Cambridge IGCSE Mathematics - Additional 0606 — 2019 May/June Paper 2 · Variant 3

1 Find the values of x for which 9x2 + 18x - 1 1 x + 1. [3]

Mark scheme: Question Answer Marks Partial Marks 1 For attempting to solve M1 where * may be any inequality sign or = 9 x 2 + 17 x − 2[*0] 1 A1 Critical values , − 2 9 1 A1 FT their critical values from ax 2 + bx + c < 0 with a −<2 x < final answer 9 > 0

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3 The points A, B and C have coordinates (4, 7), (-3, 9) and (6, 4) respectively. (i) Find the equation of the line, L, that is parallel to the line AB and passes through C. Give your answer in the form ax + by = c, where a, b and c are integers. [3] (ii) The line L meets the x-axis at the point D and the y-axis at the point E. Find the length of DE. [2]

Mark scheme: 3(i) 7 − 9 2 M1 oe or − seen 4 −−( 3) 7  2  M1 y − 4 = their  −  ( x − 6)  7   2  or y = their  −  x + c  7  40 and their c = oe 7 2 x + 7 y = 40 oe A1 3(ii) 2 2 M1 FT their equation from part (i)  40   40  their   + their    2   7  20.8[00…] A1

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4 The function f is defined, for 0° G x G 360°, by f (x) = 4 + 3 sin 2x . (i) Sketch the graph of y = f(x) on the axes below. y 8 00° 60° 120° 180° 240° 300° 360° x –8 [3] (ii) State the period of f. [1] (iii) State the amplitude of f. [1]

Mark scheme: 4(i) Correct graph B3 B1 Correct sinusoidal shape with midline at y = 4 B1Two cycles B1 Correct amplitude Maximum of 2 marks if not fully correct 4(ii) 180° B1 4(iii) 3 B1

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5 (a) Given that A = 0 - 1 and that A + O = A, 6 4 (i) state the order of the matrix A, [1] (ii) write down the matrix O. [1] 1 - 1 0.4 0.2 (b) B = and C = . e3 2o e- 0.6 0.2o Find the matrix product BC and state a relationship between B and C. [2] a 4 a (c) D = , where a is a positive integer. Find D-1 in terms of a. [2] e- 1 5o

Mark scheme: 5(a)(i) 3 by 2 or 3 × 2 B1 5(a)(ii)  0 0  B1   0 0      0 0  5(b)  1 0  B2 B1 for either   and B is C−1 oe  0 1  5(c) 1  5 −4 a  B2 1    5 − 4 a    B1 for   or ... ×  9 a  1 a  9 a    1 a 

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6 A curve has equation y = 3x - 5 - 2x . dy d2 y (i) Find and 2 . [4] dx dx (ii) Find the exact value of the x-coordinate of each of the stationary points of the curve. [2] (iii) Use the second derivative test to determine the nature of each of the stationary points. [2]

Mark scheme: 6(i) 9(3 x − 5) 2 − 2 isw B2 B1 for k (3 x − 5) 2 k ≠ 9 seen 54(3 x − 5)[1] isw B2 B1 for k (3 x − 5)[1] k ≠ 54 seen 6(ii) Solves their 9(3 x − 5) 2 − 2 = 0 M1 5 2 A1 [ x = ] ± or exact equivalent 3 9 6(iii) 5 2 M1 Substitutes their + 3 9 5 2 or their − into 3 9 their 54(3 x − 5)[1] and considers sign of result 5 2 d 2 y A1 When x = + > 0 3 9 d x 2 so minimum 5 2 d 2 y and when x = − < 0 3 9 d x 2 so maximum

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7 A B 50 cm D C 4r rad 9 O The diagram shows a company logo, ABCD. The logo is part of a sector, AOB, of a circle, centre O and radius 50 cm. The points C and D lie on OB and OA respectively. The lengths AD and BC are equal and 4r AD : AO is 7 : 10. The angle AOB is radians. 9 (i) Find the perimeter of ABCD. [5]

Mark scheme: 7(i) [ AD = BC = ] 35 soi B1 Valid method for finding DC M1 [ DC = ]19.2836... A1 4π M1 50 × oe 9 4π A1 35 + 35 + 19.2836… + 50 × 9 = 159 or awrt 159 isw 7(ii) Sector – triangle: M1 or Segment + trapezium : 1 2 4π 1 2  4π 4π  × 50 × × 50  − sin  2 9 2  9 9   1 2  4π   M1  1  −  × their15 × sin    oe +  ( 64.2787... + 19.2836... ) × 26.81155   2  9    2  1630 or 1634.538… rot to 4 or more A1 figs, isw

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8 (a) (i) Given that x 2 - = x 16 - 4x 13 + qx 10 + rx 7 + f, find the value of each of the constants p, px q and r. [3] 8 2 1 (ii) Explain why there is no term independent of x in the binomial expansion of x - . [1] e px o n x (b) In the binomial expansion of 1 - , where n is a positive integer, the coefficient of x is 30. Form e 2 o an equation in n and hence find the value of n. [4]

Mark scheme: 8(a)(i) p = 2, q = 7, r = −7 B3 B1 for each or M1 for any two of 2 6  1  2 7  1  8 × 7 2 x x 8 −  , ( )  −  , ( )   px  2  px  3 8 × 7 × 6 2 x −  ( 5  1  )  3 × 2  px  or better 8(a)(ii) Valid explanation B1 8(b) 2 B1 n ( n − 1)  1   −  [ x ] seen or implied 2  2   n ( n − 1)  1  2  M1 their   −   = 30  2  2     n 2 − n − 240 = 0 A1 16 A1

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9 y y = 16 - x 2 y = 7 O x The diagram shows the curve y = 16 - x2 and the straight line y = 7. Find the area of the shaded region. You must show all your working. [6]

Mark scheme: 9 Roots of curve: (4, 0) or (−4, 0) oe B1 Intersections: (−3, 7) or (3, 7) oe B1 Correct strategy for finding area B1 2 x 3 M1 (16 − x )dx = 16 x − ∫ 3 2 x 3 (9 − x )dx = 9 x − or ∫ 3 F(b) – F(a) M1 148 1 A1 or 49 or 49.3[33…] 3 3

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10 O q p Q P R B A The diagram shows a triangle OAB. The point P is the midpoint of OA and the point Q lies on OB such that 1 OQ = OB . The position vectors of P and Q relative to O are p and q respectively. 4 (i) Find, in terms of p and q, an expression for each of the vectors PQ, QA and PB. [3] (ii) Given that PR = mPB and that QR = n QA , find an expression for PQ in terms of m, n, p and q. [2] (iii) Using your expressions for PQ, find the value of m and of n. [4]

Mark scheme: JJJG 10(i) PQ = q − p B3 B2 for anyJJJG two correctJJJG or JJJG B1 for OA = 2p or OB = 4q soi QA = 2p − q JJJG PB = 4q − p JJJG JJJG JJJG JJJG 10(ii) PQ = λ(4q − p ) − µ(2p − q ) oe isw B2 B1 for PQ = λPB − µQA soi 10(iii) For equating the coefficients of p or q in M1 q − p = λ(4q − p ) − µ(2p − q ) 4λ+ µ= 1 oe A1 FT their (ii) provided in terms of λ, µ, p and q λ+ 2µ= 1 oe Solves their equations in λand µ M1 1 3 A1 λ= , µ= 7 7

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11 A particle travelling in a straight line passes through a fixed point O. The displacement, x metres, of the particle, t seconds after it passes through O, is given by x = 5t + sin t . (i) Show that the particle is never at rest. [2] (ii) Find the distance travelled by the particle between t = r and t = r . [2] 3 2 (iii) Find the acceleration of the particle when t = 4. [2] (iv) Find the value of t when the velocity of the particle is first at its minimum. [2] Question 12 is printed on the next page.

Mark scheme: 11(i)  d x  B1 v = = 5 + cos t    d t  5 + cos t ≠ 0 (and so never at rest) oe B1 11(ii)  π   π  M1 x = 5   + sin   or  3   3   π   π  x = 5   + sin   seen  2   2  2.75 to 2.752 A1 11(iii)  d v  M1 FT their v provided of the form k ± cos t a = = − sin t    d t  [ t = 4, a = − sin 4 = ] 0.757 or A1 0.7568[024…] 11(iv) Valid method soi e.g. M1 their ( − sin t ) = 0 or cos t = −1 sketch of v = 5 + cos t t = π A1

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12 Do not use a calculator in this question. The line y = 4x - 6 intersects the curve y = 10x3 - 19x2 - x at the points A, B, and C. Given that C is the point (2, 2), find the coordinates of the midpoint of AB. [10]

Mark scheme: 12 Eliminates y: M1 10 x 3 − 19 x 2 − x = 4 x − 6 oe Collects terms: A1 10 x 3 − 19 x 2 − 5 x + 6 = 0 x – 2 is a factor soi B1 Correct quadratic factor: B2 B1 for any two correct elements of quadratic factor (10x2 + x – 3) Factorises their 3-term quadratic or M1 solves their 3-term quadratic = 0: (5x + 3)(2x – 1) oe (−0.6, −8.4), (0.5, −4) A2 A1 for x = −0.6, x = 0.5 or y = −8.4, y = −4 nfww or for one correct (x, y) pair  their ( − 0.6 + 0.5) their ( − 8.4 + − 4)  M1  ,  oe  2 2  ( −0.05, −6.2) oe A1

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