Hua Yi Sec Prelim 2024 A Math Paper 2 MS

[Turn Over HUA YI SECONDARY SCHOOL PRELIMINARY EXAM 2024 4-G3 / 5-G2 NAME CLASS INDEX NUMBER ADDITIONAL MATHEMATICS 4049/02 PAPER 2 (XX) 26 August 2024 2 hour 15 minutes Candidates answer on the Question Paper No Additional Materials is required. ANSWER SCHEME Setter : Mdm Suzanne Lye

1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0 , x = -b ± b2 - 4ac 2a Binomial expansion a + b( ) n = an + n 1 æ èç ö ø÷ an-1b + n 2 æ èç ö ø÷ an-2b2 + ...+ n r æ èç ö ø÷ an-rbr + ...+ bn , where n is a positive integer and n r æ èç ö ø÷ = n! r!(n - r)! = n(n -1)...(n - r +1) r! 2. TRIGONOMETRY Identities sin2 A+ cos2 A = 1 sec2 A = 1+ tan2 A cosec2 A = 1+ cot2 A sin( A ± B) = sin Acos B ± cos Asin B cos( ) cos cos sin sinA B A B A B= tan tantan( ) 1 tan tan ABAB AB = sin2A = 2sin Acos A cos2A = cos2 A- sin2 A = 2cos2 A-1= 1- 2sin2 A tan 2 A = 2 tan A 1- tan2 A Formulae for DABC a sin A = b sin B = c sinC a2 = b2 + c2 - 2bccos A Cabsin2 1=

3 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 2 1 Show that 𝑥 = 1 2 is a solution of the equation 2𝑥3 + 𝑥2 − 3𝑥 + 1 = 0 and hence solve the equation completely. [5] 𝑓(𝑥) = 2𝑥3 + 𝑥2 − 3𝑥 + 1 𝑓 ( 1 2) = 2 ( 1 2) 3 + ( 1 2) 2 − 3 ( 1 2) + 1 ---------------------M1 = 0 𝑓(𝑥) = (2𝑥 − 1)(𝑥2 + 𝑏𝑥 − 1) ------------------------M1 Compare coeff. of 𝑥2, 1 = 2𝑏 − 1 b= 1 ---------------------------M1 𝑓(𝑥) = (2𝑥 − 1)(𝑥2 + 𝑥 − 1)-------------------------M1 𝑥 = 1 2, −1+√5 2 , −1−√5 2 --------------------------------------A1 Alternative Method : Using long division to find (𝑥2 + 𝑥 − 1).

4 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 2 2 (a) By considering the general term in the binomial expansion of (𝑝𝑥 + 1 𝑥3) 9 , where p is a constant, explain why there are no even powers of x in this expansion. [3] General Term =(9 𝑟) (𝑝𝑥)9−𝑟 ( 1 𝑥3) 𝑟 -----------------M1 = (9 𝑟) 𝑝9−𝑟𝑥9−4𝑟 -------------------M1 Since 9 − 4𝑟 = 1 + 4(2 − 𝑟), one added to any even number will give an odd value, hence there are no even powers of x in this expansion. --------A1

5 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 2 (b) Given that the coefficient of 𝑥8 is equal to the coefficient of 𝑥 in the expansion of (2𝑥3 + 1) (𝑝𝑥 + 1 𝑥3) 9 , find the value of p. [4] (2𝑥3 + 1) (… … (9 1) (𝑝𝑥)8 ( 1 𝑥3) 1 + (9 2) (𝑝𝑥)7 ( 1 𝑥3) 2 ) 9 ---------------------M1 = (2𝑥3 + 1) (… + 9𝑝8𝑥5 + 36𝑝7𝑥 + ⋯ ) -----------------------------------M1 2(9𝑝8) = 36𝑝7 -------------------------------------------------M1 𝑝 = 2 ------------------------------------------------------------A1 (c) Using the value of p in (b), find the term independent of x in the expansion of (2𝑥3 + 1) (𝑝𝑥 + 1 𝑥3) 9 . [2] Term independent of 𝑥 = (2) (9 3) (26) ----------------------M1 =10752 ---------------------------A1

6 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 2 3 (a) Given that 𝑦 = 2𝑥 (3𝑥+1) 1 2 , show that 𝑑𝑦 𝑑𝑥 = 3𝑥+2 (3𝑥+1) 3 2 . [4] 𝑑𝑦 𝑑𝑥 = 2(3𝑥+1) 1 2−2𝑥(1 2)(3)(3𝑥+1)−1 2 (3𝑥+1)1 -------------------M2 (numerator and denominator) = 2(3𝑥+1)−3𝑥 (3𝑥+1) 3 2 ----------------------------------M1 = 3𝑥+2 (3𝑥+1) 3 2 -----------------------------------------A1 (b) Hence find the value of ∫ 𝑥 (3𝑥+1) 3 2 2 0 𝑑𝑥. [5] 𝐹𝑟𝑜𝑚 𝑑𝑦 𝑑𝑥 = 3𝑥+2 (3𝑥+1) 3 2 , 𝑑𝑦 𝑑𝑥 = 3𝑥 (3𝑥 + 1) 3 2 + 2 (3𝑥 + 1) 3 2 3𝑥 (3𝑥+1) 3 2 = 𝑑𝑦 𝑑𝑥 − 2 (3𝑥+1) 3 2 ∫ 3𝑥 (3𝑥+1) 3 2 𝑑𝑥 2 0 = 2𝑥 (3𝑥+1) 1 2 − ∫ 2 (3𝑥+1) 3 2 2 0 𝑑𝑥 -----------------------------------M1 ∫ 𝑥 (3𝑥+1) 3 2 𝑑𝑥 2 0 = 1 3 [ 2𝑥 (3𝑥+1) 1 2 ] 0 2 − 1 3 ∫ 2 (3𝑥+1) 3 2 2 0 𝑑𝑥 = 1 3 [ 2𝑥 (3𝑥+1) 1 2 ] 0 2 − 1 3 [ 2(3𝑥+1)−1 2 (−1 2)(3) ] 0 2 -----------------------M1, M1 = 1 3 ( 4 √7 − 0) + ( 4 9) ( 1 √7 − 1) ------------------------------M1 = 0.227 -------------------------------------A1

7 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 2 4 Show that the equation 5𝑒𝑥 = 1 𝑒𝑥 − 4 has only one solution and find its value correct to 2 decimal places. [4] 5𝑒𝑥 = 1 𝑒𝑥 − 4 5𝑒2𝑥 = 1 − 4𝑒𝑥 5𝑒2𝑥 + 4𝑒𝑥 − 1 = 0 ------------------------M1 5(𝑒𝑥)2 + 4𝑒𝑥 − 1 = 0 (5𝑒𝑥 − 1)(𝑒𝑥 + 1) = 0 --------------------M1 𝑒𝑥 = 1 5 or 𝑒𝑥 = −1 𝑥 = 𝑙𝑛 ( 1 5) (no solution, reject)----------A1 = −1.61 (2 dp) ------A1

8 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 2 5 The equation of a curve is 𝑦 = −2𝑥2 + 3𝑥 + 5. (a) Find the set of values for x for which the curve lies below the line 𝑦 = 3 and represent this set of values on a number line. [4] −2𝑥2 + 3𝑥 + 5 < 3 -----------------------M1 2𝑥2 − 3𝑥 − 2 > 0 (2𝑥 + 1)(𝑥 − 2) > 0 ----------------------M1 𝑥 < − 1 2 or 𝑥 > 2 ------------------------A1 A1 -0.5 0 2

9 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 2 The line 𝑦 = 𝑥 + 𝑘 is a tangent to the curve at the point Z. (b) Find the value of the constant k. [3] −2𝑥2 + 3𝑥 + 5 = 𝑥 + 𝑘 2𝑥2 − 2𝑥 + 𝑘 − 5 = 0 -------------------------------M1 Line is tangent to curve => 𝑏2 − 4𝑎𝑐 = 0 (−2)2 − 4(2)(𝑘 − 5) = 0 ----------------------------M1 4 − 8𝑘 + 40 = 0 𝑘 = 11 2 or 5 1 2 -------------------------------------------A1 (c) Find the coordinates of Z. [2] 𝑦 = 𝑥 + 11 2 --------------eqn 1 Sub eqn 1 into eqn of curve, −2𝑥2 + 3𝑥 + 5 = 𝑥 + 11 2 ----------------M1 −4𝑥2 + 6𝑥 + 10 = 2𝑥 + 11 −4𝑥2 + 4𝑥 − 1 = 0 4𝑥2 − 4𝑥 + 1 = 0 (2𝑥 − 1)2 = 0 𝑥 = 1 2 , 𝑦 = 6 𝑍 = ( 1 2 , 6) -------------------------------A1

10 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 2 6 (a) The speed V m/s of a vehicle, t s after passing a fixed point O, is given for 𝑡 ≥ 0, 𝑉 = 1 + 𝑝𝑒𝑞𝑡, where p and q are constants. Explain how a straight line can be drawn to represent the formula, and state how the value of p and q can be obtained from the line. [4] 𝑉 − 1 = 𝑝𝑒𝑞𝑡 ln(𝑉 − 1) = ln(𝑝𝑒𝑞𝑡) ln(𝑉 − 1) = ln 𝑝 + 𝑞𝑡 --------------------M1 Draw ln(𝑉 − 1) against t ---------------M1 y- intercept = ln 𝑝 --------------------A1 gradient = q -------------------------- A1 (b)(i) Data of the speeds of the vehicle was collected. The table below shows the corresponding values of V and t. t 2 4 6 8 10 V 10.35 8.40 6.70 5.42 5.95 Using (a), draw the straight line graph on the next page. t 2 4 6 8 10 ln(V-1) 2.24 2.00 1.74 1.49 1.60 M1 - Calculate ln (𝑣 − 1), M1 – correct points plotted, M1 – Best fit line [3] (ii) Estimate the values of p and q. 𝑞 = 2.5−1.5 0−8 --------------------M1 = −0.125 ---------------------A1 𝑙𝑛𝑝 = 2.5 𝑝 = 12.18 ----------------A1 [3]