OPSS AMPrelim 2024 4E P1 MS

1 ORCHID PARK SECONDARY SCHOOL Preliminary Examination 2024 Marker 1 Chan Ho Lun Marker 2 Marker 3 Marker 4 Marker 5 MATHEMATICS Paper 1 Secondary 4 Express / 5 Normal (Academic) Setter: Mr Chan Ho Lun 4049/01 20 October 2024 2 hours 15 minutes 90 Marks Additional Materials: NIL Answer Cover Page Note: -2m maximum for not giving 1 dp for angles for whole paper. -2m maximum for not giving 3 sf for inaccurate answers for whole paper.

2 Qn Solution Ma rks Remarks 1 𝑦 = 𝑝𝑥 − 5____________(1) 𝑦 = 3𝑥2 + 4𝑥 − 2______(2) Sub (1) into (2): 3𝑥2 + 4𝑥 − 2 = 𝑝𝑥 − 5 3𝑥2 + (4 − 𝑝)𝑥 + 3 = 0 Line meets curve => Discriminant ≥ 0: (4 − 𝑝)2 − 4(3)(3) ≥ 0 16 − 8𝑝 + 𝑝2 − 36 ≥ 0 𝑝2 − 8𝑝 − 20 ≥ 0 (𝑝 − 10)(𝑝 + 2) ≥ 0 𝑝 ≤ −2 𝑜𝑟 𝑝 ≥ 10 M1 M1 M1 A1 2 Let 5𝑥2−6𝑥+13 (𝑥−1)(𝑥2+3) = 𝐴 𝑥−1 + 𝐵𝑥+𝐶 𝑥2+3 5𝑥2 − 6𝑥 + 13 = 𝐴(𝑥2 + 3) + (𝐵𝑥 + 𝐶)(𝑥 − 1) Method 1: When 𝑥 = 1, 12 = 4𝐴 𝐴 = 3 When 𝑥 = 0, 13 = 3(3) + C (-1) 𝐶 = −4 When 𝑥 = −1, 24 = 3(4) + (−𝐵 − 4)(−2) 𝐵 = 2 Hence, 5𝑥2−6𝑥+13 (𝑥−1)(𝑥2+3) = 3 𝑥−1 + 2𝑥−4 𝑥2+3 M1 M1 M1 M1 A1 3 RHS = 𝑐𝑜𝑠𝑒𝑐 𝜃 sec 𝜃 − 2 tan 𝜃 = 1 sin 𝜃 ( 1 cos 𝜃) − 2 sin 𝜃 cos 𝜃 = 1 sin 𝜃 cos 𝜃 − 2 sin2 𝜃 sin 𝜃 cos 𝜃 = 1 − 2 sin2 𝜃 sin 𝜃 cos 𝜃 = cos 2𝜃 1 2 sin 2𝜃 = 2 cos 2𝜃 sin 2𝜃 = 2 cot 2𝜃 = 𝐿𝐻𝑆 M1 M1 M2 A1 for changing either cosec, sec or tan correctly for common denominator 1 mark for cos 2𝜃, 1 mark for sin 2𝜃

3 4 (a) 5𝑒2𝑥+1 + 10𝑥𝑒2𝑥+1 B2 1 mark for each term (b) 5𝑥𝑒2𝑥+1 = 5 2 𝑒2𝑥+1 + 10 ∫ 𝑥 𝑒2𝑥+1 𝑑𝑥 + 𝑐1 M1 M1 where 𝑐1 is an arbitrary constant 5𝑥𝑒2𝑥+1 − 5 2 𝑒2𝑥+1 − 𝑐1 = 10 ∫ 𝑥 𝑒2𝑥+1 𝑑𝑥 ∫ 𝑥 𝑒2𝑥+1 𝑑𝑥 = 1 2 𝑥𝑒2𝑥+1 − 1 4 𝑒2𝑥+1 + 𝑐 where 𝑐 = − 1 10 𝑐1 M1 A1 Qn Solution Marks Remarks 5 1 + 3(1 − cos2 𝜃) = 4 cos 𝜃 −3 cos2 𝜃 − 4 cos 𝜃 + 4 = 0 Let 𝑦 = cos 𝜃 −3𝑦2 − 4𝑦 + 4 = 0 3𝑦2 + 4𝑦 − 4 = 0 (3𝑦 − 2)(𝑦 + 2) = 0 𝑦 = 2 3 or 𝑦 = −2 cos 𝜃 = 2 3 𝑜𝑟 cos 𝜃 = −2 (𝑟𝑒𝑗) 𝜃 = 0.841 𝑜𝑟 − 0.841 (3 s.f.) M1 M1 M2 A2 6 (a) 𝑓′(𝑥) = (𝑥 − 2𝑘)(2𝑥) − 𝑥2(1) (𝑥 − 2𝑘)2 = 𝑥2 − 4𝑘𝑥 (𝑥 − 2𝑘)2 M1 A1

4 6 (b) 𝑔′(𝑥) = 𝑥2 − 4𝑘𝑥 = 𝑥(𝑥 − 4𝑘) g decreases => 𝑔′(𝑥) < 0 𝑥(𝑥 − 4𝑘) < 0 0 < 𝑥 < 4𝑘 Since 𝑥 > 2𝑘, then 2𝑘 < 𝑥 < 4𝑘______(1) But it is also given that 𝑘 < 𝑥 < 6, i.e. 2𝑘 < 𝑥 < 12___(2) By (1) and (2), 4𝑘 = 12 𝑘 = 3 M1 M1 M1 A1 7 For the function to be always positive, 𝑘 > 0 (so that the graph is U-shaped)___(1) We also require discriminant < 0 (so that graph never cuts x-axis) 22 − 4(𝑘)(−2𝑘 − 3) < 0 2𝑘2 + 3𝑘 + 1 < 0 (𝑘 + 1)(2𝑘 + 1) < 0 −1 < 𝑘 < − 1 2 _____________________(2) But (1) and (2) cannot happen at the same time (k cannot be positive but yet also be between - 1 and − 1 2) ∴ There is no value of k for which the function is positive (proven) M1 M1 M1 M1 M1 A1

5 Qn Solution Marks Remarks 8 𝑦 − 𝑥 = 2 𝑦 = 𝑥 + 2 ________________________(1) 𝑦2 = 4(2𝑥 + 1) ___________________(2) Sub (1) into (2): (𝑥 + 2)2 = 4(2𝑥 + 1) 𝑥2 + 4𝑥 + 4 = 8𝑥 + 4 𝑥2 − 4𝑥 = 0 𝑥(𝑥 − 4) = 0 𝑥 = 0 𝑜𝑟 𝑥 = 4 Sub into (1): 𝑦 = 2 𝑜𝑟 𝑦 = 6 The two points are (0, 2) and (4, 6) M1 M1 M1 M1 A1 9 (a) 𝑑𝑦 𝑑𝑥 = 3𝑥2 + 𝑚 When 𝑥 = 2, 𝑑𝑦 𝑑𝑥 = 0 (𝑔𝑖𝑣𝑒𝑛): 0 = 3(2)2 + 𝑚 𝑚 = −12 M1 A1 (b) 𝑑𝑦 𝑑𝑥 = 3𝑥2 − 12 For stationary points, 𝑑𝑦 𝑑𝑥 = 0 3𝑥2 − 12 = 0 𝑥2 = 4 𝑥 = 2 𝑜𝑟 − 2 When 𝑥 = −2, 𝑦 = (−2)3 − 12(−2) − 15 𝑦 = 1 B is (-2, 1) M1 A1

6 (c) For gradient to be a min, 𝑑2𝑦 𝑑𝑥2 = 0 6𝑥 = 0 𝑥 = 0 When 𝑥 = 0, 𝑦 = −15 P is (0, -15) M1 M1 A1 (d) 𝑑3𝑦 𝑑𝑥3 = 6 Since 𝑑3𝑦 𝑑𝑥3 is positive, the gradient is a minimum. M1 A1 Qn Solution Marks Remarks 10 (a) Method 1: (𝑥 − 2)2 + (𝑦 − 3)2 − 22 − 32 − 12 = 0 (𝑥 − 2)2 + (𝑦 − 3)2 = 25 Method 2: 2𝑔 = −4 2𝑓 = −6 𝑔 = −2 𝑓 = −3 𝑔2 + 𝑓2 − 𝑐 = 4 + 9 + 12 = 25 Centre = (2, 3) Radius = 5 units M1 M1 M1 M1 A1 A1 (b) When 𝑦 = 0, 𝑥2 − 4𝑥 − 12 = 0 (𝑥 − 6)(𝑥 + 2) = 0 𝑥 = 6 𝑜𝑟 𝑥 = −2 The points are (6, 0) and (-2, 0) M1 A2 (c) Centre = (-2, 3), radius = 5 units Eqn: (𝑥 + 2)2 + (𝑦 − 3)2 = 25 M1 A1 /B2 (d) Since the centre is only 2 units away from the y- axis and the radius is 5 units, the circle will cut the y- axis and does not lie entirely in the 2nd quadrant. M1 A1 Accept any similar answer (e.g. centre is 3 units away from x-axis)

7 11 (a) 𝐴 = 𝜋𝑟2 + 2𝜋𝑘𝑟 + 𝑐 where c is an arbitrary constant 𝑊ℎ𝑒𝑛 𝑟 = 0, 𝐴 = 0 : (when there is no radius, there is no area) 𝑐 = 0 Hence, 𝐴 = 𝜋𝑟2 + 2𝜋𝑘𝑟 M1 A1 (b) Curved surface area: 2𝜋𝑟ℎ = 2𝜋𝑘𝑟 ℎ = 𝑘 B1 (c) 𝑑𝑟 𝑑𝑡 = 5 2 ln(2𝑡 + 1) + 𝑐1 where 𝑐1 is an arbitrary constant 𝑊ℎ𝑒𝑛 𝑡 = 0, 𝑑𝑟 𝑑𝑡 = 3: 3 = 𝑐1 𝑑𝑟 𝑑𝑡 = 5 2 ln(2𝑡 + 1) + 3 M1 A1 (d) 𝑑𝐴 𝑑𝑡 = 𝑑𝐴 𝑑𝑟 × 𝑑𝑟 𝑑𝑡 𝑑𝐴 𝑑𝑡 = 2𝜋(𝑟 + 𝑘) × [5 2 ln(2𝑡 + 1) + 3] When 𝑟 = 15, 𝑡 = 4, 𝑘 = 10: 𝑑𝐴 𝑑𝑡 = 2𝜋(15 + 10) × (5 2 ln 9 + 3) 𝑑𝐴 𝑑𝑡 = M1 M1 A1 Qn Solution Marks Remarks 12 (a) 4𝑥 + 4𝑦 = 100 𝑦 = 25 − 𝑥 M1 A1 (b) 𝐴 = 𝑥2 + 𝑦2 = 𝑥2 + (25 − 𝑥)2 = 2𝑥2 − 50𝑥 + 625 = 2 [(𝑥 − 25 2 ) 2 − ( 25 2 ) 2 + 625 2 ] = 2 (𝑥 − 25 2 ) 2 + 625 2 M1 M1 M1 A1

8 (c) Min area = 625 2 when 𝑥 = 25 2 B1 B1 (d) 25 − 𝑥1 (E.g. if 𝑥1 = 10, then 25 − 𝑥1 = 15 will give the same area) B1 13 (a) (3 − 7 2 , 2ℎ − 10 2 ) = (−2, ℎ − 5) M1 A1 (b) 2ℎ + 10 3 + 7 = ℎ + 5 5 M1 A1 (c) Let the equation of the perpendicular bisector be 𝑦−(ℎ−5) 𝑥−(−2) = − 5 ℎ+5 Sub (h, 3): 3−(ℎ−5) ℎ−(−2) = − 5 ℎ+5 ℎ2 − 8ℎ − 50 = 0 ℎ = 8±√82−4(1)(−50) 2(1) ℎ = 12.1 𝑜𝑟 − 4.12 (3 𝑠. 𝑓. ) M2 M1 M1 M1 A2 1 mark for substituting M 1 mark for gradient