NASS 2026 AM-O S4E5N Prelim P2 (MR)

Name: .............................................................. Register no: ........... Class: ........... NGEE ANN SECONDARY SCHOOL O PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS 4049/02 Paper 2 31 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 25 printed pages and 1 blank page. MARKERS’ REPORT

2 NAS/2026/Prelim/AM-O/P2 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 2cos2 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/P2 1 The simultaneous equations 2 6y x px   and 32 2y x  have a common solution for which 3x . Find the value of p, and hence solve the simultaneous equations. [5] SOLUTIONS: Given the common solution has x = 3, 3 152(3) 2 2y   Subst. into 2 6y x px   : 215 3 3 62 3 2 p p     Hence    2 2 3 3 6 22 2 2 15 0 2 5 3 0 53 or 2 x x x x x x x x x             Using 32 2y x  , 153, 2 5 7,2 2 x y x y     Conceptual gaps Some students did not know that given x = 3 is a common solution (from question), we can sub in x = 3 to the equations to find the value of p. Many students did not find the values of y after solving for x.

4 NAS/2026/Prelim/AM-O/P2 2 (a) (i) In an architectural design, a support beam is modelled by the line 2y x k  and a parabolic arch is modelled by 2 4 7y x x   . The beam is to touch, but not cut, the arch. Find the value of k. [3] SOLUTIONS:   2 2 4 7 2 6 7 0 x x x k x x k         For tangency, discriminant = 0.      2 6 4 1 7 0 36 28 4 0 2 k k k         Algebraic manipulation mistakes Most students knew to equate discriminant to 0, but a few students made algebraic mistakes related to the signs. (ii) The beam is then raised vertically by 3 units from this tangent position. Without drawing or referring to a graph, determine the number of points at which the beam intersects the arch. [2] SOLUTIONS: If raised vertically by 3 units, 1k  . Discriminant =      2 6 4 1 6 12 0    Hence, there will be 2 points of intersection. Did not read question properly/ understand the question Some students made reference to graph(s) to answer this question when it was explicitly mentioned in question not to. Conceptual gaps Some students did not know that shifting beam up vertically by 3 units means the equation of the beam is now y = 2x + 1.

5 NAS/2026/Prelim/AM-O/P2 (b) Represent the solution set of   2 2 1 4 8x x   on a number line. [3] SOLUTIONS:      2 2 2 2 1 4 8 4 4 1 4 8 4 7 0 2 7 2 7 0 x x x x x x x x             7 7 or 2 2x x  Conceptual gaps Some students did not know how to solve this quadratic inequality. In particular, they write 2 7 2x  which is wrong. (What you should do) For quadratic inequalities, we typically make RHS = 0, then factorise LHS to identify the roots (i.e. x-intercepts). 7 2 7 2

6 NAS/2026/Prelim/AM-O/P2 3 A cubic polynomial is given as   3 2P x x ax bx c    . The graph of  Py x cuts the y-axis at  0, 2 . The remainder when  P x is divided by 1x is 3 , and 2x is a factor of  P x . (a) Find the values of a, b and c. [4] SOLUTIONS: Since the graph cuts the ݕ-axis at  0, 2 ,  P 0 2 2 c   Since 2x is a factor,  P 2 0 8 4 2 2 0 2 5 --- (1) a b a b        The remainder when  P x is divided by 1x is 3 , so  P 1 3 1 2 3 4 --- (2) a b a b          Solving (1) and (2), 3, 1, 2a b c   . Overall Performance Most students were able to apply the remainder and factor theorem correctly and finding the values of a, b and c. Common Mistakes Mostly careless mistakes including taking remainder as 3 instead of 3 , manipulation errors or computational errors.

7 NAS/2026/Prelim/AM-O/P2 (b) Hence solve the equation  P 0x  , leaving your answers in exact form. [3] SOLUTIONS:   3 2P 3 2x x x x    Since 2x is a factor,      2P 2 1x x x dx    . Comparing coefficients of x, 2 1 1 1 d d          2 P 2 1x x x x     Hence,    22 1 0x x x      22 0 or 1 0x x x     x = 2 or 1 5 1 52, , 2 2x   Overall Performance Most students were able to factorise P(x) by applying long division method to find the quadratic factor before factorising completely. Common Mistakes Mostly error carried forward from (a) leading to wrong factors and solutions. A couple of students left the answers in significant figures instead of exact form.      21 1 4 1 1 2 1 5 2 x      

8 NAS/2026/Prelim/AM-O/P2 4 The amount of medicine remaining in a patient’s bloodstream t hours after it is taken is M mg. It is modelled by the equation ln ln16 ln 25 3 M t      . (a) Show that the model can be written in the form ln 2380 t M e   . [2] SOLUTIONS: ln 23 ln 23 ln 23 ln 23 ln ln16 ln 25 3 ln ln16 ln 2 ln5 3 ln ln16 ln5 ln ln 165 165 80 (shown) t t t t M t M t e M e M e M e M e                                         Overall Performance Many students were able to show how to convert from logarithmic form to exponential form but some did not apply the concept ln xe x  . Common Mistakes Most students were unable to show the conversion clearly, there were several trials and errors or incomplete solutions.

9 NAS/2026/Prelim/AM-O/P2 (b) Find the time taken for the medicine to decrease to 20 mg. [3] SOLUTIONS: When 20M  , ln 23 ln 23 ln 23 80 20 1 4 1ln ln 4 1ln 2 ln3 4 13ln 4 ln 2 6 t t t e e e t t t                           Time taken is 6 hours. Overall Performance Most students were able to determine the value of t correctly. Common Mistakes There were a small number of computational errors. (c) A second dose should be taken only when the amount of medicine remaining in the bloodstream has fallen to less than 15 mg. A patient takes the medicine at 07 00. The patient asks wh