Queensway Sec Prelim 2024 Add Math P2 Solutions

1 4E AM Prelim 2024 P2 Solutions 1. The Singapore government issued a savings bond in January 2024 with a yield of 2.75% per year. Mr Tan invested $15 000 in the bond. The total amount he will receive, after t years, is given by 𝐴 = 15000(1.0275)𝑡. (a) Calculate the total amount he will receive in January 2030, correct to the nearest dollar. 𝐴 = 15000(1.0275)6 = 17651.52 𝐴𝑚𝑜𝑢𝑛𝑡 = $ 17652 (b) In which year will the amount first exceed $22 000? 22000 = 15000(1.0275)𝑡 1.0275𝑡 = 22 15 𝑡 = ln (22 15) / ln 1.0275 𝑡 = 14.117 Year 2039.

2 2. (a) Without using a calculator, evaluate the value of 6𝑥 given that 22𝑥+6 × 35𝑥−1 = 27𝑥+1 22𝑥+6 × 35𝑥−1 = 33𝑥+3 22𝑥 × 26 × 35𝑥 × 3−1 = 33𝑥 × 33 22𝑥 × 35𝑥 33𝑥 = 33 26 × 3−1 22𝑥 × 32𝑥 = 34 26 62𝑥 = 81 64 6𝑥 = 9 8 (b) Solve the equation 𝑙𝑜𝑔𝑥9 = 5𝑙𝑜𝑔3𝑥, giving your answers correct to 2 significant figures. 𝑙𝑜𝑔𝑥9 = 5𝑙𝑜𝑔3𝑥 𝑙𝑜𝑔39 𝑙𝑜𝑔3𝑥 = 5𝑙𝑜𝑔3𝑥 2𝑙𝑜𝑔33 = 5(𝑙𝑜𝑔3𝑥)2 (𝑙𝑜𝑔3𝑥)2 = 2 5 𝑙𝑜𝑔3𝑥 = ±√2 5 𝑥 = 3 √2 5 𝑜𝑟 3 −√2 5 𝑥 = 2.0 𝑜𝑟 0.50

3 3. (a) A curve has the equation 𝑦 = (𝑝 − 1)𝑥2 + 2(𝑝 − 3), where p is a constant. A line has the equation 𝑦 = 6𝑥 + 3. Find the range of values of p if the curve lies completely above the line. (𝑝 − 1)𝑥2 + 2𝑝 − 6 = 6𝑥 + 3 (𝑝 − 1)𝑥2 − 6𝑥 + 2𝑝 − 9 = 0 𝑏2 − 4𝑎𝑐 < 0 (−6)2 − 4(𝑝 − 1)(2𝑝 − 9) < 0 36 − (4𝑝 − 4)(2𝑝 − 9) < 0 36 − 8𝑝2 + 8𝑝 + 36𝑝 − 36 < 0 −8𝑝2 + 44𝑝 < 0 8𝑝2 − 44𝑝 > 0 4𝑝(2𝑝 − 11) > 0 𝑝 < 0 𝑜𝑟 𝑝 > 5.5 (rejected) (b) By expressing 𝑦 = −2𝑥2 + 10𝑥 − 5 in the form 𝑦 = 𝑎(𝑥 + 𝑏)2 + 𝑐, where a, b and c are constants, find the maximum value of y. 𝑦 = −2(𝑥2 − 5𝑥) − 5 = −2 [𝑥2 − 5𝑥 + (− 5 2) 2 − (− 5 2) 2 ] − 5 = −2[(𝑥 − 2.5)2 − 2.52] − 5 = −2(𝑥 − 2.5)2 + 7.5 Max value of y = 7.5

4 4. (a) In the binomial expansion of (1 − 2 7 𝑥) 𝑛 , the sum of the coefficient of the second and third term is zero. Calculate the value of 𝑛 and hence, find the sixth term. 𝑇2 = (𝑛 1) (1)𝑛−1 (− 2 7 𝑥) 1 = − 2 7 𝑛𝑥 𝑇3= (𝑛 2) (1)𝑛−2 (− 2 7 𝑥) 2 = 𝑛(𝑛 − 1) 2 ( 4 49 𝑥2) − 2 7 𝑛 + 2𝑛(𝑛 − 1) 49 = 0 −14𝑛 + 2𝑛2 − 2𝑛 = 0 2𝑛2 − 16𝑛 = 0 2𝑛(𝑛 − 8) = 0 𝑛 = 0 𝑜𝑟 𝑛 = 8 (rejected) 𝑇6 = (8 5) (1)8−5 (− 2 7 𝑥) 5 = − 256 2401 𝑥5

5 (b) Write down the general term in the binomial expansion of ( 1 𝑥3 − 2𝑥) 8 . Hence, find the value of the constant term in the expansion of (3 + 𝑥2 2 ) 2 ( 1 𝑥3 − 2𝑥) 8 . 𝑇𝑟+1 = (8 𝑟) ( 1 𝑥3) 8−𝑟 (−2𝑥)𝑟 = (8 𝑟) 𝑥−24+3𝑟(−2)𝑟𝑥𝑟 = (8 𝑟) (−2)𝑟𝑥4𝑟−24 (3 + 𝑥2 2 ) 2 = 9 + 3𝑥2 + 𝑥4 4 𝑥4𝑟−24 = 𝑥0 𝑜𝑟 𝑥4𝑟−24 = 𝑥−2 𝑜𝑟 𝑥4𝑟−24 = 𝑥−4 𝑟 = 6 𝑜𝑟 𝑛𝑜 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑣𝑎𝑙𝑢𝑒 𝑜𝑟 𝑟 = 5 𝑇7= (8 6) (−2)6 = 1792 𝑇6= (8 5) (−2)5 = −1792 Constant term = (1792 × 9) + ( 1 4 × −1792) = 15680

6 5. (a) The function f is defined as 𝑓(𝑥) = 𝑝 − 𝑞 sin(𝑟𝑥), for −𝜋 ≤ 𝑥 ≤ 𝜋, where p, q and r are positive integers. Given that the amplitude of the function is 6, the period is  and the maximum value of is 9. (i) State the values of p, q and r. 𝑝 = 3 𝑞 = 6 𝑟 = 2 (ii) Hence, sketch the graph of 𝑓(𝑥). 9 3 −𝜋 − 𝜋 2 𝜋 2 𝜋 −3 - Shape of curve and range - Points plotted correctly

7 (b) The acute angles A and B are such that cot(𝐴 − 𝐵) = 1 3 and cot 𝐴 = 1 5. Without using a calculator, find the exact value of cos 𝐵. cot 𝐴 = 1 5 tan 𝐴 = 5 cot(𝐴 − 𝐵) = 1 3 tan(𝐴 − 𝐵) = 3 tan 𝐴 − tan 𝐵 1 + tan 𝐴 tan 𝐵 = 3 tan 𝐴 − tan 𝐵 = 3 + 3 tan 𝐴 tan 𝐵 5 − tan 𝐵 = 3 + 15 tan 𝐵 16 tan 𝐵 = 2 tan 𝐵 = 1 8 cos 𝐵 = 8 √65 = 8√65 65

8 6. A circle passes through the points (−5, 12) and (9, 14). The centre of the circle lies on the line 2𝑦 + 𝑥 = 15. (a) Find the equation of the circle. 𝑚 = 14 − 12 9 − (−5) = 1 7 𝑚⊥ = −7 𝑚𝑖𝑑𝑝𝑜𝑖𝑛𝑡 = (−5 + 9 2 , 12 + 14 2 ) = (2, 13) 13 = −7(2) + 𝑐 𝑐 = 27 Eqn of perpendicular bisector 𝑦 = −7𝑥 + 27 2(−7𝑥 + 27) + 𝑥 = 15 −14𝑥 + 54 + 𝑥 = 15 −13𝑥 = −39 𝑥 = 3 𝑦 = 6 Centre of circle (3,6) Radius = √(9 − 3)2 + (14 − 6)2 = 10 𝑢𝑛𝑖𝑡𝑠 Equation of circle (𝑥 − 3)2 + (𝑦 − 6)2 = 100 Or 𝑥2 − 6𝑥 + 𝑦2 − 12𝑦 − 55 = 0

9 (b) Explain why the line 𝑦 = 𝑚𝑥 + 6 intersects the circle at 2 distinct points for all values of 𝑚. (𝑥 − 3)2 + (𝑚𝑥 + 6 − 6)2 = 100 𝑥2 − 6𝑥 + 9 + 𝑚2𝑥2 − 100 = 0 (1 + 𝑚2)𝑥2 − 6𝑥 − 91 = 0 𝑏2 − 4𝑎𝑐 = (−6)2 − 4(1 + 𝑚2)(−91) = 400 + 364𝑚2 400 + 364𝑚2 > 0 for all values of m ∴line cuts circle at 2 distinct points 7. (a) A curve has the equation 𝑦 = 3𝑥−5 4𝑥+1 for 𝑥 > 0. Explain, with working, why the curve has no stationary points. 𝑑𝑦 𝑑𝑥 = 3(4𝑥 + 1) − 4(3𝑥 − 5) (4𝑥 + 1)2 = 12𝑥 + 3 − 12𝑥 + 20 (4𝑥 + 1)2 = 23 (4𝑥 + 1)2 Since (4𝑥 + 1)2 ≥ 0 for all x, 𝑑𝑦 𝑑𝑥 > 0 Curve has no stationary point as 𝑑𝑦 𝑑𝑥 ≠ 0.

10 (b) It is given that 𝑓(𝑥) is such that 𝑓′(𝑥) = cos 4𝑥 − 3 sin 2𝑥. Given also that 𝑓(𝜋) = 0, show that 𝑓′′(𝑥) + 4𝑓(𝑥) = −3(sin 4𝑥 + 2). 𝑓(𝑥) = 1 4 𝑠𝑖𝑛4𝑥 + 3 2 𝑐𝑜𝑠2𝑥 + 𝑐 𝑓(𝜋) = 0 3 2 + 𝑐 = 0 𝑐 = − 3 2 𝑓(𝑥) = 1 4 sin 4𝑥 + 3 2 cos 2𝑥 − 3 2 𝑓′′(𝑥) = −4 sin 4𝑥 − 6 cos 2𝑥 𝑓′′(𝑥) + 4𝑓(𝑥) = −4 sin 4𝑥 − 6 cos 2𝑥 + 4 (1 4 sin 4𝑥 + 3 2 cos 2𝑥 − 3 2) = −4 sin 4𝑥 − 6 cos 2𝑥 + sin 4𝑥 + 6 cos 2𝑥 − 6 = −3 sin 4𝑥 − 6 = −3(sin 4𝑥 + 2) (𝑠ℎ𝑜𝑤𝑛) (c) Given that 𝑑 𝑑𝑥 ( 2−𝑥 √1−2𝑥) = 𝑎𝑥+𝑏 √(1−2𝑥)3, find the value of a and b. 𝑑 𝑑𝑥 ( 2 − 𝑥 √1 − 2𝑥 ) = (1 − 2𝑥) 1 2(−1) − (2 − 𝑥) (1 2) (1 − 2𝑥)−1 2(−2) (1 − 2𝑥) = (1 − 2𝑥)−1 2[(−1)(1 − 2𝑥) − (2 − 𝑥) (1 2) (−2) (1 − 2𝑥) = (1 − 2𝑥)−1 2[−1 + 2𝑥 + 2 − 𝑥] (1 − 2𝑥) = 1 + 𝑥 (1 − 2𝑥) 3 2 𝑎 = 1, 𝑏 = 1