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 =