ACSI 2023 Y3 Core Mathematics Paper 2 Solution

Anglo - Chinese School (Independent) FINAL EXAMINATION 2023 YEAR 3 INTEGRATED PROGRAMME CORE MATHEMATICS PAPER 2 Thursday 6th October 2023 1 hour 30 minutes ADDITIONAL MATERIALS: Answer Paper (6 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 5 printed pages. [Turn over

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2023/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 t herefore 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) Make y the subject of the formula 23 5 xy y xx + = . [2] (b) Simplify 2 12 1 111 a aa a   −−  +−   , expressing your answer as a single fraction in its simplest form. [4] 𝑥𝑥𝑥𝑥 + 2𝑥𝑥 𝑥𝑥 = 3𝑥𝑥 5 5𝑥𝑥𝑥𝑥 + 10𝑥𝑥 = 3𝑥𝑥2 𝑥𝑥(5𝑥𝑥 + 10) = 3𝑥𝑥2 𝑥𝑥 = 3𝑥𝑥2 5𝑥𝑥 + 10 � 1 𝑎𝑎 + 1 − 2𝑎𝑎 (𝑎𝑎 − 1)(𝑎𝑎 + 1)� �1 − 𝑎𝑎 𝑎𝑎 � = � 𝑎𝑎 − 1 − 2𝑎𝑎 (𝑎𝑎 − 1)(𝑎𝑎 + 1)� �1 − 𝑎𝑎 𝑎𝑎 � = � −1 − 𝑎𝑎 (𝑎𝑎 − 1)(𝑎𝑎 + 1)� �−(𝑎𝑎 − 1) 𝑎𝑎 � = 1 + 𝑎𝑎 𝑎𝑎(𝑎𝑎 + 1) = 1 𝑎𝑎 Do not leave answers in fraction over a fraction for example 𝑥𝑥 = 3 5𝑥𝑥2 (𝑥𝑥+2)

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2023/FinalExamination 3 (c) Evaluate 2lg[(6.82 1.55 10 )] 32.8 −÷× , leaving your answer correct to 2 significant figures. [1] 2. [Maximum mark: 8] Julian paid $168 for x tickets to a theme park. (a) Write down an expression, in terms of x , for the cost of a ticket in dollars. [1] (b) During the peak season, he could buy 7 less tickets with the same amount of money. Write down an expression, in terms of x , for the cost of one ticket during the peak session. [1] (c) Given that the increase in the cost of one ticket during the peak season is $2, form an equation in x and show that it reduces to 2 7 588 0xx−− = . [3] 168 𝑥𝑥 168 𝑥𝑥 − 7 168 𝑥𝑥 − 7 − 168 𝑥𝑥 = 2 168𝑥𝑥 − 168(𝑥𝑥 − 7) 𝑥𝑥(𝑥𝑥 − 7) = 2 168𝑥𝑥 − 168𝑥𝑥 + 1176 = 2𝑥𝑥2 − 14𝑥𝑥 2𝑥𝑥2 − 14𝑥𝑥 − 1176 = 0 𝑥𝑥2 − 7𝑥𝑥 − 588 = 0 −0.041 Order of calculation should be from left to right: 6.82 divide by 1.55 then multiply by 10−2. Many students divide 6.82 by (1.55 × 10−2), treating it as standard form.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2023/FinalExamination 4 (d) How many tickets can he purchase with $200 during the peak season. [3] 𝑥𝑥2 − 7𝑥𝑥 − 588 = 0 𝑥𝑥 = −(−7) ± �(−7)2 − 4(1)(−588) 2(1) 𝑥𝑥 = 7 ± √2401 2 𝑥𝑥 = 7 ± 49 2 𝑥𝑥 = 28 Price of a ticket during peak season = 168 28−7 Price of a ticket during peak season = 8 Number of tickets = 200 8 = 25

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2023/FinalExamination 5 3. [Maximum mark: 4] (a) Find the range of values of k for which t he expression ( 6) 2 34 6x x px p+− +− is always positive for all real values of x . [4] (b) Solve the equation 2 11 12 0xx− += [4] 𝑥𝑥2 + 6𝑥𝑥 − 2𝑝𝑝𝑥𝑥 + 34 − 6𝑝𝑝 = 𝑥𝑥2 + (6 − 2𝑝𝑝)𝑥𝑥 + 34 − 6𝑝𝑝 𝐷𝐷 < 0 (6 − 2𝑝𝑝)2 − 4(1)(34 − 6𝑝𝑝) < 0 36 − 24𝑝𝑝 + 4𝑝𝑝2 − 136 + 24𝑝𝑝 < 0 4𝑝𝑝2 − 100 < 0 𝑝𝑝2 − 25 < 0 (𝑝𝑝 − 5)(𝑝𝑝 + 5) < 0 −5 < 𝑝𝑝 < 5 2𝑥𝑥 + 12 = 11√𝑥𝑥 (2𝑥𝑥 + 12)2 = �11√𝑥𝑥� 2 4𝑥𝑥2 + 48𝑥𝑥 + 144 = 121𝑥𝑥 4𝑥𝑥2 − 73𝑥𝑥 + 144 = 0 𝑥𝑥 = −(−73) ± �(−73)2 − 4(4)(144) 2(4) 𝑥𝑥 = 73 ± 55 8 𝑥𝑥 = 16 or 𝑥𝑥 = 2.25 Students are still unclear about what the Discriminant should be. When they read always positive, they assume that D > 0. DO NOT TAKE SQUARE ROOT LEFT SIDE AND RIGHT SIDE WHEN SOLVING QUADRATIC INEQUALITY Common mistake includes squaring every single term such as: 4𝑥𝑥2 + 144 = 121𝑥𝑥

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2023/FinalExamination 6 (c) Given that 2 77xx+=+ , find the exact values of x , leaving your answers in simplest form. [4] 𝑥𝑥2 − 7 + 𝑥𝑥 − √7 = 0 𝑥𝑥2 − �√7� 2 + 𝑥𝑥 − √7 = 0 (𝑥𝑥 − √7)(𝑥𝑥 + √7) + 𝑥𝑥 − √7 = 0 �𝑥𝑥 − √7��𝑥𝑥 + √7 + 1� = 0 𝑥𝑥 − √7 = 0 or 𝑥𝑥 + √7 + 1 = 0 𝑥𝑥 = √7 𝑥𝑥 = −√7 − 1 Common mistake includes squaring every single term to get rid of √7 such as: 𝑥𝑥4 + 𝑥𝑥2 = 49 + 7

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2023/FinalExamination 7 4. [Maximum mark: 6] During a workshop, water dispenser in the shape of a right circular cylinder of radius 12 cm, is provided for 250 participants. The initial height of water in the container is 90 cm. If each participant drinks once from the dispenser using a conical cup of diameter 5 cm and water level at 7 cm, determine if the water the dispenser has is enough for all the participants. Show your working clearly. 12 cm 90 cm Volume of water in container = 𝜋𝜋(12)2(90) = 40715.0408 Volume of each cup = 1 3 × 𝜋𝜋 × 52 × 7 Volume of each cup = 45.8208 Volume of 250 cups = 11455.208 Volume of 250 cups < Volume in container, Water is sufficient for all participants.

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2023/FinalExamination 8 5. [Maximum mark: 9] (a) Find the equation of the straight line that passes through the points (2, 5) and (0, 1). [2] (b) A quadratic curve has a maximum point at (2, 5) and it passes through the point (0, 1). Find the equation of the quadratic curve. [2] Gradient = 5−1 2−0 Gradient = 2 Equation of line: 𝑥𝑥 − 5 = 2(𝑥𝑥 − 2) 𝑥𝑥 − 5 = 2𝑥𝑥 − 4 𝑥𝑥 = 2𝑥𝑥 + 1 Equation of quadratic curve: 𝑥𝑥 = 𝑎𝑎(𝑥𝑥 − 2)2 + 5 Let 𝑥𝑥 = 0, 𝑥𝑥 = 1 1 = 𝑎𝑎(0 − 2)2 + 5 𝑎𝑎 = −1 Equation of quadratic curve: 𝑥𝑥 = −(𝑥𝑥 − 2)2 + 5

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2023/FinalExamination 9 (c) In the diagram, ABC is a right angled triangle and D is a point on AC such that BD is perpendicular to AC . Given that AD k= , 2DC k= and 32BD= cm, find the length of BC . [5] A B C D k 2k 32 cm Small right angle triangles: 𝐴𝐴𝐴𝐴2 = 𝑘𝑘2 + 322 𝐴𝐴𝐵𝐵2 = (2𝑘𝑘)2 + 322 Big triangle: 𝐴𝐴𝐵𝐵2 = 𝐴𝐴𝐴𝐴2 + 𝐴𝐴𝐵𝐵2 𝐴𝐴𝐵𝐵2 = 𝐴𝐴𝐵𝐵2 − 𝐴𝐴𝐴𝐴2 (2𝑘𝑘)2 + 322 = (3𝑘𝑘)2 − 𝑘𝑘2 − 322 4𝑘𝑘2 + 1024 = 9𝑘𝑘2 − 𝑘𝑘2 − 1024 4𝑘𝑘2 = 2048 𝑘𝑘2 = 512 𝑘𝑘 = 22.63

ACS(Independent)MathDept/Y3IPCoreMathPaper2/2023/FinalExamination 10 6. [Maximum mark: 11] (a) Solve the equation 7 10 37xxee −+= . [5] 7𝑒𝑒𝑥𝑥 + 10 𝑒𝑒𝑥𝑥 = 37 Let 𝑥𝑥 = 𝑒𝑒𝑥𝑥, 7𝑥𝑥 + 10 𝑥𝑥 = 37 7𝑥𝑥2 + 10 = 37𝑥𝑥 7𝑥𝑥2 − 37𝑥𝑥