SCSS 2023 4NA AM Prelim P2 MS

Preliminary Examination 2023 Solution Level: Sec 4G2 Additional Mathematics Paper 2 (70 marks) Qn. # Solution Mark Allocation 1 3𝑥2 + 7𝑥 − 4 2 26xx−+ 433 50 24x x x+ + − −(3𝑥4 − 6𝑥3 + 18𝑥2) 7𝑥3 − 18𝑥2 + 50𝑥 −(7𝑥3 − 14𝑥2 + 42𝑥) −4𝑥2 + 8𝑥 − 24 −(−4𝑥2 + 8𝑥 − 24) 0 43 2 3 50 24 26 x x x xx + + − −+ = 3𝑥2 + 7𝑥 − 4 M1: Attempt to do long division A1: Obtain 3𝑥2 + 7𝑥 in the quotient A1: Obtain completely correct solution 2 31 3Vx = 𝑑𝑉 𝑑𝑥 = 𝜋𝑥2 When 0.9x= , 𝑑𝑉 𝑑𝑥 = 𝜋(0.9)2 = 0.81𝜋 𝑑𝑉 𝑑𝑡 = 𝑑𝑉 𝑑𝑥 × 𝑑𝑥 𝑑𝑡 0.2 = 0.81𝜋 × 𝑑𝑥 𝑑𝑡 𝑑𝑥 𝑑𝑡 = 0.2 0.81𝜋 = 0.0786 The depth of water in the tank is increasing at a rate of 0.0786 cm/min. B1: 𝑑𝑉 𝑑𝑥 = 𝜋𝑥2 M1: Find required 𝑑𝑉 𝑑𝑥 B1: Used 𝑑𝑉 𝑑𝑡 = 0.2 M1: Chain rule used correctly A1 (with units) 3i 𝑥2 − 6𝑥 + 14 = (𝑥 − 6 2) 2 + 14 − (− 6 2) 2 = (𝑥 − 3)2 + 5 B1: Obtain correct value for a M1: Show correct process for completing square A1: Obtain correct answer 3ii (3, 5) B1 for each coordinate 4 32 4 3 8y x x x= + − −

Preliminary Examination 2023 Solution Level: Sec 4G2 Qn. # Solution Mark Allocation 𝑑𝑦 𝑑𝑥 = 3𝑥2 + 8𝑥 − 3 At 2x=− , 𝑑𝑦 𝑑𝑥 = 3(−2)2 + 8(−2) − 3 = −7 𝑦 = (−2)3 + 4(−2)2 − 3(−2) − 8 = 6 At (−2, 6), 6 = (−7)(−2) + 𝑐 𝑐 = −8 Equation of tangent is 𝑦 = −7𝑥 − 8 M1: Attempt to differentiate A1: All correct A1: 𝑑𝑦 𝑑𝑥 = −7 and 𝑦 = 6 M1: Substitute into 𝑦 = 𝑚𝑥 + 𝑐 A1 5i 32p( ) 6 18x ax x bx= + + − 𝑝(−2) = 0 𝑎(−2)3 + 6(−2)2 + 𝑏(−2) − 18 = 0 −8𝑎 − 2𝑏 + 6 = 0 4𝑎 + 𝑏 = 3 ------ (1) 𝑝(3) = 75 𝑎(3)3 + 6(3)2 + 𝑏(3) − 18 = 75 27𝑎 + 3𝑏 = 39 9𝑎 + 𝑏 = 13 ------ (2) (2) − (1): 5𝑎 = 10 𝑎 = 2 (shown) Sub 𝑎 = 2 into (1): 4(2) + 𝑏 = 3 𝑏 = −5 M1: Any attempt at using 𝑝(−2) = 0 AND 𝑝(3) = 75 to form two equations A1: Both equations correct, need not be simplified DM1: Solving their simultaneous equations, must be 2 unknowns in each A1: Obtain correct answers 5ii p(𝑥) = 2𝑥3 + 6𝑥2 − 5𝑥 − 18 𝑝 (1 2) = 2 (1 2) 3 + 6 (1 2) 2 − 5 (1 2) − 18 = −18 3 4 M1: Attempt to find 𝑝 ( 1 2) for their cubic A1: Obtain correct answer 6a 3 2 13 xy x −= + 𝑑𝑦 𝑑𝑥 = (1 + 3𝑥)(3𝑥2) − (𝑥3 − 2)(3) (1 + 3𝑥)2 = 6𝑥3 + 3𝑥2 + 6 (1 + 3𝑥)2 = 3(2𝑥3 + 𝑥2 + 2) (1 + 3𝑥)2 M1: Attempt to differentiate a quotient or appropriate product A1: Correct denominator A1: Correct numerator

Preliminary Examination 2023 Solution Level: Sec 4G2 Qn. # Solution Mark Allocation 6b ( ) 423 2 1y x x=+ 𝑑𝑦 𝑑𝑥 = 3𝑥2[4(2𝑥 + 1)3(2)] + (2𝑥 + 1)4(6𝑥) = 6𝑥(2𝑥 + 1)3[4𝑥 + (2𝑥 + 1)] = (2𝑥 + 1)3(36𝑥2 + 6𝑥) M1: Attempt to differentiate