Hua Yi Sec Prelim 2024 A Math Paper 1 MS

1 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 1 1 A cuboid has a base area of ( 7 + 4√5)cm2 and a volume of ( 16 + 18√5)cm3. Find, without using a calculator, the height of the cuboid, in cm, in the form (𝑎 + 𝑏√5), where 𝑎 and 𝑏 are integers. [3] ℎ = 16+18√5 7+4√5 × 7−4√5 7−4√5 = 112−64√5+126√5−360 −31 = 8 − 2√5

2 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 1 2 The curve 5𝑥 − 𝑥𝑦 = 20 and the line 𝑥 − 2𝑦 − 3 = 0 intersects at the points 𝐴 and 𝐵. Find the 𝑦 −coordinate of 𝐴 and of 𝐵. 𝑠𝑢𝑏 𝑥 = 2𝑦 + 3 𝑖𝑛𝑡𝑜 𝑓𝑖𝑟𝑠𝑡 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 5(2𝑦 + 3) − 𝑦(2𝑦 + 3) = 20 2𝑦2 − 7𝑦 + 5 = 0 (2𝑦 − 5)(𝑦 − 1) = 0 or by quadratic formula 𝑦 = 2.5 𝑜𝑟 1 [3]

3 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 1 3 (a) Express 12𝑥 − 13 − 3𝑥2 in the form 𝑎(𝑥 + 𝑏)2 + 𝑐 and hence state the coordinates of the turning point of the curve 𝑦 = 12𝑥 − 13 − 3𝑥2. [4] −3(𝑥2 − 4𝑥) − 13 or −3(𝑥2 − 4𝑥 + 13 3 ) ------ =−3[(𝑥 − 2)2 − 4] − 13 or =−3[(𝑥 − 2)2 − 4 + 13 3 ] = −3(𝑥 − 2)2 − 1 Turning point (2, −1) (b) State the range of 𝑘 such that 𝑦 = 𝑘 will intersect the curve 𝑦 = 12𝑥 − 13 − 3𝑥2. 𝑘 ≤ −1 [1]

4 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 1 4 Integrate 4 5𝑥+1 − 6 𝑥3 with respect to 𝑥. ∫ 4 5𝑥+1 − 6𝑥−3𝑑𝑥 = 4 ln(5𝑥+1) 5 + 3 𝑥2 + 𝑐 [2] 5 Express 7𝑥2−17𝑥+1 (𝑥2+1)(2−3𝑥) in partial fractions. 𝐴𝑥+𝐵 𝑥2+1 + 𝐶 2−3𝑥 M1 7𝑥2 − 17𝑥 + 1 = (𝐴𝑥 + 𝐵)(2 − 3𝑥) + 𝐶(𝑥2 + 1) A = -4, B = 3 C = -5 3−4𝑥 𝑥2+1 − 5 2−3𝑥----------------- [5]

5 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 1 6 Given that 𝑥5 + 𝑎𝑥3 + 𝑏𝑥2 − 3 ≡ (𝑥2 − 1)𝑄(𝑥) − 𝑥 − 2, where 𝑄(𝑥) is a polynomial. (a) State the degree of 𝑄(𝑥). Degree of 𝑄(𝑥) = 3 [1] (b) Show that 𝑎 = −2 and 𝑏 = 1. Sub 𝑥 = 1, 𝑎 + 𝑏 = −1 ------equation Sub 𝑥 = −1, − 𝑎 + 𝑏 = 3 -------equation Solve simultaneous equations, 𝑎 = −2 and 𝑏 = 1 --------------- [3] (c) Find the polynomial 𝑄(𝑥). 𝑄(𝑥)(𝑥2 − 1) = 𝑥5 − 2𝑥3 + 𝑥2 − 3 + 𝑥 + 2 𝑄(𝑥) = (𝑥5 − 2𝑥3 + 𝑥2 − 3 + 𝑥 + 2) ÷ (𝑥2 − 1)--- By long division or comparing terms, 𝑄(𝑥) = 𝑥3 − 𝑥 + 1 [3]

6 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 1 7 The sketch above shows part of the graph of 𝑦 = 𝑎 sin ( 𝑥 𝑏) + 𝑐, where 𝑥 is in degrees. (a) Explain why 𝑐 = 1. amplitude is 3. So the maximum value of y is 4, 3+c =4, means that c = 1. for showing how to get c [2] (b) State the value of 𝑏. b = 2. [1] (c) Find the value of 𝑚 and explain how you get it. The period is 720°. ---------------------- Hence m = 180 + 720 = 900 [2] ( 𝑚, 4)

7 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 1 8 The figure shows the curve 𝑦 = 𝑥2 and the point 𝑅(1,0). The variable point 𝑃(𝑝, 0) moves along the 𝑥 axis and 𝑃𝑄 is vertical. p is decreasing at the rate of 1.2 units per second. (a) Show that he area of the triangle 𝑃𝑄𝑅, 𝐴 units2, is 𝐴 = 1 2 𝑝2 − 1 2 𝑝3. getting y coordinate of Q = 𝑝2 ---- A = 1 2 (1 − 𝑝)(𝑝2) = 1 2 𝑝2 − 1 2 𝑝3( Shown ) [2] (b) Find the rate at which 𝐴 is increasing at the instant when 𝑝 = −7 units. 𝑑𝐴 𝑑𝑡 = 𝑑𝐴 𝑑𝑝 × 𝑑𝑝 𝑑𝑡 𝑑𝐴 𝑑𝑡 = 𝑝 − 3 2 𝑝2------ 𝑑𝐴 𝑑𝑡 = 𝑑𝐴 𝑑𝑝 × 𝑑𝑝 𝑑𝑡----- When p = -7, 𝑑𝐴 𝑑𝑡 = −80.5 × −1.2 = 96.6 unit2 per second [4]

8 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 1 9 From a rectangular piece of metal of width 2m and length 6m, two squares of side x m and two rectangles of sides x m and ( x + y) m are removed as shown. The metal is then folded about the dotted lines. To give a closed box with height x m. (a) Show that the volume of the box, V m3, is given by 𝑉 = 2𝑥3 − 8𝑥2 + 6𝑥. Length = y , breadth = 2-2x , height = x 𝑥 + 𝑦 + 𝑥 + 𝑦 = 6---→ 𝑦 = 3 − 𝑥 V = 𝑥𝑦(2 − 2𝑥) = 𝑥(3 − 𝑥)(2 − 2𝑥) = 2𝑥3 − 8𝑥2 + 6𝑥 ( 𝑠ℎ𝑜𝑤𝑛 ) − − − − [3]

9 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 1 (b) Given that x can vary, find the stationary value of V. 𝑑𝑣 𝑑𝑥 = 6𝑥2 − 16𝑥 + 6 -- 𝑑𝑣 𝑑𝑥 = 0 → 6𝑥2 − 16𝑥 + 6 = 0 Solve using quadratic formula ( you need to show working) 𝑥 = 2.215 ( 𝑟𝑒𝑗𝑒𝑐𝑡) 𝑜𝑟 0.4514 Stationary value of V = 1.26 m3 [4] (c) Shows that this value of V is the maximum. Show either by first or second derivative test [2]

10 2024_4-G3 / 5-G2_PRELIMS_ ADDITIONAL MATHEMATICS _PAPER 1 10 The diagram shows a triangle 𝐴𝐵𝐶 with vertices at A ( 0, 6 ) , B ( 8, 15 ) and C ( 20, k ). (a) Given that AB = BC, find the value of 𝑘. Since AB = BC, it means 82 + 92 = 122 + (15 − 𝑘)2---- (15 − 𝑘 )2= 1 ( 15 – k ) = 1 or ( k – 15 ) = -1 𝑘 = 14 ----- [4] A ( 0, 6 ) B ( 8, 15) C ( 20, k) x y