Juying 4E5N AM Prelims P1 2024 Solutions

1 CANDIDATE NAME CENTRE NUMBER S INDEX NUMBER ADDITIONAL MATHEMATICS 4049/01 Paper 1 22 August 2024 2 hours 15 minutes Candidates answer on the Question Paper. READ THESE INSTRUCTIONS FIRST Write your Centre number, index number and name on all the work you hand in. Write in dark blue or black pen. You may use a pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. If working is needed in any question it must be shown with the answer. Omission of essential working will result in loss of marks. The total number of marks for this paper is 90. The use of an approved scientific calculator is expected, where appropriate. If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place. For 𝜋, use either your calculator value or 3.142. This document consists of 18 printed pages. Set by: Mr Albert Lui Vetted by: Mdm Norhafiani Bte Abdul Majid General Certificate of Education Ordinary Level JUYING SECONDARY SCHOOL, SINGAPORE Secondary Four Express/Five Normal Academic Preliminary Examination

2

3 Answer ALL the questions 1 (a) The function 𝑓 is defined, for all values of 𝑥, by 𝑓(𝑥) = (2𝑥 − 𝑥2)𝑒𝑥. Find the range of values of 𝑥 such that 𝑓(𝑥) is a decreasing function. [4] 𝑓′(𝑥) = (2 − 2𝑥)𝑒𝑥 + (2𝑥 − 𝑥2)𝑒𝑥 = 𝑒𝑥(2 − 𝑥2) B1 Decreasing Function: 𝑓′(𝑥) < 0 𝑒𝑥(2 − 𝑥2) < 0 M1 Since 𝑒𝑥 > 0, 2 − 𝑥2 < 0 𝑥2 − 2 > 0 (𝑥 + √2)(𝑥 − √2) > 0 M1 𝑥 < −√2 𝑥 > √2 A1 (b) The gradient function of the curve is 2(𝑝 + 1)𝑥 + 2, where 𝑝 is a constant. Given that the tangent to the curve at (2 , −2) is parallel to 𝑦 + 2𝑥 − 5 = 0, find the value of 𝑝. [3] 𝑑𝑦 𝑑𝑥 = 2(𝑝 + 1)𝑥 + 2 2(𝑝 + 1)𝑥 + 2 = −2 M1 (𝑝 + 1)𝑥 = −2 M1 When 𝑥 = 2, 2𝑝 + 2 = −2 𝑝 = −2 A1

4 2 The diagram shows a chocolate bar in the form of a triangular prism and the cross- section of the chocolate bar is an isosceles triangle with 𝐴𝐵 = 𝐴𝐶. 𝑀𝐶 = (√2 + 1 2) cm and ∠𝐴𝐶𝐵 = 45°. (a) Find the exact length of 𝑨𝑪. [3] 𝐜𝐨𝐬 𝟒𝟓° = √𝟐+𝟏 𝟐 𝑨𝑪 M1 𝑨𝑪 = 𝟐(√𝟐+𝟏 𝟐) √𝟐 = 𝟐√𝟐+𝟏 √𝟐 × √𝟐 √𝟐 M1 = 𝟐 + √𝟐 𝟐 or 𝟒+√𝟐 𝟐 A1 (b) Given that the volume of the chocolate bar is (𝟐𝟓 + 𝟐𝟐√𝟐)𝐜𝐦𝟑, find the length of 𝑨𝑫 in the form (𝒂 + 𝒃√𝟐) cm, where 𝒂 and 𝒃 are integers. [4] Vol = 𝟏 𝟐 × ( 𝟒+√𝟐 𝟐 ) × ( 𝟒+√𝟐 𝟐 ) × 𝑨𝑫 M1 𝟐𝟓 + 𝟐𝟐√𝟐 = 𝟗+𝟒√𝟐 𝟒 𝑨𝑫 𝑨𝑫 = 𝟐𝟓+𝟐𝟐√𝟐 𝟗+𝟒√𝟐 𝟒 = 𝟏𝟎𝟎+𝟖𝟖√𝟐 𝟗+𝟒√𝟐 × 𝟗−𝟒√𝟐 𝟗−𝟒√𝟐 M1 = 𝟗𝟎𝟎−𝟒𝟎𝟎√𝟐+𝟕𝟗𝟐√𝟐−𝟕𝟎𝟒 𝟒𝟗 = 𝟏𝟗𝟔+𝟑𝟗𝟐√𝟐 𝟒𝟗 M1 = 𝟒 + 𝟖√𝟐 A1 C M B D A cm

5 3 The diagram shows a circle, centre O, with diameter AB. The points D and F lie on the circle. The point E is such that EB and EF are tangents to the circle. (a) Given that the points 𝐶 and 𝐷 are midpoints of 𝐵𝐸 and 𝐴𝐸 respectively, prove that angle 𝐷𝐶𝐸 = 90°. [3] ∠𝐴𝐵𝐶 = 90° ( tangent perpendicular radius) M1 𝐷𝐶 parallel 𝐴𝐵 (mid point theorem) M1 Angle 𝐷𝐶𝐸 = 90° (corresponding angles) A1 (b) Given that triangle BEF is equilateral, prove that ∠𝐵𝐸𝐹 = ∠𝐵𝐴𝐹. [2] ∠𝐸𝐵𝐹 = ∠𝐵𝐴𝐹 (alternate segment theorem) M1 Since ∠𝐸𝐵𝐹 = ∠𝐵𝐸𝐹, ∠𝐵𝐸𝐹 = ∠𝐵𝐴𝐹 (shown) A1 4 (a) Find the remainder when 6𝑥3 − 13𝑥2 + 17𝑥 − 6 is divided by 2𝑥 − 1. [2] When 𝑥 = 1 2, Remainder = 6 ( 1 2) 3 − 13 ( 1 2) 2 + 17 ( 1 2) − 6 M1 = 0 A1 A B O F E C D

6 (b) Show that there is only one real root of the equation 6𝑥3 − 13𝑥2 + 17𝑥 − 6 = 0. [3] (2𝑥 − 1)(6𝑥2 − 10𝑥 + 12) = 0 B1 𝑥 = 1 2 6𝑥2 − 10𝑥 + 12 = 0 3𝑥2 − 5𝑥 + 6 = 0 Discriminant: 𝑏2 − 4𝑎𝑐 = 25 − 4(3)(6) = −47 B1 Since −47 < 0, 3𝑥2 − 5𝑥 + 6 = 0 has no real roots, hence equation has only 1 real root which is 𝑥 = 1 2. B1 5 Solve the following equations. (a) 5𝑥 − 5 𝑥 2+1 = 6, [3] Let 𝑦 = 5 𝑥 2 𝑦2 − 5𝑦 − 6 = 0 M1 𝑦 = 6 𝑦 = −1 (reject) A1 𝑥 2 lg 5 = lg 6 𝑥 = 2.23 B1

7 (b) 2 lg(𝑥 − 3) − lg(𝑥 + 7) = 1 log100 10. [4] lg (𝑥−3)2 𝑥+7 = lg 100 lg 10 M2 100 = (𝑥−3)2 𝑥+7 M1 𝑥2 − 106𝑥 − 691 = 0 𝑥 = 112 or 𝑥 = −6.16 (rej) A1 6 (a) State the values between which the principal value of sin−1 𝑥 must lie. [1] −90° ≤ sin−1 𝑥 ≤ 90° − 𝜋 2 ≤ sin−1 𝑥 ≤ 𝜋 2 (b) Find the principal value of tan−1 1 in radian in exact form. [1] Principal value = 𝜋 4 7 Given that cot 𝜃 = − 3 4 and that tan 𝜃 and cos 𝜃 have opposite signs, without evaluating 𝜃, find the exact values of each of the following. (a) cos(−𝜃), [2] tan 𝜃 = − 4 3 , lies in 4th quadrant M1 cos(−𝜃) = cos 𝜃 = 3 5 A1 (b) sin 2𝜃 [2] = 2 sin 𝜃 cos 𝜃 = 2 (− 4 5) ( 3 5) M1 = − 24 25 A1

8 8. The approximate mean distance x (in millions of kilometres) from the centre of the Sun and the period of the orbit T (in Earth years) are recorded in the table. Mercury Venus Mars Uranus x 58 108 228 2871 T 0.24 0.62 1.88 84.11 It is believed that the planets orbiting around the Sun obey a law of the form nkxT = , where k and n are constants. (a) Express the equation in a form suitable for drawing a straight line graph and draw the graph using appropriate scaling on both axes. [4] lg 𝑇 = 𝑙𝑔𝑘 + 𝑛𝑙𝑔𝑥

9 (b) Use your graph to estimate the value of k and of n, to two significant figures. [3] Lg k = -3.2 k = 0.00063 (0.00063 to 0.00079) n = 2−9−3.2) 3.5−0 = 1.49 = 1.5 (1.4 to 1.6) (c) Using the graph, find the orbital period of the Earth, if the distance between the Earth and the Sun is about .106.149 6 km Give your answer correct to the nearest integer. [2] lg 149.6 = 2.17 = lg x lg T = 0 ⇒ T = 1 (0.79 to 1.25) (d) If the orbital period of the Jupiter is 11.86 Earth years, estimate the distance of the Jupiter from the Sun in km using your graph. [2] lg 11.86 = 1.07 lg x = 2.9 = 794000000 km (631000000 to 1000000000 km)

10 9. (a) Express 2𝑥3+2𝑥2−7𝑥+4 𝑥(𝑥−1)2 in partial fractions. [5] = 2 + 6𝑥2−9𝑥+4 𝑥(𝑥−1)2 B1 Let 6𝑥2−9𝑥+4 𝑥(𝑥−1)2 = 𝐴 𝑥 + 𝐵 𝑥−1 + 𝐶 (𝑥−1)2 6𝑥2 − 9𝑥 + 4 = 𝐴(𝑥 − 1)2 + 𝐵𝑥(𝑥 − 1) + 𝐶𝑥 When 𝑥 = 0, 4 = 𝐴 B1 When 𝑥 = 1, 6 − 9 + 4 = 𝐶 𝐶 = 1 B1 When 𝑥 = −1, 6 + 9 + 4 = 4(4) + 2𝐵 − 1 2𝐵 = 4 𝐵 = 2 B1 2𝑥3+2𝑥2−7𝑥+4 𝑥(𝑥−1)2 = 2 + 4 𝑥 + 2 𝑥−1 + 1 (𝑥−1)2 B1