MSHS 2025 AMATH PRELIM P2 MS

[Turn over For Examiners’ Use Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 / 7 / 7 / 9 / 7 / 12 / 9 / 8 / 11 Q9 Q10 SUBTOTAL / 9 / 11 Statement Presentation Units Rounding Off Class/ Index Number Centre Number/ ‘O’ Level Index Number Name / / MARIS STELLA HIGH SCHOOL PRELIMINARY EXAMINATION SECONDARY FOUR ADDITIONAL MATHEMATICS 4049/2 Paper 2 Solution 26 August 2025 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 ax 2 + bx + c = 0, x = a ac b b 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 rn r − = ! ) 1 )...( 1 ( r r n n n+ − − 2. TRIGONOMETRY Identities sin 2 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) = B A B A tan tan1 tan tan  ± sin2 2sin cosA AA= 22 2 2cos 2 cos sin 2cos 1 1 2sinA AA A A= − = −=− 2 2 tantan 2 1 tan AA A= − Formulae for ∆ABC A a sin = B b sin = C c sin 2 22 2 cosa b c bc A=+− 1 sin2 bc A∆=

3 1 (a) A polynomial f (x) has a remainder of −1 when divided by (2x + 3). (i) F ind the remainder when f(x) – 1 is divided by (2x+3) [1] ( ) ( ) ( ) () 2 3 () 1 () 1 2 3 () 1 1 2 3 () 2 Remainder 2 f x x hx f x x hx x hx = +− −= + −− = +− =− (ii) Find in terms of f (x), a polynomial which is divisible by (2x + 3). [1] () 1Polynomial f x= + (b) The cubic polynomial g( x) is such that the coefficient of x3 is 2 and the roots of g(x) = 0 are −1, m and 2m, where m is an integer. It is given that g(x) has a remainder of 12 when divided by x – 1. Find an expression of g( x) in descending powers of x. [5] ( )( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 3 22 32 32 () 2 1 ( 2 ) (1) 2(2)(1 )(1 2 ) 12 4(1 3 2 ) 12 2 3 13 2 3 20 2 1 20 12 () 2 ( ) 2 1 2 ( 4) 2 24 2 4 428 2 5 28 2 10 4 16 gx x x m x m g mm mm mm mm mm m or reject gx x x x xx x x xx xx xxx x xx =+− − = −−= −+ = − += − −= + −= = − =+−− = −− − = − −+−+ = − ++ = − ++

4 2 The equation of a curve is ln xy x= . (i) Find the coordinates of the stationary point of this curve. Leave your answer in terms of e. [4] 2 2 ln 1 ln 1 ln 0 1 ln 0 1 ln 1 xy x xxdy x dx x x x dy dx x x xe y e =  −= −= = −= = = = Coordinates of stationary point are 1,e e   (ii) Determine the nature of the stationary point. [3] ( ) ( ) 2 2 2 24 4 2 23 1 ln 1 2 1 ln 2 1 ln 1 0 1Therefore , is a maximum point. dy x dx x x xxdy x dx x xx x x xe dy dx e e e −= −− −= −− −= = = −<   Alternative method : First derivative test. x 2.6 e 2.8 dy/dx >0 0 <0 sign +ve 0 -ve Therefore it is a maximum point.

5 3 The decay of a certain radioactive isotope can be modelled by the exponential equation 0 atN Ne −= after t weeks, where N represents the amount of radioactive isotope, N0 and a are constants. A sample of this radioactive isotope has a mass of 100.9 g initially. (i) After 2 weeks, it is found that the amount of this sample left is 84.6 g. Calculate the value of a. [3] 2 2 84.6 100.9 84.6 100.9 84.62 ln 100.9 1 84.6ln2 100.9 0.088098 0.0881 a a e e a a − − = = −=  =−   ≈ = (ii) What percentage of this sample has decayed after 5 weeks? [3] 5(0.088098) 5 100.9 64.951 100.9 64.951Percentage decayed 10 0%100.9 35.6% t Ne − = = = −= × = (iii) Find the number of weeks when the amount of radioactive isotope decayed first exceeds 60 g. Give your answer correct to the nearest week. [3] (0.088098) (0.088098) 100.9 60 100.9 40.9 40.9 100.9 40.9(0.088098) ln 100.9 1 40.9ln0.088098 100.9 10.24 t t N e e t t − − <− < < −<  >−   > The nearest week that the amount of radioactive isotope decayed exceeds 60 g = 11

6 4 In the diagram, AB is the diameter of the circle with centre O. DE and BF are tangents to the circle at C and B respectively. DCE and BEF are straight lines. Prove that (i) Triangle ABC and triangle AFB are similar. [3] 90 (Angles in semicircle) 90 (Radius perpendicular to tangent) (common angle) Triangle ABC and triangle AFB are similar (AA similarity) ACB ABF ACB ABF CAB BAF ∠= ° ∠= ° ∠= ∠ ∠= ∠ (ii) Show that EC = EF. [4] Triangle ABC and triangle AFB are similar. . (same angle) (tangent chord theorem) (vertically opposite angle) FCE, triangle CFE is i ABC AFB CFE AFB ABC DCA DCA FCE FCE ABC AFB Therefore CFE ∠= ∠ ∠= ∠ ∠= ∠ ∠= ∠ ∠= ∠= ∠ ∠= ∠ sosceles, EC = EF Alternative solution: ( ) Triangle ECB is isosceles. . . 180 90 (Adjacent angles in a straight line ) 90 (tangent chord theorem) 180 90 (angle sum of triangle) 90 T EB = EC common tangent Let EBC x ECB x FCE x x CAB ECB x CFE x x FCE ∠= ∠= ∠ = −− = − ∠= ∠ = ∠ = −− = − =∠ herefore triangle CFE is isosceles, EC = EF

7 5 (a) Find the integer a and b which satisfy the equation ( ) 1 1 32 21 428 abyyy −  ÷= [3] ( ) ( ) ( ) ( ) ( ) 1 11 13 2 33 2222 33 24 1 42 28 22 2 2, 4 yy y y yy y ab − −− − −  ÷= ÷ = ÷ = = =− (b) Solve the equation ( )3 1log 8 2 log 3x x−= − . [5] ( ) ( ) ( ) ( ) ( ) ( )( ) 3 33 33 3 2 2 1log 8 2 log 3 log 8 2 log log 8 log 2 log 8 2 83 8 90 9 10 9, 1( , 8 0) x x xx xx xx xx xx xx x rej x −= − −= − −+ = −= −= − −= − += =− −> (c) Solve the equation ( ) 14 72 2xx+ += . [4] ( ) ( ) ( ) ( ) ( ) ( )( ) 1 2 2 2 4 72 2 4 47 22 0 42 72 2 0 2 4 7 20 41 20 1, 2( )4 122 4 2 xx xx xx x x Let y yy yy y rej x + − += + −= + −= = + −= − += = − = = =−

8 6 (a) Given that 2cos sin 2 sin cos 2y xx x x= − , show that cos cos2dy kx xdx = , stating the value of k. [4] ( )( ) ( )( ) 2cos sin 2 sin cos 2 2 cos 2cos 2 ( sin )(sin 2 ) sin 2sin 2 cos cos 2 4cos cos 2 2sin sin 2 2sin sin 2 cos cos 2 3cos cos 2 y xx x x dy x x x x x x xxdx x x xx xx x x xx = − = +− − − + = −+− = k = 3 (b) Hence, show that ( ) 24 0 727cos cos2 sec 13x x x dx π += +∫ [5] ( ) [ ] [ ] ( ) 224 44 0 00 4 4 00 4 0 7cos cos2 sec 7cos cos2 sec 7 3cos cos2 tan3 7 2cos sin2 sin cos2 tan 034 722 10132 72 13 x x x dx x x dx x dx x x dx x xx x x π ππ π π π π += + = + = − +−  = −+ = + ∫ ∫∫ ∫ 7 Two points A( −3, −6) and B (7, −6) lie on a circle centre G. The equation of the tangent to the circle at the point Q (1, −12) is x + 5y +59 = 0. (a) Show that the coordinates of G is (2, −7). [5] Mid-point of AB = 37 ,62 −+ − = (2, − 6) The x- coordinate of G = 2 Gradient of tangent at Q = 1 5− Gradient of normal at Q = 5 Equation of normal at Q: ( )( 12) 5 1 5 5 12 5 17 yx yx yx −− = − = −− = − The normal at Q passes through the center of the circle. When x= 2, y = 5(2) – 17 = −7 Therefore G = (2, −7)

9 Alternative solution: Mid-point of AB = 37 ,62 −+ − = (2, − 6)