Cambridge IGCSE Mathematics - Additional 0606 — 2020 Oct/Nov Paper 2 · Variant 3

2 Solve the simultaneous equations. x 2 + 3xy = 4 2x + 5y = 4 [5]

Mark scheme: 2 2  4 − 2 x  M1 eliminate x or y x + 3 x   = 4  5  x 2 − 12 x + 20 ( = 0 ) A1 3 terms on one side if eliminating y 5 y 2 + 16 y ( = 0 ) if eliminating x ( x − 2 )( x − 10 ) ( = 0 ) M1 or y ( 5 y + 16 ) ( = 0 ) x = 2 or x = 10 nfww A1 or correct pair 16 A1 y = 0 or y = − nfww 5

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3 Find the values of k for which the equation x 2 + ( k + 9) x + 9 = 0 has two distinct real roots. [4]

Mark scheme: 3 use b 2 − 4 ac ( k + 9 ) 2 −×4 9 ( > 0) M1 k 2 + 18 k + 45 ( > 0 ) A1 k =−15 k =−3 A1 k < −15 or k > −3 no isw A1 not ‘and’ mark final answer A0 if combined as one statement

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4 It is given that y = ln ( 1 + sin x) for 0 1 x 1 r. d y. [2] (a) Find d x d y r 1(b) Find the value of when x = , giving your answer in the form , where a is an integer. d x 6 a [2] d y (c) Find the values of x for which = tan x . [5] d x

Mark scheme: 4(a) dy 1 M1 = dx 1 + sinx cosx A1 × cosx = 1 + sinx 4(b) π dy M1 insert into their 6 dx 1 A1 3 not 3 3 4(c) cos x sin x M1 sinx their = replace tan x with 1 + sin x cos x cos x use cos 2 x = 1 − sin 2 x M1 earned when equation reduced to a 2 quadratic in sinx 2sin x + sin x −=1 0 ( ) ( 2sin x − 1)( sin x + 1) = 0 M1 solve three term quadratic in sinx π A1 or 0.524 or better radians only x = 6 if M0 M0 M0 and (a) and (b) correct, allow SC2 for 1 π tanx = , x = 3 6 5 π A1 or 2.62 or better radians only x = A0 if extra solution(s) in range 6

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5 Solve the following simultaneous equations. 3 x # 9 y - 1 = 243 1 2 x + 1 y - 2 2 8 # 2 = [5] 4 2

Mark scheme: 5 express an equation correctly in powers M1 of 3 or powers of 2 x + 2 y − 2 = 5 oe (x + 2y = 7) A1 accept unsimpified 2 x + 1 − 2.5 = 3 + y − 0.5 oe (2x – y = 4) A1 accept unsimplified solve correct equations for x or y M1 x = 3 and y = 2 A1

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6 A 4-digit code is to be formed using 4 different numbers selected from 1, 2, 3, 4, 5, 6, 7, 8 and 9. Find how many different codes can be formed if (a) there are no restrictions, [1] (b) only prime numbers are used, [1] (c) two even numbers are followed by two odd numbers, [2] (d) the code forms an even number. [2]

Mark scheme: 6(a) 3024 B1 6(b) 24 B1 6(c) 4P2 × 5P2 M1 4×3 × 5×4 240 no isw A1 6(d) 4P1 × 8P3 M1 4 × 8×7×6 1344 no isw A1

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7 A curve has equation y = x cos x . d y. [2] (a) Find d x (b) Find the equation of the normal to the curve at the point where x = r , giving your answer in the form y = mx + c . [4] r (c) Using your answer to part (a), find the exact value of x sin x d x . [5] 6y0

Mark scheme: 7(a) − x sin x + cos x isw B2 accept unsimplified if incorrect allow B1 for d ( cos x ) = − sin x clearly seen d x 7(b) x = π, y = −π B1 or –3.14 or better d y B1 d y x = π, = − 1 from correct d x d x gradient of normal =1 M1 use m1m2 = –1 with their grad of tangent y = x −π2 cso A1 or y = x − 6.28 or better fully correct solution their ( a ) = x cos x7(c)  M1 * − sin x + cos xd x = x cos x (  ) B1 clearly seen anywhere cos xdx = sin x  − x cos x + sin x A1 implies previous marks if (a) is correct π M1 * dep insert into their integral 6 1 3 A1 reject decimals −π 2 12

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8 DO NOT USE A CALCULATOR IN THIS QUESTION. log 2 ( y + 1) = 3 - 2 log 2 x log 2 ( x + 2) = 2 + log 2 y (a) Show that x 3 + 6x 2 - 32 = 0 . [4] (b) Find the roots of x 3 + 6x 2 - 32 = 0 . [4] (c) Give a reason why only one root is a valid solution of the logarithmic equations. Find the value of y corresponding to this root. [2]

Mark scheme: 8(a) x 2 ( y + 1) = 8 oe B1 x + 2 = 4 y oe B1 2  x + 2  M1 eliminate y from correct equations x  + 1  = 8  4  x 3 + 6 x 2 − 32 = 0 A1 answer given 8(b) x = 2 or x = −4 seen B1 or ( x − 2 ) or ( x + 4 ) seen find quadratic factor M1 x2 and 16 or long division to x2 + kx or x2 and –8 or long division to x2 + kx not from expanding two linear factors 2 2 A1 x + 8 x + 16 or x + 2 x − 8 ( ) ( ) 2 A1 answer only without working earns B1 ( x − 2)( x + 4) and x = 2, − 4, − 4 above only 8(c) no real value for log 2 ( −4 ) B1 must identify specific term in one of original equations and use x = –4 or log 2 ( −+4 2 ) y = 1 B1

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9 x m B C D 300 m A E 400 m The rectangle ABCDE represents a ploughed field where AB = 300 m and AE = 400 m . Joseph needs to walk from A to D in the least possible time. He can walk at 0.9 ms -1 on the ploughed field and at 1.5 ms -1 on any part of the path BCD along the edge of the field. He walks from A to C and then from C to D. The distance BC = x m . (a) Find, in terms of x, the total time, T s, Joseph takes for the journey. [3] (b) Given that x can vary, find the value of x for which T is a minimum and hence find the minimum value of T. [6]

Mark scheme: 9(a) 2 2 B1 ( AC = ) 300 + x seen isw 2 2 M1 using clearly indicated value for their 300 + x time for AC = oe AC or their CD 0.9 400 − x or time for CD = oe 1.5 2 2 A1 300 + x 400 − x T = + oe seen 0.9 1.5 isw 9(b) − 1 B2 accept unsimplified; 2 2 if incorrect allow B1 for correct dT 1 ( 300 2 + x 2 ) = × 2 x − oe 1 dx 2 0.9 3 2 2 ± 2 differentiation of 300 + x ( ) dT M1 d T set their = 0 must be a function of x dx d x 2 2 2 A1 equation in x2 with square root removed 25 x = 9 300 + x oe ( ) x = 225 (m) A1 T = 533 (s) or 1600/3 (exact value) A1 or 8 min 53 s

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10 (a) The sum of the first 4 terms of an arithmetic progression is 38 and the sum of the next 4 terms is 86. Find the first term and the common difference. [5] (b) The third term of a geometric progression is 12 and the sixth term is - 96 . Find the sum of the first 10 terms of this progression. [6] Question 11 is printed on the next page.

Mark scheme: 10(a) use S4 or S8 M1 4 A1 accept unsimplified S 4 = [ 2 a + 3d ] = 38 ( 2 a + 3d = 19 ) 2 8 A1 accept unsimplified S 8 = [ 2 a + 7 d ] = 38 + 86 ( 2 a + 7 d = 31) 2 or 8 4 S8 – S4 = [ 2 a + 7 d ] − [ 2 a + 3 d ] = 86 2 2 ( 4 a + 22 d = 86 ) solve correct equations for a or d M1 a = 5 and d = 3 A1 10(b) ar 2 = 12 soi B1 ar 5 = −96 soi B1 solve correct equations for a or r M1 r = − 2 and a = 3 A1 insert their a and r into S10 M1  10  3 1 −−( 2 )  ( )  S10 = )    1 −−( 2 )    –1023 A1

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11 DO NOT USE A CALCULATOR IN THIS QUESTION. Solve the quadratic equation 7 - 2 x 2 - 4x + 7 + 2 = 0 , giving each of your answers in the form ` j ` j a + b 7 , where a and b are constants. [7]

Mark scheme: 7 − 2 7 + 2 = 3 soi B1 seen anywhere11 ( )( ) use quadratic formula to solve for x M1 A1 4 ± 16 − 4 7 − 2 7 + 2 ( )( ) x = 2 7 − 2 ( ) 4 ± 2 A1 or 4 ± 4 in numerator x = 2 7 − 2 ( ) rationalise one of their solutions B1 full rationalisation statement must be shown 7 + 2 4 + 2 ( ) e.g. × 2 7 − 2 7 + 2 ( ) ( ) x = 2 + 7 nfww A1 2 1 A1 2 + 7 x = + 7 nfww accept 3 3 3

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