Cambridge IGCSE Mathematics - Additional 0606 — 2018 May/June Paper 2 · Variant 1

1 A, B and C are subsets of the same universal set. (i) Write each of the following statements in words. (a) A 1Y B [1] (b) A + C = Q [1] (ii) Write each of the following statements in set notation. (a) There are 3 elements in set A or B or both. [1] (b) x is an element of A but it is not an element of C. [1]

Mark scheme: Question Answer Marks Partial Marks 1(i)(a) A is not a [proper] subset of B oe B1 1(i)(b) A and C are mutually exclusive oe B1 or A intersection C is the empty set oe 1(ii)(a) n ( A ∪ B ) = 3 B1 1(ii)(b) x ∈ ( A ∩ C′ ) oe B1

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2 The variables x and y are such that y = ln 3x - 1 for x 2 . 3 dy (i) Find . [2] dx (ii) Hence find the approximate change in x when y increases from ln 1.2 to ln 1.2 + 0. 125 . [3] ^ h ^ h

Mark scheme: 2(i) 1 M1 k × 3 x − 1 1 A1 3 × 3 x − 1 2(ii) 11 B1 x = soi 15 d y M1 0.125 ≈ their × δx oe d x x = their 11 15 0.05 nfww A1

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3 A 7-character password is to be selected from the 12 characters shown in the table. Each character may be used only once. Characters Upper-case letters A B C D Lower-case letters e f g h Digits 1 2 3 4 Find the number of different passwords (i) if there are no restrictions, [1] (ii) that start with a digit, [1] (iii) that contain 4 upper-case letters and 3 lower-case letters such that all the upper-case letters are together and all the lower-case letters are together. [3]

Mark scheme: 3(i) ( 12 P7 = ) 3991680 B1 3(ii) 11 B1 (4 × 6P = ) 1 330 560 3(iii) 4! × 4! × 2 oe M2 M1 for 4! × 4! oe only or 4 P4 × 4 P3 oe only 1152 A1

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4 Do not use a calculator in this question. It is given that x + 4 is a factor of p x = 2x 3 + 3x 2 + ax - 12 . When x is divided by x - 1 the ^ h p^ h remainder is b. (i) Show that a =- 23 and find the value of the constant b. [2] (ii) Factorise x completely and hence state all the solutions of p x = 0 . [4] p^ h ^ h

Mark scheme: 4(i) 2( − 4) 3 + 3 ( − 4 )2 − 4 a − 12 = 0 with one B1 Note: = 0 must be seen or may be implied by e.g. −92 = 4a or 92 = −4a correct interim step leading to a = −23 or convincingly showing that 2( − 4) 3 + 3 ( − 4 )2 − 4( − 23) − 12 = 0 or correct synthetic division at least as far as −4 2 3 a −12 −8 20 −4 a − 80 2 −5 a + 20 0 then a = −23 or correct long division to, e.g. verify −23, at least as far as 2 x 2 − 5 x − 3 x + 4 2 x 3 + 3 x 2 − 23 x − 12 2 x 3 + 8 x 2 − 5 x 2 − 23 x − 5 x 2 − 20 x − 3 x − 12 −3 x − 12 0 p(1) = 2 + 3 − 23 − 12 B1 b = −30 4(ii) finds a correct quadratic factor B2 B1 for quadratic factor with 2 correct terms e.g. (2x2 − 5x – 3) OR B1for finding (x – 3) using factor theorem B1for convincingly finding (2x + 1) as third factor Product of three linear factors M1 (2x + 1)(x – 3)(x + 4) 1 A1 If M0 then SC1 if quadratic factorised correctly but x = − , x = 3, x = − 4 nfww does not show full factorisation but does give all 3 2 solutions correctly

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5 The function f is defined by f x = for x 2 2.5 . 2x - 5 (i) Find an expression for f -1 x [2] ^ h. (ii) State the domain of f -1 x [1] ^ h. ax + b (iii) Find an expression for f x giving your answer in the form , where a, b, c and d are 2^ h, cx + d integers to be found. [3]

Mark scheme: 5(i) Putting y = f(x), changing subject to x M1 and swopping x and y or vice versa − 1 1  1  5 x + 1 A1 f ( x ) =  + 5  or oe isw 2  x  2 x 5(ii) x > 0 oe B1 5(iii) 1 B1  1 2  − 5  2 x − 5  1 M1 FT if expression of equivalent difficulty oe 2 − 5(2 x − 5) 1 e.g. 2 x − 5  1  − 5  2 x − 5  2 x − 5 A1 Completes to oe −10 x + 27 final answer

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6 40 cm A D x rad O C B 16 cm In the diagram AOB and DOC are sectors of a circle centre O. The angle AOB is x radians. The length of the arc AB is 40 cm and the radius OB is 16 cm. (i) Find the value of x. [2] (ii) Find the area of sector AOB. [2] (iii) Given that the area of the shaded region ABCD is 140 cm2, find the length of OC. [3]

Mark scheme: 6(i) 16x = 40 oe M1 x = 2.5 oe (radians) A1 6(ii) 1 M1 (16 )2 (2.5) oe 2 320 A1 6(iii) 1 2 M1 FT provided their 320 > 140 r ( their 2.5 ) = (their 320) − 140 oe 2 correct simplification to r2 = … M1 dep on first M1 12 A1

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7 Differentiate with respect to x (i) 4x tan x , [2] e 3x + 1 (ii) 2 . [3] x - 1

Mark scheme: 7(i) 4tan x + 4 x sec 2 x isw B2 Fully correct B1 for one correct term as part of e.g. a sum of 2 terms 3 x +1 3 x +1 B17(ii) d ( e ) = 3e dx 3 x +1 3 x +1 M1 ( x 2 − 1)( their 3e ) − their (2 x )e 2 2 ( x − 1) 3 x +1 3 x +1 A1 ( x 2 − 1)( 3e ) − 2 xe oe isw 2 2 ( x − 1)

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8 An experiment was carried out recording values of y for certain values of x. The variables x and y are thought to be connected by the relationship y = axn , where a and n are constants. (i) Transform the relationship y = axn into straight line form. [2] The values of ln y and ln x were plotted and a line of best fit drawn. This is shown in the diagram below. ln y 5 4 3 2 1 0 0 2 4 6 8 ln x (ii) Use the graph to find the value of a and of n, stating the coordinates of the points that you use. [3] (iii) Find the value of x when y = 50. [2]

Mark scheme: 8(i) Takes logs of both sides M1 ln y = ln a + n ln x A1 or lg y = lg a + n lg x 8(ii) n = −0.2 to −0.3 nfww B1 attempts to equate y-intercept to ln a M1 or forms their ln equation with their gradient and a point on the line or uses two points on the line to form a pair of simultaneous equations a = e4.7 isw or 110 or 109.9[47…] A1 maximum of 2 marks if no coordinates stated 8(iii) use of ln(50) and lnx = 3 to 3.2 M1 50 their n or for = x or better theira or for ln50 = ln(theira ) + (their n )ln x oe awrt 22 or 23 to 2 significant figures A1 implies M1

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9 (i) Express 5x 2 - 14x - 3 in the form p x + q 2 + r , where p, q and r are constants. [3] (ii) Sketch the graph of y = 5x 2 - 14x - 3 on the axes below. Show clearly any points where your graph meets the coordinate axes. [4] y O x (iii) State the set of values of k for which 5 x 2 - 14 x - 3 = k has exactly four solutions. [2]

Mark scheme: 9(i) 2 B3 B1 for each of p, q, r correct in correct format;  7  64 5  x −  − allow correct equivalent values.  5  5 If B0, then  7  64 SC2 for 5  x −  −  5  5 or SC1 for correct values but incorrect format 9(ii) B4 B2 for fully correct shape in correct position or B1 for fully correct shape translated parallel to the x-axis B1 for y-intercept at (0, 3) marked on graph 3 B1 for roots marked on graph at −0.2 and 3 −0.2 O 3 9(iii)  64  B2 FT their (i) 0 < k < their −   64  5  B1 for any inequality using their or max y 5 value is their 12.8soi

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10 A particle moves in a straight line such that its displacement, s metres, from a fixed point O at time t seconds, is given by s = 4 + cos 3t , where t H 0 . The particle is initially at rest. (i) Find the exact value of t when the particle is next at rest. [2] r r (ii) Find the distance travelled by the particle between t = and t = seconds. [3] 4 2 (iii) Find the greatest acceleration of the particle. [2]

Mark scheme: 10(i) ds B1 v = = −3sin3t dt π B1 When v = 0, t = 3 10(ii) Finding s when M1 π π t = and t = 4 2 Finding s when M1 Using their (i) correctly π t = their and correct plan 3 1.29 nfww A1 10(iii) dv B1 a = = −9cos3t dt 9 B1 FT their k cos3t

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11 (a) Solve 10 cos 2 x + 3 sin x = 9 for 0° 1 x 1 360° . [5] (b) Solve 3 tan 2y = 4 sin 2 y for 0 1 y 1 r radians. [5]

Mark scheme: 11(a) 10 (1 − sin 2 x ) + 3sin x = 9 M1 Solves 10sin 2 x − 3sin x −=1 0 oe M1 dep on first M1 Solves their three term quadratic in sin ݔ 1 1 A1 sin x = , sin x = − 2 5 30º, 150º and 191.5º, 348.5º awrt A2 A1 for any two correct solutions 11(b) sin 2 y M1 3 = 4sin 2 y oe cos2 y Solves 3sin 2 y − 4sin 2 y cos2 y [ = 0] M1 dep on first M1 3 A1 sin2 y = 0 cos2 y = 4 Any two of A1 π, 0.72273…, 5.56045… nfww π A1 SC: cancels out sin2y after M1M0 , 0.361, 2.78 awrt nfww allow SC1 for 0.72273… and 5.56045... and 2 SC1for 0.361 and 2.78

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12 In this question all lengths are in metres. x A B 30° h 5 C A water container is in the shape of a triangular prism. The diagrams show the container and its cross-section. The cross-section of the water in the container is an isosceles triangle ABC, with angle ABC = angle BAC = 30°. The length of AB is x and the depth of water is h. The length of the container is 5. (i) Show that x = 2 3 h and hence find the volume of water in the container in terms of h. [3] (ii) The container is filled at a rate of 0.5 m3 per minute. At the instant when h is 0.25 m, find (a) the rate at which h is increasing, [4] (b) the rate at which x is increasing. [2]

Mark scheme: 12(i) h M1 tan30 = oe x 2 Correct completion to given answer A1 V = 5 3 h 2 isw B1 12(ii)(a) dV 5 3 B1 FT theirV = k h 2 = their10 3 h or dh 2 dh dh dV M1 = × soi dt dV dt dh 1 M1 = × 0.5 dt  d V  their    dh  0.115 or 0.11547 to 0.1155 oe A1 12(ii)(b)  dx dx dh  1 M1  = × =  2 3 × their  dt dh dt  5 3 2 A1 5

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