Cambridge IGCSE Mathematics - Additional 0606 — 2022 May/June Paper 2 · Variant 3

1 Solve the equation 4 7x - 3 - 5 = 9 . [3]

Mark scheme: Question Answer Marks Partial Marks 1 13 B1 x = oe 14 7 x  3 3.5 oe, soi M1 or 28 x  12 14 oe, soi 1 A1 x =  oe 14 Alternative method 196 x 2  168 x  13  0 oe (B1) factorising e.g. 14 x  13 14 x  1 (M1) 13 1 (A1) x = ,  14 14

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2 DO NOT USE A CALCULATOR IN THIS QUESTION. Variables x and y are related by the equation y = kx2 . When x = 1 + 2 , y = 1 - 2 . Find the value of k, giving your answer in the form a + b c , where a, b and c are integers. [4]

Mark scheme: 2  2  1 2  3  2 2 M1  1  2   leading to   3  2 2  1  2 3  2 2 M1 FT their 3  2 2 if of equivalent  difficulty 3  2 2 3  2 2 3  2 2  3 2  4 DM1 FT their 3  2 2 if of equivalent Correctly expands 9  8 difficulty 7  5 2 A1 Alternative method (M1) 1  2 1  2 1  2     1  2 1  2 1  2 1  2       (M1) 3  2 2 1  2    1  2  2 Correctly expands (DM1) FT their 3  2 2 if of equivalent 3  3 2  2 2  4 difficulty [(  1) 2 ] 7  5 2 (A1)

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3 The points A, B and C have coordinates (2, 6), (6, 1) and (p, q) respectively. Given that B is the mid‑point of AC, find the equation of the line that passes through C and is perpendicular to AB. Give your answer in the form ax + by = c , where a, b and c are integers. [5]

Mark scheme: 3 C(10, 4) B1  1 5 M1  m AC  6 oe or oe 2  6 4 4 M1 1 m FT 5 5 their 4 4 A1 FT their coordinates of C providing y – (4) = ( x  10) oe that one coordinate is correct and their 5 perpendicular gradient 4x – 5y = 60 oe A1

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4 (a) Find the range of values of x satisfying the inequality ( 5x - 1)( 6 - x) 1 0 . [2] 2 1 (b) Show that the equation ( 2k + 1) x - 4kx + 2k - 1 = 0 , where k !- , has distinct, real roots. 2 [3]

Mark scheme: 4(a) 1 M1 CVs , 6 5 1 A1 mark final answer x < , x > 6 5 4(b) (4k)2 – 4(2k + 1)(2k – 1) M1 16k2 – 4(4k2 – 1) A1 or 16 k 2  16 k 2  8k  8k  4 or better 4 > 0 A1

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5 y P ( 0, 3) Q ( 2r, 7) 0 x x =- 4r x = 4r The diagram shows part of the graph of y = a tan bx + c . The graph has vertical asymptotes at x =- 4 r and x = 4 r and passes through the points P and Q. (a) Write down the period of a tan bx + c . [1] (b) Find the values of a, b and c. [4]

Mark scheme: 5(a) 8π B1 5(b) c = 3 B1  π  1 B1  8π b     b  8  π  M1 7  a tan    3 oe  4  a = 4 A1

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6 The polynomial p ( x) is such that p ( x) = 6x 3 + ax 2 - 52x + b , where a and b are integers. It is given that p ( x) is divisible by 2x - 3 and that p l ( 1) = 4 . (a) Find the values of a and b. [5] DO NOT USE A CALCULATOR IN THIS PART OF THE QUESTION. (b) Using your values of a and b, factorise p ( x) fully. [3]

Mark scheme: 6(a) p  x : 18 x 2  2 ax  52 B1 18 + 2a – 52 = 4 M1 FT if at least 2 terms correct in derivative and has a term in a a  19 A1 Correct method to find b M1 FT their integer value of a if used 6  27 9(19) 52  3    b  0 9(19)  4b  231 8 4 2 81 171 or   78  b  0 oe 4 4 or correct elimination of a using 9 a  4b  231 oe and 2a – 34 = 4 oe b  15 A1 If 0 scored, SC1 for 9 a  4b  231 or 81 9 a 9 231   78  b  0 or a  b  4 4 4 4 oe 6(b) 2 M2 M1 for two terms correct in quadratic 3 x  14 x  5 p x  2 x  3   factor  2 x  3  3 x  1 x  5  A1

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7 (a) (i) Write down the set of values of x for which lg ( 5x - 3) exists. [1] (ii) Solve the equation lg ( 5x - 3) = 1. [1] 1 (b) It is given that log y x = 4 + log y 64 + log y 162 , where y 2 0 . Find an expression for y in 2 terms of x. Simplify your answer. [5]

Mark scheme: 7(a)(i) x  0.6 B1 7(a)(ii) B1 nfww  x  13 or 2.6 5 7(b) 1 B1 log y 64  log y 8 soi 2 or 4 = log y y 4 soi x B2 B1 for one further correct application Combines log terms e.g. log y  4 of a relevant log law 1296 e.g. log y x  4  log y 1296 4 1296 or log y x  log y 1296 y or log y  4 4 x or log y x  log y y  log y 8  log y 162 4 x 4 M1 y  or 1296y  x oe 1296 1 A1 mark final answer 4 x x 4 y  or y  6 6

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8 (a) Differentiate y = 2xe 4 x with respect to x. [2] (b) Hence find xe 4 x dx . [4] y

Mark scheme: 8(a) d 4 x 4 x B1 (e )  4e d x 8 xe 4 x  2e 4 x isw B1 FT their ke4x 8(b) Use of part (a) B1 FT part (a) providing of form 2e 4 x dx oe kxe 4 x  2e 4 x  8 xe 4 x dx  2 xe 4 x   4 x 1 4 x 1 4 x M1 FT part (a) providing of form e d x oe xe dx  xe   kxe 4 x  2e 4 x 4 4  xe 4 x e 4 x A2 A1 for any 2 terms correct   c oe 4 16

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9 (a) Find the unit vector in the direction of 40i - 9 j . [2] (b) The position vectors of points P and Q relative to an origin O are p and q respectively. The point R PR lies on the line PQ and is between P and Q such that = k . PQ (i) Write down the set of all possible values of k. [1] (ii) Given that the position vector of R relative to O is mp + nq show that m + n = 1. [3]

Mark scheme: 9(a) 2 2 M1 40 ( 9) soi 40 9 A1 mark final answer i  j oe 41 41 9(b)(i) 0 < k < 1 B1  9(b)(ii) OR  p  k (q  p ) M1  or OR  q  (1  k )(p  q )  OR  (1  k )p  kq A1  +  = 1 – k + k = 1 A1

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10 y A y = 3 + 2x - x 2 B O C x The diagram shows part of the curve y = 3 + 2 x - x 2 . The point A lies on the curve and has an x‑coordinate of 1.5. The tangent to the curve at A meets the x‑axis at B. The curve meets the x‑axis at C. Find the area of the shaded region. [10]

Mark scheme: 10 A(1.5, 3.75) soi B1 Factorises or solves 3  2 x  x 2  0 M1 C(3, 0) soi A1 implies M1 dy d y B1  2  2 x , when x = 1.5  1 dx d x y – 3.75 = (x – 1.5) oe B2 B1 for y – 3.75 = (x – 1.5) oe leading to B(5.25, 0) soi 1 B1 1  (5.25  1.5)  3.75 OR  (5.25  3)  2.25 2 2 2 5.25 2  1.5 2  2 5.25 2  32  5.25    5.25(1.5)   5.25    5.25(3)   2  2  2  2   3 2 3  M1 OR  (3  2 x  x )dx   F( x ) 1.5   3   1.5   3 2  3 x  x )d x   G( x ) 1.5  3  1.5(2.25   2 3  2 x x   3 x     3 x 3 2 x 3   2 3 1.5  2.25 x    oe  2 3 1.5  225 M1 dep on at least 2 correct terms in the  their   F(3)  F(1.5)  integration  32   81  OR  G(3)  G(1.5)   their   32  117 A1 or 3.65625 32

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11 (a) The sum of the first 20 terms of an arithmetic progression is 1100. The sum of the first 70 terms is 14 350. Find the 12th term. [6] 1(b) The first three terms of a geometric progression are x + 6 , x - 9 , ( x + 1) . Show that x 2 satisfies the equation x 2 - 43x + 156 = 0 . Hence show that a sum to infinity exists for each possible value of x. [7]

Mark scheme: 11(a) 20 B1  S 20    2 a  19 d   1100 oe 2 70 B1  S 70    2 a  69 d   14350 oe 2 Solves their linear equations in a and d M1 dep on at least B1 and an attempt to form the other equation using the sum formula a = 2, d = 6 A2 A1 for each  u12 2 11  6   64 B1 11(b) 1 M2 1 ( x  1) ( x  1) x  9 2 x  9 2  M1 for either or ; may x  6 x  9 x  6 x  9  x  9  1 be embedded or ( x  9)    ( x  1)  x  6  2  x  9  2 1 ( x  6)    ( x  1) oe  x  6  2 Correct simplification to given equation A1 x 2  43 x  156  0 Factorises or solves: M1 (x – 4)(x – 39) = 0 x = 4, x = 39 A1 1 2 A2 r =  r = A1 for either value of r 2 3 |r| < 1 for each progression [and so the sum to infinity exists]

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12 In this question all lengths are in centimetres. x y A container is a half‑cylinder, open as shown. It has length y and uniform cross‑section of radius x. The volume of the container is 25 000. Given that x and y can vary and that the outer surface area, S, of the container has a minimum value, find this value. [8]

Mark scheme: 12 πx 2 y M1  25000 oe 2 50000 50000 A1 y  or πxy  πx 2 x  50000  2 M1 k S  πx  2   πx FT their y providing of the form 2  πx  π x 50000 2 A1 S   πx x d S  2 M1 k 2  50000 x  2π x oe FT from expression of form  mx d x x Equates to zero and solves for x: A1 25000 x  3 or 19.9647.... π 2 DM1 dep on previous M1; 50000  25000  S min   π  3  FT their positive value of x used in a 25000  π  correct expression for S 3 π 3760 or 3756 to 3757 A1

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