NASS 2025 AMATH PRELIM P1 MS

MARK SCHEME Register no: .............................. Class: ........... NGEE ANN SECONDARY SCHOOL O PRELIMINARY EXAMINATION ADDITIONAL MATHEMATICS 4049/01 Paper 1 1 September 2025 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 24 printed pages.

2 NAS/2025/Prelim/AM-O/P1 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 02 = + +c bx ax, a ac b bx 2 42 − ± −= Binomial Expansion ( ) 1 22 ... ...12 n n n n nr r nnn nab a a b a b a b b r −− −  + = + + ++ ++     , where n is a positive integer and ( ) ( ) ( ) 1 ... 1 ! ! ! ! n nn nrn r r nr r − −+ = = − 2. TRIGONOMETRY Identities 22sin cos 1AA+= 22sec 1 tanAA= + 22cosec 1 cotAA= + ( ) B A B A B Asin cos cos sinsin ±= ± ( )cos cos cos sin sinAB A B A B±=  ( ) B A B AB A tan tan1 tan tantan  ±= ± A A Acos sin2 2sin = 22 2 2cos2 cos sin 2cos 1 1 2sinA AA A A= − = −=− 2 2tantan2 1 tan AA A= − Formulae for ΔABC sin sin sin abc ABC= = 222 2 cosa b c bc A=+− 1 sin2 bc A∆=

3 NAS/2025/Prelim/AM-O/P1 1 Determine with working, whether the function ( ) ( ) 7f 31x x= + , 0x> , is an increasing or decreasing function. [3] [ ] 2 2 2 2 df() f () d 7( 1)(3 1) (3) M1 21 (3 1) For 0, (3 1) 0 21 0 M1(3 1) Since f ( ) 0, for 0,f( ) is a decreasing fu nction. A1 xx x x x x x x x xx − ′ = = − + −−−−− =− + > +> − < −−−−−+ ′ < > −−−−−

4 NAS/2025/Prelim/AM-O/P1 2 The equation of a quadratic curve is given by 2 17y x px= −+ − and it has a maximum point ( )4,q where p and q are constants. (a) By expressing y in the form of ( ) 2 xa b−− + , find the value of p and of q. [4] ( ) 2 2 22 22 2 2 17 17 17 M1 for completing the square 22 1722 17 24 Since maximum point is (4, ), 4 & 2 y x px x px ppx ppx ppx q p = −+ − = −−−   =− − − − − −−−−−       = − − +− −     = −− + −  = 2 2 17 M14 8 8 A1 17 4 1 A1 pq p = − −−−−− = −−−−− = − =− −−−−− Alternatively, ( ) 2 2 Since maximum point is (4, ), 4 M1 8 16 M1 8 A1 & 17 16 1 A1 q yx q xx q pq q =− − + −−−−− = − + − + −−−−− = −−−−− − =− + =− −−−−− (b) Hence, write down the nature of the turning point of 2 5 17y x px= −+ − and state the coordinates of the turning point. [2] Minimum point B1 at ( )4, 5 B1− −−−−−

5 NAS/2025/Prelim/AM-O/P1 3 A line that is not parallel to the axisx− has the equation 2 33ax y a b+= −− , where a and b are constants. A curve has the equation 2223xyb+= + . Given that the line and the curve intersect at the point ( )0,a , find the value of a and of b. [5] 2 2 22 2 2 33 3 3 (1) 2 3 (2) When 0, , From (1), 3 3 M1 for substituting point in to line 2 3 (3) From (2), 3 M1 for substituting point int o curve 3 ax y a b y ax a b xyb x ya a ab ba ab ab += −− =− + − − −−−−− + = + −−−−− = = = − − −−−−− = − −−−−− = + −−−−−− = + ( ) 2 2 (4) Substitute (3) into (4), 2 3 3 M1 for solving simultaneously 20 20 0 (rej, since not parallel) or 2 A1 with rejection 1 A1 aa aa aa aa b −−−−− = − + −−−−− −= −= = = −−−−− = −−−−−

6 NAS/2025/Prelim/AM-O/P1 4 The first three terms in the expansion of ( )1 n ax+ are 23313 8xx−+ , where a and n are constants. Find the value of a and of n. [5] ( ) ( ) ( ) ( ) ( ) 2 22 22 2 11 112 2 By comparing the coefficients of: : 3 M11 3 3 (1) 33: M12 8 1 33 (2)28 Sub. (1) into (2): 1 3 2 n nn nax ax ax nax a x nxa na a n nxa nn a nn n    + = + + += + + +        =− −−−−− =− =− −−−−−  = −−−−−  − = −−−−− − −  ( ) 2 2 2 22 2 2 33 M18 9 9 33 28 72 72 66 6 72 0 12 0 12 0 0 (rej, since 0) or 12 0 12 A1 Sub. 12 into (1): 3 12 1 4 1 and 12 A1 with rejection4 nn n n nn nn nn nn n nn n n a a an = −−−−− − = −= −= −= −= = ≠ −= = −−−−− = =− =− ∴ =− = −−−−− Note: if solve for a first, no need for rejection of value as there is only one value of n.

7 NAS/2025/Prelim/AM-O/P1 5 (a) Prove that 1 cos sin tan1 cos sin 2 xx x xx −+ =++ . [4] 2 2 2 1 cos sin 1 cos sin 2sin 2sin cos2 22 B1 for 1 cos 2sin and B1 for sin 2sin cos2 222cos 2sin cos2 22 sin sin cos22 2 M1 for factorising cos sin cos22 2 sin 2 A cos 2 xx xx x xx x xxxxx xx xx x xx x x x −+ ++ + = −−−−− − = = +  += −−−−− +  = −−−−− 1 tan [AG]2 x= −−−−−

8 NAS/2025/Prelim/AM-O/P1 (b) It is given that ( )f 1 2sin 3xx= − . (i) State the least and greatest values of ( )f x . [2] ( ) ( ) ( ) ( ) Least value of f 1 2 1 1 B1 Greatest value of f 1 2 1 3 B1 x x = − =− −−−−− = − − = −−−−− (ii) State the period of ( )f x . [1] ( ) 360 2Period of f 120 or B133x π° = = ° −−−−− (iii) Solve 1 2sin 3 0x−= for 0 x π≤≤ . [3] 1 1 2sin 3 0, 0 1sin 3 0 3 32 1Basic angle, sin M 12 6 3 , , 2 , 3 5 13 173 , , , 66 6 6 4 correct answers: 2 marks 5 13 17, , , A 2 2 or 3 correc18 18 18 18 xx xx x x x π π α πα απα πα πα ππ π π ππ π π − − = ≤≤ = ≤≤ = −−−−− = = −+− = = −−−−− t answers: 1 mark otherwise: 0 mark   

9 NAS/2025/Prelim/AM-O/P1 Correct Shape with 1.5 cycle for ( )fyx= – B1 y-axis readings ( )1, 1 and 3y=− – B1 Critical points readings (x = 5,, ,62 6 ππ π π ) – B1 (iv) Sketch the graph of ( )fyx= for 0 x π≤≤ . [3]

10 NAS/2025/Prelim/AM-O/P1 6 (a) It is given that 1sin 5 A=− and 3tan 4B=− , where A and B are in the same quadrant. Without using a calculator, find the exact value of (i) ( )tan A− , [2] ( )tan tan M1 1 A12 A A − =− −−−−− = −−−−− (ii) ( )sin AB+ , [2] ( )sin sin cos cos sin 14 2 3 M15555 10= 55 2 5 25 A15 AB A B A B+= +      = − + − −−−−−          − =− =− −−−−−