ACSI 2020 Y3 Core Mathematics Paper 2 Solution

FINAL EXAMINATION 2020 YEAR 3 INTEGRATED PROGRAMME CORE MATHEMATICS PAPER 2 MONDAY xxth October 2020 1 hour 30 minutes ADDITIONAL MATERIALS: Answer Paper (7 sheets) Graph Paper ( 1 sheet) INSTRUCTIONS TO STUDENTS Do not open this examination paper until instructed to do so. A calculator is required for this paper. Answer all the questions on the answer sheets provided. At the end of the examination, fasten the answer sheets together. Unless otherwise stated in the question, all numerical answers must be given exactly or correct to three significant figures. Answers in degrees are to be given to one decimal place. INFORMATION FOR STUDENTS The maximum mark for this paper is 80. _____________________________________________________ This question paper consists of 4 printed pages. [Turn over

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2020/FinalExamination 2 Full marks are not necessarily awarded for a correct answer with no working. Answers must be supported by working and/or explanations. Where an answer is incorrect, some marks may be given for correct method, provided this is shown by written working. You are therefore advised to show all working. Answer all the questions on the answer sheets provided. Begin each question on a new page. 1. [Maximum mark: 6] (a) Evaluate 2 4 (0.3578) 7.647 43.96  , leaving your answer correct to 3 significant figures. [2] 0.354019 2.57492 = 0.137 (b) Express 2 1 1 10 3 2 2 4 x x x x ++++ − − as a single fraction in its simplest form. [4] 1 𝑥 + 2 + 1 2 − 𝑥 + 10 + 3𝑥 𝑥2 − 4 = 1 𝑥 + 2 − 1 𝑥 − 2 + 10 + 3𝑥 (𝑥 − 2)(𝑥 + 2) = 𝑥 − 2 − 𝑥 − 2 + 10 + 3𝑥 (𝑥 − 2)(𝑥 + 2) = 6 + 3𝑥 (𝑥 − 2)(𝑥 + 2) = 3 (𝑥 − 2) Some student take square of the expression to get rid of the square root first. They need to know that this is an expression, hence they cannot do that! Very careless 1 2 − 𝑥 = − 1 𝑥 − 2 Some students did not leave it in the simplest form

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2020/FinalExamination 3 2 [Maximum mark: 7] (a) Solve the equation 923 2x x−= , giving your answers correct to two decimal places. [4] 2𝑥(2𝑥) − 9 = 3(2𝑥) 4𝑥2 − 9 = 6𝑥 4𝑥2 − 6𝑥 − 9 = 0 𝑥 = −(−6) ± √(−6)2 − 4(4)(−9) 2(4) 𝑥 = −0.93 or 𝑥 = 2.43 (b) Expand and simplify 2 1 2 1 2 4 3 3 3 3 3 3p q q q p p  + − +      . [3] 𝑝 2 3𝑞 2 3 − 𝑞 1 3𝑝 4 3 + 𝑝2 + 𝑞 − 𝑞 2 3𝑝 2 3 + 𝑞 1 3𝑝 4 3 = 𝑝2 + 𝑞 3 [Maximum mark: 9] (a) Simplify ( ) ( ) 32 12 44 2 3 24 (4 ) xx xx − −  , expressing your answer in positive indices. [3] 8𝑥−6 × 4𝑥12 16𝑥−8 × 𝑥12 = 32𝑥6 16𝑥4 = 2𝑥2 The question specifically asked for 2 dp. Some student left it as 3 sf. Easy question, but surprising very poorly done. Many tried to factorize the expression instead of manually expanding it. A lot of misconceptions such as: (2𝑥−2)3 = 2𝑥−6 = 1 2𝑥6 (4𝑥−4)2 = 4𝑥−8

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2020/FinalExamination 4 (b) Given that 1 2 2 21 16 2(2 ) 28 xx xx k + −+ + = , find the value of k . [3] 16𝑥(16) + 2(24𝑥) 2𝑥2−223𝑥23 = 24𝑥(16) + 2(24𝑥) 2(24𝑥) = 24𝑥(16 + 2) 2(24𝑥) = 18 2 = 9 (c) Find the value of 1 1 1 log log logpqrpqr pqr pqr++ . [3] log 𝑝𝑝𝑞𝑟 + log 𝑞𝑝𝑞𝑟 + log 𝑟𝑝𝑞𝑟 = log (𝑝𝑞𝑟)𝑝𝑞𝑟 = 1 4 [Maximum mark: 15] The diagram shows four towns A, B, C and D. Given that Town C and Town D lie west of B, AB = 8.2 km, BC = 8.8 km, CD = 9.3 km and 118oACD= , calculate A B C D 8.2 8.8 9.3 118o Very careless. Mistakes such as the following are very common: 16𝑥+1 = 24𝑥+1 2(22𝑥)2 = 2(44𝑥) = 84𝑥 Poorly attempted for a very simple question. Many just split the denominator into 3 separate terms or just tried to combine the 3 fractions using common denominator.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2020/FinalExamination 5 (a) ABC , [4] ∡𝐴𝐶𝐵 = 180 − 118 ∡𝐴𝐶𝐵 = 62 sin 62 8.2 = sin ∡𝐶𝐴𝐵 8.8 sin ∡𝐶𝐴𝐵 = 0.947554 ∡𝐶𝐴𝐵 = 71.4 ∡𝐴𝐵𝐶 = 180 − 71.4 − 62 ∡𝐴𝐵𝐶 = 46.6 (b) the bearing of B from A, [2] 90 − 46.6 = 43.4 180 − 43.4 = 136.6 (c) the distance of AD. [3] 𝐴𝐷2 = 18.12 + 8.22 − 2 ∗ 18.1 ∗ 8.2 ∗ cos 46.6 𝐴𝐷2 = 190.895 𝐴𝐷 = 13.8 Alternative methods such as finding the length of AC first, then use consine rule and then sine rule again is quite common. Students should try to use the easiest method to solve a problem. Alternative: 90 + 46.6 = 136.6 Well attempted for students who has gotten part (a) and (b) correct. Very direct and straight forward.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2020/FinalExamination 6 A cyclist starts from D at 1030 and travels towards A at a constant speed of 15 km/h. (d) Find the time, to the nearest minute, when he will be nearest to Town C. [6] 𝐴𝐶 sin 46.6 = 8.2 sin 62 𝐴𝐶 = 6.74 Area of triangle ACD = 1 2 × 9.3 × 6.74 × sin 118 Area of triangle ACD = 27.7 1 2 × 13.8 × ℎ = 27.7 ℎ = 4.02 Distance cycled = √9.32 − 4.022 Distance cycled = 8.39 Time cycled = 8.39 15 = 0.559 hr = 34 minutes Time nearest = 1104 5 [Maximum mark: 13] (a) The equation of a curve is 23 3 12y qx px q= − + where p and q are positive integers. Show that the x− axis is the tangent to the curve if 22 p q = . [4] (−3𝑝)2 − 4(3𝑞)(12𝑞) = 0 9𝑝2 − 144𝑞2 = 0 𝑝2 𝑞2 = 144 9 𝑝 𝑞 = 12 3 𝑝 2𝑞 = 2 • This is a 6 mark question, some students just used answer in part (c) to divide by speed. • Many students failed to find the perpendicular distance. • GOOD: Alternative method includes finding angle ADC and then use cosine to find the distance travelled. • Question is asking for the time, not just the time taken. Some students are unable to explain why they need to let D = 0. Alternative method includes attempting to complete the square and show that the turning point is at (2, 0), hence x-axis is a tangent.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2020/FinalExamination 7 (b) Explain why the line 2y kx=− will always intersect the curve 3 3y x= − . [4] 𝑘𝑥 − 2 = 3 𝑥 − 3 𝑘𝑥2 − 3𝑘𝑥 − 2𝑥 + 6 = 3 𝑘𝑥2 − 3𝑘𝑥 − 2𝑥 + 3 = 0 D = [−(3𝑘 + 2)]2 − 4(𝑘)(3) = 9𝑘2 + 12𝑘 + 4 − 12𝑘 = 9𝑘2 + 4 Since 𝐷 > 0, hence the line always intersect the curve. (c) The roots of ( 1)(3 )x x m− − = are  and  . (i) Find the value of + and  in terms of m . (ii) Given that ( ) 2 1 1 4   += , find the value of m . [5] 3𝑥 − 𝑥2 − 3 + 𝑥 = 𝑚 −𝑥2 + 4𝑥 − 3 − 𝑚 = 0 𝑥2 − 4𝑥 + 3 + 𝑚 = 0 𝛼 + 𝛽 = 4 𝛼𝛽 = 3 + 𝑚 Very careless in manipulation. Many students are not able to expand (−3𝑘 − 2)2 correctly! Hence, leading to the wrong conclusion or unable to explain why the two lines always intersect. They expanded it as: (−3𝑘 − 2)2 = 9𝑘2 − 12𝑘 + 4 Very poorly attempted. Students must expand and ensure that the right hand side of the equation is 0 before reading off the a, b and c of the quadratic equation.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2020/FinalExamination 8 1 𝛼 + 1 𝛽 = 4 (𝛼𝛽)2 𝛼 + 𝛽 𝛼𝛽 = 4 (𝛼𝛽)2 𝛼 + 𝛽 = 4 𝛼𝛽 4 = 4 3 + 𝑚 3 + 𝑚 = 1 𝑚 = −2 6 [Maximum mark: 18] (a) Evaluate 32log 2 log 3+ . [2] lg 2 lg 3 + lg 3 lg 2 = 2.215 (b) Find the value of k if 2 5 kee + = . [3] 𝑒𝑘+2 = ln 5 𝑘 + 2 = ln(ln 5) 𝑘 = ln(ln 5) − 2 𝑘 = −1.52 Simple question but many failed to see that they can only obtain the value by changing to base 10 or base e. Hence, the question is poorly attempted. Surprisingly, this was much better than part (a).