Cambridge IGCSE Mathematics - Additional 0606 — 2018 Oct/Nov Paper 2 · Variant 3

1 Solve the equation 5x - 3 =- 3x + 13 . [3]

Mark scheme: Question Answer Marks Partial Marks 1 x = 2 B1 3 − 5 x = −3 x + 13 oe M1 x = –5 A1

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2 On each of the Venn diagrams below, shade the region indicated. A B C (A B C)' A A B B C C A B C' (A B) C' [3]

Mark scheme: 2 3 B1 for each correct diagram

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3 (i) Write 8 + 7x - x 2 in the form a - (x - b) 2 , where a and b are constants. [3] (ii) Hence state the maximum value of 8 + 7 x - x 2 and the value of x at which it occurs. [2] (iii) Using your answer to part (i), or otherwise, solve the equation 8 + 7z 2 - z 4 = 0 . [3]

Mark scheme: 3(i) 2 3 7 81  7  B1 b = −  x −  2 4  2   7  2 M1 ± 8 ±   seen  2  or expand given form and equate for 8 or 7 A1 fully correct 3(ii) 81 2 maximum their B1 4 7 when x = their B1 2 from their correct form 3(iii)  2 7  2 81 M1 replace x by z 2 in their (i) and  z −  = oe  2  4 equate to zero. 7 9 M1 z2 = ± 2 2 z = ± 8 A1

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4 2 = 2x + 4 dx (x + 1) dy dy (i) Find , given that = 1 when x = 1. [3] dx dx (ii) Find y in terms of x, given that y = 3 when x = 1. [3]

Mark scheme: 4(i) integrate: increase in powers of at least one term M1 * dy 2 1 A1 = x − + ( C ) 3 dx ( x + 1) 1 A1 C = 8 4(ii) integrate their (i): increase in powers of at least one M1 Dep* term 1 3 1 1 A1 two correct terms in x y = x + + x + ( D ) 2 3 2 ( x + 1) 8 29 A1 D = 12

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5 Given that A = and B = , find 1 4 - 2 5 (i) A-1 , [2] (ii) the matrix C such that CA = B , [2] (iii) the matrix D such that A-1 D + B = I . [3]

Mark scheme: 5(i) 1  4 −3  2  4 − 3    B1   5  −1 2   − 1 2  1 B1 5 5(ii) post multiply by A–1 M1 C = BA–1 1  0 5  A1   5  −13 16  5(iii)  0 − 4   − 4 23  B1 I − B =   or AB =    2 − 4   − 7 24  D = A ( I − B ) or D = A – AB M1  6 − 20  A1 D =    8 − 20 

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6 Solve the simultaneous equations log 2 (x + 2y) = 3 , log 2 3x - log 2 y = 1. [5]

Mark scheme: 6 3 x B1 implied by one correct equation log28 = 3 or log3 x − log y = log (any base) y or log22 = 1 soi x + 2 y = 8 B1 3 x B1 = 2 y solve correct equations for x or y M1 x = 2 and y = 3 A1

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7 A squad of 20 boys, which includes 2 sets of twins, is available for selection for a cricket team of 11 players. Calculate the number of different teams that can be selected if (i) there are no restrictions, [1] (ii) both sets of twins are selected, [2] (iii) one set of twins is selected but neither twin from the other set is selected, [2] (iv) exactly one twin from each set of twins is selected. [2]

Mark scheme: 7(i) 167 960 1 7(ii) evidence of selecting from 16 M1 [16 C7 =] 11 440 A1 7(iii) 2 × n Cr with n = 16 or r = 9 M1   2 ×16 C 9 =  22880 A1 7(iv) 4 × n Cr with n = 16 or r = 9 M1   4 ×16 C 9 =  45760 A1

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8 Variables x and y are such that when y2 is plotted against e2x a straight line is obtained which passes through the points (1.5, 5.5) and (3.7, 12.1). Find (i) y in terms of e2x, [3] (ii) the value of y when x = 3 , [1] (iii) the value of x when y = 50 . [3]

Mark scheme: 8(i) 12.1 − 5.5 B1 correct expression for gradient [= 3] 3.7 − 1.5 y 2 − 5.5 M1 = their grad e 2 x − 1.5 or correctly use y2 = (their m) e2x + c with one point to find c 2 x A1 y = [ ± ] 3e + 1 8(ii) [±]34.8 1 8(iii) 2 x B1 * 50 = ( their 3 ) e + their1 or 2500 = ( their 3 ) e 2 x + their1  2499  M1 Dep* 2 x = ln   obtain 2x explicitly  3  3.36 cao A1

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9 (a) Solve 2 sin x + r = 3 for 0 1 x 1 r radians. [3] 4 (b) Solve 3 sec y = 4 cosec y for 0° 1 y 1 360° . [3]

Mark scheme: 9(a) π π M1 x + = 4 3 π 5π A2 A1 for one correct and (0.262 and 1.31) 12 12 9(b) 1 1 M1 correctly use sec y = and cosec y = cos y sin y 4 A1 obtain expression for tany or y tan y = explicitly 3 53.1° and 233.1° A1 9(c) correctly rewrite equation in terms of sinz and cosz M1 use sin 2 z = 1 − cos 2 z M1 appropriate use of pythagorean identity for forming an equation in one trig ratio 8cos 2 z − 2cos z −=1 0 oe A1 ( 4cos z + 1)( 2cos z − 1) = 0 M1 solve 3 term quadratic in cosz 60° and 300° and 104.5° and 255.5° A2 A1 for any two correct

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10 The equation of a curve is y = x 2 3 + x for x H- 3 . dy (i) Find . [3] dx (ii) Find the equation of the tangent to the curve y = x 2 3 + x at the point where x = 1. [3] (iii) Find the coordinates of the turning points of the curve y = x 2 3 + x . [4]

Mark scheme: 10(i) d 1 − 1 B1 2 3 + x = ( 3 + x ) d x 2 1 − 1 M1 2 correctly substitute their ( 3 + x ) 2 and their 2x into product rule d y 2 1 − 1 1 A1 2 = x × ( 3 + x ) 2 + 2 x ( 3 + x ) d x 2 10(ii) y = 2 B1 d y 17 B1 = d x 4 y − 2 17 17 9 B1 17 = ( y = x − ) oe FT on their 2 and their from x − 1 4 4 4 4 or use y = mx + c and find c d y their d x 10(iii) d y M1 set their = 0 d x obtain correct quadratic equation A1 5x2 + 12x [= 0] soi (0, 0) and (–2.4, 4.46) A2 A1 for one point or two correct values of x

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11 A line with equation y =- 5x + k + 5 is a tangent to a curve with equation y = 7 - kx - x 2 . (i) Find the two possible values of k. [5] (ii) Find, for each of your values of k, • the equation of the tangent • the equation of the curve • the coordinates of the point of contact of the tangent and the curve. [5] (iii) Find the distance between the two points of contact. [2]

Mark scheme: 11(i) −5 x + k + 5 = 7 − kx − x 2 M1 * 2 2 M1 Dep* b − 4 ac ( = 0 ) → ( k − 5 ) − 4 ( k − 2 ) ( = 0 ) k 2 − 14 k + 33 ( = 0 ) A1 ( k − 11)( k − 3 ) ( = 0 ) M1 Dep dep * solve quadratic in k k = 11 and k = 3 A1 11(ii) y = –5x + 16 and y = 7 – 11x – x2 B2 FT their k B1 for any two correct y = –5x + 8 and y = 7 – 3x – x2 solve one tangent/curve pair for one variable from M1 quadratic equation with repeated root (–3, 31) and (1, 3) A2 A1 for one correct point or two correct x values 11(iii) find distance between any two points found in (ii) M1 800 oe A1

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