CEDAR 2025 AMATH PRELIM P1 MS

CEDAR GIRLS’ SECONDARY SCHOOL PRELIMINARY EXAMINATION 2025 SECONDARY FOUR CANDIDATE NAME Mark Scheme CLASS INDEX NUMBER ADDITIONAL MATHEMATICS 4049/01 Paper 1 25 August 2025 2 hours 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. 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, highlighters, 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 90 This document consists of 21 printed pages and 1 blank page. [Turn over]

Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 0 2 =++ cbxax , a acbbx 2 4 2 −−= Binomial expansion nrrnnnnn bbar nbanbanaba ++     ++     +     +=+ −−− ......21)( 221 , where n is a positive integer and ! )1(...)1( )!(! ! r rnnn rnr n r n +−−=−=      2. TRIGONOMETRY Identities 1cossin 22 =+ AA AA 22 tan1sec += 22cosec 1 cotAA=+ BABABA sincoscossin)sin( = BABABA sinsincoscos)cos( = BA BABA tantan1 tantan)tan(  = AAA cossin22sin = AAAAA 2222 sin211cos2sincos2cos −=−=−= A AA 2 tan1 tan22tan −= Formulae for ABC C c B b A a sinsinsin == Abccba cos2 222 −+= 1 sin2 bc A=

3 Cedar Girls’ Secondary School 4049/01/S4/Prelim/2025 [Turn over Answer all the questions. 1 (a) Express 6 2 9 21 − − in the form of 2ab+ , where a and b are integers. [2] 6 2 9 2 1 2 1 2 1 12 6 2 9 2 9 21 3 3 2 −+ −+ + − −= − =− (b) 6 2 9 21 − − is a root of the equation 2 0x px q+ + = where p and q are integers. Using the result in part (a), find the value of p and of q. [4] 2(3 3 2) (3 3 2) 0 9 18 2 18 3 3 2 0 (18 3 ) 2 27 3 0 18 3 0 (1) 27 3 0 (2) 6 9 pq p p q p p q p pq p q − + − + = − + + − + = − + + + + = + = −−− + + = −−− =− =− ( ) 2 2 2 2 2 2 4 2 43 3 2 2 6 6 2 4 Comparing, 6 4 6 2 4 36 2 6 4 72 9 p p qx p p q p p q p pq pq q q −  −= −  −−= − =−  − =− − − =− − =  − − = =−

4 Cedar Girls’ Secondary School 4049/01/S4 Prelim/2025 2 Find the range of values of the constant m for which the curve 2( 3)y m x mx m= − + + lies completely above the x-axis. [5] 𝑦 = (𝑚 − 3)𝑥2 + 𝑚𝑥 + 𝑚 𝑏2 − 4𝑎𝑐 < 0 𝑚2 − 4(𝑚 − 3)(𝑚) < 0 𝑚2 − 4𝑚2 + 12𝑚 < 0 −3𝑚2 + 12𝑚 < 0 −3𝑚(𝑚 − 4) < 0 𝑚 < 0 or 𝑚 > 4 𝑚 − 3 > 0 𝑚 > 3 Therefore 𝑚 > 4

5 Cedar Girls’ Secondary School 4049/01/S4/Prelim/2025 [Turn over 3 Express 3 2 25 ( 1) x xx − − in partial fractions. [6] 2𝑥3 − 5 𝑥2(𝑥 − 1) = 2 + 2𝑥2 − 5 𝑥2(𝑥 − 1) 2𝑥2 − 5 𝑥2(𝑥 − 1) = 𝐴 𝑥 + 𝐵 𝑥2 + 𝐶 𝑥 − 1 2𝑥2 − 5 = 𝐴𝑥(𝑥 − 1) + 𝐵(𝑥 − 1) + 𝐶𝑥2 Let 𝑥 = 1, −3 = 𝐶 Let 𝑥 = 0, −5 = 𝐵(−1) 𝐵 = 5 Let 𝑥 = −1, −3 = −𝐴(−2) + 5(−2) − 3(1) 𝐴 = 5 2𝑥3 − 5 𝑥2(𝑥 − 1) = 2 + 5 𝑥 + 5 𝑥2 − 3 𝑥 − 1 2𝑥3 − 5 𝑥2(𝑥 − 1) = 2 + 2𝑥2 − 5 𝑥2(𝑥 − 1) 2𝑥2 − 5 𝑥2(𝑥 − 1) = 𝐴𝑥 + 𝐵 𝑥2 + 𝐶 𝑥 − 1 2𝑥2 − 5 = (𝐴𝑥 + 𝐵)(𝑥 − 1) + 𝐶𝑥2 Let 𝑥 = 1, −3 = 𝐶 Let 𝑥 = 0, −5 = 𝐵(−1) 𝐵 = 5 Let 𝑥 = −1, −3 = −𝐴(−2) + 5(−2) − 3(1) 𝐴 = 5 2𝑥3 − 5 𝑥2(𝑥 − 1) = 2 + 5𝑥 + 5 𝑥2 − 3 𝑥 − 1 = 2 + 5 𝑥 + 5 𝑥2 − 3 𝑥 − 1

6 Cedar Girls’ Secondary School 4049/01/S4 Prelim/2025 4 (a) State the range of principal values of 1tan x− . [1] (a) 1tan22 x −−   (b) The curve cosy a b cx=+ is shown below for 3π0 2x radians. (i) Find the values of a, b and c. [3] (b)(i) 2 2 7 1 4 3 4 3cos 2 period c c ab ab a b yx  == = −= += = =− =− y x

7 Cedar Girls’ Secondary School 4049/01/S4/Prelim/2025 [Turn over (ii) On the same axes, draw a straight line to determine the number of solutions for the equation cos 5 2a b cx x  + = − for 30 2x  . [3] (b)(ii) cos 5 2 2cos 5 a b cx x a b cx x     + = − + = − There are 3 solutions.

8 Cedar Girls’ Secondary School 4049/01/S4 Prelim/2025 5 When f(x) is divided by 1x− , the remainder is 61. The remainder is 123− when f(x) is divided by 3x+ . Find the remainder when f(x) is divided by 2 23xx+− . [5] 2 2 3 ( 1)( 3) ( ) ( 1)( 3) ( ) ( ) (1) 61 61 ( 3) 123 3 123 46 15 x x x x f x x x Q x ax b f ab f ab a b + − = − + = − + + + = += − =− − + =− = = The remainder is 46 15x+ .

9 Cedar Girls’ Secondary School 4049/01/S4/Prelim/2025 [Turn over 6 A curve is defined by the equation 2 eln 10 x y x =  + . Show that the curve has no turning points for all real values of x. [5] 2 2 2 2 2 2 2 2 2 2 ln 10 ln ln( 10) ln( 10) 21 10 10 2 10 ( 1) 10 1 10 ( 1) 9 10 x x ey x y e x y x x dy x dx x dy x x dx x dy x dx x dy x dx x =  + = − + = − + =− + +−= + − + −= + −+= + ( ) ( ) 22 2 Since 1 0 then 1 9 0 and 10 0 xx x −  − +  + therefore 2 2 ( 1) 9 010 x x −+ + and 0dy dx  The curve has no turning points. 2 2 2 2 2 2 2 2 2 ln 10 ln ln( 10) ln( 10) 21 10 10 2 010 10 2 0 2( 2) ( 2) 4(1)(10) no solutions or 4 02 x x ey x y e x y x x dy x dx x xx x xx x b ac =  + = − + = − + =− + +− =+ + − = − −  − −= −  0dy dx  . The curve has no turning points.

10 Cedar Girls’ Secondary School 4049/01/S4 Prelim/2025 7 Solutions to this question by accurate drawing will not be accepted. The diagram above shows a triangle ABC with vertices B ( )7, 2 and C ( )6, 2−− . AB is perpendicular to AC and is parallel to the line 2yx=− . M and N are the mid-points of AB and BC respectively. (a) Find the coordinates of A. [6] 1 1 2 2 gradient of 2 gradient of gradient of 1 1gradient of 2 Equation of line : 2 subt(8,0) 16 Equation of line : 2 16 1Equation of line : 2 subt( 6, 2) 1 Equation of line : AB AC AB AC AB y x c c AB y x AC y x c c AC y =−  =− = =− + = =− + =+ −− = 1 12 x=+ 12 16 1 2 6 2(6) 16 4 xx x y − + = + = =− + = A (6 , 4) M A B y x N O