MSHS 2025 AMATH PRELIM P1 MS

This document consists of 17 printed pages and 1 blank page. [Turn over For Examiners’ Use Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 / 5 / 6 / 5 / 6 / 5 / 6 / 7 / 7 / 7 Q10 Q11 Q12 Q13 SUBTOTAL / 8 / 8 / 10 / 10 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 4049/1 Paper 1 21 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 The curve 22( 1) ( 3) 5xy−+− = intersects the line 35yx−= at two points. Find the coordinates of these two points. [5] 22( 1) ( 3) 5xy−+− = ------(1) 35yx−= 35yx= + -----------(2) Subst (2) to (1), 22( 1) (3 5 3) 5xx− + +− = M1 22 2 1 9 12 4 5xx x x− ++ + += 210 10 0xx+= M1 10 ( 1) 0xx += M1 x = 0 or x = –1 y = 5 or y = 2 the coordinates are (0, 5) and (–1, 2) A2

4 2 A stone is thrown vertically upwards such that its height, h metres from the ground at time t seconds after being thrown is given by the formula 29 12 1ht t= −++ . (a) Explain the meaning of the constant term in the formula. [1] The initial height of the stone is 1m above the ground. B1 (b) Express h in the form 2()at b c++ , where a, b and c are constants to be determined. [3] 2 491 3ht = − −+ 22 2 42 291 33 3tt   = − −+ − +     2 295 3t= −− + . a = – 9, b = 2 3− , c = 5 B3 for each of the value (c) Hence state the maximum height attained by the stone and the time at which this occurs. [2] Maximum height = 5m when time = 2 3 s √B2 FT from (b)

5 3 Find the range of values of the constant p such that 2 43y px x p= − +− is always positive for all real values of x. [5] for 2 43y px x p= − +− to be always positive, p>0 and discriminant < 0, (– 4)2 – 4 (p)(p – 3) < 0 M1 for using discriminant < 0 16 – 4p 2 +12p < 0 p2 – 3p – 4 > 0 M1 for quadratic eqn ( p + 1) (p – 4) >0 M1 for factorising p < – 1 or p > 4 M1 since p >0, p > 4 A1

6 4 Express 2 2 15 13 18 ( 2)(3 1) xx xx −+ −+ in partial fractions. [6] Let 2 22 15 13 18 ( 2)(3 1) 2 3 1 x x A Bx C xx x x −+ + = +−+− + M1 2215 13 18 (3 1) ( )( 2)x x A x Bx C x− + = ++ + − M1 when x = 2, 13A = 52 A = 4 M1 when x = 0, –2C + 4 = 18 C = – 7 M1 when x = 1, 16 – B + 7 = 20 B = 3 M1 2 22 15 13 18 4 3 7 ( 2)(3 1) 2 3 1 xx x xx x x −+ − = +−+− + A1

7 5 (a) Find the first 4 terms in the expansion of 6 23a xx  + in ascending power of x, simplifying each term. [3] ( ) ( ) ( ) 66 5 4 3 232 22 266 63 3 3 3 ...12 3 a a a a ax xx xx xx x x        +=+ + + +                 B1 if at least 2 terms correct and B2 if all 4 terms correct = 56 4 33 63 18 135 540 ...aa a axxx++ + + B1 (b) Given that there is no term in 3x in the expansion of 6 32(2 1) 3 axx x −+  , find the value of the positive constant a. [2] ( ) 56 3 4 33 63 182 1 135 540 ...aax a axxx − ++ + +  since there is no x3 term, ( ) 432 135 540 0aa −= M1 43270 540 0aa−= ( ) 3270 2 0aa −= a = 0 (rej), a = 2 A1

8 6 A particle moves along the curve 2 5 (3 1)y x= − , where 1 3x≠ , in such a way that the x-coordinate is decreasing at a rate of 0.1 units per second. (a) Find the rate of change of the y -coordinate when x = 1. [4] 25(3 1)yx −= − 35( 2)(3 1) (3)dy xdx −= −− M1 3 30 (3 1)x −= − M1 when x = 1, 30 8 dy dx −= 30( 0.1)8 dy dt −= − M1 = 3 8 units/s A1 (b) Find the value of x when y increases at the rate of 3 125 units per second. [2] 3 33 125 (3 1)x= − (3x – 1)3 = 125 M1 3x – 1 = 5 x = 2 A1

9 7 A calculator must not be used in this question. (a) Show that tan15 2 3°= − . [4] tan 60 tan 45tan(60 45 ) 1 tan 60 tan 45 °− °°− ° = +°° M1 31 31 1 3 31 −−= × +− M 1 for ratios, M1 for rationalization 3 23 1 31 −+= − 4 23 2 −= = 23− (shown) A1 (b) Use the result from part (a) to find an expression for 2cosec 15°, in the form 3ab+ where a and b are integers. [3] 22cosec 15 1 cot 15°= + ° ( ) 2 11 23 = + − M1 11 4433 = + −+ 1 7431 743 743 += +× −+ M1 7431 49 48 += + − 843= + A1

10 8 (a) Prove the identity sin 1 cos 2cosec 1 cos sin xx xxx −+=− . [4] LHS= sin 1 cos 1 cos sin xx xx −+− ( ) ( ) 22sin 1 cos 1 cos sin xx xx +−= − M1 ( ) 22sin 1 2cos cos 1 cos sin x xx xx +− += − ( ) 1 1 2cos 1 cos sin x xx +−= − M1 ( ) 2(1 cos ) 1 cos sin x xx −= − M1 = 2cosec x A1 = RHS (b) Hence solve the equation sin 1 cos 51 cos sin xx xx −+=− for 0 ≤ x ≤ 2π. [3] 2cosec x = 5 5cosec 2x= 2sin 5x= M1 Basic angle = 0.41151 M1 x = 0.412, 2.73 (3s.f) A1