Cambridge IGCSE Mathematics - Additional 0606 — 2024 May/June Paper 2 · Variant 1

1 (a) On the axes, sketch the graph of y = 4x - 6 , showing the points where the graph meets the axes. [2] y O x (b) Solve the equation 4x - 6 = 2 x . [3]

Mark scheme: Question Answer Marks Partial Marks 1(a) Fully correct graph with intercepts marked B2 B1 for a graph of correct shape with vertex on x-axis y (0, 6) O (1.5, 0) x 1(b) 4x – 6 = 2x and 4x – 6 = –2x oe M1 x = 3 x = 1 A2 A1 for either correct Alternative method 12x2 – 48x + 36 = 0 oe (B1) Factorises or solves (M1) x = 1, x = 3 (A1)

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2 (a) Write 3 + 4x - 2 x 2 in the form a + b ( x + c) 2 , where a, b and c are integers. [3] (b) Hence write down the range of the function f ( )x = 3 + 4x - 2x 2 , where x ! R . [1]

Mark scheme: 2(a) 5 – 2(x  1)2 B3 Mark final expression B2 for – 2(x  1)2 or B1 for (x  1)2 or b = –2, c = –1 and B1 for 5 + b(x + c)2 oe with numerical values of b and c or a = 5 2(b) f  their 5 B1 STRICT FT of their 5 from (a)

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4 DO NOT USE A CALCULATOR IN THIS QUESTION. (a) Find the exact distance between the two points where the curve 9 ( x - 1) 2 + 4 ( y - 3) 2 = 36 cuts the y-axis. [4] (b) Find the coordinates of the points where the curve with equation 2x 2 + 83 xy = x 3 y - 20 x 1 intersects the curve with equation y = . Give each of your answers in the form a + b c , where x a and b are rational and c is the smallest integer possible. [6]

Mark scheme: 4(a) 4( y  3) 2  36  9 or 4 y 2  24 y  9[  0] M1 27 A1 y  3  or exact equivalent, soi 4 27  27  M1 FT their a  b providing b is not a 3    3   oe 4  4  square number 27 or 3 3 or exact equivalent, nfww A1 4(b) Eliminates one unknown and simplifies M1 terms: 2 x 2  83  x 2  20 x oe, soi x 2  20 x  83  0 A1 Applies quadratic formula or completes the M1 FT their 3-term quadratic 20  20 2  4[1](83) square: x  2 x 10  17 A1 1 10  17 M1 FT their x = a  b providing y   10  17 10  17 previous M1 awarded 10 17 A1 dep on all marks previously awarded x 10  17 , y   83 83

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5 There are 3 women, 2 men and 4 children in a choir. (a) The choir stands in a single straight line. (i) Find the number of possible arrangements if the first person and last person are both women. [2] (ii) Find the number of possible arrangements if all the children stand next to each other. [2] (b) Four of the choir are selected to sing in a group. (i) Find the number of different selections if no man is chosen. [2] (ii) Find the number of different selections if at least 2 women are chosen. [2]

Mark scheme: 5(a)(i) 30 240 2 M1for 3  7!  2 or 3 P1  7 P7  2 P1 oe 5(a)(ii) 17 280 2 M1for 4!  6! oe or 4 P4  6 P5 oe 5(b)(i) 35 2 M1 for 7C4 or 1 + 4 + 18 + 12 5(b)(ii) 51 2 M1 for 3C2  6C2 + 3C3  6C1 oe or 18 + 4 + 3 + 2 + 24

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6 Variables x and y are such that y = cos x sin 2 x . Use differentiation to find the approximate change in y as x increases from 3 to 3+ h , where h is small. [5]

Mark scheme: 6 d 2 B1 (sin x )  2sin x cos x soi d x cos x  their (2sin x cos x )  M1 d 2 FT their (sin x ) (their  sin x )  sin 2 x d x cos x  their (2sin x cos x ) ( sin x )  sin 2 x A1 d 2 FT their (sin x ) isw d x δy 2 3 M1 FT their derivative  their (2sin3cos 3  sin 3) or better h 0.274h or A1 dep on correct derivative seen 0.2738[08...]h where the coefficient of h is rot to 4 or more sf

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7 It is given that y = mx 2 + + n , where m and n are non-zero constants. It is also given that 2 2 2 d y d y 3 2 = - y for all values of x. Find the values of m and n. [4] f d x p e d x o

Mark scheme: 7 d y 1 B1  2 mx  d x 2 d 2 y B1  2 m d x 2 1 5 2 M1 for m  and n  and no other values 2 4 4  1   2 x  3(2 m )   2 mx     mx   n   2   2  soi

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8 (a) In an arithmetic progression, the sum of the first 30 terms is - 1065 . The sum of the next 20 terms is - 2210 . Find the first term and the common difference. [5] (b) A geometric progression is such that the first term is 4 and the sum of the first three terms is 7. Find the two possible values of the common ratio and find the sum to infinity for the convergent progression. [5]

Mark scheme: 8(a) 30 B1 S 30   2 a  29 d  1065 2 S 50  S 30  B1 50  30   2 a  49 d     2 a  29 d   2210 2  2  50 or S 50   2 a  49 d  2210  1065 2 Solves their linear equations in a and d as M1 dep on an attempt to form their far as a = .. or d = … equations using at least one sum formula Some correct pairs are e.g. 150a + 2175d = –5325 150a + 3675d = –9825 or 30a + 435d = –1065 50a + 1225d = –3275 or 2a + 49d = –131 2a + 29d = –71 a = 8, d = 3 A2 A1 for each 8(b) 4(1  r 3 ) B1 4 + 4r + 4r2 = 7 or  7 oe 1  r 4r 2  4r  3   0  oe B1 Solves or factorises their 3-term quadratic M1 oe e.g. (2r – 1)(2r + 3) [= 0] r = 0.5 , –1.5 A1  4  A1  8 only, nfww    1  0.5 

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9 The functions f and g are defined by 3x 2 f ( )x = for x 1 0 4x - 1 1 g ( )x = 2 for x 1 0 . x (a) Explain why the function fg does not exist. [1] (b) Given that the function gf does exist, find and simplify an expression for gf ( )x . [2] -1 px - x ( qx + r) (c) Show that f ( )x can be written as where p, q and r are integers. [4] 3

Mark scheme: 9(a) Valid explanation: Range of g is g > 0 oe B1 9(b) 1 M1 2   3 x 2    4 x  1  (4 x  1) 2 A1 or simplified equivalent, isw 9 x 4 9(c) 3 x 2  4 xy  y   0  or 3 y 2  4 xy  x   0  B1 2 M1 FT their expression providing it has at ( 4 y )  ( 4 y )  4(3)( y )  x   oe most one sign error 2(3) or ( 4 x )  ( 4 x ) 2  4(3)( x )  y   oe 2(3) Justifies the negative square root B1 A1 1 2 x  x (4 x  3) f ( x )  3

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10 (a) Show that ( tan x + sec x) 2 can be written as . [4] 1- sinx (b) Hence solve the equation ( tan 3 i + sec 3i ) 2 = 6 for 0° G i G 180 ° . [4]

Mark scheme: 10(a) tan 2 x  2tan x sec x  sec 2 x M1  sin x 1  2 or     cos x cos x  sin 2 x sin x 1 1 A1  2    cos 2 x cos x cos x cos 2 x 1 2 or factorises sin x  1 oe 2  cos x (1  sin x ) 2 A1 1  sin 2 x (1  sin x )(1  sin x ) 1  sin x A1 must be fully justified  (1  sin x )(1  sin x ) 1  sin x or (1  sin x ) 2 1  sin x  (1  sin x )(1  sin x ) 1  sin x 10(b) 7sin3 5 B1 One correct value for 3 soi e.g. M1 45.58… 134.4… 405.5… 494.4… 15.2 or 15.19 to 15.195 A2 with no extras in range 44.8 or 44.80 to 44.81 135.2 or 135.19 to 135.195 A1 for any 2 correct, ignoring extras 164.8 or 164.80 to 164.81

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11 In this question all lengths are in centimetres. A B x D C The diagram shows a rectangle ABCD with BC = x . The area of the rectangle is 400 cm2. x Two identical quarter-circles of radius 2, with centres A and C, are removed from the rectangle to make the shaded shape. Given that x can vary, find the value of x that gives the minimum value of the perimeter of the shaded shape and hence find this minimum value. [7]

Mark scheme: 11  πx   400 x  B2 400 P = 2    x  2    oe B1 for rectangle length = soi  4   x 2  x Correct first derivative B1 FT their P providing it is of the form π 800 a   bx oe with a an integer and b a 2 x 2 x constant d P M1 FT provided one term of their Equates their to 0 and solves for x derivative is correct d x 40 A1 x = or 22.6 π or 22.56[75...] π  40 800 M1 FT their value of x provided it is P =    greater than 0 2  π  40 π P = 70.9 or 70.89[81...] A1

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12 O A D P B C The diagram shows a triangle OBC. OA : OB = 4 : 7 and OD : OC = 4 : 7. OB = b and OC = c The point P is the point of intersection of AC and BD such that AP = m AC and BP = n BD where m and n are scalars. (a) Find two expressions for OP, each in terms of b, c and a scalar, and hence show that P divides both AC and DB in the ratio 4 : 7. [7] 2 2(b) The point Q is such that OQ = b + c . 7 7 Use a vector method to show that O, Q and P are collinear. Justify your answer. [2]

Mark scheme: 12(a)  4 4  B3  4 4  OP  b   c  b  and B2 for OP  b   c  b  or 7  7  7  7   4   4  OP  b   c  b  OP  b   c  b  oe  7   7  or B1 for   4   4   OP   their  b   c   their  b   7    7   or   4   OP  b    their  c  b  or   7   Equates components e.g.: M1 FT providing at least B1 awarded and   4  4 two expressions for OP in terms of b,  their  (1  )  1   or  their   7  7 c,  and  found 4 4 A1 (1  )  1     oe 7 7 4 7 A2 4 7  and  oe A1 for  or  oe 11 11 11 11 and conclusion AP : AC = 4 : 11 therefore AP : PC = 4 : 7 BP : BD = 7 : 11 therefore DP : PB = 4 : 7 oe 12(b)  4  4 4 2 M1 for OP = c b   b  c  or OP =   4 4 4 11 11 11 OP = = c  b  c  or OP b  and 11 11 11   OP and OQ are scalar multiples of each other and have a point in common oe

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