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 b