OPSS_Prelims_2024_4E_Math_P2_Marking_Scheme

Paper 2 Qn No. Solutions 1ai 2𝑥2 − 6𝑥 − 12 = 2(𝑥2 − 3𝑥 − 6) = 2[(𝑥 − 1.5)2 − 2.25 − 6] = 2[(𝑥 − 1.5)2 − 8.25] =2(𝑥 − 1.5)2 − 16.5 1aii 2𝑥2 − 6𝑥 − 12 = 0 2(𝑥 − 1.5)2 − 16.5 = 0 2[(𝑥 − 1.5)2 = 16.5 (𝑥 − 1.5)2 = 8.25 𝑥 − 1.5 = 2.87 𝑜𝑟 𝑥 − 1.5 = −2.87 𝑥 = 4.37 𝑜𝑟 𝑥 = −1.37 1b 3𝑥+1 2𝑥2+11𝑥+12 − 1 𝑥+4 = 3𝑥+1 (2𝑥+3)(𝑥+4) − 1 𝑥+4 = 3𝑥+1 (2𝑥+3)(𝑥+4) − (2𝑥+3) (2𝑥+3)(𝑥+4) = 3𝑥+1−2𝑥−3 (2𝑥+3)(𝑥+4) = 𝑥−2 (2𝑥+3)(𝑥+4) 2 2𝑥 + 𝑦 = 3𝑥 + 5𝑦 + 3 𝑥 = −4𝑦 − 3 − (1) 4𝑥 + 5𝑦 – 7 = 𝑥 + 𝑦 3𝑥 + 4𝑦 = 7 − (2) Sub (1) into (2) 3(−4𝑦 − 3) + 4𝑦 = 7 −12𝑦 − 9 + 4𝑦 = 7 −8𝑦 = 16 𝑦 = −2 Sub 𝑦 = −2 into (1) 𝑥 = −4(−2) − 3

𝑥 = 5 3a (𝑥 − 3)(2𝑥 + 8) = −12 2𝑥2 + 8𝑥 − 6𝑥 − 24 + 12 = 0 2𝑥2 + 2𝑥 − 12 = 0 𝑥2 + 𝑥 − 6 = 0 (𝑥 + 3)(𝑥 − 2) = 0 𝑥 = −3 𝑜𝑟 𝑥 = 2 3b 4 2𝑥−3 − 3 𝑥+2 = 1 4(𝑥+2)−3(2𝑥−3) (2𝑥−3)(𝑥+2) = 1 4(𝑥 + 2) − 3(2𝑥 − 3) = (2𝑥 − 3)(𝑥 + 2) 4𝑥 + 8 − 6𝑥 + 9 = 2𝑥2 + 𝑥 − 6 2𝑥2 + 3𝑥 − 23 = 0 𝑥 = −3±√32−4(2)(−23) 2(2) 𝑥 = −3±√193 4 𝑥 = 2.72 𝑜𝑟 − 4.22 4a TSA of cone = 𝜋𝑟2 + 𝜋𝑟𝑙 = 𝜋(2𝑦)2 + 𝜋(2𝑦)(𝑙) = 2𝜋𝑦(2𝑦 + 𝑙) 4b TSA of hemisphere = 𝜋𝑟2 + 1 2 (4𝜋𝑟2) = 𝜋(3𝑦)2 + 2𝜋(3𝑦)2 = 9𝜋𝑦2 + 18𝜋𝑦2 = 27𝜋𝑦2 2𝜋𝑦(2𝑦 + 𝑙) = 27𝜋𝑦2 2𝑦 + 𝑙 = 27𝑦 2 𝑙 = 27𝑦 2 − 2𝑦 𝑙 = 23𝑦 2 𝑜𝑟 11.5𝑦 4c Vol of hemisphere 1 2 ( 4 3 𝜋𝑟3) = 729 2 3 𝜋(3𝑦)3 = 729 18𝜋𝑦3 = 729 𝑦 = 2.3448

height of cone = √(𝑙2 − 𝑟2) = √(( 23 2 (2.3448)2) 2 − (2 × 2.3448)2) = 26.554 cm Vol of cone = 1 3 𝜋𝑟2ℎ = 1 3 (𝜋)(2 × 2.3448)2(26.554) = 611.547 ≈ 612𝑐𝑚3 (𝑡𝑜 3𝑠. 𝑓. ) 5a 2 ∶ 100000 1 ∶ 50000 5b 1 𝑐𝑚 ∶ 50000 𝑐𝑚 1 𝑐𝑚 ∶ 0.5 𝑘𝑚 Area 1 cm2 : 0.25 km2 456 cm2 : 114 km2 6ai sin ABE = sin EBC = 2 √8 6aii cos ABE = −cos EBC =− 3 √8 6b Area = ( 1 2) (𝐴𝐵)(𝐵𝐸)𝑠𝑖𝑛𝐴𝐵𝐸 = ( 1 2) (4)(√8) ( 2 √8) = 4 cm2 6c 𝐴𝐵 𝐴𝐶 = 4 7 𝐴𝑟𝑒𝑎 𝐴𝐵𝐸 𝐴𝑟𝑒𝑎 𝐴𝐶𝐷 = 16 49 Area ABE = 4 ÷ 16 × 49 = 12.25 cm2 7a When 𝑥 = −10, 𝑦 = −10 − 4 (−10)2 = −10.0

7b 7ci 7cii -1.07 or 1.07 (±0.1) 7ciii 2𝑦 − 2𝑥 = −7 : (equation 1) 𝑦 = 𝑥 − 4 𝑥2 : (equation 2)

Sub (2) into (1): 2 (𝑥 − 4 𝑥2) − 2𝑥 = −7 2𝑥 − 8 𝑥2 − 2𝑥 = −7 − 8 𝑥2 = −7 8 = 7𝑥2 7𝑥2 − 8 = 0 𝐴 = 7, 𝐵 = −8 8a ∠𝑃𝑂𝑄 = 180∘ − 60∘ = 120∘ ∠𝑂𝑄𝑃 = 180∘−120∘ 2 = 30∘ (base ∠ of iso △) ∠𝑇𝑄𝑃 = 90∘ − 30∘ = 60∘ 8b ∠𝑃𝑅𝑄 = 120∘ 2 = 60∘ (∠ at centre = 2 ∠ at circumference) 8c 𝑂𝑃𝑅 = 180∘ − 60∘ − 20∘ − (2 × 30∘) = 40∘ (sum of ∠ of △) 8d Obtuse ∠𝑄𝑂𝑃 = 360∘ − 120∘ = 240∘ Area of major sector = 240∘ 360∘ × π(5)2 = ( 50 3 𝜋) 𝑐𝑚2 Area of △ 𝑃𝑂𝑄 = 1 2 (5)(5) sin 1 20∘=10.825 Total Area = 63.2 𝑐𝑚2 9a 𝐿𝑒𝑛𝑔𝑡ℎ = √[−6 − (−4)]2 + [−2 − (−7)]2 = √(−2)2 + 52 = 5.39 9b 𝑚𝐴𝐵 = 𝑚𝐶𝐷 = −2−(−7) −6−(−4) = 5 −2 𝑂𝐶⃗⃗⃗⃗⃗ = 𝐵𝐶⃗⃗⃗⃗⃗ + 𝑂𝐵⃗⃗⃗⃗⃗ = ( 8 −2) + (−4 −7) = ( 4 −9) 𝐶(4, −9) 𝑦 = −2.5𝑥 + 𝑐 −9 = −2.5(4) + 𝑐 𝑐 = 1 𝑦 = − 5 2 𝑥 + 1 𝑜𝑟 2𝑦 = −5𝑥 + 2 9ci 𝑋𝐶⃗⃗⃗⃗⃗ = 1 2 𝐴𝐶⃗⃗⃗⃗⃗ = 1 2 [𝑂𝐶⃗⃗⃗⃗⃗ − 𝑂𝐴⃗⃗⃗⃗⃗ ]

= 1 2 [( 4 −9) − (−6 −2)] = ( 5 −7/2) 9cii 𝑂𝑋⃗⃗⃗⃗⃗ = 𝑂𝐶⃗⃗⃗⃗⃗