ACSI 2017 Y3 Core Mathematics Paper 2

FINAL EXAMINATION 2017 YEAR 3 INTEGRATED PROGRAMME CORE MATHEMATICS PAPER 2 MONDAY 9th October 2017 1 hour 30 minutes 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 5 printed pages. [Turn over

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2017/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: 10] (a) It is given that 2 2 1 atuts += . (i) Express a in terms of s , u and t . [2] (ii) Find the value of a when 4.12=s , 256.0=u and 25.0=t , leaving your answer corrected to the nearest whole number. [2] (b) Solve the equation 8 1 1 4 −=− y y , leaving your answers in the simplest surds. [3] (c) A man bought m pencils at r dollars per dozen. He sold them for k cents each. Find an expression, in terms of m , r and k , for the profit, in cents, that he made. [3] 2 [Maximum mark: 6] Find the value(s) of k for which the line kyx =+ is a tangent to the curve 53 22 =+− yxx . [6] 3 [Maximum mark: 11] Solve the following equations: (a) 7323 6482 +−+ = xxx [3] (b) 3 122loglog 82 =+ xx [4] (c) Solve the equation t t ee 721 =+− , giving your answer correct to 2 decimal places. [4]

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2017/FinalExamination 3 4 [Maximum mark: 7] (a) The sum S of the first n integers is given by the formula )1(2 1 += nnS . What is the minimum number of integers required in order for the sum to exceed 325? [3] (b) Find the values of the integers a and b such that 52 525 525 5 + += + + ba . [4] 5 [Maximum mark: 10] In the diagram, the points A, B, C and D are on level ground. The point D is equidistant from A and C and ∠ADC = 90o. C is due north of A, AC = 37 m, BC = 40 m and ∠BCA = 55o. Calculate (a) the area of △ 𝐴𝐵𝐶, [2] (b) ∡𝐶𝐵𝐴, [4] (c) the distance AD, [2] (d) the bearing of A from B. [2] D North A B C 40 m 37 m 55o

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2017/FinalExamination 4 6 [Maximum mark: 12] The diagram below shows the curve of a quadratic function that cuts the y-axis at C and the x-axis at B and A where 3=x . It has a maximum point at (1, 8). A straight line passes through C and A. (a) Find the equation of the curve. [4] (b) Find the equation of the perpendicular bisector of AC. [4] (c) The perpendicular bisector of AC intersects the line AC at D and the y-axis at E. Find the area of triangle CDE. [4] 7 [Maximum mark: 7] Given that the equation 042 =++ hxx has two real roots, (i) find the range of possible values of h . [2] (ii) find the value of h if the two roots are  and  where 34=− . [5] C B A 3 (1, 8) y x

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2017/FinalExamination 5 8 [Maximum mark: 13] A closed rectangular box of height h cm has a horizontal rectangular base of sides 3x cm and 2x cm, where 61  x . (a) If the volume of the box is 200 cm3, express h in terms of x and show that the total surface area, A cm2, of the box is given by xxA 3 100012 2+= . [4] The table below shows some values of x & the corresponding values of A (correct to the nearest integer). x 1 2 3 4 5 6 A 345 215 a 275 367 488 (b) Calculate the value of a. [1] (c) Taking 2 cm to represent 1 unit on the horizontal axis and 2 cm to represent 50 units on the vertical axis, draw the graph of xxA 3 100012 2+= for 61  x . [4] (d) From the graph, estimate (i) the minimum value of A, and the corresponding height of the box, (ii) the range of values of x for which 300A . [4] 9 [Maximum mark: 4] The roots,  and  , of the equation 02 =++ cbxax are in the ratio 1 : n . Show that 22)1( nbacn =+ . [4] --------------------------------------------------------END OF PAPER---------------------------------------------------

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2017/FinalExamination 6 1ai) 𝑠 = 𝑢𝑡 + 1 2 𝑎𝑡2 𝑠 − 𝑢𝑡 = 1 2 𝑎𝑡2 2𝑠 − 2𝑢𝑡 = 𝑎𝑡2 𝑎 = 2𝑠 − 2𝑢𝑡 𝑡2 = 2(𝑠 − 𝑢𝑡) 𝑡2 1aii) 𝑎 = 2(12.4) − 2(0.256)(0.25) 0.252 𝑎 = 395 1b) 4 𝑦 − 1 = 𝑦 − 1 8 (𝑦 − 1)2 = 32 𝑦 − 1 = ±4√2 𝑦 = 1 ± 4√2 1c) Cost of each pencil = 𝑚𝑟100 12 Selling price = 𝑚𝑘 Profit = 𝑚𝑘 − 𝑚𝑟100 12 = 𝑚(𝑘 − 25 3 𝑟) 2) 𝑥 = 𝑘 − 𝑦 (𝑘 − 𝑦)2 − 3(𝑘 − 𝑦) + 𝑦2 = 5 𝑘2 − 2𝑘𝑦 + 𝑦2 − 3𝑘 + 3𝑦 + 𝑦2 − 5 = 0 2𝑦2 + 3𝑦 − 2𝑘𝑦 − 5 − 3𝑘 + 𝑘2 = 0 2𝑦2 + (3 − 2𝑘)𝑦 − 5 − 3𝑘 + 𝑘2 = 0 (3 − 2𝑘)2 − 4(2)(𝑘2 − 3𝑘 − 5) = 0 9 − 12𝑘 + 4𝑘2 − 8𝑘2 + 24𝑘 + 40 = 0 −4𝑘2 + 12𝑘 + 49 = 0 4𝑘2 − 12𝑘 − 49 = 0 𝑘 = 5.31 or 𝑘 = −2.31 3a) 2𝑥+3 × 26𝑥−9 = 26𝑥+42 27𝑥−6 = 26𝑥+42

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2017/FinalExamination 7 7𝑥 − 6 = 6𝑥 + 42 𝑥 = 48 3b) log2 𝑥 + log8 2𝑥 = 2 1 3 log8 𝑥3 + log8 2𝑥 = 2 1 3 log8 2𝑥4 = 2 1 3 2𝑥4 = 128 𝑥4 = 64 𝑥 = 2.83 3c) 𝑒1−𝑡 + 2 = 7 𝑒𝑡 𝑒 𝑒𝑡 + 2 = 7 𝑒𝑡 𝑒 + 2𝑒𝑡 = 7 𝑒𝑡 = 7 − 𝑒 2 ln 𝑒𝑡 = ln (7 − 𝑒 2 ) 𝑡 = 0.761 4a) 1 2 𝑛(𝑛 + 1) > 325 𝑛2 + 𝑛 − 650 > 0 (𝑛 − 25)(𝑛 + 26) > 0 𝑛 < −26 or 𝑛 > 25 Hence, minimum number of integers = 26 4b) 𝑎 + 𝑏√5 5 + 2√5 = 5 + 2√5 2 + √5 𝑎 + 𝑏√5 = (5 + 2√5)2 2 + √5 𝑎 + 𝑏√5 = 45 + 20√5 2 + √5

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2017/FinalExamination 8 𝑎 + 𝑏√5 = 45 + 20√5 2 + √5 × 2 − √5 2 − √5 𝑎 + 𝑏√5 = 90 − 45√5 + 40√5 − 100 −1 𝑎 + 𝑏√5 = 10 + 5√5 5a) Area of triangle = 1 2 𝑎𝑏 sin 𝐶 Area of triangle = 1 2 (37)(40) sin 55 Area of triangle = 606 5b) 𝐴𝐵2 = 372 + 402 − 2(37)(40) cos 55 𝐴𝐵2 = 1271.213 𝐴𝐵 = 35.65 sin 55 35.65 = sin ∡𝐶𝐵𝐴 37 sin ∡𝐶𝐵𝐴 = 0.85017 ∡𝐶𝐵𝐴 = 58.2𝑜 5c) cos 45𝑜 = 𝐴𝐷 37 𝐴𝐷 = 26.2 5d) Alternate Angle = 55𝑜 Bearing = 360 − 55 + 58.2 Bearing = 246.8𝑜 7a) 𝑦 = 𝑎(𝑥 − 1)2 + 8 When 𝑥 = 3 and 𝑦 = 0, 0 = 𝑎(3 − 1)2 + 8 −8 = 4𝑎 𝑎 = −2 𝑦 = −2(𝑥 − 1)2 + 8 7b) When 𝑥 = 0, 𝑦 = 6 Coordinate of C(0, 6)

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2017/FinalExamination 9 Midpoint of AC = ( 0+3 2 , 6+0 2 ) Midpoint of AC = ( 3 2 , 3) Gradient of AC = − 6 3 Gradient of AC = −2 Gradient of perpendicular bisector = 1 2 Equation is: 𝑦 − 3 = 1 2 (𝑥 − 3 2) 𝑦 = 1 2 𝑥 + 9 4 7c) D ( 3 2 , 3) When 𝑥 = 0, 𝑦 = 9 4 Area of triangle CDE = 1 2 (6 − 9 4)( 3 2) Area of triangle CDE = 2.81 8ai) 42 − 4(1)(ℎ) > 0 16 − 4ℎ > 0 −4ℎ > −16 ℎ < 4 8aii) 𝛼 + 𝛽 = −4 𝛼𝛽 = ℎ 𝛼 − 𝛽 = √(𝛼 + 𝛽)2 − 4𝛼𝛽 4√3 = √(−4)2 − 4ℎ 16(3) = 16 − 4ℎ 4ℎ = −32 ℎ = −8 9a) 3𝑥 × 2𝑥 × ℎ = 200 ℎ = 200 6𝑥2 Area = 2(3𝑥)(2𝑥) + 2(2𝑥)(ℎ) + 2(3𝑥)(ℎ) 𝛼 + 𝛽 = −4 𝛼2 + 𝛽2 + 2𝛼𝛽 = 16 ------(1) 𝛼 − 𝛽 = 4√3 𝛼2 + 𝛽2 − 2𝛼𝛽 = 48 ----- (2) (1) – (2): 4𝛼𝛽 = −32 𝛼𝛽 = −8

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2017/FinalExa