Cambridge IGCSE Mathematics - Additional 0606 — 2019 Feb/March Paper 2 · Variant 2

1 A band can play 25 different pieces of music. From these pieces of music, 8 are to be selected for a concert. (i) Find the number of different ways this can be done. [1] The 8 pieces of music are then arranged in order. (ii) Find the number of different arrangements possible. [1] The band has 15 members. Three members are chosen at random to be the treasurer, secretary and agent. (iii) Find the number of ways in which this can be done. [1]

Mark scheme: Question Answer Marks Partial Marks 1(i) 1 081 575 B1 1(ii) 40 320 B1 1(iii) 2730 B1

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2 Variables x and y are related by the equation y = . ex d y l - x ln x (i) Show that = x . [4] d x xe (ii) Hence find the approximate change in y as x increases from 2 to 2 + h, where h is small. [2]

Mark scheme: 2(i) x B2 B1 for each e d ( ln x ) 1 d ( ) x = , = e soi dx x dx x 1 x M1 e × their − ( ln x ) × their e d y x = dx x 2 e ( ) correct completion to given answer, A1 d y 1 − x ln x = d x xe x 2(ii)  1 − 2ln 2  M1 δy = × h soi  2   2e  −0.0261[…]h isw A1

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3 The function f is defined, for 0c G x G 360c, by f (x) = a + b sin cx , where a, b and c are constants with b 2 0 and c 2 0 . The graph of y = f ( x) meets the y-axis at the point (0, -1), has a period of 120° and an amplitude of 5. (i) Sketch the graph of y = f ( x) on the axes below. [3] y 6 0 0° 60° 120° 180° 240° 300° 360° x – 6 (ii) Write down the value of each of the constants a, b and c. [2] a = ..................................... b = ..................................... c = .....................................

Mark scheme: 3(i) Fully correct curve B3 B1 for correct shape for sine with y- intercept at −1 y 6 B1 for curve with period 120° B1 for curve with amplitude 5 Maximum of 2 marks if not fully x correct. 60 120 180 240 300 360 -6 3(ii) a = −1 b = 5 c = 3 B2 B1 for any 2 correct

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4 (a) Find the values of x for which 2x + 1 G 3x + 4 . [3] x 2 2 (b) Show that, whatever the value of k, the equation + kx + k + 1 = 0 has no real roots. [3] 4

Mark scheme: 4(a) Expands, rearranges to form a M1 3-term quadratic on one side 4 x 2 + x − 3 [*0 ] 3 A1 Critical values and −1 4 3 A1 FT their critical values − 1 - x - final answer 4 4(b) 2  1  2 M1 k − 4 k + 1   ( )  4  −1 A1 discriminant independent of k and negative oe A1 FT their −1

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5 Solutions to this question by accurate drawing will not be accepted. The points A(3, 2), B(7, -4), C(2, -3) and D(k, 3) are such that CD is perpendicular to AB. Find the equation of the perpendicular bisector of CD. [6]

Mark scheme: 5 2 + 4 3 M1 [ m AB = ] oe or − soi 3 − 7 2 2 M1 [ mCD = ] their oe, soi 3 2 3 + 3 M1 their = oe or 3 k − 2 2 3 + 3 = their ( x − 2) oe 3 k = 11 nfww A1  (their 11) + 2 3 + −3  M1  ,  oe  2 2  3 A1 FT their mAB and (their 6.5, 0) y = − ( x − 6.5 ) oe isw 2

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6 The relationship between experimental values of two variables, x and y, is given by y = Abx , where A and b are constants. (i) Transform the relationship y = Abx into straight line form. [2] The diagram shows ln y plotted against x for ten different pairs of values of x and y. The line of best fit has been drawn. ln y 8 7 6 5 4 3 2 1 0 1 2 3 4 x (ii) Find the equation of the line of best fit and the value, correct to 1 significant figure, of A and of b. [4] (iii) Find the value, correct to 1 significant figure, of y when x = 2.7. [2]

Mark scheme: 6(i) Takes logs, to any base, of both sides and applies M1 the addition/multiplication law for logs ln y = ln( Ab x ) ⇒ ln y = ln A + ln b x ⇒ ln y = ln A + x ln b A1 6(ii) ln y = 1.4x + 2.2 oe B2 B1 for either m = 1.4 or ln b = 1.4 or c = or ln y = xln 4 + ln 9 oe 2.2 or ln A = 2.2 [ A = e their 2.2 = ]9 and B2 FT their 2.2 and their 1.4 [b = e their 1.4 = ] 4 their 2.2 their 1.4 B1 FT for A = e or b = e or correct FT decimal rounded to more than 1 sf 6(iii) ln y = 6 M1 or y = their 9(their 4 2.7 ) or y = e their 2.2 (e their 1.4 × 2.7 ) or ln y = their1.4(2.7) + their 2.2 or ln y = (2.7)ln(their 4) + ln( their 9) awrt 400 correct to 1 sf A1

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7 (i) Given that y = x x 2 + 1 , show that = p , where a, b and p are positive constants. [4] dx 2 x + 1 ` j (ii) Explain why the graph of y = x x 2 + 1 has no stationary points. [2]

Mark scheme: 7(i) B2 2 = x 2 + 1 2 × 2 x = kx x 2 + 1 B1 for d ( x d ( x 2 + 1 ) 2 + 1 ) 1 ( ) − 1 ( ) − 1 dx 2 dx where k ≠ 1 2 M1 x + 1 −   1 2 1 2 x + 1 + x × their  × 2 x  ( )   2    d y  2 x 2 + 1 A1 =   1  d x  2 2 x + 1 ( ) 1 or a = 2, b = 1, p = nfww 2 7(ii) Complete argument B2 FT their positive a and b d y e.g. For stationary points = 0 and when a and B1 FT for a partially correct argument d x 2 d y b are positive, ax + b cannot be 0 e.g. Because cannot be 0. d x or 2x2 cannot be −1

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8 Relative to an origin O, the position vectors of the points A and B are 2i + 12j and 6i - 4j respectively. (i) Write down and simplify an expression for AB. [2] The point C lies on AB such that AC : CB is 1 : 3. (ii) Find the unit vector in the direction of OC. [4] The point D lies on OA such that OD : DA is 1 : m. (iii) Find an expression for AD in terms of m, i and j. [2]

Mark scheme: 8(i) 6 i − 4 j − ( 2 i + 12 j ) oe M1 4i − 16 j oe, isw A1 8(ii) JJJG JJG 1 JJJG M1  OC =  OA + AB oe   4 JJJG JJJG JJJG 3 or  OC =  OB − AB oe   4 JJJG JJJG JJG 1 3 or  OC =  OB + OA oe   4 4 or 3( x − 2) = 6 − x and 3( y − 12) = −−4 y 3i + 8 j oe A1 JJJG 2 2 M1 OC = their 3 + their 8 3i + 8 j A1 FT their 3i + 8 j and their 73 their 73 8(iii) λ B2 λ − ( 2 i + 12 j ) oe, isw B1 for ( 2 i + 12 j ) seen or 1 + λ 1 + λ JJJG 1 OD = ( 2 i + 12 j ) oe 1 + λ

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9 (a) It is given that g ()x = 6x 4 + 5 for all real x. (i) Explain why g is a function but does not have an inverse. [2] (ii) Find g2 ()x and state its domain. [2] It is given that h ()x = 6x 4 + 5 for x G k . (iii) State the greatest value of k such that h-1 exists. [1] (iv) For this value of k, find h-1(x). [3] (b) The function p is defined by p ()x = 3ex + 2 for all real x. (i) State the range of p. [1] (ii) On the axes below, sketch and label the graphs of y = p (x) and y = p -1 (x) . State the coordinates of any points of intersection with the coordinate axes. [3] y y = x O x (iii) Hence explain why the equation p (x) = p -1 (x) has no solutions. [1]

Mark scheme: 9(a)(i) Valid explanation e.g. B2 B1 for either each x is mapped to a Each x is mapped to a unique value of y [and so g unique value of y oe or for inverse does is a function] but the inverse does not exist not exist because it is many to one oe because it is many to one oe 9(a)(ii)  g 2 ( x ) =  6(6 x 4 + 5) 4 + 5 isw B2 B1 for  g 2 ( x ) =  6(6 x 4 + 5) 4 + 5 isw     for all real x B1 for correct domain 9(a)(iii) [k = ] 0 B1 9(a)(iv) 4 y − 5 M1 4 x − 5 x = soi or y = 6 6 y − 5 A1 x − 5 x = ± 4 or y = ± 4 6 6 −1 x − 5 A1 If M1 A0 A0, allow SC1 for an answer h ( x ) = − 4 x − 5 x − 5 6 of h −1 ( x ) = 4 or y = 4 6 6 9(b)(i) p > 2 B1 9(b)(ii) For p: B2 B1 for each Correct exponential shape tending to y = 2 passing through (0, 5) For the inverse function: B1 Approximate reflection of p in the dotted line passing through (their 5, 0) 9(b)(iii) Valid explanation e.g. B1 The graphs do not intersect and so there are no solutions oe

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10 y B y = 1 + x + 5 x y - 3x = 3 A O D C x The diagram shows the curve y = 1 + x + 5 x and the straight line y - 3x = 3 . The curve and line intersect at the points A and B. The lines BC and AD are perpendicular to the x-axis. (i) Using the substitution u2 = x, or otherwise, find the coordinates of A and of B. You must show all your working. [6]

Mark scheme: 10(i) Eliminates x or y e.g. M1 3 x + 3 = x + 5 x + 1 or 3 + 3u 2 = u 2 + 5u + 1 Rearranges to a 3-term quadratic e.g. A1 0 = 2 x − 5 x + 2 or 0 = 2u 2 − 5u + 2 Factorises or solves 0 = 2 x − 5 x + 2 oe M1 or 0 = 2u 2 − 5u + 2 oe x = 0.5 , x = 2 A1 or u = 0.5 , u = 2 A(0.25, 3.75) B (4, 15) oe A2 A1 for each or for x = 0.25 and x = 4 10(ii) Method 1: Finding the area of the trapezium and subtracting Valid method to find the area of the trapezium soi M1 1125 5 A1 or 35 or 35.2 or 35.15625 rot to 4 or 32 32 more figs, soi Attempts to integrate M1 their 4 ∫their 0.25( x + 5 x + 1)dx [ −their 35.2] their 4 3 A1    x 2 5 x 2  + + x [ − their 35.2] oe   2 3  2    their 0.25 F(their 4) – F(their 0.25) [ −their 35.2] M1 45 13 A1 or or 2.8125 isw 16 216 or 2.81, or 2.812 Method 2: Finding the difference of two integrals Attempts to integrate M2 M1 for an attempt to form the their 4 difference with at most one error and ∫their 0.25( x + 5 x + 1 − (3 + 3 x ))dx attempts to integrate their 4 or ∫their 0.25( −2 x + 5 x − 2)dx oe their 4 A1 FT 3       −2 x 2 5 x 2   dep on at least M1 already awarded; their + − 2 x oe  must be at least 3 terms and, if FT, must     2 3    2   be of equivalent difficulty   their 0.25 F(their 4) – F(their 0.25) M1 45 13 A2 or or 2.81, 2.812 or 2.8125 16 216

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11 (a) Find y 6 dx . [3] x i - 5 ) d i . [2] (b) (i) Find y cos (4 2 i - 5 ) d i . [2] (ii) Hence evaluate y cos (4 .125

Mark scheme: 11(a) x 2 ( x 6 + 1) 2 1 B1 = x + soi x 6 x 4 x 3 x −3 B2 B1 for any two out of three terms + + c oe, isw correct 3 −3 11(b)(i) k sin(4θ− 5) where M1 1 k > 0 or k = − 4 sin(4θ− 5) A1 ( +c ) 4 11(b)(ii) sin(4(2) − 5) sin(4(1.25) − 5) M1 FT their (b)(i), dep on M1 awarded in − (b)(i) 4 4 sin(3) sin(0) or − 4 4 0.0353 or 0.03528[…] oe, cao A1

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