MSHS 2026 Prelim AM P2 Solutions

This document consists of 22 printed pages. [Turn over For Examiners’ Use Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 /4 /5 /6 /7 /4 /9 /10 /9 Q9 Q10 Q11 Q12 SUBTOTAL /8 /9 /9 /10 Statement Presentation Units Rounding Off Class/ Index Number Centre Number/ ‘O’ Level Index Number Name SOLUTIONS / / MARIS STELLA HIGH SCHOOL PRELIMINARY EXAMINATION SECONDARY FOUR ADDITIONAL MATHEMATICS 4049/2 Paper 2 25 August 2026 Candidates answer on the Question Paper. 2 hours 15 minutes READ THESE INSTRUCTIONS FIRST Write your class, index number and name 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 of the marks for this paper is 90. 90

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x = a acbb 2 42 −− Binomial expansion (a + b)n = an +       1 n an − 1b +       2 n an − 2b2 + ... +       r n an − r br + ... + bn, where n is a positive integer and       r n = ! !( )! n r n r− = ! )1)...(1( r rnnn +−− 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A sin(A  B) = sin A cos B  cos A sin B cos(A  B) = cos A cos B  sin A sin B tan(A  B) = BA BA tantan1 tantan   sin 2 2sin cosA A A= 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 ABC A a sin = B b sin = C c sin 2 2 2 2 cosa b c bc A= + − 1 sin2 bc A=

3 1 Do not use a calculator in this question. In triangle ABC, ( )65AB=+ cm and 30ABC =  . Given that the triangle has an area of ( ) 1 32 5 54 − cm2, find the exact length of BC. Give your answer in the form ( )5ab+ , where a and b are integers. [4] ( )( ) ( ) ( )( ) ( ) 116 5 sin 30 32 5 524 1 1 1 6 5 32 5 52 2 4 32 5 5 6 5 6 5 6 5 192 32 5 30 5 25 31 217 62 5 31 BC BC BC + = − + = −  −−= +− −−+= −= ( ) 7 2 5 cm=−

4 2 In a sports meet, the height, h metres, of a javelin thrown by an athlete is modelled by the formula, ( ) 23 36 76100h w w=− − − , where w is the horizontal distance travelled by the javelin in metres. (a) By expressing h in the form 2()A w B C−+ , where A, B and C are constants, find the maximum height reached by the javelin and state the corresponding horizontal distance when this occurs. [3] ( ) ( ) ( ) 2 2 2 2 3 36 76100 3 18 18 76100 3 18 12100 h w w w w =− − − =− − − − =− − + Maximum height of javelin = 12 m when the horizontal distance is at 18 m. (b) Find the horizontal distance covered by the javelin before it hits the ground. [2] When ℎ = 0, ( ) ( ) 2 2 3 18 12 0100 18 400 18 20 38 or 2 (rejected) w w w w − − + = −= − = =− OR ( )( ) 2 2 3 27 57 0100 25 25 36 76 0 38 2 0 38 or 2 (rejected) ww ww ww ww − + + = − − = − + = = =− The javelin covered 38 m before hitting the ground.

5 3 Solve the equation 2 cosec 5sin 3cot  −= for 0 360   . [6] ( ) ( ) ( ) ( )( ) ( ) 2 2 2 2 2cos ec 5sin 3cot 2 5sin 3cos 2 5 1 cos 3cos 5cos 3cos 3 0 3 3 4 5 3cos 25 3 69 10 1.13066 (rejected) or 0 .530662        −= −= − − = − − = − −  − − −= = = =− 57.9499 122.1 , 237.9   = =  

6 4 A curve passes through the point ( )0, 1A − and the gradient of the curve at this point is parallel to the line 21yx+ =− . Given 2 22 2 d 2e 6e 9cos3d xxy xx −= + + , find y in terms of x. [7] ( ) 22 22 1 1 1 22 22 22 2 2 d 2e 6e 9cos3 dd e 3e 3sin 3 dWhen 0, 2, d 2 1 3 0 0 d e 3e 3sin 3d e 3e 3sin 3 d 13 e e cos322 When 0, 1, 13 1 1 22 xx xx xx xx xx y xxx xc yx x c c y xx y x x xc xy c − − − − − = + + = − + + = =− − = − + + = = − + = − + = + − + = =− − = + − +   2 22 2 13e e cos3 222 xx c yx − =− = + − −

7 5 (a) Given that 10 3 f ( ) d 15xx = and f ( )x lies above the x-axis for 3x , find the exact value of 7 10 2 37 5 3f ( ) d 3f ( ) dx x x x x −− . [3] ( ) ( ) 7 10 2 37 7 7 102 3 3 7 7 7 102 3 3 7 7 10 2 33 7 3 3 33 5 3f ( ) d 3f ( ) d 5 d 3f ( ) d 3f ( ) d 5 d 3 f ( ) d f ( ) d 5 d 3 f ( ) d 5 3 153 5 7 3 453 2 1445481 or 33 x x x x x x x x x x x x x x x x x x x x x x  −− = − − = − +  =− =− = − − =         (b) The diagram shows part of the graph of g( )yx= . Given that the curve passes through the points ( )0,1 , ( )1,0− and ( )2,9 , find the value of 29 01 d dy x x y+ . [1] 29 01 d d 29 18 y x x y+ = =  (0, 1) (–1, 0) 0 • (2, 9) y x g( )yx=

8 6 (a) Find the value of a for which 23x− is a factor of ( ) 32P 2 5 6x x x ax= − + + , where a is an integer. Hence, determine whether 2x+ is a factor of ( )P x . [4] 32 3 P 02 3 3 32 5 6 02 2 2 27 45 3 6 04 4 2 33 22 1 a a a a  =      − + + =           − + + = =− =− ( ) 32P 2 5 6x x x x = − − + ( ) ( ) ( ) ( ) 32 Since P 2 2 2 5 2 2 6 = 28 0 − = − − − − − + − Hence, 2x+ is not a factor of ( )P x . (b) Factorise ( )P x completely. [2] ( ) 32 2 By observation, P 2 5 6 (2 3)( 2) Comparing terms, 3 4 1 1 x x x x x x bx xb b = − − + = − + − − − =− =− OR By Long division, ( ) ( ) ( ) 2 32 32 2 2 2 2 3 2 5 6 2 3 6 2 3 4 6 4 6 0 xx x x x x xx xx xx x x −− − − − + −− − − + − − + −+ − − + ( ) ( )( ) ( )( )( ) 22 3 2 2 3 2 1 P x x x x x x x = − − − = − − +

9 (c) Using your answers in (b), solve the equation ( ) 3 1 22 5 2 2 6y y y+ − = − . [3] ( ) ( ) ( ) ( )( )( ) 3 1 2 32 2 5 2 2 6 2 2 5 2 2 6 0 2 3 2 2 2 1 0 32 or 2 2 or 2 1 (rejected, 2 0)2 3ln 2 1ln 2 0.585 (to 3s.f.) y y y y y y y y y y y y y yy + − = − − − + = − − + =  = = =−  == =

10 7 In the diagram, A, B, C and D lie on a circle. XT is the tangent to the circle at A and the line DB extended meets the tangent at T. Points E and F lie on X