KRSS AM Prelim P1 2024 MS

Kent Ridge Secondary School Secondary 4 Express/5 Normal Academic Preliminary Examination 2024 Add Math Prelim 2024 P1 Mark scheme Qn Solutions Marks 1a tan 2𝐴 = 2 tan 𝐴 1 − tan2 𝐴 1 = 2 tan 𝐴 1 − tan2 𝐴 1 − tan2 𝐴 = 2 tan 𝐴 tan2 𝐴 + 2 tan 𝐴 − 1 = 0 tan 𝐴 = −2 ± √4 + 4 2 = −1 + √2 M1 A1 M1 A1 1b sec2 𝐴 = 1 + tan2 𝐴 = 1 + (−1 + √2) 2 = 1 + 1 − 2√2 + 2 = 4 − 2√2 M1 A1 2a Gradient 𝐿1 = −2 Gradient 𝐿2 = 1 2 −3 = 1 2 (2) + 𝑐 𝑐 = −4 𝐵(0, −4) M1 M1 A1 2b (4 − 2)2 + (𝑘 + 3)2 = 25 (𝑘 + 3)2 = 21 𝑘 = ±√21 − 3 M1 M1 or apply quad formula to general eqn A1 3a Min -2 and max 2 B1 3b Min -3 + 1 = -2, max = 3+1 = 4 B1 3c f(x) period or shape correct g(x) period or shape correct f(x) fully correct g(x) fully correct B1 B1 B1 B1 3d 3 B1 4a (𝑎 − 𝑥2)6 = 𝑎6 − 6𝑎5𝑥2 + (6 2) 𝑎4𝑥4 +… = 𝑎6 − 6𝑎5𝑥2 + 15𝑎4𝑥4 +… B1,B1,B1 4b ( 1 𝑥2 + 2 + 𝑥2) (𝑎 − 𝑥2)6 ( 1 𝑥2 + 2 + 𝑥2) (𝑎6 − 6𝑎5𝑥2 + 15𝑎4𝑥4) Term independent of x: −6𝑎5 + 2𝑎6 = 0 −6 + 2𝑎 = 0 𝑎 = 3 Term in x2: 𝑏 = 15𝑎4 − 12𝑎5 + 𝑎6 = −972 M1 M1 A1 M1,A1 5a f ′(𝑥) = (𝑥 − 3) − (𝑥 + 1) (𝑥 − 3)2 M1

Qn Solutions Marks f ′(𝑥) = −4 (𝑥 − 3)2 Since (𝑥 − 3)2 > 0, 𝑥 ≠ 3 𝐟 ′(𝒙) is always negative or Gradient of f is always negative, so it is a decreasing function A1 B1 B1 5b f(𝑥) = 1 + 4 𝑥 − 3 ∫ 1 + 4 𝑥 − 3 𝑑𝑥 = 𝑥 + 4 ln(𝑥 − 3) + 𝑐 M1 M1, M1, A1 6a 75℃ B1 6b 75𝑒−0.02𝑡 = 65 𝑒−0.02𝑡 = 65 75 −0.02𝑡 = ln ( 65 75) 𝑡 = ln(65 75) −0.02 = 7.16 min M1 M1 A1 6c 63𝑒15𝑘 = 54.2 𝑘 = ln(54.2 63 ) 15 = −0.01003 = −0.01 M1 M1, A1 6d 75𝑒−0.02𝑡 = 63𝑒−0.01𝑡 𝑒−0.02𝑡 𝑒−0.01𝑡 = 63 75 𝑒−0.01𝑡 = 63 75 𝑡 = ln(63 75) −0.01 = 17.4 minutes M1 M1 A1 7a 𝑥 (1 𝑥) + ln 𝑥 = 1 + ln 𝑥 M1 – diff ln x correctly A1 7b f(𝑥) = ln 𝑥 − 𝑥 + 𝑐 0 = ln 1 − 1 + 𝑐 c = 1 f(𝑥) = ln 𝑥 − 𝑥 + 1 M1 M1 A1 8a 2𝑥2 − 8𝑥 = −𝑥2 − 4𝑥 − 3 3𝑥2 − 4𝑥 + 3 = 0 Discriminant = (−4)2 − 4(3)(3) = −20 Since discriminant < 0, there are no real roots to the simultaneous equations. The 2 curves do not intersect M1 M1 A1 8b 2(𝑥2 − 4𝑥) = 2[(𝑥 − 2)2 − 4] = 2(𝑥 − 2)2 − 8 −(𝑥2 + 4𝑥 + 3) = −[(𝑥 + 2)2 − 1] = −(𝑥 + 2)2 + 1 Sketch min curve with TP (2,-8) passing through O Sketch max curve with TP (-2,1) passing through (0,-3) Non intersecting M1 M1 A1 B1 B1

Qn Solutions Marks 9a 𝑟 𝑥 = 5 25 𝑟 = 𝑥 5 Volume of liquid = 1 3 𝜋(5)2(25) − 1 3 𝜋 ( 𝑥 5) 2 (𝑥) = 1 3 𝜋(625 − 𝑥3 25) M1 B1 9b 𝑑𝑉 𝑑𝑥 = 1 3 𝜋(− 3𝑥2 25 ) 𝑑𝑉 𝑑𝑡 = 𝑑𝑉 𝑑𝑥 × 𝑑𝑥 𝑑𝑡 = 1 3 𝜋(− 3𝑥2 25 ) (− 1 2 𝑡) 𝑥 = ∫ − 1 2 𝑡 𝑑𝑡 𝑥 = − 1 2 (𝑡2 2 ) + 𝑐 Sub t = 0, x = 25 𝑐 = 25 𝑥 = − 𝑡2 4 + 25 Sub x = 2 − 𝑡2 4 + 25 = 2 𝑡2 = 92 𝑡 = √92 𝑑𝑉 𝑑𝑡 = 1 3 𝜋 (− 3(2)2 25 ) (− 1 2 √92) = 2.41 𝑐𝑚3/𝑠 M1 M1 M1 A1 A1 A1 10a 𝐵𝐶 = 8 sin 𝜃 𝐴𝐶 = 8 cos 𝜃 Area = 1 2 (8 sin 𝜃)(8 cos 𝜃) = 32 sin 𝜃 cos 𝜃 = 16 sin 2𝜃 M1 – either BC or AC found M1 A1 10b Max = 16 B1 10c Perimeter = AB + BC + CA = 8 + 8 sin 𝜃 + 8 cos 𝜃 = 8(1 + sin 𝜃 + cos 𝜃) B1 10d sin 𝜃 + cos 𝜃 = √2 sin(𝜃 + 45°) Max perimeter = 8(1 + √2) B1, B1 B1 11(a) 𝑓(0) = −4 + 𝑏 = 0 𝑏 = 4 𝑓(3) = 1 3 (3)3 + 𝑎(3)2 − 20(3) − 4 + 4 = −51 𝑎 = 0 M1 A1 M1 A1 11(b) 𝑓(𝑥) = 1 3 𝑥3 − 20𝑥 𝑓 ′(𝑥) = 𝑥2 − 20 = 0 𝑥 = ±√20 = ±2√5 M1 A1

Qn Solutions Marks 𝑓 ′′(𝑥) = 2𝑥 When 𝑥 = 2√5, 𝑓 ′′(𝑥) > 0, it is a minimum point When 𝑥 = −2√5, 𝑓 ′′(𝑥) < 0, it is a maximum point M1 A1 A1 12(a) sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵 sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵 = 2 3 3 sin 𝐴 cos 𝐵 − 3 cos 𝐴 sin 𝐵 = 2 sin 𝐴 cos 𝐵 + 2 cos 𝐴 sin 𝐵 sin 𝐴 cos 𝐵 = 5 cos 𝐴 sin 𝐵 sin 𝐴 cos 𝐵 cos 𝐴 cos 𝐵 = 5 cos 𝐴 sin 𝐵 cos 𝐴 cos 𝐵 tan 𝐴 = 5 tan 𝐵 M1 M1 B1 12(b) sin(45° − 𝜃) sin(45° + 𝜃) = 2 3 𝐴 = 45°, 𝐵 = 𝜃 tan 45° = 5 tan 𝜃 tan 𝜃 = 1 5 𝛼 = tan−1 (1 5) = 11.30° 𝜃 = 11.30, 180 + 11.30 = 11.3°, 191.3° M1 M1 A1 A1