BBSS 4E AM Prelim P2 2021 Marking Scheme

Habits of Mind: Striving for Accuracy and Precision Bukit Batok Secondary School GCE ‘O’ Level Preliminary Examination 2021 Secondary 4 Express ADDITIONAL MATHEMATICS (Marking Scheme) 4049/02 1 (i) LHS = 1 – 𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 + 𝑠𝑖𝑛𝑠𝑖𝑛 𝜃 1+ 𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 + 𝑠𝑖𝑛𝑠𝑖𝑛 𝜃 = 1 – (1 − 2 𝜃 2 ) + 𝜃 2 𝜃 2 1 + (2 𝜃 2 − 1) + 𝜃 2 𝜃 2 = 2 𝜃 2 + 𝜃 2 𝜃 2 2 𝜃 2 + 𝜃 2 𝜃 2 = 𝜃 2 (𝜃 2 +𝜃 2 ) 𝜃 2 (𝜃 2 +𝜃 2 ) = 𝑡𝑎𝑛 𝑡𝑎𝑛 𝜃 2 = RHS M1 for converting at least one cos θ correctly M1 for converting sin θ correctly M1 for factorizing A1 (ii) 1 + sin θ = 2 cos θ ------ (1) 1 − 𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 + 𝑠𝑖𝑛𝑠𝑖𝑛 𝜃 1+ 𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 + 𝑠𝑖𝑛𝑠𝑖𝑛 𝜃 =𝑡𝑎𝑛 𝑡𝑎𝑛 𝜃 2 --------- (2) Sub (1) into (2): 2𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 − 𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 𝜃 + 𝑐𝑜𝑠𝑐𝑜𝑠 𝜃 =𝑡𝑎𝑛 𝑡𝑎𝑛 𝜃 2 1 3 =𝑡𝑎𝑛 𝑡𝑎𝑛 𝜃 2 (since cos θ ≠ 0) Let β be a basic angle. tan β = 1 3 β ≈ 18.435o 𝜃 2 ≈ 18.435o θ ≈ 36.9o M1 for substitution M1 for simplifying A1 [7] 2 (i) x2 + y2 – 6x + 8y = 0 Centre of circle = (3, – 4) Radius of circle = √32 + (−4)2 = 5 units Diameter of circle = 10 units B1 for centre M1 for radius A1 for diameter (ii) tangent at R: 3x + 4y = 18 y = − 3 4 x + 9 2 Gradient of normal = 4 3 Normal will pass through centre of circle (3, – 4). M1 for making y the subject M1 for gradient of normal

2 BBSS/Prelim/2020/Sec 4NA/AMaths/Paper1 Habits of Mind: Striving for Accuracy and Precision Equation of normal: – 4 = 4 3 (3) + c c = – 8 Equation of normal: y = 4 3 x – 8 A1 (iii) Since the origin (0, 0) satisfies the equation, x2 + y2 – 6x + 8y = 0, Thus the origin lies on the circle. B1 [7] 3 (i) 2x3 + cx2 + dx – 6 = (x2 – 2x – 3) Q(x) + 23x + 21 2x3 + cx2 + dx – 6 = (x – 3) (x + 1) Q(x) + 23x + 21 When x = 3, 2(3)3 + c(3)2 + d(3) – 6 = 23(3) + 21 9c + 3d = 42 3c + d = 14 ----- (1) When x = – 1, 2(– 1)3 + c(– 1)2 + d(– 1) – 6 = 23(– 1) + 21 c – d = 6 ----- (2) (1)+ (2): 4c = 20 c = 5 Sub c = 5 into (1): d = 14 – 15 = – 1 M1 for factorising (x2 – 2x – 3) correctly M1 for sub x = 3 or sub x = – 1 correctly A1 for correct c A1 for correct d (ii) f(x) = 2x3 + 5x2 – x – 6 f(1) = 2(1)3 + 5(1)2 – (1) – 6 = 0 (x – 1) is a factor of f(x). f(x) = (x – 1)(2x2 + 7x + 6) = (x – 1) (2x + 3) (x + 2) M1 for finding one factor M1 for finding another factor by long division or comparing coefficients. A1 [7] 4 (a) (3𝑛) = (3𝑛) √27 = 3 + 𝑛 3 23 = (1+𝑚) 3 M1 for change of base M1 for either numerator or denominator correct A1 (b) √𝑥 + 3 = 1 − 1 2 (1 − 2𝑥) 1 2 (𝑥 + 3) + 1 2 (1 − 2𝑥) = 1 (𝑥 + 3) (1 − 2𝑥) = 2 (x + 3)(1 – 2x) = 22 – 2x2 – 5x – 1 = 0 2x2 + 5x + 1 = 0 x = −5 ±√(5)2−4(2)(1) 2(2) = −5 ±√17 4 ≈ – 0.219 or – 2.28 M1 combining log M1 for removing log M1 for quadratic eq. M1 for formula A1 [8] 5 (i) Height of triangle = √(13𝑥)2−(10𝑥 2 ) 2 M1

3 BBSS/Prelim/2020/Sec 4NA/AMaths/Paper1 Habits of Mind: Striving for Accuracy and Precision = 12x cm Area of triangle = 1 2 (10𝑥)(12𝑥) = 60x2 cm2 Volume of box, 3840 = 60x2 h h = 3840 60𝑥2 = 64 𝑥2 Surface area, A = (13𝑥 + 13𝑥 + 10𝑥)ℎ + 60x2 = (36𝑥) (64 𝑥2) + 60x2 = 2304 𝑥 + 60x2 M1 M1 A1 (ii) A = 2304 𝑥 + 60x2 𝑑𝐴 𝑑𝑥 = − 2304 𝑥2 + 120x − 2304 𝑥2 + 120x = 0 x3 = 2304 120 x = √ 96 5 3 𝑑2𝐴 𝑑𝑥2 = 4608 𝑥3 + 120 When x3 = 2304 120 , 𝑑2𝐴 𝑑𝑥2 = 4608 2304 120 + 120 = 360 > 0 Minimum internal surface area, A = 2304 √96 5 3 + 60(√ 96 5 3 ) 2 ≈ 1290 cm2 M1 A1 M1 M1 A1 [9]

4 BBSS/Prelim/2020/Sec 4NA/AMaths/Paper1 Habits of Mind: Striving for Accuracy and Precision 6 (i) d 1 2 3 4 5 lg P 1.623 1.708 1.792 1.914 1.959 C1 for all points plotted correctly, C1 for straight line. (ii) lg P = 2d lg b + lg a Gradient = 1.959 −1.623 5 − 1 2lg b = 0.084 (acceptable range: 0.07 to 0.1) b = 10 0.042 ≈ 1.10 Vertical intercept = 1.54 lg a = 1.54 a = 10 1.54 M1 A1 M1 for vertical intercept (1.53 to 1.55) A1 lg P d 0 1 2 3 4 5 2.0 1.9 1.8 1.7 1.6 1.5 1.4 x x x x x

5 BBSS/Prelim/2020/Sec 4NA/AMaths/Paper1 Habits of Mind: Striving for Accuracy and Precision ≈ 34.7 (acceptable range: 33.9 to 35.5)

6 BBSS/Prelim/2020/Sec 4NA/AMaths/Paper1 Habits of Mind: Striving for Accuracy and Precision (iii) Incorrect P = 82 When d = 4, lg P = 1.875 P ≈ 75.0 (acceptable range: 73.2 to 76.7) B1 B1 [8] 7 (i) 𝑥3 + 4𝑥2+ 3 (𝑥2+3) (𝑥−1) = 𝑥3 + 4𝑥2+ 3 𝑥3− 𝑥2+ 3𝑥−3 = 1 + 5𝑥2 − 3𝑥 + 6 (𝑥2+3) (𝑥−1) Let 5𝑥2 − 3𝑥 + 6 (𝑥2+3) (𝑥−1) = 𝐴𝑥+𝐵 (𝑥2+3) + 𝐶 (𝑥−1) 5𝑥2 − 3𝑥 + 6 = (Ax + B)(x – 1) + C(x2 + 3) When x = 1, 5 – 3 + 6 = 0 + 4C C = 2 When x = 0, 6 = – B + 3(2) B = 0 When x = 2, 5(2)2 – 3(2) + 6 = 2A + 2(22 + 3) A = 3 𝑥3 + 4𝑥2+ 3 (𝑥2+3) (𝑥−1) = 1 + 3𝑥 (𝑥2+3) + 2 (𝑥−1) M1 for expansion M1 for division M1 M1 for one correct A, B or C Another M1 for one correct A, B or C A1 (ii) 𝑑 𝑑𝑥 [𝑙𝑛 𝑙𝑛 (𝑥2 + 3) ] = 2𝑥 𝑥2 + 3 B1 (iii) ∫ 𝑥3 + 4𝑥2+ 3 (𝑥2+3) (𝑥−1) dx = ∫ [1 + 3𝑥 (𝑥2+3) + 2 (𝑥−1)] dx = x + 3 2 𝑙𝑛 𝑙𝑛 (𝑥2 + 3) + 2 𝑙𝑛 𝑙𝑛 (𝑥 − 1) + c, where c is an arbitrary constant. M1 A1 for 3 2 𝑙𝑛 𝑙𝑛 (𝑥2 + 3) A1 for 2 𝑙𝑛 𝑙𝑛 (𝑥 − 1) A1 for all final correct answer [11] 8 (i) 𝑑2𝑦 𝑑𝑥2 =𝑐𝑜𝑠 𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 𝑐𝑜𝑠 2𝑥 –𝑠𝑖𝑛 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 𝑠𝑖𝑛 2𝑥 – 4 𝑠𝑖𝑛 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑐𝑜𝑠 𝑥 = cos (2x + x) – 2 sin 2x = cos 3x – 2 sin 2x M1 for sin 2x or cos 3x A1 (ii) 𝑑𝑦 𝑑𝑥 = ∫ (𝑐𝑜𝑠 3𝑥 – 2 𝑠𝑖𝑛 2𝑥) 𝑑𝑥 = 1 3 sin 3x + cos 2x + c, where c is an arbitrary constant 𝑦 = ∫ ( 1 3 𝑠𝑖𝑛 3𝑥 + 𝑐𝑜𝑠 2𝑥 + 𝑐) 𝑑𝑥 = − 1 9 cos 3x + 1 2 sin 2x + cx + d, where c and d are arbitrary constants M1 for integration M1 for integration of c M1 for all integration correct

7 BBSS/Prelim/2020/Sec 4NA/AMaths/Paper1 Habits of Mind: Striving for Accuracy and Precision At origin, 0 = − 1 9 cos 0 + 1 2 sin 0 + c(0) + d, d = 1 9 At