BVSS 2024 Prelim EM Paper 2 (Marking Scheme)

Answer all the questions. 1. (a) Factorise completely 2 44x x xy y− − + . Answer ………………………… [2] (b) Given that 11 1244 8 2 x− += , find the exact value of x. Answer x = …………………… [3] (c) Given that 13 12 − += h hk , express h in terms of k. Answer h = …………………… [2] (d) Solve 4 73 2 12 3 2 ++ xxx . Answer ………………………… [3] S4E/5N EM Prelim 2024 Paper 2 Marking Scheme 2 44x x xy y− − + = ( 4) ( 4)x x y x− − − ------ M1 (1st level) = ( )( 4)x y x−− ------ A1 11 1244 8 2 x− += 112( ) 3( ) 1242 2 2 x− += ------ M1 (base 2 for at least one term) 3 ( 1)1 422 x−+− = 311 4 x− = − − ------ M1 (compare power, no ECF) 3 4x= ------ A1 13 12 − += h hk (3 1) 2 1k h h− = + 3 2 1kh k h− = + ------ M1 (linear and expand correctly) 3 2 1kh h k− = + (3 2) 1h k k− = + 1 32 kh k += − ------ A1 accept 1 32 kh k −−= −+ 2 2 1 32 xx + and 2 1 3 7 24 xx++ 4 6 3xx+ ------- M1 (linear) 8 4 6 14xx+  + ------- M1 (linear) 23x− 2 10x 3 2x− 5x 3 52 x −   ------- A1

2 2. The diagram shows three points, A, B and C on the ground. AB = 174 m and BC = 195 m. The bearing of A from B is 238°.The bearing of C from A is 108°. A point D lies on the path AC such that it is equidistant to A and to B (a) Show that angle BAC = 50. [3] (b) Find the angle BCA. Answer ………….………….. [2] North A B C 174 195 360 238x = − (angles at a point) ---------- M1 (with reason) = 122 y = 180 – 122 (int angles) ---------- M1 (with reason) = 58 108 58BAC = − ---------- A1 (shown) = 50 (MUST label angles clearly to be awarded full marks) 238 x sin sin 50 174 195 BCA= ---------- M1 1sin (0.683547)BCA −= = 43.12147 = 43.1 ---------- A1 D y

3 (c) Find the area of triangle ABC. Answer ………….………… m2 [3] (d) Find distance from C to D. Answer ………….………… m [4] ---------------------------------------------------------------------------------------------------------------- 180 43.12147 50ABC = − − ---------- M1 = 86.87853 Area of triangle = 1 174 195 sin86.878532   ---------- M1 (using ‘their’ ABC) = 16939.82 = 16900 m2 ---------- A1 (or any accurate higher approximation) Since AD = BD, ABD is isosceles triangle. BDC = 50 + 50 (ext  of ) ---------- M1 = 100 DBC = 180 – 100 – 43.12147 ---------- M1 = 36.87853 Using sine rule 195 sin 36.87853 sin100 CD =  ---------- M1 (using their DBC) 118.8287782CD= = 119 m (3 sf) ---------- A1 (accept more accurate approximations) North A B C 174 195 238 x y or 180 – 50 – 50 – 43.12147 = 36.87853 gets 2 marks Alternative Using cosine rule AC= -- M2 AD = M1 CD = AC – AD ---- A1

4 3. The variables x and y are connected by the equation 3 432 xyx= − + . Some corresponding values of x and y are given in the table below . x . . 3.5− . . 3− . . 2− . . 1− . . 0 . . 1 . . 2 . . 2 .5. . 3 . . y . 4.4− . . 1.5 . . p . . 6.5 . . 3 . . 0.5− . . 1− . 0.8 . 4.5 . (a) Find the value of p . Answer p = ………………… [1] (b) On the graph paper found in the next page, draw the graph of 3 432 xyx= − + for 33.5 x−  . [3] Points – 2 M smooth curve 1 M (c) By drawing a tangent, find the gradient of the curve at x = 2.5. Answer …………………… [2] (d) (i) On the same axes, draw the line 1yx=− − for 33.5 x−  . Answer [on graph] [1] (ii) Write down the x-coordinate of the point where this line intersects the curve. Answer x = ………………… [1] (e) State the minimum value of 3 432 xyx= − + for 03 x . Answer ………….………… [1] p = 7 ------- B1 Tangent line --- M1 Gradient = 5.4 ------- A1 (accept 4.9 to 5.8) –2.9 ------- B1 (accept –2.95 to –2.8) –1.35 ------- B1 (accept –1.5 to –1.2)

5 (f) The equation 3 4 5 02 x x− + = has only one solution. Explain how this can be seen from your graph. Answer : [2] –6 –5 –4 –3 –2 –1 0 1 2 8 6 –4 –3 –2 –1 0 1 2 3 1 2 3 4 5 6 x y –2 4 –4 2 3 4 5 02 x x− + = 3 245 2 02 x x −+= −− 3 2432 x x− + = − ------ M1 By drawing the line y = –2, there is one point of intersection. ------ A1 (no need to draw, but must mention y = -2)

6 4. The diagram shows a circle with centre O. ABCDE is a straight line. AC = CD and line OD meets the circle at F. It is given that OB = 7.5 cm, CD = 12 cm and OBD = 90. (a) Find the length of OD. Answer ………….………… cm [1] (b) Without the use of a calculator, find the value of cos ODE in its simplest form. Answer ………….…………… [1] (c) Show that angle COD is approximately 0.501 rad. Answer : [3] A B C D E O 7.5 cm 12 cm 2 2 2 7.5 18OD =+ OD = 19.5 cm ---------- B1 18 12cos 19.5 13ODE =− =− ---------- B1 18tan 7.5BOD= ---------- M1 = 1.176005 = 1.176 rad 6tan 7.5BOC= ---------- M1 = 0.67474 rad COD= 1.176005 – 0.67474 = 0.501264 = 0.501 rad ---------- A1 F

7 (d) Find the perimeter of shaded region. Answer ………………………. cm [3] ---------------------------------------------------------------------------------------------------------------- Radius OC = 227.5 6+ = 9.60468 cm ---------- M1 Length arc CE = 9.60468 × 0.501264 ---------- M1 (using ‘their’ radius) = 4.81448 cm Perimeter = 7.5 + 6 + 9.60458 + 4.81448 = 27.919 = 27.9 cm ---------- A1

8 5. A lead technician working with his trainee, can repair a machine together. If each of them worked alone, the lead technician would take x hours, while the trainee will need 3.5 hours more. (a) Find in terms of x, (i) the fraction of work done by the lead technician in one hour, Answer ………….………… [1] (ii) the fraction of work done by the trainee in one hour. Answer ………….………… [1] (b) In one hour, both the lead technician and his trainee will complete 5 21 of the repairs for the machine. Form an equation and show that it reduces to 210 49 147 0xx− − = . [3] 1 x ------- B1 1 3.5x+ ------- B1 1 1 5 3.5 21xx+= + ------- M1 accept 1 1 1 3.5 4.2xx+= + Or 42 42 13.5xx+= + 2 3.5 5 3.5 21 xx xx ++ =+ ------- M1 (single fraction from their first equ of same difficulty level) 242 73.5 5 17.5x x x+ = + 20 5 24.5 73.5xx= − − ------- A1 210 49 147 0xx− − =

9 (c) Solve 210 49 147 0xx− − = . Answer x = …………. or x = ….……… [3] (d) Hence, find the number of hours needed to repair the machine by two trainees. Answe