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 using the product rule A2: A1 for each term B1: Manipulate the answer to the correct format 7i ( )( ) 5f( ) 3 2 1 xx xx= −+ 5𝑥 (3𝑥 − 2)(𝑥 + 1) = 𝐴 3𝑥 − 2 + 𝐵 𝑥 + 1 5𝑥 = 𝐴(𝑥 + 1) + 𝐵(3𝑥 − 2) Sub 𝑥 = −1: −5 = −5𝐵 𝐵 = 1 Sub 𝑥 = 2 3: 5 (2 3) = 5 3 𝐴 𝐴 = 2 5𝑥 (3𝑥 − 2)(𝑥 + 1) = 2 3𝑥 − 2 + 1 𝑥 + 1 B1: Correct partial fraction seen M1: Attempt to find A and B A2: Obtain correct answers 7ii 𝑓′(𝑥) = − 2(3) (3𝑥 − 2)2 − 1 (𝑥 + 1)2 = − 6 (3𝑥 − 2)2 − 1 (𝑥 + 1)2 M1: Attempt chain rule for either term A2: Obtain correct answers 7iii Since (3𝑥 − 2)2 > 0, − 6 (3𝑥−2)2 < 0 and (𝑥 + 1)2 > 0, − 1 (𝑥+1)2 < 0, 𝑓′(𝑥) = − 6 (3𝑥−2)2 − 1 (𝑥+1)2 < 0, therefore the gradient of every point on the curve is negative. B1 B1 8a 𝐿𝐻𝑆 = cot2 𝜃 − 1 cosec2 𝜃 = cosec2 𝜃 − 1 − 1 cosec2 𝜃 = 1 − 2 cosec2 𝜃 = 1 − 2 sin2 𝜃 = cos 2𝜃 B1: Use cot2 𝜃 = cosec2 𝜃 − 1 B1: Use 1 cosec2 𝜃 = sin2 𝜃 B1: Use cos 2𝜃 = 1 − 2 sin2 𝜃 8bi Let 5cos 2sinxx− = ( )cosRx +

Preliminary Examination 2023 Solution Level: Sec 4G2 Qn. # Solution Mark Allocation 5 cos 𝑥 − 2 sin 𝑥 = 𝑅 cos 𝑥 cos 𝛼 − 𝑅 sin 𝑥 sin 𝛼 𝑅 cos 𝛼 = 5 and 𝑅 sin 𝛼 = 2 𝑅 = √52 + 22 = √29 α = tan−1 (2 5) = 21.801° 5 cos 𝑥 − 2 sin 𝑥 = √29cos(𝑥 + 21.8°) M1 M1: Attempt to solve 2 equations in R and α A2: Obtain correct answers 8bii Least value of 8 5cos 2sinxx+− = 8 − √29 = 2.61 Least value occurs when cos(𝑥 + 21.8°) = −1 𝑥 + 21.801° = 180° 𝑥 = 158.2° M1: −√29 A1: Obtain correct answer M1: 𝑥 + 21.801° = 180° A1: Obtain correct answer 9i Point B lies on the x-axis → 𝑦 = 0 𝑥 = 2𝑦 + 14 𝑥 = 2(0) + 14 = 14 Coordinates of B are (14, 0) B1 9ii x-coordinate of X= 2 + 14−2 4 × 3 = 11 y-coordinate of X= −4 + 0−(−4) 4 × 3 = −1 Coordinates of X are (11, −1) M1 A1 9iii Gradient of AX= −1 ÷ 1 2 = −2 At X(11, −1), −1 = (−2)(11) + 𝑐 𝑐 = 21 Equation of AX is 𝑦 = −2𝑥 + 21 ------- (1) Gradient of OA= 1 2 Equation of OA is 𝑦 = 1 2 𝑥 ------- (2) Sub (1) into (2): 1 2 𝑥 = −2𝑥 + 21 2 1 2 𝑥 = 21 𝑥 = 8 2 5 M1 M1 M1 M1 M1

Preliminary Examination 2023 Solution Level: Sec 4G2 Qn. # Solution Mark Allocation Sub 𝑥 = 8 2 5 into (2): 𝑦 = 1 2 (8 2 5) = 4 1 5 Coordinates of A are (8 2 5, 4 1 5) A1 9iv Area of the trapezium OABC= 1 2 |0 2 0 −4 14 8 2 5 0 0 4 1 5 0 | = 1 2 |14 × 4 1 5 − 14 × (−4)| = 1 2 |114 4 5| = 57 2 5 units2 M1 M1 A1 10 Area of shaded region = ∫ (1 2 𝑥 + 14) − (𝑥2 − 4𝑥 + 5) d𝑥 6 0 = ∫ (−𝑥2 + 4 1 2 𝑥 + 9) d𝑥 6 0 = [− 𝑥3 3 + 4 1 2 (𝑥2 2 ) + 9𝑥] 0 6 = [− 63 3 + 4 1 2 (62 2 ) + 9(6)] = 63 units2 M1: Attempt to integrate ∫ ((𝑥2 − 6 0 4𝑥 + 5)) d𝑥 (at least one term correct) A1: Correct result M1: Correct use of limits in their expression A1: 30 units2 B1: When 𝑥 = 0, 𝑦 = 5 or ∫ ( 1 2 𝑥 + 14) d𝑥 6 0 B1: Area of trapezium or substituting limits and getting 93 A1: 63 units2