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

1 Variables x and y are such that y = sin x + e -x . Use differentiation to find the approximate change in y as x increases from r to r + ,h where h is small. [4] 4 4

Mark scheme: Question Answer Marks Partial Marks 1 dy − x B2 B1 for cos x or –e–x = cos x − e dx dy M1 δy = their × h dx x = π 4 0.251h A1

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2 DO NOT USE A CALCULATOR IN THIS QUESTION. 10 + 2 5 The point 1 - 5, p lies on the curve y = 2 . Find the exact value of p, simplifying your ` j x answer. [5]

Mark scheme: 2 2 B1 or rationalises Squares: (1 − 5 ) = 1 − 5 − 5 + 5 10 + 2 5 (1 + 5 ) 2 × (1 − 5 ) 2 (1 + 5 ) 2 Rationalises, e.g. B1 or squares 10 + 2 5 6 + 2 5 (1 + 5 ) 2 = 1 + 5 + 5 + 5 × 6 − 2 5 6 + 2 5 Multiplies out, e.g. M1 Multiplies out 60 + 20 5 + 12 5 + 4(5)  10 + 2 5 6 + 2 5   2 × 2 =  36 − 20  (1 − 5 ) (1 + 5 )  60 + 20 5 + 12 5 + 4(5) (1 − 5 ) 2 5 + 2 5 A2 A1 for k + 2 5 or 5 + k 5

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3 Find the values of k for which the line y = x - 3 intersects the curve y = k 2 x 2 + 5kx + 1 at two distinct points. [6]

Mark scheme: 3 x −=3 k 2 x 2 + 5kx + 1 M1 k 2 x 2 + ( 5k − 1) x + 4 = 0 soi A1 (5 k − 1) 2 − 4( k 2 )(4) M1 9k 2 − 10k + 1 ∗ 0 M1 Critical values: A1 1 and 1 soi 9 1 A1 k < or k > 1 9

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4 The three roots of p ( )x = 0 , where p ( )x = 2x 3 + ax 2 + bx + c are x = , x = n and x =- n, where 2 a, b, c and n are integers. The y-intercept of the graph of y = p ( )x is 4. Find p ( x), simplifying your coefficients. [5]

Mark scheme: 4 Factorised form: B1 ( x + n)( x − n)(2x −1) oe Multiplies out correctly M1 FT their factorised form provided of equivalent difficulty Correct expanded form in terms of n: A1 2 x 3 − x 2 − 2n 2 x + n 2 Uses (their n2) = 4 in their expression M1 2 x3 − x 2 − 8 x + 4 A1 If A0A0 then SC1 for ( x + n)( x − n)( x − 0.5) giving n 2 = 8 3 1 2 leading to x − x − 8 x + 4 2 Alternative method: B1 for factorised form: ( x + n)( x − n)(2x − 1) M1 for their n 2 = 4 A1 for n = 2 M1 for multiplying out ( x + their 2)( x − their 2)(2 x − 1) A1 for 2 x3 − x 2 − 8 x + 4 If A0A0 then SC1 for ( x + n)( x − n)( x − 0.5) giving n 2 = 8 3 1 2 leading to x − x − 8 x + 4 2

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5 Solutions to this question by accurate drawing will not be accepted. The points A and B are (4, 3) and (12, −7) respectively. (a) Find the equation of the line L, the perpendicular bisector of the line AB. [4] (b) The line parallel to AB which passes through the point (5, 12) intersects L at the point C. Find the coordinates of C. [4]

Mark scheme: 5(a) Finds coordinates of mid-point B1 (8, –2) 3 + 7  5  B1 m AB =  = −  oe soi 4 − 12  4  −1 M1 mL = oe − 54 4 A1 y + 2 = ( x − 8) oe isw 5 5(b) 5 B1 y − 12 = − ( x − 5) 4 Attempts to solve their equations M1 (13, 2) A2 A1 for x = 13 or y = 2

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6 (a) Find the equation of the tangent to the curve 2y = tan 2x + 7 at the point where x = r . 8 Give your answer in the form ax - y = r + c , where a, b and c are integers. [5] b (b) This tangent intersects the x-axis at P and the y-axis at Q. Find the length of PQ. [2]

Mark scheme: 6(a) dy 2 B1 = sec 2 x dx dy B1 d y their = their 2 FT their dx x = π d x 8 π B1 x = , y = 4 8  π  M1 y − their 4 = ( their 2 )  x −  oe  8  π A1 2 x − y = − 4 4 6(b) 2 2 M1  π   π   − 2  +  4 −  oe  8   4  3.59 or 3.59[03…] rot to four or more A1 figs

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7 Giving your answer in its simplest form, find the exact value of 4 10 (a) dx, [4] y 0 5 x + 2 nl 2 2 4 x + 2 (b) e d.x [5] y 0 ` j

Mark scheme: 7(a) 2ln(5x + 2) B2 B1 for kln (5x + 2) 2 ( ln(22) − ln(2) ) oe soi M1 2 A1 2ln11 or ln121 or ln11 e 8 x + 4 d x M17(b)  ln 2 M1  1 8 x+ 4  e    8  0 1 ln2 8 4 4 M2 1 ln2 ( e × e − e ) oe M1 for ( e 8 + 4 − e 4 ) 8 8 A1 255e 4 or exact equivalent 8

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8 (a) Solve 3 cot 2 x - 14 cosec x - 2 = 0 for 0° 1 x 1 360°. [5] sin 4 y - cos 4 y (b) Show that = tan y - 2 cos y sin y. [4] cot y

Mark scheme: 8(a) 3(cosec 2 x − 1) − 14cosec x − 2 [ = 0 ] M1 3cosec2 x − 14cosec x −=5 0 A1 (cosecx – 5)(3cosecx + 1) M1 1 A1 sinx = nfww 5 11.5 and 168.5 nfww A1 8(b) Correct use of sin 2 y + cos 2 y = 1 B1 Factorises using the difference of 2 B1 squares 1 cos y B1 Uses = tan y or cot y = cot y sin y correctly Full and correct completion to given B1 answer: tan y − 2cos y sin y

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9 (a) Solve the equation x - 2 = 243. [3] 27 1 (b) log a b - = log b a, where a 2 0 and b 2 0. 2 Solve this equation for b, giving your answers in terms of a. [5]

Mark scheme: 9(a) 310 x B1 3 x− 6 [ = 243] oe or 3 log 9 5 x − log 27 x− 2 = log 243 oe 37 x+ 6 = 35 soi oe or M1 5 x ( log9 ) − ( x − 2 ) log27 = log243 1 A1 x = − 7 9(b) 1 1 1 B2 1 log a b − = B1 for bringing down the power of 2 2 log a b 2 1 1 e.g. log a b or for a change of base 1 2 2 or − = log b a log b a 2 1 e.g. loga b Clears the fraction and rearranges M1 1 2 1 ( log a b ) − log a b = 1 oe 2 2 ( log a b ) 2 − log a b − 2 = 0 oe or let x = log a b x 2 − x − 2 = 0 oe or 1 1 2 − log b a = (log b a ) 2 2 0 = 2(log b a ) 2 + log b a − 1 oe or let y = log b a 2 y 2 + y −=1 0 (log a b − 2)(log a b + 1) oe or M1 (2logb a − 1)(logb a + 1) [log a b = 2, log a b = −1 or A1 1 log b a = , log b a = −1 2 leading to ] b = a2, b = oe

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10 (a) The first 5 terms of a sequence are given below. 4 - 2 1 - .05 0.25 (i) Find the 20th term of the sequence. [2] (ii) Explain why the sum to infinity exists for this sequence and find the value of this sum. [2] (b) The tenth term of an arithmetic progression is 15 times the second term. The sum of the first 6 terms of the progression is 87. (i) Find the common difference of the progression. [4] (ii) For this progression, the nth term is 6990. Find the value of n. [3] Question 11 is printed on the next page.

Mark scheme: 10(a)(i) 4 × ( − 0.5)19 M1 1 6 A1 − or −7.63 ×10− or 131072 −7.62939... ×10−6 rot to four or more figs 10(a)(ii) Valid explanation e.g. B1 the common ratio is between −1 and 1 4 8 B1 = 1 −−( 0.5) 3 10(b)(i) a + 9d = 15( a + d ) B1 6 B1 { 2 a + 5 d } = 87 2 Solves their equations for d e.g. M1  3  2  − d  + 5d = 29  7  d = 7 A1 10(b)(ii) a = − 3 soi B1 6990 = their ( −3) + ( n − 1)(their 7) M1 n = 1000 A1

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11 A C1 C2 B The circles with centres C1 and C2 have equal radii of length r cm. The line C1C2 is a radius of both circles. The two circles intersect at A and B. (a) Given that the perimeter of the shaded region is 4r cm, find the value of r. [4] (b) Find the exact area of the shaded region. [4]

Mark scheme: 11(a) 4 B2 2 [perimeter =] πr soi B1 for angle ACB = π 3 3  4 M1  t h e ir π r = 4π oe  3  r = 3 A1 11(b) 1 2 2π M1 × their 3 × their oe 2 3 1 2 2π M1 × their 3 × sin their oe 2 3 For subtracting and doubling: M1 2 2π their 3 × their − 3 2 2π their 3 × sin their 3 9 A1 6π − 3 or exact equivalent 2

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