggested Solutions Post-mortem 1 [3] For stationary points, →d𝑦d𝑥=e𝑥cos𝑥−e𝑥sin𝑥=e𝑥(cos𝑥−sin𝑥)=0 →cos𝑥=sin𝑥 (∵e𝑥>0) →tan𝑥=1 ∴𝑥=14𝜋+𝑛𝜋, where 𝑛∈ℤ Substituting this result into the curve equation, →𝑦=e(14𝜋+𝑛𝜋)cos(14𝜋+𝑛𝜋) →𝑦=e(14𝜋+𝑛𝜋)[cos14𝜋cos𝑛𝜋−sin14𝜋sin𝑛𝜋] →𝑦=e(14𝜋+𝑛𝜋)[√22cos𝑛𝜋−√22(0)] →𝑦=e(14𝜋+𝑛𝜋)[√22(−1)𝑛+1] ∴𝑦=−√22e14𝜋(−e𝜋)𝑛 Letting 𝑛=𝑘+1 and 𝑛=𝑘 to find ratio of the 𝑦-coordinates of two consecutive turning points, →−√22e14𝜋(−e𝜋)𝑘+1 −√22e14𝜋(−e𝜋)𝑘=−e𝜋, which is constant. ∴ The 𝑦-coordinates of the turning points follow a geometric progression. Required common ratio =−e𝜋 This question assesses on differentiation and geometric series. The keyword “turning points” should sufficiently hint at the use of calculus, which will reveal that tan𝑥=1 at all turning points. Finding the required progression afterwards requires candidates to appreciate that there are infinitely many solutions to tan𝑥=1, not just 𝑥=14𝜋. With the general formula for the 𝑥-coordinates of the turning points, candidates can substitute it into the curve equation and reorganise the terms to obtain the required ratio. Note the negative sign in the ratio, which may elude the unwary. This could easily be prevented by cross-checking using GC which will confirm the curve’s oscillation about the 𝑥-axis.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 1 6 2 (a) [2] Given that (𝑢𝑛+1+𝑢𝑛+2+⋯+𝑢2𝑛) is a constant multiple of (𝑢1+𝑢2+⋯+𝑢𝑛), →𝑢𝑛+1+𝑢𝑛+2+⋯+𝑢2𝑛=𝑘(𝑢1+𝑢2+⋯+𝑢𝑛), where 𝑘∈ℝ. →𝑛2[(1+𝑛𝑑)+(1+(2𝑛−1)𝑑)]=𝑘⋅𝑛2[(2(1)+(𝑛−1)𝑑)] →2+(3𝑛−1)𝑑=𝑘(2+(𝑛−1)𝑑) →2+(3𝑛−1)𝑑2+(𝑛−1)𝑑=𝑘, which is constant. This question mainly tests on arithmetic series, in particular the expression for the arithmetic sum. 2 (b) [2] Let 2+(3𝑛−1)𝑑2+(𝑛−1)𝑑=𝑘 →2+(3𝑛−1)𝑑=2𝑘+(𝑛−1)𝑘𝑑 →2+3𝑛𝑑−𝑑=2𝑘+𝑘𝑛𝑑−𝑘𝑑 →3𝑛𝑑−𝑘𝑛𝑑=2𝑘−𝑘𝑑−2+𝑑 →𝑛𝑑(3−𝑘)=𝑘(2−𝑑)−(2−𝑑) →𝑛𝑑(3−𝑘)=(𝑘−1)(2−𝑑) 𝑑 and 𝑘 are constants → (𝑘−1)(2−𝑑) is constant →𝑛𝑑(3−𝑘) is constant for all integers 𝑛≥1 → 𝑛𝑑(3−𝑘)=0 ∴𝑘=3 or 𝑑=0 (rej. ∵ sequence is non-constant) Given 𝑛𝑑(3−𝑘)=0 and 𝑘=3, →0=(2−𝑑)(3−1) →𝑑=2 This question may prove challenging. The keyword “hence” hints the use of the expression in (a), from which candidates are expected to gain further insight. This suggested solution yields a factorised form for a convenient deduction. Most importantly, this question calls for candidates to appreciate that an expression containing constants and variables can only be constant when the expression itself evaluates to 0. Upon recognising this, along with the fact that that the sequence is non-constant, candidates should be able to deduce 𝑑.

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 1 7 3 (a) [2] Substituting 𝑦=−𝑎 to 𝑦2=𝑎𝑥, →(−𝑎)2=𝑎𝑥 →𝑎2=𝑎𝑥 →𝑥=𝑎 Substituting 𝑦=−𝑎 to 𝑎𝑦=𝑥2−2𝑎𝑥, →𝑎(−𝑎)=𝑥2−2𝑎𝑥 →−𝑎2=𝑥2−2𝑎𝑥 →0=𝑥2−2𝑎𝑥−𝑎2=(𝑥−𝑎)2 →𝑥=𝑎 Both curves have a point in common at (𝑎,−𝑎). This question assesses candidates on graph sketching, and tangentially on graph transformation. Candidates may encounter difficulties visualising the effects of the constant 𝑎 to each graph simultaneously. Substituting 𝑎 with a real value to obtain a sketch using GC would be most helpful, but graph transformations may also prove to be a handy tool here: • Consider 𝑦2=𝑥 with 𝑦2=𝑎𝑥. Observe that the two graphs are obtainable by scaling the other graph parallel to the 𝑥-axis. • Consider 𝑦=𝑥2 with 𝑎𝑦=𝑥2−2𝑎𝑥. Observe that the second quadratic equation can be expressed as 𝑦=1𝑎(𝑥−𝑎)2−𝑎. This expression gives insight to not only the transformations between these two graphs, but also the coordinates of the turning point. Marks would be awarded each for the correct shape of both curves, and the coordinates of the turning point of 𝑎𝑦=𝑥2−2𝑎𝑥, which must lie on the curve 𝑦2=𝑎𝑥. This part also aims to confirm two of the four 𝑥-coordinates at which the two curves intersect (𝑥=0 and 𝑥=𝑎). These, along with other information deduced in this question, are relevant in the next part, which may prove much more challenging. 𝑦 𝑥 𝑂 𝑎𝑦=𝑥2−2𝑎𝑥 𝑦2=𝑎𝑥 (𝑎,−𝑎)

X Junior College Preparatory Examinations 9758 Mathematics Suggested Solutions and Post-mortem © X Junior College Set III – Paper 1 8 3 (b) [3] Letting 𝑎=1 yields 𝑦2=𝑥 and 𝑦=𝑥2−2𝑥. Using the sketch in (a), the following are deduced: • 𝑦=√𝑥 is the modulus transformation of 𝑦2=𝑥. • Hence, 𝑦=√𝑥 and 𝑦=|𝑥2−2𝑥| meet at the same 𝑥-coordinate as do 𝑦2=𝑥 and 𝑦=𝑥2−2𝑥. • Hence, 𝑦=√𝑥 and 𝑦=|𝑥2−2𝑥| meet at 𝑥=0 and 𝑥=1. Finding the remaining principal values for the inequality using the graph in (a) as proxy, →𝑥=(𝑥2−2𝑥)2 →0=𝑥4−4𝑥3+4𝑥2−𝑥 →0=𝑥(𝑥3−4𝑥2+4𝑥−1) →0=𝑥(𝑥−1)(𝑥2−3𝑥+1), by considering root 𝑥=1 (from sketch) and comparing coefficient →𝑥=0 or 𝑥=1 or 𝑥=3±√52 ∴𝑥=0 or 3−√52≤𝑥≤1 or 𝑥≥3+√52 This question may prove challenging. Success in this part is highly dependent on the correctness of the answers in part (a). With 𝑎=1, we have the two curves 𝑦2=𝑥 and 𝑦=𝑥2−2𝑥 pertinent to the given inequality. This question requires candidates to appreciate that modulus transformation preserves 𝒙-coordinates. Note that 𝑦2=𝑥 can also be written as 𝑦=±√𝑥, which upon modulus t