Cambridge IGCSE Mathematics - Additional 0606 — 2019 Oct/Nov Paper 2 · Variant 1

1 (i) On the axes below, draw the graph of y = 2x - 3 . y 8 6 4 2 -2 0 2 4 x -2 -4 [2] (ii) Solve the equation 7 - 2x - 3 = 0 . [3]

Mark scheme: Question Answer Marks Partial Marks 1(i) 6 y B2 B1 shape 4 B1 Correct intersection with axes. 2 x −2 2 4 −2 −4 1(ii) 7 = 2x – 3 → x = 5 B1 Uses 7 = 3 – 2x oe M1 x = –2 A1

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2 y 6 4 2 0 r 2r x -2 The figure shows part of the graph of y = p + q cos rx . Find the value of each of the integers p, q and r. p = q = r = [3]

Mark scheme: 2 p = 2 B3 B1 for each q = 4 r = 3

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3 (a) Solve e 2x + 1 = 3e 4 - 3x . [3] (b) Solve lg (y - 6) + lg (y + 15) = 2 . [5]

Mark scheme: 3(a) obtain e5x – 3 = 3 M1 OR Take logs → 2x + 1 = ln3 + 4 – 3x take logs correctly M1 OR Collect like terms → 5x = 3 + ln3 → 5x – 3 = ln3 3 + ln3 A1 x = or x = 0.820 5 3(b)` Use of laws of logs M1 → lg(y – 6)(y + 15) = 2 Uses 10ଶ= 100 B1 → [(y – 6)(y +15)] = 100 Obtain correct quadratic A1 → y2 + 9y – 190 = 0 Solve a three term quadratic M1 y = 10 only A1

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4 Do not use a calculator in this question. Solve the following simultaneous equations, giving your answers for both x and y in the form a + b 2 , where a and b are integers. 2x + y = 5 3x - 2 y = 7 [5]

Mark scheme: 4 Eliminate x or y M1 7 + 5 2 1 A1 x = or y = 3 + 2 2 3 + 2 2 Multiply numerator and denominator by M1 3 − 2 2 x = 1 + 2 A1 y = 3 − 2 2 A1

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5 A particle is moving in a straight line such that t seconds after passing a fixed point O its displacement, s m, is given by s = 3 sin 2 t + 4 cos 2 t - 4 . (i) Find expressions for the velocity and acceleration of the particle at time t. [3] (ii) Find the first time when the particle is instantaneously at rest. [3] (iii) Find the acceleration of the particle at the time found in part (ii). [2]

Mark scheme: 5(i) Differentiate M1 Obtain 2cos2t or − 2sin2t v = 6cos2t − 8sin2t A1 a = −12sin2t − 16cos2t A1 5(ii) Equate v to 0 and attempt to solve M1 tan2t = 0.75 A1 or sin2t = 0.6 or cos2t = 0.8 t = 0.32(2) A1 Must be in radians 5(iii) Insert value of t into expression for a M1 Radians or degrees a = −20 A1 Must have used radians

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6 Do not use a calculator in this question. The curve xy = 11x + 5 cuts the line y = x + 10 at the points A and B. The mid-point of AB is the point C. Show that the point C lies on the line x + y = 11. [7]

Mark scheme: 6 Eliminate y M1 x 2 − x − 5 = 0 A1 Use formula M1 1 ± 21 A1 x = 2 21 ± 21 A1 y = 2 Find mid-point M1 ( 0.5 ,1 0.5 ) Show that mid-point lies on x + y = 11 A1

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7 (a) (i) Use the factor theorem to show that 2x - 1 is a factor of p ()x , where p ()x = 4x 3 + 9x - 5 . [1] (ii) Write p ()x as a product of linear and quadratic factors. [2] (b) (i) Show that 13 tan x sec x - 4 sin x - 5 sec 2 x = 0 can be written as 4 sin 3 x + 9 sin x - 5 = 0 . [3] (ii) Using your answers to part (a)(ii) and part (b)(i) solve the equation 13 tan x sec x - 4 sin x - 5 sec 2 x = 0 for 0 1 x 1 2 r radians. [4]

Mark scheme: 7(a)(i) f ( 0.5 ) = 0.5 + 4.5 − 5 = 0 B1 7(a)(ii) Factorise to obtain 2 x 2 and 5 M1 2 x + x + 5 ( 2 x − 1)( 2 A1 ) 7(b)(i) sin x 1 M1 sin x 5 Replace tan x by and sec x by 13 − 4sin x − = 0 cos x cos x cos 2 x cos 2 x Uses cos 2 x = 1 − sin 2 x M1 13sin x − 4sin x 1 − sin 2 x − 5 = 0 ( ) 4sin 3 x + 9sinx − 5 = 0 A1 Completed correctly 7(b)(ii) 2sin 2 x + sinx + 5 = 0 no real roots B1 Suitable statement seen 2sinx −=1 0 M1 Attempt to solve π A1 x = 6 5 π A1 x = 6

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8 The equation of a curve is given by y = xe -2x . dy (i) Find . [3] dx (ii) Find the exact coordinates of the stationary point on the curve y = xe -2x . [2] -2x 1(iii) Find, in terms of e, the equation of the tangent to the curve y = xe at the point ,1 [2] e e2 o. (iv) Using your answer to part (i), find xe -2x d x . [3] y

Mark scheme: 8(i) –2e–2x seen B1 Product rule M1 Clear attempt e −2 x (1 − 2 x ) A1 8(ii) dy M1 Must have two terms Set = 0 and attempt to solve dx  1 1  A1  ,   2 2e  8(iii) d y M1 Attempt to find at x = 1 d x 1 −1 1 2 A1 y − = x + 2 2 ( x − 1) or y = − 2 e e e e 2 8(iv) Integrate part(i) M1 xe −2 x = ∫− 2 xe −2 x + e −2 x d x ( ) Integrate e −2 x and make ∫ xe −2 x dx the M1 subject − xe − 2 x e −2 x A1 − + c 2 4

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9 Given that A = and B = , find - 9 - 3 6 5 (i) A -1 , [2] (ii) B2, [2] (iii) the matrix C, where B -1 C + A = B , [3] (iv) the matrix D, where B -2 DA = I . [3]

Mark scheme: 9(i) 1 B1 3  − 3 − 2  B1 ×   9 5  9 (ii) 2  10 7  B2 Minus one each error B =    42 31  9(iii) C = B 2 − BA M1  1 1  A1 BA =    − 15 − 3   9 6  A1 C =    57 34  9(iv) D = B2A–1 M1 1  33 15  A2 Minus one each error D =   3  153 71 

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10 (i) Expand (3 + )x 4 evaluating each coefficient. [3] p 4 In the expansion of x - (3 + x) the coefficient of x is zero. b x l (ii) Find the value of the constant p. [2] (iii) Hence find the term independent of x. [1] (iv) Show that the coefficient of x2 is 90. [2]

Mark scheme: 10(i) 81 + 108 x + 54 x 2 + 12 x 3 + x 4 B3 B1 for coefficients B1 for powers B1 for all Correct 10(ii) Identify and select two terms in x and M1 81 – 54p = 0 equate to zero p = 1.5 A1 10(iii) Constant term = –108p = –162 A1 FT using their p 10(iv) Correctly identify two terms in x2 M1 x2 term = 108 – 12p 108 – 18 = 90 A1

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11 A plane, which can travel at a speed of 300 km h–1 in still air, heads due north. The plane is blown off course by a wind so that it travels on a bearing of 010° at a speed of 280 km h–1. (i) Find the speed of the wind. [3] (ii) Find the direction of the wind as a bearing correct to the nearest degree. [3]

Mark scheme: 11(i) Uses correct triangle with vw opposite 10° M1 Sides of 300 and 280 include 10° Use cosine rule M1 vw 2 = 300 2 + 280 2 − 2 × 300 × 280cos10 vw = 54.3 A1 11(ii) Use sine rule M1 280 54.3 = sinα sin10 ° α= 63° or 64° A1 Bearing 117° or 116° A1

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