Cambridge IGCSE Mathematics - Additional 0606 — 2020 May/June Paper 2 · Variant 3

1 Solutions to this question by accurate drawing will not be accepted. Find the equation of the perpendicular bisector of the line joining the points (4, - 7 ) and ( - 8 , 9). [4]

Mark scheme: Question Answer Marks Partial Marks 1 Coordinates of mid-point B1 ( −2,1) 9 −−7  16  B1 m AB =  = −  −−8 4  12  −1 M1 m⊥ = −1612 3 A1 y −=1 ( x + 2) oe 4

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3 (a) On the axes below, sketch the graph of y =- ( x + 2)( x - 1)( x - 6), showing the coordinates of the points where the graph meets the coordinate axes. y O x [2] (b) Hence solve - ( x + 2)( x - 1)( x - 6) G 0. [2]

Mark scheme: 3(a) Correct sketch B2 B1 for correct shape y B1 for correct coordinates (−2, 0), (1, 0), (6, 0) and (0, −12) (-2, 0) O (1, 0) (6, 0) x (0, -12) 3(b) –2 ⩽ x ⩽1 and x ⩾ 6 B2 B1 for –2 ⩽ x ⩽1 or x ⩾ 6 with no contradictions

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4 (a) (i) Find how many different 5-digit numbers can be formed using five of the eight digits 1, 2, 3, 4, 5, 6, 7, 8 if each digit can be used once only. [2] (ii) Find how many of these 5-digit numbers are greater than 60 000. [2] (b) A team of 3 people is to be selected from 4 men and 5 women. Find the number of different teams that could be selected which include at least 2 women. [2]

Mark scheme: 4(a)(i) 6720 B2 B1 for 8 × 7 × 6 × 5 × 4 or 8 5P 4(a)(ii) 2520 B2 B1for 3 × 7 × 6 × 5 × 4 or 3 P1 × 7 P4 4(b) 4C1 × 5C2 + 5C3 M1 50 A1

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5 DO NOT USE A CALCULATOR IN THIS QUESTION. 128 (a) Simplify . [2] 72 1 3 (b) Simplify - , giving your answer as a fraction with an integer denominator. [4] 1 + 3 3 + 2 3

Mark scheme: 5(a) 128 64 × 2 M1 = 72 36 × 2 128 16 or simplifies to 72 9 4 A1 correct completion to 3 5(b) M1 3 + 2 3 − 3 1 + 3 ( ) 1 + 3 3 + 2 3 ( )( ) 3 M1 3 + 2 3 + 3 3 + 6 3 9 − 5 3 M1 × 9 + 5 3 9 − 5 3 9 3 − 15 A1 or equivalent 6 Alternative method M1 for 1 − 3 3 ( 3 − 2 3 ) − (1 + 3 )(1 − 3 ) ( 3 + 2 3 )( 3 − 2 3 ) 1 − 3 3 3 − 6 M1 for − 1 − 3 9 − 12 M1 for writing with a common denominator 9 3 − 15 A1 for or equivalent 6

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6 (a) The curve y = a sin bx + c has a period of 180°, an amplitude of 20 and passes through the point (90°, - 3 ). Find the value of each of the constants a, b and c. [3] x (b) The function g is defined, for - 135° G x G 135°, by g ( x) = 3 tan - 4. Sketch the graph of 2 y = g ( x) on the axes below, stating the coordinates of the point where the graph crosses the y-axis. [2] y - 135° 0 135° x

Mark scheme: 6(a) a = 20 B3 B1 for each b = 2 c = −3 6(b) Correct sketch: B2 B1 for correct tan shape with one y continuous section only B1 for correct y-intercept (0, −4) x -135 135 (0, -4)

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7 Variables x and y are connected by the relationship y = Axn , where A and n are constants. (a) Transform the relationship y = Axn to straight line form. [2] When ln y is plotted against ln x a straight line graph passing through the points (0, 0.5) and (3.2, 1.7) is obtained. (b) Find the value of n and of A. [4] (c) Find the value of y when x = 11. [2]

Mark scheme: 7(a) ln y = ln( Ax n ) and so M1 ln y = ln A + ln x n ln y = ln A + n ln x A1 7(b) lnA = 0.5 M1 A = e 0.5 or 1.6 A1 n = 1.7 − 0.5 M1 3.2 − 0 3 A1 n = oe 8 7(c) their 3 M1 y = their e 0.5 (11) 8 oe 4.05 or 4.05200... rot to four or more figs A1

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8 (a) Differentiate y = tan( x + 4) - 3 sin x with respect to x. [2] ln( 2x + 5) (b) Variables x and y are such that y = 3 x . Use differentiation to find the approximate 2e change in y as x increases from 1 to 1 + h, where h is small. [6]

Mark scheme: 8(a) sec 2 ( x + 4) − 3cos x B2 B1 for each 8(b) d(ln(2x +5)) 2 B1 = dx 2 x + 5 d(2e 3 x ) 3 x B1 = 6e d x d y M1 FT their derivatives of ln(2x + 5) and 2e3x = d x  2e 3 x 3 x ln(2 x + 5)  their 2 − their 6e  2 x + 5  4e 6 x d y A1 = d x  2e 3 x 3 x ln(2 x + 5)  2 − 6e  2 x + 5  4e 6 x d y M1 δy = their × h d x x =1 −0.138h A1

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9 DO NOT USE A CALCULATOR IN THIS QUESTION. 1 6 (a) Find the term independent of x in the binomial expansion of 3x - . [2] b x l x n (b) In the expansion of 1 + the coefficient of x4 is half the coefficient of x6. Find the value of b 2 l the positive constant n. [6]

Mark scheme: 9(a) –540 B2 3 6 × 5 × 4 3  1  B1 for ( 3 x )  −  oe 3!  x  9(b) 6 B1 n ( n − 1)( n − 2)( n − 3)( n − 4)( n − 5)  1  ×  6!  2  4 B1 n ( n − 1)( n − 2)( n − 3)  1  ×  4!  2  Forms a correct equation with their M1 coefficients in terms of n Simplifies their equation to M1 (n – 4)(n – 5) = 240 or better Factorises or attempts to solve their 3-term M1 quadratic n = 20 A1

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10 Solve the equation (a) 5 sec 2 A + 14 tan A - 8 = 0 for 0° G A G 180°, [4] (b) 5 sin 4B - r + 2 = 0 for - r G B G r radians. [4] b 8 l 4 4

Mark scheme: 10(a) 5(1 + tan2A) + 14 tan A – 8 = 0 soi B1 Solves or factorises their 3-term quadratic M1 in tanA oe 11.3 and 108.4 A2 with no extras in range; or not from clearly wrong working but 11.30[99...] and 108.43[49...] rot to four or allow recovery from minor slips more decimal places or A1 for either, ignoring extras 10(b) π −1 2  B1 4 B − = sin  −  soi 8  5  −0.411[516...] rot to three or more figs M1 −0.00470[444...] A1 rot to three or more figs −0.584[344...] rot to three or more figs A1

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11 In this question all lengths are in centimetres. 1 2 The volume, V, of a cone of height h and base radius r is given by V = r r h. 3 R w 90 180 The diagram shows a large hollow cone from which a smaller cone of height 180 and base radius 90 has been removed. The remainder has been fitted with a circular base of radius 90 to form a container for water. The depth of water in the container is w and the surface of the water is a circle of radius R. (a) Find an expression for R in terms of w and show that the volume V of the water in the container is r 3 given by V = w + 180 - 486000r . [3] 12 ` j (b) Water is poured into the container at a rate of 10 000 cm3s−1. Find the rate at which the depth of the water is increasing when w = 10. [4]

Mark scheme: 11(a) 1 B1 R = ( w + 180) 2 1 2 M1 V = π ( their R ) ( w + 180 ) 3 1 2 − π ( 90 ) (180 ) 3 Correct completion to given answer: A1 π V = ( w + 180 )3 − 486000π 12 11(b) dV π B1 = 3 ( w + 180 )2 oe dw 12 dw dw dV M1 = × soi dt dV dt d w 1 M1 = × 10000 d t  d V  their    dw  w =10 0.353 [cms−1] A1 or 0.3526[97...] [cms−1] rot to four or more figs

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12 (a) (i) Given that f( x) = , show that f l ( x) = tan x sec x. [3] cos x (ii) Hence find y `3 tan x sec x - 4 e 3 xj dx. [3] 5 p (b) Given that dx = ln 2, find the value of the positive constant p. [5] y 2 px + 10

Mark scheme: 12(a)(i) −−( sin x ) B2 − sin x oe B1 for oe cos 2 x cos 2 x Correct completion to given answer: B1 dep on all previous marks having been tanxsecx awarded 12(a)(ii) 3 x B1 4 3e x = e 4 oe 3 x 3 x M1 3 4 3 4 e d x = − ke oe cos x − cos x 3 x A1 3 4 4 − e + c oe cos x 3 = ln 2 12(b) [ ln( px + 10) ]52 M1 ln(5 p + 10) − ln(2 p + 10) = ln 2 M1 ln  5 p + 10 = ln 2 M1    2 p + 10  5 p + 10 = 2(2 p + 10) M1 p = 10 A1

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