Peirce Prelim 2021 S4E Add Math P1 MS

Peirce Secondary School 07 Sept 2021 Preliminary Exam UPDATED 13 Sept Sec 4E/5N Add Math 4049/I Question Answer Marks Partial Marks Guidance 1 Solve the equation . 1 102𝑥. 10 − 80(10𝑥) = −70 6 M1 For splitting Let u = 10𝑥 M1 For substitution 𝑢2 − 8𝑢 + 7 = 0 M1 For quadratic equation. Accept 10𝑢2 − 80𝑢 + 70 = 0 [M2 for (10𝑥)2 − 8(10𝑥) + 7 = 0 𝑖𝑓 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑛𝑜 𝑠𝑢𝑏𝑠𝑖𝑡𝑢𝑡𝑖𝑜𝑛 𝑢𝑠𝑒𝑑] (u – 7) (u – 1) = 0 u = 7 or u = 1 M1 For u or 10𝑥 values x = lg 7 or x = 0 x = 0.845 (3 sf) or x = 0 A2 Accept 0.84509804. Reject lg 7 Question Answer Marks Partial Marks Guidance 2 𝑓(𝑥) = 𝑥2 − 5𝑥 + 7 2(i) 𝑏2 − 5𝑏 + 7 = 𝑐2 − 5𝑐 + 7 3 M1 f(b) = f(c) or f(b)– f(c) = 0 or f(c) – f(b)=0 (𝑏 − 𝑐)(𝑏 + 𝑐) = 5(𝑏 − 𝑐) M1 For factorisation Since b≠c, b-c ≠0 Therefore 𝑏 + 𝑐 = 5 (shown) A1 For conclusion 2(ii) 𝑏 = 5 − 𝑐 4(5 − 𝑐)(𝑐) = 21 4 M1 For substitution (2𝑐 − 3)(2𝑐 − 7) = 0 M1 c = 1.5 or c = 3.5 b = 3.5 or b = 1.5 M1 Reject pre-mature rejection of values

Since b > c, b = 3.5 and c = 1.5 A1 Accept fractions Question Answer Marks Partial Marks Guidance 3(i) = 38 + (8 1 )37 (− 𝑥 2) 1 + (8 2 )36 (− 𝑥 2) 2 + (8 3 )35 (− 𝑥 2) 3 + ⋯ 3 M1 For expansion = 6561 – 8748𝑥 + 5103𝑥2 − 1701𝑥3 + ⋯ A2 For 4 terms [A1 for 2-3 terms; A0 if 0-1 term correct.] 3(ii) (𝑥 − 5 𝑥) 2 (3 − 𝑥 2) 8 = (𝑥2 − 10 + 25 𝑥2) (6561 – 8748𝑥 + 5103𝑥2 − 1701𝑥3 + ⋯ ) 3 M1 Expansion of (𝑥 − 5 𝑥) 2 = 𝑥2 − 10 + 25 𝑥2 = … + 87480x – 42525x + … = … + 44955x + … M1√ √ For overall expansion if expansion of (𝑥 − 5 𝑥) 2 Coefficient of x = 44955 A1 Reject 44955x Question Answer Marks Partial Marks Guidance 4(i) When 𝑡 = 0, 𝑀 = 24 𝑔 When 𝑡 = 35, 𝑀 = 12 𝑔 3 M1 For M = 12 g 𝑒−𝑘(35) = 1 2 or 𝑒𝑘(35) = 2 −35 𝑘 = 𝑙𝑛 1 2 or 35 𝑘 = 𝑙𝑛 2 M1 For isolating e or taking ln k = 0.0198042 = 0.0198 (3 s f) A1 Reject 𝑘 = − 1 35 𝑙𝑛 1 2 or 1 35 𝑙𝑛 𝑙𝑛 2 or0.020 Accept 0.01980

4(ii) 𝑀 = 24𝑒−0.0198042(365) 2 M1√ Accept t = 366 days , √ for k = 0.017414028 = 0.0174 g (3 s f) A1 Question Answer Marks Partial Marks Guidance 5 5(i) 𝑑𝑦 𝑑𝑥 = (2𝑥 − 3)(3) − 3𝑥(2) (2𝑥 − 3)2 5 M1 For Quotient or Product Rule applied correctly = −9 (2𝑥−3)2 = −9 4 M1√ √ for equating 𝑑𝑦 𝑑𝑥 to gradient of tangent x = 2.5 or x = 0.5 A1√ √ for both x values y = 3.75 or y = -0.75 A1√ √ for both y values Points are (2.5, 3.75) and (0.5, -0.75) A1 For both points 5(ii) 𝑑𝑦 𝑑𝑥 = −9 (2𝑥−3)2 2 Since (2𝑥 − 3)2 > 0, 𝑥 ≠ 1.5 −9 (2𝑥−3)2 < 0 M1 Reject using one/two values of x to show 𝑑𝑦 𝑑𝑥 < 0. Curve is a decreasing function as 𝑑𝑦 𝑑𝑥 < 0. A1√ √ For correct conclusion 𝑑𝑦 𝑑𝑥 < 0 𝑜𝑟 𝑖𝑚𝑝𝑙𝑖𝑒𝑑 𝑘𝑛𝑜𝑤𝑙𝑒𝑑𝑔𝑒 𝑑𝑦 𝑑𝑥 < 0 .

Question Answer Marks Partial Marks Guidance 6(i) a = 2 2 B1 b = 2 B1 6(ii) 𝑚 = −𝜋 + 𝑝 1 B1 6(iii) 𝑘 < 1, 𝑘 > 3 1 B1 6(iv) 𝑥 − 𝜋 𝑐𝑜𝑠 𝑐𝑜𝑠 𝑏𝑥 = 𝜋𝑎 − 𝜋 𝑥 𝜋 + 1 = 𝑎 +𝑐𝑜𝑠 𝑐𝑜𝑠 𝑏𝑥 3 M1 By adding/drawing the straight line 𝑦 = 𝑥 𝜋 + 1 A1√ √ for correctly plotted the line Number of solutions = 2 A1√ √ implied from the graph

Question Answer Marks Partial Marks Guidance 7(i) 𝑑𝑦 𝑑𝑥 = 𝑒𝑥(−𝑠𝑖𝑛 𝑥 − 𝑐𝑜𝑠 𝑥) + (𝑐𝑜𝑠 𝑐𝑜𝑠 𝑥 − 𝑠𝑖𝑛 𝑥) 𝑒𝑥 2 M1 For either one term correct = -2𝑒𝑥 𝑠𝑖𝑛 𝑠𝑖𝑛 𝑥 (shown) A1 7(ii) ∫ 𝜋 3 0 𝑒𝑥 𝑠𝑖𝑛 𝑠𝑖𝑛 𝑥 𝑑𝑥 = − 1 2 [𝑒𝑥(𝑐𝑜𝑠 𝑐𝑜𝑠 𝑥 − 𝑠𝑖𝑛 𝑥)] 0 𝜋 3 4 M1 = − 1 2 [𝑒 𝜋 3(𝑐𝑜𝑠 𝑐𝑜𝑠 𝜋 3 − 𝑠𝑖𝑛 𝜋 3) − 𝑒0(𝑐𝑜𝑠 𝑐𝑜𝑠 0 − 𝑠𝑖𝑛 𝑠𝑖𝑛 0 ) ] M1√ √ for substitution of boundary values = − 1 2 [𝑒 𝜋 3 (1 2 − √3 2 ) − (1)] = (√3 − 1 4 ) 𝑒 𝜋 3 + 1 2 a = 1, b = 1 2 A2 Answers given correctly

Question Answer Marks Partial Marks Guidance 8 8(i) 𝑑𝑦 𝑑𝑥 = 2𝑥 9 + 1 6 6 M1 At 𝑥 = 6, 𝑑𝑦 𝑑𝑥 = 3 2 M1√ √ for correct substitution of 𝑥 = 6 𝑖𝑛𝑡𝑜 𝑑𝑦 𝑑𝑥 Gradient of normal = − 2 3 M1√ √ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑛𝑜𝑟𝑚𝑎𝑙 ∶ 𝑦 = − 2 3 𝑥 + 11 M1√ √ for equation of normal Substitute 𝑥 = 6, 𝑦 = 5 + 𝑘: OR 5 + 𝑘 = − 2 3 (6) + 11 62 9 + 6 6 + 𝑘 = − 2 3 (6) + 11 M1√ 𝑘 = 2 (shown) A1 8(ii) Shaded area = ∫ 6 0 [(− 2 3 𝑥 + 11) − ( 𝑥2 9 + 𝑥 6 + 2)] 𝑑𝑥 4 M1√ √ for applying equation of normal = [− 𝑥3 27 − 5𝑥2 12 + 9𝑥] 0 6 M1√ √ for correct integration = (−8 − 15 + 54) − (0) M1√ √ for correct substitution of limits = 31 sq units A1 OR Area of trapezium = 1 2 (7 + 11)(6) = 54 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠 M1 Area under graph = ∫ 6 0 ( 𝑥2 9 + 𝑥 6 + 2) 𝑑𝑥 = [ 𝑥3 27 + 𝑥2 12 + 2𝑥] 0 6 M1 For integration

= (23) − (0) M1 For substitution of limits Shaded area = 54 – 23 = 31 sq units A1 Question Answer Marks Partial Marks Guidance 9(i) 𝑆𝑃 = 5 𝑐𝑜𝑠 𝑐𝑜𝑠 𝜃 𝑆𝑄 = 12 𝑠𝑖𝑛 𝑠𝑖𝑛 𝜃 4 M1 M1 𝑡𝑎𝑛 𝑡𝑎𝑛 𝜃 = 5 12 M1 𝜃 = 22.61986° = 22.6° (1 𝑑 𝑝) A1 Reject 22.62 9(ii) PQ = 12 𝑠𝑖𝑛 𝑠𝑖𝑛 𝜃 − 5 𝑐𝑜𝑠 𝑐𝑜𝑠 𝜃 4 M1√ √ for PQ = SQ - SP R = 13 α = 22.61986◦ M1√ M1√ √ for 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 √𝑎2 + 𝑏2 √ for 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑡𝑎𝑛−1 = 𝑏 𝑎 PQ = 13 𝑠𝑖𝑛 𝑠𝑖𝑛 (𝜃 − 22.6°) A1 9(iii) Range: 22.6° < 𝜃 < 90° 1 B1√ √ Realises range is between part (i) 𝜃 value and 90°.

Question Answer Marks Partial Marks Guidance 10 v = t2 – 6t + 5 10(i) 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 = ∫ (𝑡2 – 6𝑡 + 5 ) 𝑑𝑡 = 𝑡3 3 − 6𝑡2 2 + 5𝑡 + 𝑐 2 M1 When t = 0, displacement = 0 => c = 0 Displacement = 𝑡3 3 − 3𝑡2 + 5𝑡 A1 10(ii) At A and B, v = 0 => t = 1 and t = 5 4 M1 𝐹𝑜𝑟 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑡 At 𝑡 = 1, displacement = 2 1 3 𝑐𝑚 At t = 5, displacement = −8 1 3 𝑐𝑚 𝐴𝐵 = ∫ 5 1 (𝑡2 – 6𝑡 + 5 ) 𝑑𝑡 = −10 2 3 𝑐𝑚 M1√ M1√ √ for 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑡 = 1 √ for 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑡 = 5 [M1 for correct integration and M1substitution of limits] AB = 10 2 3 cm A1 Accept 10.7 cm; reject −10 2 3 𝑐𝑚 10(iii) Average speed = 102 3 4 2 M1√ √ for applying AB in part (ii) = 2.67 cm/s A1 Accept 2 2 3 10(iv) 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑑𝑣 𝑑𝑡 = 2𝑡 − 6 4 M1 𝑡 = 3 M1 OC = 3 cm M1√ √ For applying t value when acceleration=0 OB = 8 1 3 cm OC < 1 2 𝑂𝐵 A1√ Correct conclusion with comparison

BC = 5 1 3 cm > OC Reject if no comparison between OC, BC and/or 1 2 𝑂𝐵 is made. Therefore, C is nearer to O Question Answer Marks Partial Marks Guidance 11 11(i) Centre (5,6), radius = 3 units 3 M1 A1A1 Any correct method Correct answers given 11(ii) radius =3, centre (5,6) and y = 6 is a horizontal line 5-3 = 2 𝑥 = 2 is a vertical line and is a tangent to the circle at (2,6). 2 M1 A1 Any correct explanation implying tangent perpendicular to radius Accept proof of discriminant = 0 11(iii) Centre (-1,6) 2 M1 (𝑥 + 1)2 + (𝑦 − 6)2 = 9 A1 Accept General Form Question Answer Marks Partial Marks Guidance 12(i) 𝑥 + 𝑦 = 20 𝑦 = 20 − 𝑥 3 M1 For y in terms of x 𝐴𝑟𝑒𝑎 = 𝑥𝑦 − 𝜋(1 2 𝑥)2 M1 For Area = Rectangle - Circle = 20𝑥 − ( 4+𝜋 4 )𝑥2 (𝑠ℎ𝑜𝑤𝑛) A1 12(ii) 𝑑𝐴 𝑑𝑥 = 20 − (4 + 𝜋 4 ) (2𝑥) 3 M1 For correct differentiation 𝑆𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑣𝑎𝑙𝑢𝑒 => 𝑑𝐴 𝑑𝑥 = 0 𝑑𝐴 𝑑𝑥 = 20 − (4 + 𝜋 4 ) (2𝑥) = 0 M1√ √ for 𝑑𝐴 𝑑𝑥 = 0, sets to 0 and solves

𝑥 = 40 4 + 𝜋 = 5.60099 = 5.60 A1 Reject 5.6 Question Answer Marks Partial Marks Guidance 12(iii) 𝑑2𝐴 𝑑𝑥2 = − (4 + 𝜋) 2 2 M1√ √ Correct differentiation Accept 1st Derivative Test with clear presentation 𝑑2𝐴 𝑑𝑥2 = − (4 + 𝜋) 2 < 0 → 𝐴 𝑜𝑟 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑒𝑟𝑦 𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 𝑎 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 The landscape designer might be delighted as he could optimise/make full use of the length/dimension of the fencing to obtain a maximum area for his flower beds. A1√ √ For correct conclusion Reject x is maximum and making area maximum.