(Bedok South) AM4 P2 (2025) Students

1 BEDOK SOUTH SECONDARY SCHOOL PRELIMINARY EXAMINATION 2025 4E5N CANDIDATE NAME CLASS REGISTER NUMBER ADDITIONAL MATHEMATICS Paper 2 4049 / 02 2 hours 15 minutes 90 Answer all the questions. 1 (a) Solve the equation 232(4 ) 8 11(2 ) 2x x x +− = − , giving your answer correct to 3 significant figures. [5] (b) Show that the solution from part (a) may be written in the form lognmp− where m, n and p are integers to be determined. [2] 2 The mass, M grams, of a radioactive substance, present at time t years after first being observed, is given by the formula 200 ktMe −= , where k is a constant. The mass of the substance was 123.7 g after being observed for 2 years. (a) (i) State the initial mass of the substance. [1] (ii) Show that k is approximately 0.240, correct to 3 significant figures. [1] (iii) Find the mass of the substance when t = 5, [1] (iv) Find the value of t when the mass of the substance is 15% of its initial mass. Give your answers correct to three significant figures. [2] (b) Explain, with clear working, why the mass of the substance can never be more than 200 grams. [1] (c) Sketch the graph of M against t. [2] 3 (a) It is given that 2 25yx x= − , where 𝑥 ≥ ℎ. Show that d( 25 2) d yx x kx x − − = , where k is an integer and determine the value of h and of k. [4] (b) Hence evaluate   7 3 10 ( 2 2 ) 6 d . 5 x x x x − − + [4] 4 (a) (i) Factorise completely 3 64x + . [2] (ii) Hence, express 2 3 3 64 x x + in partial fractions. [5] (b) Using the results in part (a), or otherwise, find 32 3 2 3 128 d64 xx xx ++ + . [3]

2 5 A particle moves in a straight line so that, at time t seconds after passing a fixed point O, its velocity is v m/s, where 4 8cos 2vt=+ . Find (a) the velocity of the particle at the instant it passes O, [1] (b) the least value of the particle’s acceleration, [1] (c) the values of t, in terms of π, when the particle is at rest for 03 t , [4] (d) the distance travelled in the first 2 seconds. [4] 6 The diagram shows right angled trapezium OCDF inside a semicircle with centre O and radius 10 cm such that angle BOC is θ radians, and angle CDF and angle OFD are right angles. (a) Show that the perimeter, P cm, of trapezium OCDF is given by 10 30 cos 10 sinP = + + [2] (b) Find the value of R when 10 sin 30 cos+ is expressed as cos( )R − , where R and α are constants, and hence state the maximum perimeter of the trapezium. [3] (c) Show that the area, A cm2, of trapezium OCDF is given by 75 sin 2A = [2] (d) The area of the trapezium varies with the value of θ. Find the value of θ for which the area has a stationary value and determine whether this area is a maximum or a minimum. [4] 7 Solutions to this question by accurate drawing will not be accepted. In the diagram, P, Q and R are points on the circle. (a) Explain, with geometrical reason, why PR is the diameter of the circle. [3] (b) Find the equation of the circle in the form 22 0x y ax by c+ + + + = , where a, b and c are integers. [3] y x O P (−13, 11) Q (1, 9) R (−1, −5) B A C E O D 10 θ F

3 (c) Find the equation of the perpendicular bisector of PQ. [3] (d) The point S lies on the circle such that it is furthest from the point Q. Show that the coordinates of S are (−15, −3) and hence calculate the area of the quadrilateral PQRS. [3] 8 The diagram shows part of the curve 10 112y x=− − passing through the point P(2k, k −1), where k is a constant. The curve meets the x-axis at the point X. The tangent and normal at P meet the x-axis at the points T and N respectively. (a) Find the equation of the normal at P. [6] (b) Find the exact area of the shaded region. [6] 9 The table below shows experimental values of two variables x and y. x 1 2 3 4 5 6 y 2.20 1.74 1.71 1.83 1.87 1.96 The variables x and y are related by the equation 1y bxax=+ , where a and b are constants. One value of y has been recorded incorrectly. (a) Show how 1y bxax=+ is transformed to plot a graph of xy against xx . [1] (b) Draw a straight line graph of xy against xx for the given data. [2] (c) Using your graph, (i) find an approximate value of y to replace the incorrect value, [2] (ii) estimate the value of a and of b, [3] (iii) find the value of y when 2.52x= . [2] (d) A pair of values of x and y is considered acceptable only if the xy value is within 2% vertical difference from the straight line. A student recorded a pair of values such that 4.50x= and 1.86y= . Verify whether the recorded values by the student are acceptable. [2] x y O P T N X

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5 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax bx c2 0+ + = , x b b ac a= −  −( )2 4 2 Binomial Expansion ( ) ... ...a b a n a b n a b n r a b bn n n n n r r n+ = +     +     +     + +− − − 1 2 1 2 2 , where n is a positive integer and ( ) ( ) ! 1...1 ! ! )( ! r rnnn rrn n r n +−−=−=      2. TRIGONOMETRY Identities BABABA BABABA AAec AA AA sinsincoscos)(cos sincoscossin)(sin cot1cos tan1sec 1cossin 22 22 22 = = += += =+ tan (A  B) = tan tan tan tan A B A B  1  sin sin cos2 2A A A= cos cos sin cos sin2 2 1 1 22 2 2 2A A A A A= − = − = − tan tan tan2 2 1 2A A A= − . Formulae for  ABC a A b B c Csin sin sin= = a b c bc A2 2 2 2= + − cos Cabsin2 1=

6 ANSWERS (Bedok South) 4E5N AM P2 (Prelim 2025) (w Ans) 1. (a) 𝑥 = 0.678 (to 3 sig. fig.) (b) 𝑚 = 3, 𝑛 = 2, 𝑝 = 5 2. (a) (i) 𝑀0 = 200 g (ii) 𝑘 = 0.240 (to 3 sig. fig.) (iii) 𝑀 = 60.2 g (to 3 sig. fig.) (iv) 𝑡 = 7.90 years (to 3 sig. fig.) (c) 3. (a) ℎ = 5 2 , 𝑘 = 5 (b) 300 4. (a) (i) (𝑥 + 4)(𝑥2 − 4𝑥 + 16) (ii) 1 𝑥 + 4 + 2𝑥 − 4 𝑥2 − 4𝑥 + 16 (b) 2𝑥 + ln(𝑥3 + 64) + 𝑐 5. (a) 𝑣 = 12 m/s (b) Least 𝑎 = −16 m/s2 (c) 𝑡 = 1 3 𝜋, 2 3 𝜋 (d) 10.3 m (to 3 sig. fig.) 6. (a) Hint: Draw a line CG ⊥ BE (b) 𝑅 = 10√10 or 31.6 (to 3 sig. fig.) Max 𝑃 = 41.6 cm (to 3 sig. fig.) (c) Hint: Apply Double Angle formula (d) 𝜃 = 𝜋 4 7. (a) Hint: Right angle in semicircle (b) 𝑥2 + 𝑦2 + 14𝑥 − 6𝑦 − 42 = 0 (c) 𝑦 = 7𝑥 + 52 (d) Hint: The furthest two points on a circle is its diameter. 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑃𝑄𝑅𝑆 = 200 units2 8. (a) 𝑦 = − 2 5 𝑥 + 8 (b) 10 ln 5 + 12 9. (a) Hint: Multiply equation with 𝑎𝑥 to get 𝑥𝑦 = 𝑎𝑏(𝑥√𝑥) + 𝑎 (b) (c) (i) 𝑦 = 1.78 (to 3 sig. fig.) (ii) 𝑎 = 1.50 (to 3 sig. fig.) 𝑏 = 0.467 (to 3 sig. fig.) (iii) 𝑦 = 1.71 (to 3 sig. fig.) (d) Unacceptable, Vertical difference = 2.073% 7.10 8.20 9.55

7 MARKING SCHEME (Bedok South) 4E5N AM P2 (Prelim 2023) (w Ans) Qn Solution Testing 1. Exponential Equations (a) 2(4𝑥) − 8 = 11(2𝑥) − 22𝑥+3 2(22𝑥) − 8 = 11(2𝑥) − (22𝑥)23 2(2𝑥)2 − 8 = 11(2𝑥) − 8(2𝑥)2 Let 𝑢 = 2𝑥 2𝑢2 − 8 = 11𝑢 − 8𝑢2 10𝑢2 − 11𝑢 − 8 = 0 (2𝑢 + 1)(5𝑢 − 8) = 0 𝑢 = − 1 2 or 𝑢 = 8 5 2𝑥 = − 1 2 (reject) 2𝑥 = 8 5 lg 2𝑥 = lg 8 5 𝑥 = lg 8 5 lg 2 = 0.678072 = 𝟎. 𝟔𝟕𝟖 (to 3 sig. fig.) Convert to base 2 Apply substitution correctly Solving for u Apply logarithm appropriatel y (b) From (a), 2𝑥 = 8 5 log2 2𝑥 = log2 8 5 𝑥 log2 2 = log2 8 − log2 5 𝑥 = log2 23 − log2 5 = 3 log2 2 − log2 5 = 3 − log2 5 = 𝑚 − log𝑛 𝑝 𝑚 = 𝟑, 𝑛 = 𝟐, 𝑝 = 𝟓 Apply log2 onto equation Power law Quotient law Identity property 2. Expone