Cambridge IGCSE Mathematics - Additional 0606 — 2020 Feb/March Paper 2 · Variant 2

1 Find the values of x for which 12x 2 - 20x + 5 1 ( 2 x + 1)( x - 1). [4]

Mark scheme: Question Answer Marks Partial Marks 1 Expands right hand side and attempts to M1 collect terms Factorises or solves their 3-term quadratic M1 2 3 A1 correct CVs , 5 2 2 3 A1 FT their CVs, provided both M marks < x < mark final answer awarded 5 2

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2 Variables x and y are such that, when 1g y is plotted against x3 , a straight line graph passing through the points (6, 7) and (10, 9) is obtained. Find y as a function of x. [4]

Mark scheme: 2 Valid method to find m M1 9 − 7  1  m =  =  10 − 6  2  Valid method to find c M1 FT their m 1 e.g. 7 = their × 6 + c 2  1  3 M1 lg y =  their  x + their 4  2  1 x 3 + 4 A1 y = 10 2 oe, isw

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3 Find the exact solution of 3 2 x - 3 x + 1 - 4 = 0. [4]

Mark scheme: 3 Rewrites in quadratic form soi M1 e.g. y = 3x then y 2 − 3 y − 4 = 0 or (3 x ) 2 − 3(3 x ) − 4 = 0 Factorises or solves their 3-term quadratic M1 e.g. (y + 1)(y – 4) [= 0] or (3x + 1)(3x – 4) [= 0] 3x = 4 A1 ignore 3x = −1 ln 4 A1 x = log 3 4 or oe, only ln3

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4 The position vectors of three points, A, B and C, relative to an origin O, are , and - 7 - 4 y respectively. Given that AC = 4BC, find the unit vector in the direction of OC. [5]

Mark scheme:     4 OC − OA = 4 ( OC − OB ) soi B1   15  B2 B1 for [x = ] 15 or [ y = ] −3 [ OC = ]    −3   2 2 M1 OC = their15 + their ( − 3) 1  15  A1  15    oe FT their   and their 234 234  −3   − 3 

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5 (a) On the axes below, sketch the graph of y = 5x - 7 , showing the coordinates of the points where the graph meets the coordinate axes. [3] y O x (b) Solve 5 5x - 7 - 1 = 14 . [3]

Mark scheme: 5(a) Correct V shape with vertex on positive x- B1 axis (0, 7) B1  7  B1  , 0   5  5(b) x = 2 B1 5 x − 7 = their ( − 3) oe, soi M1 or 25 x − 35 = their ( − 15) oe, soi 4 A1 x = oe 5 Alternative method 25 x 2 − 70 x + 40 = 0 oe (B1 factorising e.g. ( 5 x − 4 )( x − 2 ) M1 4 A1) x = 2, 5

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6 (a) A circle has a radius of 6 cm. A sector of this circle has a perimeter of 2 6 + 5r cm. Find the area of this sector. [4] (b) A 7 cm O r rad 4 B The diagram shows the sector AOB of a circle with centre O and radius 7 cm. Angle AOB = r radians. Find the perimeter of the shaded region. [3] 4

Mark scheme: 6(a) 2(6) + 6θ = 2(6 + 5π) oe M1 5 A1 θ = π oe, soi 3 1 2 5π  M1 × 6 × their  2  3  94.2 or 30π A1 Alternative method arc AB = 10π (M1 10π 5 B1 sector is = of the circle 12π 6 5 M1 × 36π 6 94.2 or 30π A1) 6(b)  π 7π M2  π   7π  2  7sin + oe, soi M1 for 2  7sin  + their   or  8  4  8   4  their 2  7sin π +  7π   8   4 10.9 or 10.85 to 10.86 A1

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8 f(x) 2 1 0 x 2r 4r 2r 8r - 1 3 3 3 - 2 - 3 - 4 - 5 - 6 8r The diagram shows the graph of f( x) = a cos bx + c for 0 G x G radians. 3 (a) Explain why f is a function. [1] (b) Write down the range of f. [1] (c) Find the value of each of the constants a, b and c. [4]

Mark scheme: 8(a) Valid explanation e.g. B1 Each value of x is mapped to a unique value of y. 8(b) −5 f 1 B1 8(c) a = 3, b = 0.75 oe, c = − 2 B4 B1 for a = 3 B1 for c = −2 2π 8π M1 for = oe b 3 A1 for b = 0.75 oe

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9 Variables x and y are such that y = . Use differentiation to find the approximate change in y x2 as x increases from 0.5 to 0.5+ ,h where h is small. [6]

Mark scheme: 9 d(e 3 x ) 3 x B1 = 3e soi d x Applies product rule to e.g. numerator: M1 or to x −2 sin x : x −2 cos x + ( −2 x −3 )sin x their(3e3x)sinx + e3x cosx 3 x 2 or to e × x− : e 3 x × ( −2 x −3 ) + their (3e 3 x ) × x −2 Correct quotient rule: M1 − 2 x ( e 3 x sin x ) x 2 ( their ( 3e 3 x sin x + e 3 x cos x ) ) 4 or applies product rule for a second x time e.g. : x −2 (their ( 3e 3 x ) sin x + e 3 x cos x ) + ( −2 x −3 )( e 3 x sin x ) Fully correct derivative; isw A1  d y  M1 δ y = their   × h  d x x =0.5  7.14h A1 or 7.137[66...]h with coefficient rot to 4 or Answer only, without working, scores more figs SC1 isw

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10 (a) g( )x = 3 + for x H 1. x (i) Find an expression for g -1 ( x). [2] (ii) Write down the range of g -1 . [1] (iii) Find the domain of g -1 . [2] 2(b) h( )x = 21n( 3x - 1) for x H . 3 The graph of y = h( )x intersects the line y = x at two distinct points. On the axes below, sketch the graph of y = h( )x and hence sketch the graph of y = h -1 ( x). [4] y O x

Mark scheme: 10(a)(i) Correct method to find inverse M1 −1 1 A1 g ( x ) = oe x − 3 10(a)(ii) g−1 ⩾ 1 or [1, ∞) B1 10(a)(iii) 3 < x  4 or (3, 4] B2 B1 for 3 and 4 in an incorrect inequality or for x > 3 or x ⩽ 4 10(b) Correct graph for h B1 h−1 the reflection of h in y = x B1 FT their h Both graphs drawn over the correct domain B1 FT their h and h−1 h−1 B1 Correct graphs intersecting twice h 2 3 2 3

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11 h r A container is a circular cylinder, open at one end, with a base radius of r cm and a height of h cm. The volume of the container is 1000 cm3. Given that r and h can vary and that the total outer surface area of the container has a minimum value, find this value. [8]

Mark scheme: 11 1000 1000 B1 h = 2 or r = soi πr πh 2  1000  M1 S = πr + 2πr  2  oe or  πr   1000 1000 ( h ) oe S = π  + 2π  π h  πh 2  1000  A1 S = πr + 2   or better or  r  1000 1000 2 ) S = + 2π ( h 1 h π dS −2 B2 B1 FT for each term correct = 2πr − 2000 r or dr − d S −2 1 = − 1000 h + 1000π h 2 d h dS 3 1000 M1 = 0, r = oe or dr π d S 32 1000 = 0, h = oe d h π 2 M1  1000  2000 S = π  3  + or  π  1000 3 π 1 1000  1000  2 S = + 2 1000π  3  1000  π  3 π 439 or 439.3 to 439.4 A1

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12 A particle P moves in a straight line such that, t seconds after passing through a fixed point O, its acceleration, a ms -2 , is given by a =- 6. When t = 0, the velocity of P is 18 ms -1 . (a) Find the time at which P comes to instantaneous rest. [3] (b) Find the distance travelled by P in the 3rd second. [3]

Mark scheme: 12(a) v = −6t + c soi B1 v = −6t + 18 M1 −6t + 18 = 0 , t = 3 A1 12(b) − 6t 2 B1 s = + 18t soi 2 ( −3(3) 2 + 18(3) ) −−( 3(2) 2 + 18(2) ) M1 FT their s provided it is from an attempt to integrate 3 (metres) A1 Not from wrong working

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13 (a) The sum of the first two terms of a geometric progression is 10 and the third term is 9. (i) Find the possible values of the common ratio and the first term. [5] (ii) Find the sum to infinity of the convergent progression. [1] (b) In an arithmetic progression, u 1 =- 10 and u 4 = 14. Find u 100 + u 101 + u 102 + f + u 200 , the sum of the 100th to the 200th terms of the progression. [4]

Mark scheme: 13(a)(i) a + ar = 10 soi B1 ar 2 = 9 soi B1 Solves their equations M1 3 3 A2 3 3 r = − , and a = 25, 4 A1 for either r = − , or a = 25, 4 5 2 5 2 or 3 for r = − and a = 25 or 5 3 for r = and a = 4 2 13(a)(ii) 125 5 B1 or 15.625 or 158 only 8 13(b) d = 8 B1 [ S 200 − S 99 = ] M2 M1 for either sum correct or correct FT their d 200 { 2( −10) + 199(their 8)} − 2 99 { 2( −10) + 98(their 8)} oe 2 119382 cao A1 Alternative method 1 d = 8 (B1 u100 = −10 + 99 × 8 [ = 782 ] and M1 u 200 = −10 + 199 × 8 [ = 1582 ] and n = 101 1 M1 (101)(782 + 1582) 2 119382 cao A1) Alternative method 2 d = 8 (B1 u100 = −10 + 99 × 8 [ = 782 ] and M1 n = 101 1 M1 (101)(2 × 782 + (101 − 1) × 8) 2 119382 cao A1)

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