JVSS 4E A Math Prelim 2021 Paper 02 MS

JVSS/PRELIM2021/SEC4E/AM/4049/P2/MARKINGSCHEME Sec 4E A Math PRELIM Examinations 2021 P2 Marking Scheme Qn Solutions 1 𝑘2 + 4𝑘 + 4 > 13𝑘 − 16 [M1] 𝑘2 − 9𝑘 + 20 > 0 (𝑘 − 4)(𝑘 − 5) > 0 [M1] 𝑘 < 4, 𝑘 > 5 [M1] [A1] 2(a) 𝑑𝑦 𝑑𝑥 = 3( 5 6) ( 𝑥 4 + 𝑎) −1 6 ( 1 4) [M1] Gradient of normal = − 4 5 Gradient of tangent = 5 4 [M1] 5 4 = ( 15 24) ( 0.5 4 + 𝑎) −1 6 [M1] 𝑎 = − 7 64 [A1] 2(b) 𝑊ℎ𝑒𝑛 𝑥 = 0.5, 𝑦 = 3 ( 0.5 4 − 7 24) 5 6 = 3 32 [M1] 𝑦 − 3 32 = 5 4 (𝑥 − 0.5)[M1] 𝑦 = 5 4 𝑥 − 17 32 [A1] 3(i) Coefficient = (𝑛 𝑟)(−3)𝑟 [M1] When r = 2, 54 = (𝑛 2)(−3)2 → 6 = 𝑛(𝑛−1) 2 [M1] 𝑛 = 4 [A1] When r = 1, −6𝑝 = (𝑛 1)(−3)1 → 2𝑝 = 𝑛 [M1] → 𝑝 = 2 [A1] When r = 3, 𝑞 = (4 3)(−3)3 = −108 [A1] 3(ii) = (1 − 12𝑥 + 54𝑥2 − 108𝑥3)(1 + 2𝑥) = ⋯ + 108𝑥3 − 108𝑥3 + ⋯[M1] = 0 [A1] 4(i) [Removed due to CLT] ∡𝐺𝐹𝐴 = ∡𝐺𝐹𝐴 (𝑇𝑎𝑛𝑔𝑒𝑛𝑡 − 𝑐ℎ𝑜𝑟𝑑 𝑡ℎ𝑒𝑜𝑟𝑒𝑚) [M1] = ∡𝐷𝐹𝐸 (𝐴𝑙𝑡 𝑎𝑛𝑔𝑙𝑒𝑠) [M1] = ∡𝐸𝐷𝐹 (𝐼𝑠𝑜𝑠𝑐𝑒𝑙𝑒𝑠 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒) [A1] 4 5 5

JVSS/PRELIM2021/SEC4E/AM/4049/P2/MARKINGSCHEME 4(ii) ∡𝐷𝐸𝐹 = 180° − 2∡𝐸𝐷𝐹 (𝐼𝑠𝑜𝑠𝑐𝑒𝑙𝑒𝑠 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒) ∡𝐷𝐺𝐹 = 180° − 2∡𝐷𝐸𝐹 (𝑎𝑛𝑔𝑙𝑒𝑠 𝑖𝑛 𝑜𝑝𝑝 𝑠𝑒𝑔𝑚𝑒𝑛𝑡) [M1] = 2 ∡𝐸𝐷𝐹 (𝐼𝑠𝑜𝑠𝑐𝑒𝑙𝑒𝑠 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒) ∡𝐴𝐺𝐹 = 180° − ∡𝐷𝐺𝐹 = 180° − 2∡𝐸𝐷𝐹 [M1] ∡𝐺𝐴𝐹 = 180° − ∡𝐴𝐺𝐹 − ∡𝐺𝐹𝐴 ∡𝐷𝐺𝐹 = 180° − (180° − 2∡𝐸𝐷𝐹) − ∡𝐸𝐷𝐹 [M1] = ∡𝐸𝐷𝐹 = ∡𝐺𝐹𝐴 [A1] 5(i) sin 𝜃 = 𝐵𝐷 70 70sin 𝜃 = 𝐵𝐷 [M1] cos 𝜃 = 𝐵𝐸 30 30cos 𝜃 = 𝐵𝐸 [M1] ℎ = 70sin 𝜃 − 30cos 𝜃 [A1] 5(ii) 𝑅 = √702 + 302 [M1] = 76.2 (3sf) ∝ = tan−1 ( 30 70) [M1] = 23.2° ℎ = 76.2sin(𝜃 − 23.2°) [A1] 5(iii) 20 = 76.2(sin 𝜃 − 23.2°) ∝ = sin−1 ( 20 76.15773) [M1] = 15.22515° 𝜃 − 23.1985° = 15.22515° 𝜃 = 34.42365° = 34.4° [A1] 6(i) See attached B1 – ln y values B1 – axis/scale B1 – smooth curve 6(ii) 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 = 4.15−1.65 8−2 [M1] ln 𝑏 = 5 12 𝑏 = 𝑒 5 12 [M1] 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = 0.8 ln 𝑎 = 0.8 𝑎 = 𝑒0.8 [M1] 𝑦 = (𝑒0.8)(𝑒 5 12𝑥) [A1] 6(iii) See attached B1 – straight line x = 2.8 [B1]

JVSS/PRELIM2021/SEC4E/AM/4049/P2/MARKINGSCHEME 7(a) 𝑥(5𝑥 − 12) = 2𝑥 − 𝑐 [M1] 5𝑥2 − 14𝑥 + 𝑐 = 0 (−14)2 − 4(5)(𝑐) < 0 [M1] 9.8 < 𝑐 [A1] 7(b) 𝑥2 − 4𝑥 + (− 4 2) 2 − (− 4 2) 2 + 7 [M1] = (𝑥 − 2)2 + 3 [M1] Turning point = (2 , 3) [A1] 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒 = 3 7(c) 2𝑥2 − 6𝑥 − 4 + 2𝑥 = 12 [M1] (𝑥 − 4)(𝑥 + 2) = 0 𝑥 = 4, 𝑦 = 4 [A1] 𝑥 = −2, 𝑦 = 16 [A1] 8(a) log2 𝑥2 − 3 ( log2 4 log2 𝑥) = log2 2 log2 4 × log2 16 log2 4 [M1] Sub log2 𝑥 = 𝑦 2𝑦 − 6 𝑦 = 1 [M1] 2𝑦2 − 𝑦 − 6 = 0 [M1] 𝑦 = − 3 2 , 𝑦 = 2 log2 𝑥 = − 3 2 𝑥 = 2−3 2 [A1] log2 𝑥 = 2 𝑥 = 22 = 4 [A1] 8(b) 26(𝑥−1) × 23𝑥 = 24(6𝑥 3 ) [M1] 26𝑥−6+3𝑥 = 28𝑥 [M1] 9𝑥 − 6 = 8𝑥 [M1] 𝑥 = 6 [A1] 9(i) Gradient of BD = 1 [M1] Gradient of AC = – 1 [M1] 𝑦 = −𝑥 + 1 [A1] 9(ii) C (5 – 3, 4 – 5) [M1] C (2, –1) [M1] Area = 1 2 |0 1 −3 −4 2 −1 5 4 0 1| [M1] = 16 units2 [A1] 9(iii) Gradient of AB = 5 3 Gradient of p = 5 3 [M1] Accept answer with similar meaning to “same gradient”, “parallel lines”, “does not intersect” [A1]

JVSS/PRELIM2021/SEC4E/AM/4049/P2/MARKINGSCHEME 10(a) 𝑑𝑦 𝑑𝑥 = (𝑥)(2(𝑐𝑜𝑠𝑥)(−𝑠𝑖𝑛𝑥)) + (𝑐𝑜𝑠2𝑥)(1) [M2] = 𝑐𝑜𝑠2𝑥 − 2𝑥𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 [A1] 10(b) 𝑥𝑐𝑜𝑠2𝑥 + 𝑐 = ∫ 𝑐𝑜𝑠2𝑥 𝑑𝑥 − ∫ 𝑥𝑠𝑖𝑛2𝑥 𝑑𝑥 [M1] = ∫ 1+𝑐𝑜𝑠2𝑥 2 𝑑𝑥 − ∫ 𝑥𝑠𝑖𝑛2𝑥 𝑑𝑥 [M1] 𝑥𝑐𝑜𝑠2𝑥 + 𝐶1 = 1 2 𝑥 + 𝑠𝑖𝑛2𝑥 4 + 𝐶2 − ∫ 𝑥𝑠𝑖𝑛2𝑥 𝑑𝑥 [M2] ∫ 𝑥𝑠𝑖𝑛2𝑥 𝑑𝑥 = 1 2 𝑥 + 𝑠𝑖𝑛2𝑥 4 − 𝑥𝑐𝑜𝑠2𝑥 + 𝐶 [A1] 10(c) ∫ 𝑥𝑠𝑖𝑛2𝑥 𝑑𝑥 𝜋 2𝜋 3 = [ 1 2 𝑥 + 𝑠𝑖𝑛2𝑥 4 − 𝑥𝑐𝑜𝑠2𝑥]𝜋 3 𝜋 2 [M1] = 𝜋 4 − 0.47829 [M1] = 0.307 𝑟𝑎𝑑 [A1] 11(i) 𝐵𝑎𝑠𝑒 𝑎𝑟𝑒𝑎 = (60 − 𝑥)(45 − 2𝑥) [M1] Volume = (2700 − 165𝑥 + 2𝑥2)(𝑥) [M1] = (2𝑥3 − 165𝑥2 + 2700𝑥) [A1] 11(ii) 𝑑𝑉 𝑑𝑥 = 6𝑥2 − 330𝑥 + 2700 [M1] 0 = 𝑥2 − 55𝑥 + 450 [M1] (𝑥 − 45)(𝑥 − 10) = 0 [M1] 𝑥 = 45, 𝑥 = 10 [A1] (rejected) 11(iii) 𝑑2𝑉 𝑑𝑥2 = 12𝑥 − 330 [M1] = 12(10) − 330 = −210 < 0 [M1] Stationary value is maximum, best possible design [A1]