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.