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 − 𝑋𝐶⃗⃗⃗⃗⃗ = ( 4 −9) − ( 5 −7/2) = ( −1 −11/2) 9d 2 1 𝑜𝑟 2: 1 10a (35 × 20) + (45 × 39) + (55 × 16) + (65 × 20) + (75𝑥) = 50.1(20 + 39 + 16 + 20 + 𝑥) 4635 + 75𝑥 = 50.1(95 + 𝑥) 4635 + 75𝑥 = 4759.5 + 50.1𝑥 24.9𝑥 = 124.5 𝑥 = 5 10b Std Deviation = 11.6 min 10c The male participants ran faster than the females participants as their mean time was shorter. The female participants were more consistent in their running speed as their standard deviation was lesser than that of the males. 11a 𝑇5 = 72 + 17 = 66 11b The sum of 2 odd numbers or the sum of 2 even numbers will always be an even number. 11c 𝑇𝑛 = (𝑛 + 2)2 + 5 + 3(𝑛 − 1) = 𝑛2 + 4𝑛 + 4 + 5 + 3𝑛 − 3 = 𝑛2 + 7𝑛 + 6 11d 𝑇𝑝+1 − 𝑇𝑝 = (𝑝 + 1)2 + 7(𝑝 + 1) + 6 − [𝑝2 + 7𝑝 + 6] = 𝑝2 + 2𝑝 + 1 + 7𝑝 + 7 + 6 − 𝑝2 − 7𝑝 − 6 = 2𝑝 + 8 11e 2𝑝 + 8 = 4 2p = −4 ⇒ p = −2 Since 𝑝 cannot be negative, consecutive terms of the sequence cannot have a difference of 4 . 12ai Let X be North of H Angle PHX = 360o – 306o = 54o Bearing of H from P = 180o – 54o = 126o (int angles) 12aii Bearing of L from P = 126o + 124o = 250o 12aiii tan 𝜃 = 500 2500 angle of elevation = 11.3o

12b HL2 = 2.52 + 32 – 2(2.5)(3)cos124o HL = 4.86 km 12c Let X be the point where LX is the shortest distance to HPQ cos 56o = 𝑋𝑃 2.5 XP = 1.39798 HX = 1.39798 + 3 = 4.39798 Time = 4397.98 𝑚 4.5 𝑚/𝑠 = 977.33 s = 16.289 min It left harbour at 0753 hours 13a 3 4 × 0.88 = 0.66 (shown) 1 2 × 0.88 = 0.44 (shown) 13b 0.5 × 0.5 × 0.22 × 0.9 = 0.0495 13c Time Color Target Size Probability (Hit + Capture) 0s Red 2 Target 2 = 0.66 × 0.8 × 0.4 × 0.7 = 0.14784 1s Green 3 Target 3 = 0.44 × 0.95 × 0.9 × 0.7 = 0.26334 2s Yellow 4 Target 4 = 0.22 × 0.5 × 0.7 × 0.7 = 0.0539 3s Orange 1 Target 1 = 0.88 × 0.7 × 0.5 × 0.7 = 0.2156 ∴ Maximum probability happens at green Target 2 Sam should wait for 1 second for the target to change from red Target 2 to green Target 3, with maximum probability of 0.26334 ∴