ACSBR 2024 4E5N Prelim MATH P2 solutions

Mathematics Paper 2 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) Qn Steps/Answer Remarks 1 (a) USD = 10000 𝑥 (b) 10000 𝑥 − 0.030 − 10000 𝑥 = 166 10000𝑥 − 10000(𝑥 − 0.030) = 166𝑥(𝑥 − 0.030) 166𝑥2 − 4.98𝑥 − 300 = 0 83𝑥2 − 2.49𝑥 − 150 = 0 (c) 𝑥 = 2.49 ± √(−2.49)2 − 4(83)(−150) 2(83) 𝑥 = 1.359 or 𝑥 = −1.329 (d) 20000 1.359 − 0.030 = USD 15 049 Accept 15 048 2 (ai) 𝑠 = [5 − 3(−3)2] ÷ (−3)2 𝑠 = −2 4 9 or 𝑠 = − 22 9 Cannot accept 3 sf (aii) 𝑠𝑞2 = 𝑟 − 3𝑞2 𝑠𝑞2 + 3𝑞2 = 𝑟 𝑞2(𝑠 + 3) = 𝑟 𝑞 = ±√ 𝑟 (𝑠 + 3) (b)(i) (7𝑛 − 1)2 − (𝑛 − 1)2 = [(7𝑛 − 1) + (𝑛 − 1)][(7𝑛 − 1) − (𝑛 − 1)] = 6𝑛(8𝑛 − 2) = 12𝑛(4𝑛 − 1) = 12(4𝑛2 − 𝑛) is also accepted (b)(ii) 𝑝 − 3𝑛 + 12𝑛2 − 4𝑛𝑝 (7𝑛 + 1)2 − (𝑛 − 1)2 = 𝑝(1 − 4𝑛) − 3𝑛(1 − 4𝑛) 12𝑛(4𝑛 − 1) = (3𝑛 − 𝑝)(4𝑛 − 1) 12𝑛(4𝑛 − 1) = 3𝑛 − 𝑝 12𝑛

Mathematics Paper 2 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) 3 (a) One possible answer • ET = AT (tangents from external points are equal) • TO is common • OE = OA (radii of circle) Triangle TOA is congruent to triangle TOE (SSS test) Another possible answer • angle TAO = angle TEO = 90° (tangent perpendicular to radius) • TO is common • OE = OA (radii of circle) Triangle TOA is congruent to triangle TOE (RHS test) (bi) angle AOT = (180 – 90 – 32)° = 58° (tangent perpendicular to radius) angle ABF = 1 2angle AOT 1 2 (58°) = 29° (angle at centre is twice angle at circumference) angle OFG = angle ABF = 29° (alternate angles, OF //BA) (bii) angle ACF = angle ABF =29° (angles in the same segment) angle CAE = 180 – 90 – 29 = 61° (OT is perpendicular bisector of chord AE) angle CDE = 180 – 61 = 119° (angles in opposite segments) (c) As angle OET and angle OAT are right-angles, by the property of right angle in a semicircle, OT is a diameter and points E and A will lie on the circumference. OETA are thus four points on the circumference of this circle. Or Angle AOE + angle ATE = (58 × 2) + (32 × 2) = 180° Angle OET + angle OAT = 180°. By the property of angles in opposite segments, OETA are thus four points on the circumference of this circle. 4 (a)(i) 𝐴𝐵⃗⃗⃗⃗⃗ = 2𝑏 − 𝑎 (a)(ii) 𝐵𝐶⃗⃗⃗⃗⃗ = − 3 4 𝐵𝐴⃗⃗⃗⃗⃗ = − 3 4 (2𝑏 − 𝑎) 𝑂𝐶⃗⃗⃗⃗⃗ = 𝑂𝐵⃗⃗⃗⃗⃗ + 𝐵𝐶⃗⃗⃗⃗⃗ = 2𝑏 − 3 4 (2𝑏 − 𝑎) = 3 4 𝑎 + 1 2 𝑏 (b)(i) 𝑂𝑃⃗⃗⃗⃗⃗ = 𝑂𝐵⃗⃗⃗⃗⃗ + 𝐵𝑃⃗⃗⃗⃗⃗ = 2𝑏 + 3𝑎 (b)(ii) 𝑂𝑃⃗⃗⃗⃗⃗ = 3𝑎 + 2𝑏. 𝑂𝐶⃗⃗⃗⃗⃗ = 1 4 (3𝑎 + 2𝑏) As 𝑂𝐶⃗⃗⃗⃗⃗ = 1 4 𝑂𝑃⃗⃗⃗⃗⃗ , and O is a common point, O, C and P lie on a straight line.

Mathematics Paper 2 Marking Scheme Secondary 4 Express / 5 Normal Academic Preliminary Exams 2024 Anglo-Chinese School (Barker Road) (ci) 3:1 (cii) OAC : OAB : OAD 1 : 4 2 : 1 4 : 2 Therefore,