Cambridge IGCSE Mathematics - Additional 0606 — 2019 May/June Paper 2 · Variant 1

1 Find the values of x for which x(6x + 7) H 20. [3]

Mark scheme: Question Answer Marks Partial Marks 1 6 x 2 + 7 x − 20[*0] M1 where * may be any inequality sign or = 4 5 A1 Critical values , − 3 2 5 4 A1 FT their critical values using outside x ≤ − or x ≥ final answer 2 3 regions

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2 Two variables x and y are such that y = for x 2 0. x3 dy 1 - 3 ln x (i) Show that = 4 . [3] dx x (ii) Hence find the approximate change in y as x increases from e to e + h, where h is small. [2]

Mark scheme: 2(i) d 1 B1 (ln x ) = soi dx x 3  1 2 M1 x  − 3 x ln x d y  x  = d x 3 2 x ( ) −3  1 −4 or x + −3 x ln x   ( )  x  Completion to given answer: A1 dy 1 − 3ln x = dx x 4 2(ii)  1 − 3lne  M1  4  h  e  −h2 oe or −0.0366h awrt A1 e 4

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3 (i) Sketch the graph of y = 5x - 3 on the axes below, showing the coordinates of the points where the graph meets the coordinate axes. y O x [3] (ii) Solve the equation 5x - 3 = 2 - x . [3] 2

Mark scheme: 3(i) Correct shape 3 B1 correct shape must have cusp on x- 0.6 oe indicated on x-axis axis 3 indicated on y-axis B1 for each correct point There must be a sketch to award the marks for the intercepts and sketch should be continuous with one intersection only on each axis 3(ii) Solves 5 x − 3 = x − 2 oe M1 or (5 x − 3) 2 = (2 − x ) 2 1 A1 [ x = ] oe 4 5 B1 [ x = ] oe 6

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5 v ms–1 10 O 4 k k + 6 t s The velocity-time graph represents the motion of a particle travelling in a straight line. (i) Find the acceleration during the last 6 seconds of the motion. [1] (ii) The particle travels with constant velocity for 23 seconds. Find the value of k. [1] (iii) Using your answer to part (ii), find the total distance travelled by the particle. [3]

Mark scheme: 5(i) 10 B1 − oe 6 5(ii) 27 B1 5(iii) Attempts to find total area M1 1 M1 (23 + their k + 6) × 10 2 1 1 or × 4 × 10 + 23 × 10 + × 6 × 10 2 2 280 A1

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6 (a) A = 2 x x - 3 Given that A does not have an inverse, find the exact values of x. [3] 0 3 0 1 2 (b) B = - 4 1 and C = e3 - 4 5o f 5 2p (i) Write down the order of matrix B. [1] 9 - 12 15 (ii) The matrix BC = 3 - 8 - 3 Explain why CB ! BC . [2] f6 - 3 20 p.

Mark scheme: 6(a) ( x + 3)( x − 3) − 2 x ( − x ) B1 their det A = 0 M1 Can be implied by later work [ x = ] ± 3 isw A1 6(b)(i) 3 × 2 or 3 by 2 B1 6(b)(ii) BC is a 3 by 3 matrix and CB is a 2 by 2 B2 B1 for a partially correct statement e.g. matrix [so they cannot be the same] oe The orders are not the same or BC is a 3 by 3 matrix or CB is a 2 by 2 matrix  6 5  or [CB =]   [so not equal] or B1 for 3 correct elements  41 15  or finding one correct element of CB as or B1 for finding one correct element of being different from BC and CB as being different from BC, without commenting that the elements are further comment different, [the matrices cannot be the same] oe

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7 The variables x, y and u are such that y = tan u and x = u 3 + 1. (i) State the rate of change of y with respect to u. [1] (ii) Hence find the rate of change of y with respect to x, giving your answer in terms of x. [4]

Mark scheme: 7(i) sec 2 u B1 7(ii) dy dy du M1 Attempts = × dx du dx dy dy dx or = ÷ dx du du dy their sec 2 u A1 FT their (i) = dx 3u 2 u = 3 x − 1 soi B1 sec 2 ( 3 x − 1) A1 final answer cao 3( 3 x − 1) 2 If B1 only then SC1 for 2 1 − k ( x − 1) 3 sec 2 ( x − 1) 3

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8 B 8 cm E 2 r rad 9 A C D The diagram shows a right-angled triangle ABC with AB = 8 cm and angle ABC = r radians. The points D 2 and E lie on AC and BC respectively. BAD and ECD are sectors of the circles with centres A and C 2r respectively. Angle BAD = radians. 9 (i) Find the area of the shaded region. [6]

Mark scheme: 8(i) 5π B1 [angle ECD =] oe or 0.873 soi 18 Attempts to find AC and subtract 8 M1 8 e.g. AC = 2π cos 9 [ DC = ] 2.44 A1 1 2 π M2 1 2 2π × 8 × theirAC × sin M1 for × 8 × or for 2 9 2 9 1 2 5π × their 2.44 × their seen OR 2 18 1  2π  1 2 2π × 8 × 8tan   − × 8 × 2  9  2 9 1 2 5π − × their 2.44 × their 2 18 awrt 1.91 A1 8(ii) their(6.712 – 2.443) M2 M1 for either arc seen  5π   2π  + their 2.443   + 8    18   9  awrt 12.0 A1

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9 (a) Eleven different television sets are to be displayed in a line in a large shop. (i) Find the number of different ways the televisions can be arranged. [1] Of these television sets, 6 are made by company A and 5 are made by company B. (ii) Find the number of different ways the televisions can be arranged so that no two sets made by company A are next to each other. [2] (b) A group of people is to be selected from 5 women and 3 men. (i) Calculate the number of different groups of 4 people that have exactly 3 women. [2] (ii) Calculate the number of different groups of at most 4 people where the number of women is the same as the number of men. [2]

Mark scheme: 9(a)(i) 39 916 800 B1 9(a)(ii) 5! × 6! oe M1 86 400 A1 9(b)(i) 5C3 × 3C1 oe M1 30 A1 9(b)(ii) 5C 2 × 3C 2 + 5C1 × 3C1 oe M1 45 A1

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10 Solutions to this question by accurate drawing will not be accepted. The points A and B have coordinates ( p, 3) and (1, 4) respectively and the line L has equation 3x + y = 2 . 1 (i) Given that the gradient of AB is , find the value of p. [2] 3 (ii) Show that L is the perpendicular bisector of AB. [3] (iii) Given that C q, - 10 lies on L, find the value of q. [1] ` j (iv) Find the area of triangle ABC. [2]

Mark scheme: 10(i) 4 − 3 1 M1 ALT uses y = mx + c with A and B as = oe 1 −p 3 far as an equation in p only −2 A1 10(ii) Either: Finds midpoint AB B1 FT their p  their p + 1 3 + 4   ,   2 2  Verifies ( −0.5, 3.5 ) is on L B1 y = −3 x + 2 therefore m = −3 oe B1 1 and ×−=3 −1 oe 3 Or: finds midpoint AB B1 FT their p  their p + 1 3 + 4   ,   2 2  1 B1 ×−=3 −1 oe 3 y − 3.5 = − 3( x + 0.5) and completion to B1 y = −3 x + 2 10(iii) q = 4 B1 10(iv) 22.5 nfww B2 B1 for correct method to find area using correct values 1 e.g. × AB × MC where M is the 2 midpoint of AB

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11 (a) (i) Show that i i = . [4] sin i 1 + cos i cosec i - cot i 5 (ii) Hence solve = for 180° 1 i 1 360 ° . [2] sin i 2 1 r(b) Solve tan 3z- 4 =- for 0 G z G radians. [3] ` j 2 2

Mark scheme: 11(a)(i) 1  1 cosθ M2 M1 for either −   cosecθ− cotθ 1  cosθ sinθ  sinθ sinθ  = cosecθ−   sinθ sinθ sinθ  cosecθ− cotθ 1  1  or =  − cotθ sinθ sinθ sinθ  1 − cosθ M1 1 − cos 2 θ 1 − cosθ 1 A1 = (1 − cosθ)(1 + cosθ) 1 + cosθ 11(a)(ii) awrt 233.1 B2 with no extras in range 3 B1 for cosθ=− soi 5 11(b) −1 1  M1 3φ− 4 = tan  −  soi  2  awrt 0.132, 1.18 A2 with no extras in range A1 for one correct

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dy r 8 r(b) A curve is such that = 3 - 2 cos 5x . The curve passes through the point , dx b 5 5 l. (i) Find the equation of the curve. [4] ydx and hence evaluate (ii) Find r ydx . [5] r y 2y

Mark scheme: 12(a) 2e x B1 seen 2 2e a 1 M1 Uses limits correctly for their integral − = 50 and sets = 50 2 2 Rearranges and takes logs to base e: M1 Using their integral 2 a = ln101 oe 1 A1 Allow any exact equivalent a = ln101 or ln 101 final answer 2 12(b)(i) 2 B2 B1 for −k sin5 x where k > 0 [ y = ]3 x − sin5 x [ + c ] 5 8π 3π 2  π  M1 = − sin 5 × + c   5 5 5  5  2 A1 y = 3 x − sin5 x + π 5 12(b)(ii)   2   B3 2 y d x = 3 x − sin5 x + π d x B2 for cos5 x oe nfww     ∫ ∫  5    25 2 2 3 x 3 x 2 and B1FT for + … + πx [ + c ] = + cos5 x + πx [ + c ] 2 2 25  π  M1 their F(π) – their F    2  16[.0] or 15.95 to 15.96 or A1 13π 2 2 − 8 25

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