Cambridge IGCSE Mathematics - Additional 0606 — 2023 May/June Paper 2 · Variant 3

1 (a) Solve the equation = 1. [2] 7 (b) y 14 12 10 8 6 4 2 x – 10 – 8 – 6 – 4 – 2 0 2 4 6 8 10 – 2 – 4 – 6 The diagram shows the graph of y = 3 x + 9 . By drawing a suitable graph on the same diagram, solve the inequality 3x + 9 G x - 5 . [3]

Mark scheme: Question Answer Marks Partial Marks 1(a) 4x – 5 = 7 and 4x – 5 = 7 oe, soi M1 1 A1 x = 3, x =  2 1(b) Correct graph 3 B1 for correct graph y 14 and 12 10 8 6 B2 dep for 7  x   1 ; 4 dependent on a correct graph for –7 ⩽ x ⩽ 2 –1 -10 -8 -6 -4 -2 0 2 4 6 8 10 x or -2 B1 STRICT FT for their critical values -4 from the two intersections of their straight-line section of graph providing it -6 has negative gradient AND 7  x   1

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2 DO NOT USE A CALCULATOR IN THIS QUESTION. 98x 12 Write the expression in the form a b + c xd where a, b, c and d are integers. [4] 3 + 2 ` j

Mark scheme: 2 6 M2 12 7 2 x 3  2 98 x 3  2   6   M1 for 7 2x or or 3  2 3  2 3  2 3  2       98 x 6 3  2   or 3  2 3  2    their 7 2 x 6 3  2   3  2 3  2    21 2 x 6  14 x 6 7 x 6 (3 2  2) A1 or oe 9  2 9  2 (3 2  2) x 6 A1 Alternative method 12 6 (M1) 98 x 3  2 98 x 3  2  or  3  2 3  2 3  2 3  2 12 12 6 6 (M1) 3 98 x  196 x 3 98 x  196 x or 9  2 9  2 12 12 6 6 (A1) 3 98 x  196 x 3  7 2 x  14 x or oe 49 7 (3 2  2) x 6 (A1)

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3 (a) Differentiate ln ( x 3 + 3x 2 ) with respect to x, simplifying your answer. [2] x + 2 (b) Hence find dx . [2] y x ( x + 3)

Mark scheme: 3(a) 3( x  2) 3 x  6 2 mark final answer or or simplified equivalent; x ( x  3) x 2  3 x 3 x 2  6 x B1 for oe x 3  3 x 2 23(b) 1ln( x 3  3 x 2 )  c 3  3 x 2 ) B1 for 1ln( x 3 3

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4 The polynomial p is such that p ( )x = 2x 3 + 11x 2 + 22x + 40 . (a) Show that x =- 4 is a root of the equation p ( )x = 0 . [1] (b) Factorise p ( )x and hence show that p ( )x = 0 has no other real roots. [4]

Mark scheme: 4(a) 2( 4) 3  11( 4) 2  22( 4)  40  0 oe 1 4(b) ( x  4)(2 x 2  3 x  10) B2 B1 for 2 x 2  3 x  10 with two terms out of three correct Correct use of b 2  4 ac for their 3-term quadratic M1 factor 32 – 4(2)(10) < 0 isw or A1 32 – 4(2)(10) =  71 oe, cao

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5 (a) (i) A gardening group has 20 members. A committee of 6 members is to be selected. Anwar and Bo belong to the gardening group and at most one of them can be on the committee. How many different committees are possible? [2] (ii) The gate for the garden has a lock with a 6-character passcode. The passcode is to be made from Letters G A R D E N Numbers 0 1 2 3 4 5 6 7 8 9. No character may be used more than once in any passcode. Find the number of possible passcodes that have 4 letters followed by 2 numbers. [2] (b) (i) Given that n H 4 , show that ( n - 3) # n C 3 n = 4 # C 4 . [2] = 5n , where n H 3 , show that n satisfies the equation n - 3n - 28 = 0 . (ii) Given that nC 3 2 Hence find the value of n. [4]

Mark scheme: 5(a)(i) 35700 2 M1 for 20C6  18C 4 or 18C6  18C5  2C1 oe 5(a)(ii) 32400 2 M1 for 6 P4  10 P2 or ( 6  5  4  3)  (10  9) oe 5(b)(i) Correct algebraic method to show 2 n n ! n B1 for C3  or ( n  3) C3 is the same as 4nC 4 oe 3!( n  3)! n n ! C 4  4!( n  4)! 5(b)(ii) n ( n  1)( n  2) B2 n n ( n  1)( n  2)  5n or B1 for   C3   or 6 6 n(n – 1)(n – 2) = 30n n(n – 1)(n – 2) = 30n seen and completion to given answer: n 2  3 n  28  0 ( n  7)( n  4)  0 oe M1 n = 7 A1

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6 y A x O y = 5e 2 x - 3 The diagram shows the curve y = 5e 2 x - 3 . The curve meets the y-axis at the point A. The tangent to the curve at A meets the x-axis at the point B. Find the length of AB. [6]

Mark scheme: 6 dy 2 x B1  10e dx [At A, m = ] 10 B1 [At A, y = ] 2 B1 [Equation tangent is] y = 10x + 2 oe B1 2 M1 providing their 10 is derived using   their 2 2  their 2  oe differentiation AB2 =     their10  [AB = ] 2.01 or 2.009[9…] nfww, isw A1

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7 Variables x and y are such that y = . Use differentiation to find the approximate change 1– x in y as x increases from 0.1 to 0.1+ h , where h is small. [6]

Mark scheme: 7 d(4 x 3  2sin8 x ) 2 B2 B1 for 12x2+ kcos8x, where k > 0  12 x  16cos8 x soi d x Correct quotient rule: M1 or applies correct product rule to t hei r (1  x ) 12 x 2  16cos8 x  (4 x 3 + 2sin8 x )( 1) (4 x 3  2sin8 x )(1  x )1 :     (1  x ) 2 (4 x 3  2sin8 x )  (1  x ) 2 1    their (12 x 2  16cos8 x ) (1  x ) 1   Fully correct derivative; isw A1 FT their 12x2 + 16cos8x δ y  d y  M1  their   h  d x x =0.1  14.3h A1 or 14.29[54...]h with coefficient rot to 4 or more figs isw

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8 (a) The functions f and g are defined by r 3 r f ( x) = sec x for 1 x 1 2 2 g ( x) = 3 ( x 2 - 1) for all real x. (i) Find the range of f. [1] - 1 2 r (ii) Solve the equation f ( )x = . [3] 3 (iii) Given that gf exists, state the domain of gf. [1] (iv) Solve the equation gf ( )x = 1. [5] (b) The function h is defined by h ( x) = ln ( 4 - x) for x 1 4 . Sketch the graph of y = h ( x) and hence sketch the graph of y = h -1 ( x) . Show the position of any asymptotes and any points of intersection with the coordinate axes. [4] y O x

Mark scheme: 8(a)(i) f ⩽ 1 1 8(a)(ii) x = 2 nfww 3   2π    2π  M1 for x  f  sec or         3    3  1 2π sec x  3 1 A1 for  2π  cos    3  OR M1 for a complete attempt to find f x1 ( ) ; includes swapping the variables 1 1 1  A1 for f ( x )  cos    x  8(a)(iii) π 3π 1  x  2 2 8(a)(iv) gf ( x )  3(sec 2 x  1) B1 2 1 4 M1 3tan x  1 or  oe cos 2 x 3 3 M1 tan x 1 oe or cos x  oe and solves 3 4 for x, soi 5π 7π A2 A1 for one correct solution, condoning x  , and no other solutions extras 6 6 8(b) Correct diagram with intercepts indicated and 4 B1 for correct shape for h; may not be asymptotes shown. over correct domain but must have positive y-intercept and x-intercept and y appear to tend to an asymptote in the 4th quadrant h-1(x) 4 B1 for 3 and ln4 correctly marked; must 3 have attempted correct shape ln4 B1 for the position of the vertical O ln4 3 4 x asymptote indicated; must have attempted correct shape B1 for h1 the reflection of their h in the line h(x) y = x Maximum of 3 marks if not fully correct

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9 (a) Show that 3 dx = 36.6 . [3] 1 x (b) y 10y = 7 - 3x 1 y = 3x + 4 A O x 1 The diagram shows part of the line 10y = 7 - 3 x and part of the curve y = . 3x + 4 The line and curve intersect at the point A. Verify that the y-coordinate of A is 0.1 and calculate the area of the shaded region. [8]

Mark scheme: 19(a) B1  x  4 23 3  x  4 x 3 x 8 5 2 M1 FT providing one term is correct in  3 3 3  2 1  x  6 x    5 1 x 3  4 x 3 3 53 23  3 53 23  A1 (8)  6(8)   (1)  6(1)  = 36.6 5  5    9(b) 1 M2 1 10(0.1) = 7 – 3x and 0.1 = M1 for 10(0.1) = 7 – 3x and 0.1 = 3 x  4 3 x  4 and oe evaluates both expressions as x = 2 oe 1 B1 [Area trapezium =] (0.1  0.7) their 2 oe or 2 7(their 2) 3(their 2) 2   [0] oe or 0.8 10 20  1  1 B2 1 d x  ln(3 x  4) [  c ] B1 for k ln(3 x  4) k  or for     3 x  4  3 3 1ln3 x  4 3 1 1 M1 dep on at least previous B1 ln(3(2)  4)  ln(3(0)  4) 3 3 their 0.8 – 0.3054…oe M1 dep previous M1; FT their 0.8 providing the difference results in a positive value 0.495 or 0.4945[69…] rot to 4 or more sf A1

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10 An arithmetic progression, A, has first term a and common difference d. The 2nd, 14th and 17th terms of A form the first three terms of a convergent geometric progression, G, with common ratio r. (a) (i) Given that d ! 0 , find two expressions for r in terms of a and d and hence show that a =- 17d . [6] (ii) Find the value of r. [2] (b) The first term of the geometric progression, G, is q and the sum to infinity is 256. 3 Find the sum of the first 20 terms of the arithmetic progression, A. [7]

Mark scheme: 10(a)(i) a + d, a + 13d, a + 16d soi B1 a  13d a  16 d M2 FT their 3 distinct terms providing of the  oe form a  d a  13d a + kd where k ≠ 0 and at least one is correct a  13d M1 for either  r   or a  d a  16 d  r   or a  13d a  16 d  r   a  d Clears fractions and expands oe: A1 a 2  26 ad  169 d 2  a 2  17 ad  16 d 2 9ad + 153d 2 = 0 or 9 ad  153 d 2 A1 Convincingly derives a = 17d e.g. A1 9d(a + 17d) = 0 [therefore] a = 17d oe 10(a)(ii) r = 0.25 oe 2 17 d  13d 17 d  16 d M1 or 17 d  d 17 d  13d or –16d , –4d, –d 10(b) q 256 M1 FT their 0.25 providing it is between –1  oe and 1 1  0.25 3 q = 64 A1 [a + d = their 64 ] 17d + d = their 64 DM1 dep on previous M1 or a a   their 64 17 d =  4 and a = 68 oe A2 A1 for either correct OR d =  4 and S 20  150 d oe 20 M1 FT their a and d S 20   2(their 68)  19(their ( 4)) 2 or S 20  150( their d ) 600 A1

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