Cambridge IGCSE Mathematics - Additional 0606 — 2022 Oct/Nov Paper 2 · Variant 1

1 Solve the following simultaneous equations, giving your answers in the form a + b 7 where a and b are integers. x + 3y = 11 x - 7 y = 7 [5]

Mark scheme: Question Answer Marks Guidance 1 Finds by elimination 3 y + 7 y = 4 oe M1 or substitutes x = 11 − 3 y into x − 7 y = 7 oe OR Finds by elimination 3 y + 7 y = 21 + 11 7 oe 11 − x or substitutes y = into x − 7 y = 7 3 oe 4 A1 y = 3 + 7 21 + 11 7 or x = 3 + 7 4 3 − 7 M1 FT their value of x or y providing of y =  oe equivalent difficulty 3 + 7 3 − 7 21 + 11 7 3 − 7 or x =  oe 3 + 7 3 − 7 y = 6 − 2 7 and x = 6 7 − 7 A2 A1 for either and no extra values

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2 DO NOT USE A CALCULATOR IN THIS QUESTION. Find the x-coordinates of the points where the line y = 3x - 8 cuts the curve y = 2x 3 + 3x 2 - 26x + 22 . [5]

Mark scheme: 2 2 x3 + 3x 2 − 29 x + 30  = 0 B1 Uses a correct factor x – 2 or x + 5 M1 to find a quadratic factor with at least 2 terms correct ( x − 2) → (2 x 2 + 7 x − 15)  = 0  A1 or ( x + 5) → (2 x 2 − 7 x + 6)  = 0  Factorises or solves their 3-term M1 dep on previous M1 quadratic: (x + 5)(2x – 3) [= 0] or (x – 2)(2x – 3) [= 0] −7 7 2 − 4(2)( −15) or  x =  2(2) 7  ( −7) 2 − 4(2)(6) or 2(2) x = 2, −5, 1.5 A1

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3 (a) Find the coordinates of the point on the curve y = 1 + 3 x where the gradient of the normal 8 is - . [5] 3 (b) Find the equation of the normal to the curve y = 1 + 3 x at the point (8, 5) in the form y = mx + c . [3]

Mark scheme: 13(a) − dy 1 − 12 B2 dy 1 = (1 + 3 x )  3 oe, isw B1 for = (1 + 3 x ) 2  ... dx 2 dx 2 dy 1 − 12 or = (...)  3 d x 2  1 1 − dy 1  their 2  or = their (1 + 3 x )  3 d x 2 dy − 12 or = k (1 + 3 x )  3, k is a constant, dx 1 k ≠ 2 3 B1 d y mtangent = or 0.375 FT their if necessary providing at 8 d x −2 least B1 previously awarded or mnormal = 1 oe − 3(1 + 3 x ) 2  dy 3   dx   8 M1 d y FT their if necessary providing at their − = −  = or   their   d x dx   8 dy 3     least B1 previously awarded (5, 4) A1 3(b) 10 B1 mnormal = − 3 10 M1 FT their mnormal y – 5 = − (x – 8) 3 −10  −10  or y = x + c and 5 =   (8) + c 3  3  oe soi 10 95 A1 FT their mnormal y = − x + 3 3

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4 Solve the following equations, giving your answers to 3 significant figures. (a) 2 3 x + 1 = 5 x - 2 [3] 2y + 1 6 (b) e = 1 + 2 y + 1 [4] e

Mark scheme: 4(a) (3x + 1)log2 = (x – 2)log5 oe B1 (3log2 – log5)x = −log2 – 2log5 M1 FT if of equivalent difficulty x = −8.32 A1 4(b) Writes as a quadratic in e2y + 1 M1 condone one error or states u = e2y + 1 and writes as a quadratic in u oe, soi 2 y +1 ( e ) 2 − e 2 y +1 − 6  = 0 oe A1 or u 2 − u − 6 [ = 0] oe (e 2 y +1 + 2)(e 2 y +1 − 3)  = 0  leading to A1 e 2 y +1 = 3 or (u + 2)(u – 3) [=0] leading to e 2 y +1 = 3 y = 0.0493 and no other solutions A1

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5 You are given that y = . cos 2x d y k sin 2x (a) Show that = 2 where k is a constant to be found. [2] d x cos 2x d y 5 r (b) Find the values of x such that = for 0 1 x 1 . [4] d x sin 2x 2

Mark scheme: 5(a) dy −2 2sin2 x B2 B1 for − (cos2 x ) −2  m sin 2 x = −(cos2 x ) −2sin2 x = dx cos 2 2 x 0[cos2 x ] − ( m sin2 x ) or where m = 2 dy 0[cos2 x ] −−( 2sin2 x ) 2sin2 x 2 or = = cos 2 x dx cos 2 2 x cos 2 2 x or m < 0 5(b) 2tan22x = 5 M1 FT their k or 7cos22x = 2 or 7sin22x = 5 5 A1 tan2 x =  2 2 or cos2 x =  7 5 or sin2 x =  7 0.503 or 0.5034[26…] rot to 4 or more sf A2 A1 for either, ignoring extras 1.07 or 1.067[36…] rot to 4 or more sf and no extras in range

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6 (a) Write 3x 2 + 15 x - 20 in the form a ( x + b) 2 + c where a, b and c are rational numbers. [4] (b) State the minimum value of 3x 2 + 15x - 20 and the value of x at which it occurs. [2] 2 1 (c) Use your answer to part (a) to solve the equation 3y 3 + 15y 3 - 20 = 0 , giving your answers correct to three significant figures. [3]

Mark scheme: 6(a) 2 B4 2  5  155  5  3  x +  − B2 for 3  x +  or 3 ( x + 2.5 )2  2  4  2   5  2 or B1 for  x +  or ( x + 2.5 )2  2  155 B2 for c = − or −38.75 4 25 or B1 for − −3 20 oe 4 6(b) 155 5 B2 FT their c from part a and −their b from Min value − when x is − 4 2 (a) B1 for either without contradiction 6(c) 1 2 M1 FT an expression of correct form from (a)   155 3 5 3  y +  = soi  2  4 1 A1 3 5 155 Rearranges as far as: y = −  2 12 soi y = 1.31 or −226 A1

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7 The sum of the first three terms of a geometric progression is 17.5 and the sum to infinity is 20. Find the first term and the common ratio. [6]

Mark scheme: 7 a (1 − r 3 ) 2 B1 = 17.5 oe or a + ar + ar = 17.5 1 − r oe a B1 = 20 1 − r Correctly eliminates a or eliminates r M1 FT their equations providing at least B1 awarded 20(1 – r3) = 17.5 A1 or a 3 − 60a 2 + 1200a − 7000 = 0 1 A2 A1 for either r = , a = 10 2

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8 The equation of a curve is y = x sin x . d y (a) Find . [2] d x (b) Find the equation of the tangent to the curve at x = r in the form y = mx + c . [3] 2 (c) Use your answer to part (a) to find x cos x dx . [3] y r (d) Evaluate x cos x dx , giving your answer correct to 2 significant figures. [2] 4y0

Mark scheme: 8(a) Product rule attempted M1 at most one error  dy  A1 = sin x + x cos x oe    dx  8(b)  π  π B1 When x = y =    2  2  π  dy M1 FT their derivative providing at least M1 When x = = 1   awarded in (a)  2  dx y = x A1 8(c) xsinx + cosx + c B3 B2 for xsinx + cosx or B1 for  x cos xdx =  sin xdx x sin x −     8(d) π π π M1 sin + cos − ( 0 + cos0 ) 4 4 4 0.26 A1

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9 The functions f ( x) and g ( x) are defined as follows for x 2- by 3 f ( x) = x 2 + 1, g ( x) = ln ( 3 x + 2) . (a) Find fg ( x) . [1] (b) Solve the equation fg ( x) = 5 giving your answer in exact form. [3]

Mark scheme: 9(a) (ln(3x + 2))2 + 1 oe isw B1 9(b) ln(3x + 2) = [] 2 B1 e2 = 3x + 2 M1 FT ln(3x + 2) = k, where k > 0 2e − 2 A1 x = as only solution 3 9(c) ln(3ln(3x + 2) + 2) B1 their(3ln(3x + 2) + 2) = e M1 FT their gg(x) with at most one error 3ln(3x + 2) + 2 = e A1 e − 2 M1 FT their aln(3x + 2) + b = e, where a and ln(3x + 2) = or 0.239[42…] b are non-zero constants 3 e − 2 A1 3x + 2 = e 3 or 3x + 2 = 1.270[52…] awrt −0.243 A1

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= 0 the10 The acceleration, a ms -2 , of a particle at time t seconds is given by a =- 2 . When t ( t + 1 ) -1 velocity of the particle is 50 ms . (a) Find an expression for the velocity of the particle in terms of t. [4] (b) Find the distance travelled by the particle between t = 1 and t = 10 . [4]

Mark scheme: 10(a)  −45  −45(t + 1) −1 B2  −45  −1 dt = k (t + 1) dt = + C or B1 for 2   2   v =  (t + 1) v =  (t + 1)     −1 better their 45 M1 50 = + C 0 + 1 A1  v = 45 + 5 t + 1 10(b)  F(t ) =  45ln(t + 1) + 5t 101 B2 B1 for (their 45)ln(t + 1) F(10) – F(1) M1 dep on at least B1 122 (m) or 121.7[13…] rot to 4 or more sf A1 dep on all previous marks awarded

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11 A 5-digit code is to be formed using 5 different numbers selected from 1, 2, 3, 4, 5, 6, 7, 8. Find how many possible codes there are if the code forms (a) a number less than 60 000 that ends in a multiple of 3, [3] (b) an even number less than 60 000. [3]

Mark scheme: 11(a) 1080 3 M2 for a fully correct method e.g. [starts with 1, 2, 4, 5 and ends in 3, 6] 4  6  5  4  2 or 960 and [starts with 3 and ends in 6] 1  6  5  4  1 or 120 OR [ends with 6 and starts with 1, 2, 3, 4, 5] 5  6  5  4  1 or 600 and [ends with 3 and starts with 1, 2, 4, 5] 4  6  5  4  1 or 480 or M1 for a partially correct method equivalent to one of the above two steps 11(b) 2160 3 M2 for a fully correct method e.g. [starts with 1, 3, 5 and ends in 2, 4, 6, 8] 3  6  5  4  4 or 1440 and [starts with 2 or 4 and ends in 6, 8 or one of 2 or 4] 2  6  5  4  3 or 720 OR [ends with 6, 8 and starts with 1, 2, 3, 4, 5] 5  6  5  4  2 or 1200 and [ends with 2, 4 and starts with 1, 3, 5 or one of 2 or 4] 4  6  5  4  2 or 960 or M1 for a partially correct method equivalent to one of the above two steps

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