NASS 2026 AM O S4E5N Prelim P1 Mark Scheme with Marker's Report

Name: MARK SCHEME ................................... Register no: ........... Class: ........... NGEE ANN SECONDARY SCHOOL O PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS 4049/01 Paper 1 28 August 2026 2 hours 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, register number and class in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 90. For Examiner’s Use Total /90 Checked by student: ………………………………. Date: ………………… This document consists of 21 printed pages and 1 blank page.

2 NAS/2026/Prelim/AM-O/P1 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 02  cbxax , a acbbx 2 42 Binomial expansion   1 2 2 ... ...1 2 n n n n n r r nn n na b a a b a b a b b r                          , where n is a positive integer and       1 ... 1 ! ! ! ! n n n n rn r r n r r          2. TRIGONOMETRY Identities 2 2sin cos 1A A  2 2sec 1 tanA A  2 2cosec 1 cotA A    BABABA sincoscossinsin   cos cos cos sin sinA B A B A B     BA BABA tantan1 tantantan   AAA cossin22sin  2 2 2 2cos 2 cos sin 2cos 1 1 2sinA A A A A      2 2 tantan 2 1 tan AA A  Formulae for ΔABC sin sin sin a b c A B C  2 2 2 2 cosa b c bc A   1 sin2 bc A 

3 NAS/2026/Prelim/AM-O/P1 1 The percentage Gross Domestic Product (GDP) growth of a country from 2014 to 2024 can be modelled by the quadratic equation 2G( ) 0.7 7 11.2t t t   , where t is the number of years after 2014. (a) By completing the square, find the year in which the percentage growth is the lowest, and state this minimum percentage GDP growth. [4]       2 2 2 2 0.7 7 11.2 0.7 10 11.2 0.7 5 6.3 B1 for 0.7 5 and B1 for 6.3 t t t t t t            Hence, the year in which the percentage growth is the lowest is 2019 ----- A1 and the minimum percentage GDP growth is 6.3% ----- A1 Marker’s Report: 1. This part of the question was generally well done. Most candidates were able to complete the square correctly. 2. A number of candidates did not read the question carefully and left their answer as (t = 5) instead of stating the year, 2019. Some candidates also omitted the negative sign when stating the minimum percentage GDP growth. (b) Explain why this model may not be suitable for predicting the country’s percentage GDP growth in the long run. [1] The model is quadratic, hence it will eventually increase without bound as t becomes large. This is unlikely to reflect real GDP growth in the long run as economic growth is affected by many changing factors. -----B1 Marker’s Report: 1. This part of the question was generally poorly done. 2. Candidates did not fully understand the question, “Explain why this model may not be suitable,” and instead gave responses that were not based on mathematical reasoning. Candidates should also note that the increase is not exponential. For a quadratic model, the rate of increase is linear.

4 NAS/2026/Prelim/AM-O/P1 2 Do not use a calculator in this question. It is given that 2 12a b  , where a and b are positive real numbers. (a) Show that   2 2 3 216 2 6a b b    . [4]   2 2 2 2 2 12 24 144 M1 for squaring both sides 2 3 2 24 144 144 2( 12 ) M1 for factorising 144 2( 12 ( 6) ( 6) ) M1 for completing the squa re 144 2[( 6) 36] M1 for simplifying a b b b a b b b b b b b b                            2 2 144 72 2( 6) 216 2( 6) AG b b         Alternative Method:     2 2 2 12 24 144 M1 for expansion 2 3 2 24 144 216 2 6 216 2( 12 36) M1 for expansion 216 2 24 72 144 2 24 (2) M1 for simplifying Since (1) = (2), M1 for comparing 2 3 a b b b a b b b b b b b b b b a b                               2 216 2 6 AGb    Marker’s Report: 1. This question was generally poorly done. 2. Among candidates who attempted the question, some omitted relevant working, especially since this was a “show” question. As a result, they were not awarded full credit for this part of the question. (b) Hence state the least possible positive value of 1 2 3a b . [1] Maximum value of 2 3 216 1 1Least value of B12 3 216 a b a b     Marker’s Report: 1. This part of the question was either not attempted or was generally poorly done. 2. Candidates should recognise that even without successfully completing part (a), they could still have answered part (b) correctly if they understood that the least possible value occurs when the denominator is at its maximum.

5 NAS/2026/Prelim/AM-O/P1 3 (a) Express   1 1n n  in partial fractions. [3]     1Let M1 1 1 1 ( 1) When 0, 1 A1 When 1, 1 A1 1 1 1 minus 1 mark without final statement1 1 A B n n n n A n Bn n A n B n n n n                  Marker’s Report: 1. This part of the question was generally well done, although a number of candidates made careless mistakes when writing the final statement. (b) Hence show that 1 1 1 1 1 2 ...3(4) 4(5) 5(6) ( 1) ( 1) 3( 1) n n n n n n          . [3]     1 1 1 1 1 ...3(4) 4(5) 5(6) ( 1) ( 1) 1 1 1 1 1 1 1 1 1 1 ... M1 for writing specific terms3 4 4 5 5 6 1 1 1 1 M1 for simplifying3 1 1 3 M1 for same denominator3 1 2 (shown) A3 1 n n n n n n n n n n n n n                             G Marker’s Report: 1. This part of the question was generally poorly done. 2. Candidates did not fully understand what was meant by “Hence show” and, as a result, did not make use of their answers from part (a) to proceed with this part of the question. 3. A handful of candidates incorrectly thought that this was a Binomial Theorem question.

6 NAS/2026/Prelim/AM-O/P1 4 (a) Find the coefficient of 3x in the binomial expansion of 1221 .2 x x     [3]             122 2 121 1 12 2 12 3 53 121 in M1 for general term2 2 12 2 12 2 12 3 3 M1 for equating power of to 3 5 12Coefficient of 2 24 5 r r r r r r r r x xT x rx x xr xr r x r x                                                 3 A14 Marker’s Report: 1. This question was generally quite well done. 2. A number of candidates expanded the expression fully when they could have used the general term instead. 3. A number of candidates did not read the question carefully and therefore did not state the coefficient of the required term. 4. Candidates who did not arrive at the correct answer generally made algebraic slips or omi