KRSS AM Prelim P2 2024 MS

Kent Ridge Secondary School Secondary 4 Express/5 Normal Academic Preliminary Examination 2024 Add Math Prelim 2024 P2 Mark scheme Qn Solutions Marks 1a 4 (𝑥2 + 1)(𝑥 + 1) = 𝐴𝑥 + 𝐵 𝑥2 + 1 + 𝐶 𝑥 + 1 4 (𝑥2 + 1)(𝑥 + 1) = (𝐴𝑥 + 𝐵)(𝑥 + 1) (𝑥2 + 1)(𝑥 + 1) + 𝐶(𝑥2 + 1) (𝑥2 + 1)(𝑥 + 1) 4 = (𝐴𝑥 + 𝐵)(𝑥 + 1) + 𝐶(𝑥2 + 1) Sub x = -1 4 = 𝐶((−1)2 + 1) 𝐶 = 2 Sub x = 0 4 = (𝐵)(1) + 𝐶(1) 𝐵 = 2 Compare coef of x2: 𝐴 + 𝐶 = 0 𝐴 = −2 4 (𝑥2 + 1)(𝑥 + 1) = 2 − 2𝑥 𝑥2 + 1 + 2 𝑥 + 1 A1 M1 M1 M1 A1 2a Plot points of corresponding values of R and 𝟏 𝒅𝟐 Draw best fit line through points and the origin Find gradient of the line gives the value of k B1 B1 B1 B1 2bi Gradient of line = 1.5 −10 = −0.15 Lg y intercept = 3 lg 𝑦 = −0.15𝑡 + 3 𝑦 = 10−0.15𝑡+3 M1 B1 M1 A1 2bii Initial number of particles : lg 𝑦 = 3 𝑦 = 103 = 1000 Find the point on the straight line when lg 𝑦 = lg 500 = 2.69 The time taken is the t value of the point M1 M1 B1 – their t value (±0.4) 3a 𝑑𝑦 𝑑𝑥 = − sin 𝑥 + 1 2 − sin 𝑥 + 1 2 = 0 sin 𝑥 = 1 2 𝛼 = 𝜋 6 𝑥 = 𝜋 6 , 5𝜋 6 𝑥 = 5𝜋 6 M1 A1 M1 A1 3bi Gradient of tangent at x = 0 𝑑𝑦 𝑑𝑥 = − sin 0 + 1 2 = 1 2 𝑦 = cos 0 + 0 = 1 Equation of tangent 𝑦 = 1 2 𝑥 + 1 M1 M1

Qn Solutions Marks 3bii 𝐴𝑟𝑒𝑎 = ∫ 1 2 𝑥 + 1 − cos 𝑥 − 𝑥 2 5𝜋 6 0 𝑑𝑥 = [𝑥 − sin 𝑥]0 5𝜋 6 = 5𝜋 6 − sin (5𝜋 6 ) = 5𝜋 6 − 1 2 = 2.12 (3 𝑠. 𝑓. ) M1 – mtd to find area trap under tangent M1 – definite integral of curve from 0 to 𝑥𝑏 A1 – correct expr of their integrals A1 – correct sub of limits A1 4a 𝑘 2 + 8𝑥 = 1 2𝑥 + 2𝑘𝑥 𝑘𝑥 + 16𝑥2 = 1 + 4𝑘𝑥2 (4𝑘 − 16)𝑥2 − 𝑘𝑥 + 1 = 0 𝑘2 − 4(4𝑘 − 16)(1) = 0 𝑘2 − 16𝑘 + 64 = 0 𝑘 = 8 M1 M1 M1 A1 4b Discriminant: √8 2 − 4𝑎(𝑎 − 1) < 0 8 − 4𝑎2 + 4𝑎 < 0 𝑎2 − 𝑎 − 2 > 0 (𝑎 − 2)(𝑎 + 1) > 0 Since 𝑎 < 0, 𝑎 < −1 M1 – expr for D M1 – condition for D B1 A1 5a 5b 5c 𝑦 = 𝑒2𝑥 sin 3𝑥 𝑑𝑦 𝑑𝑥 = 2𝑒2𝑥 sin 3𝑥 + 3𝑒2𝑥 cos 3𝑥 𝑑2𝑦 𝑑𝑥2 = 2(2𝑒2𝑥 sin 3𝑥 + 3𝑒2𝑥 cos 3𝑥) + 3(2𝑒2𝑥 cos 3𝑥 − 3𝑒2𝑥 sin 3𝑥 𝑑2𝑦 𝑑𝑥2 = −5𝑒2𝑥 sin 3𝑥 + 12𝑒2𝑥 cos 3𝑥 2𝑒2𝑥 sin 3𝑥 + 3𝑒2𝑥 cos 3𝑥 − 5𝑒2𝑥 sin 3𝑥 + 12𝑒2𝑥 cos 3𝑥 + 𝑎𝑒2𝑥 sin 3𝑥 = 𝑏𝑒2𝑥 cos 3𝑥 2 − 5 + 𝑎 = 0 3 + 12 = 𝑏 𝑎 = 3, 𝑏 = 15 M1 either term seen A1 use of product rule and final ans M1 use of at one correct product rule of their dy/dx A1 M1 M1 M1 A1

Qn Solutions Marks 6 𝑑 𝑑𝑥 ( 𝑥 − 2 √3𝑥 + 1 ) = √3𝑥 + 1 − 3(𝑥 − 2) 2√3𝑥 + 1 3𝑥 + 1 = 2(3𝑥 + 1) 2√3𝑥 + 1 − 3(𝑥 − 2) 2√3𝑥 + 1 3𝑥 + 1 = 3𝑥 + 8 2√3𝑥 + 1(3𝑥 + 1) = 3𝑥 + 8 2√(3𝑥 + 1)3 M1 – quotient rule seen with positive sq root or product seen with negative sq root M1 – simplify with common denominator or taking out common factor M1 – all factors in denominator collected B1 6b ∫ 3𝑥 + 8 2√(3𝑥 + 1)3 𝑑𝑥 𝑥2 𝑥1 = [ 𝑥 − 2 √3𝑥 + 1 ] 𝑥1 𝑥2 ∫ 3𝑥 + 7 2√(3𝑥 + 1)3 𝑑𝑥 𝑥2 𝑥1 + ∫ 1 2√(3𝑥 + 1)3 𝑑𝑥 𝑥2 𝑥1 = [ 𝑥 − 2 √3𝑥 + 1 ] 𝑥1 𝑥2 ∫ 3𝑥 + 7 2√(3𝑥 + 1)3 𝑑𝑥 𝑥2 𝑥1 = [ 𝑥 − 2 √3𝑥 + 1 ] 𝑥1 𝑥2 − ∫ 1 2√(3𝑥 + 1)3 𝑑𝑥 𝑥2 𝑥1 = [ 𝑥 − 2 √3𝑥 + 1 ] 𝑥1 𝑥2 − 1 2 ∫ (3𝑥 + 1)−3 2𝑑𝑥 𝑥2 𝑥1 = [ 𝑥 − 2 √3𝑥 + 1 ] 0 5 − [1 2 (3𝑥 + 1)−1 2 3 (− 1 2) ] 0 5 = 3 4 − −2 1 − (− 1 3(4) + 1 3(1)) = 2 1 2 M1 – seen or implied M1 – any equivalent form To show 7 = 8-1 or 8 = 7+ 1 M1 – standard integral M1 – show the correct limits substituted into a valid integral A1 7a ∠𝐷𝐴𝐸 = ∠𝐴𝐵𝐷 (angles in alternate segment) ∠𝐴𝐷𝐸 = ∠𝐵𝐴𝐷 (alternate angles of parallel lines) triangle ABD is similar to triangle DAE (AA similarity) B1 B1 B1 7b ∠𝐵𝐴𝐷 = ∠𝐷𝐶𝐵 (corresponding angles of similar triangles) ∠𝐵𝐴𝐷 + ∠𝐷𝐶𝐵 = 180° (angles in opposite segment) B1 B1

Qn Solutions Marks ∠𝐵𝐴𝐷 = ∠𝐷𝐶𝐵 = 90° BD is diameter (angle in semicircle = 90°) B1 8a 3(3𝑥+1) = 10 − 3−𝑥 Let 𝑢 = 3𝑥 3(3𝑢) = 10 − 1 𝑢 9𝑢2 − 10𝑢 + 1 = 0 3𝑥 = 1 9 or 3𝑥 = 1 𝑥 = −2 𝑜𝑟 0 M1 – breakdown 3𝑥+1 M1 – general QE M1 – eqn in x A1 8b log100 𝑥 + lg 𝑦 = 3 lg 𝑥 lg 100 + lg 𝑦 = 3 lg 𝑥 2 + lg 𝑦 = 3 lg √𝑥 + lg 𝑦 = 3 lg √𝑥 𝑦 = 3 √𝑥𝑦 = 103 𝑦 = 1000 √𝑥 M1 – change base M1 – step before simplifying to one log term M1 – one log term A1 9a (𝑥 − 2.5)2 + (− 1 2 𝑥 + 5)2 = 365 4 𝑥2 − 5𝑥 + 6.25 + 1 4 𝑥2 − 5𝑥 + 25 = 365 4 5 4 𝑥2 − 10𝑥 + 31.25 = 365 4 5𝑥2 − 40𝑥 + 125 = 365 5𝑥2 − 40𝑥 − 240 = 0 𝑥 = 12, 𝑥 = −4 𝐴(−4, 12) M1v - substitution M1 – general QE A1 A1 9b centre of circle (2.5,5) 𝑦 = 2𝑥 + 𝑐 Sub centre of circle (2.5,5) 5 = 2(2.5) + 𝑐 𝑐 = 0 B1 M1 – grad ⊥ seen B1 A1 9c Sub y = 0 into AB 0 = − 1 2 𝑥 + 10 𝑥 = 20 D(20,0) M, Mid point AD = (8,6) Distance ME2 = (8 − 2.5)2 + (6 − 5)2 = 125 4 < 365 4 M1 M1 M1 M1 10a Sub t = 0, v = 1 B1, B1 10b 𝑎 = −8𝑒−2𝑡 + 1 = 0 𝑒−2𝑡 = 1 8 M1 M1

Qn Solutions Marks −2𝑡 = ln 1 8 −2𝑡 = ln 1 − ln 8 = − ln 8 𝑡 = 1 2 ln 8 𝑣 = 4𝑒− ln 8 + 1 2 ln 8 − 3 = −1.46 B1 A1 10c Since velocity changes from positive to negative, the particle did change its direction of motion B1 B1 10d 𝑠 = −2𝑒−2𝑡 + 𝑡2 2 − 3𝑡 + 𝑐 Sub t= 0, s = 6 6 = −2 + 𝑐 𝑐 = 8 Sub t= 2 𝑠 = −2𝑒−4 + 2 − 6 + 8 = 3.96m M1 integrate exp term M1 integrate power term M1 A1