CEDAR 2025 AMATH PRELIM P2 MS

CEDAR GIRLS’ SECONDARY SCHOOL PRELIMINARY EXAMINATION 2025 SECONDARY FOUR CANDIDATE NAME CLASS INDEX NUMBER ADDITIONAL MATHEMATICS 4049/02 Paper 2 28 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 18 printed pages. [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 4049/02/S4/Prelim/2025 [Turn over Answer all the questions. 1 An object was launched from a machine. Its height, y metres above ground, t seconds after launch, can be modelled by a quadratic function y = f( t). Four seconds after launch, the object reached its maximum height of 81 m and then dropped to a height of 54 m after a further three seconds. (a) Find the initial height of the object when it was launched. [4] ( ) 2 4 81y a t= − + Substitute (7, 54) ( ) 2 54 7 4 81a= − + a = –3 ( ) 2 3 4 81yt=− − + At t = 0 y = 33 Initial height the object was launched = 33 m (b) Without solving a quadratic equation, explain and determine how long did the object stay above 54 m? [2] Since the line of symmetry is t = 4, the object was at height 54 at t = 1. Length of time = 23 = 6 seconds

4 Cedar Girls’ Secondary School 4049/02/S4/Prelim/2025 2 Show that the solution of the equation 1212 36 4xx++= can be expressed as lg1 lg a b+ , where a and b are integers. [5] 1212 36 4xx++= 12 12 36 4 16xx =   12 4 x x = 48 3x = 48 lg3 x = lg 48 lg 3x = lg 48 lg 3x = lg(16 3) = lg16 + lg3 x = 1 + lg16 lg 3

5 Cedar Girls’ Secondary School 4049/02/S4/Prelim/2025 [Turn over 3 (a) Solve the equation 323 11 3 2 0x x x− + + = , expressing non-integer roots in surd form. [5] Let f(x) = 323 11 3 2x x x− + + 2 3f   = 32 2 2 23 11 3 23 3 3      − + +           = 0 Therefore 32x− is a factor of f(x) f(x) = 2(3 2)( 1)x x bx− + − Comparing coefficient of 2x , 32b− = –11 b = –3 f(x) = 2(3 2)( 3 1)x x x− − − When 323 11 3 2 0x x x− + + = 2(3 2)( 3 1)x x x− − − = 0 2 3x= or 23 3 4(1)( 1) 2(1)x  − −= 3 13 2x = (b) Hence, solve the equation 32 11 9 18 0x x x+ + − = , expressing non-integer roots in surd form. [2] 32 11 9 18 0x x x+ + − = 32 11 2099 xx x− − − + = 32 3 11 3 2 03 3 3 x x x     − − − + − + =           2 33 x−= or 3 13 32 x −= x = – 2 or x 9 3 13 2 −=

6 Cedar Girls’ Secondary School 4049/02/S4/Prelim/2025 4 A curve is such that 2 22 d 12 d (3 2 ) y xx = − . Given that the curve passes through the points 1 ,12   and ( ) 21,ln(8 )e , show that the y-intercept of the curve can be expressed as ln ba c   where a, b and c are constants. [8] 2 22 12 (3 2 ) dy dx x= − 2 2 2 12(3 2 )dy xdx −=− 112(3 2 ) 2( 1) dy x cdx −−=+ −− = 16(3 2 )xc−−+ 6ln(3 2 ) 2 xy cx d −= + +− = 3ln(3 2 )x cx d− − + + Since ( ) 21,ln(8 )e lies on the curve 2ln(8 )e = c + d d = 2ln(8 )e – c ----- (1) Since 1 ,12   lies on the curve 1 = –3ln2 + 2 c + d ----- (2) 1 = –3ln2 + 2 c + 2ln(8 )e – c c = 2 d = 3ln2 y = 3ln(3 2 ) 2 3ln 2xx− − + + When x = 0 y = 3ln 3 3ln 2−+ = 23ln 3  

7 Cedar Girls’ Secondary School 4049/02/S4/Prelim/2025 [Turn over 5 A tangent to a circle at the point (4, 3) intersects the x-axis at x = 7. The line with the equation 3 2 6yx=+ is normal to the circle at another point. (a) Find the equation of the circle. [8] Gradient of tangent = 30 47 − − = –1 Gradient of normal = 1 Sub (4, 3) into y x c=+ 34 c=+ c = –1 Equation of normal is 1yx=− 1yx=− -------- (1) 3 2 6yx=+ -------- (2) Sub (1) in (2) 3( 1) 2 6xx− = + 3 3 2 6xx− = + x = 9 y = 8 Centre of circle is (9, 8) Radius = 22(9 4) (8 3)− + − = 50 Equation of circle is ( ) 22( 9) 8xy− + − = 50 (b) Find, in exact form, the equations of the tangents to the circle that are parallel to the x-axis. [2] 8 50y=+ 8 50y=−

8 Cedar Girls’ Secondary School 4049/02/S4/Prelim/2025 6 A particle, P, starts moving in a straight line with a velocity of 8 m/s at a displacement of 3 m from a fixed point O. The velocity, v m/s of the particle t seconds after moving is given by 5cos 2sinv t t c= + + . (a) Find the value of c. [2] (a) 5cos 2sinv t t c= + + When t = 0 85 c=+ c = 3 (b) Find the initial acceleration of the particle. [2] (b) 5sin 2cosa t t=− + Initial acceleration = 2 m/s2

9 Cedar Girls’ Secondary School 4049/02/S4/Prelim/2025 [Turn over (c) By explaining with relevant working, show the particle changed its direction of motion during the 3rd second. [3] (c) 5cos 2sin 3v t t= + + At t = 2, v = 5cos 2 2sin 2 3++ = 2.7379 At t = 3, v = 5cos3 2sin 3 3++ = –1.6677 Since the velocity changed sign from positive to negative during t = 2 to t = 3, the particle changed its direction during the 3rd second. (d) Find displacement of P from O at t = 6. [3] Displacement 5sin 2cos 3t t t d= − + + When t = 0, s = 3 –2 + 0 + d = 3 d = 5 Therefore, 5sin 2cos 3 5s t t t= − + + When t = 6, s 5sin 6 2cos6 3(6) 5= − + + = 19.7 m Alternative When t = 0, s = 0 –2 + 0 + d = 0 d = 2 Therefore, 5sin 2cos 3 2s t t t= − + + When t = 6, s 5sin 6 2cos6 3(6) 2= − + + = 16.7 m

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