Cambridge IGCSE Mathematics - Additional 0606 — 2023 Feb/March Paper 2 · Variant 2

1 On the axes below, sketch the graph of y = 4 cos 2x for 0 G x G r , giving the coordinates of any points where the graph meets the axes. [3] y 0 r x

Mark scheme: Question Answer Marks Partial Marks 1 Correct sketch 3 B1 for correct shape, with 3 y consistent maxima, 2 cusps on the x-axes and reasonable 4 symmetry 0 π 3π π x  π   3π  4 4 B1 for  , 0  and  , 0   4   4  either seen on the graph or stated; must have attempted a graph of correct shape B1 for starts with (0, 4) and ends with (, 4) and (0, 4) either seen on the graph or stated; must have attempted a graph correct shape

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2 DO NOT USE A CALCULATOR IN THIS QUESTION. x 11 2 Expand and simplify , giving your answer with a rational denominator. [4] e 2 3 - 1 o

Mark scheme: 2 11x 2 B1 12 + 1 − 4 3 2 M1 FT their expression of 11x 13 + 4 3 ( ) equivalent difficulty 13 − 4 3 13 + 4 3 ( )( ) A1 11x 2 13 + 4 3 2 2 ( ) 143 x + 44 3 x or 169 − 48 169 − 48 A1 mark final answer x 2 13 + 4 3 2 2 ( ) 13 x + 4 3 x or 11 11 Alternative method 2 (M1)  x 11(2 3 + 1)       (2 3 − 1)(2 3 + 1)  2 2 (A1)  x 11(2 3 + 1)   2 33 x + x 11    or       12 − 1 12 − 1     11x 2 (12 + 1 + 4 3) 132 x 2 + 11x 2 + 4 363 x 2 (A1) or 121 121 2 (A1) mark final answer x 13 + 4 3 ( ) 11

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3 Solve the inequality 5x + 4 G 2x - 3 . [4]

Mark scheme: 3 (5x + 4)2 * (2x –3)2 soi M1 where * is any inequality sign or = 21x 2 + 52 x + 7*0 A1 1 7 A1 Critical values: − , − soi 7 3 7 1 A1 FT their derived critical − x − mark final answer values 3 7 Alternative method 5 x + 4*2 x − 3 oe soi (M1) and 5 x + 4*3 − 2 x oe soi where * is any inequality sign or = 1 7 (A2) 1 7 Critical values: − , − soi A1 for − or − 7 3 7 3 7 1 (A1) FT their derived critical − x − mark final answer values 3 7

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4 y = cosec 5x r Show that y = a sin bx , where a and b are integers, and hence find the value of y d x . [4] 5y0

Mark scheme: 4 1 B1 [y =] = sin5x nfww cosec5 x π B1 FT their asin5x π 5  cos5 x  5  y0 dx =  −  5  0 1  π   1  M1 FT their a(kcosbx) where − cos  5   −− cos ( 5  0 )  1 5  5   5  k < 0 or k = 5 2 A1 5

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5 DO NOT USE A CALCULATOR IN THIS QUESTION. (a) Show that x - 1 is a factor of the expression x 3 - 2 x 2 - 19x + 20 . [1] (b) Hence write x 3 - 2x 2 - 19x + 20 as a product of its linear factors. [3] (c) Hence find the exact solutions of the equation e 3 y - 2e 2 y - 19e y + 20 = 0 . [2]

Mark scheme: 5(a) 13 – 2(12) – 19 + 20 = 0 1 5(b) (x – 1)(x2 – x – 20) M2 M1 for two terms correct in the quadratic factor (x – 1)(x + 4)(x – 5) A1 5(c) e y = 1, e y = 5 M1 y = 0, y = ln5 mark final answer A1 1.61 or decimal equivalent for ln5 seen is A0 as calculator use not permitted

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6 (a) A geometric progression has first term 64 and common ratio 0.5. (i) Find the 10th term. [2] (ii) Find the sum of the first 10 terms. [2] (iii) Find the sum to infinity. [1] (b) An arithmetic progression is such that S 20 - 400 = 2 S 10 and u 1| u 6 is 1 : 5. Find the sum of the first 3 terms of this progression. [6]

Mark scheme: 6(a)(i) 1 2 M1 for 64(0.5) 9 oe or 0.125 8 6(a)(ii) 1023 2 64(1 − 0.510 ) or 127.875 M1 for oe 8 1 − 0.5 6(a)(iii) 128 1 6(b) 20 10 M2 20 2a + 19d  − 400 = 2  2a + 9d  M1 for  2 a + 19 d  or 2 2 2 oe, soi 10 2 a + 9 d  soi 2 5a = a + 5d soi M1 d = 4 A1 a = 5 A1 27 nfww B1 must have earned all previous marks

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7 (a) Variables x and y are such that y = . Use differentiation to find the approximate tan x r r change in y as x increases from to + h , where h is small. [5] 4 4 2 1 d y 1 d y ( x + 1)( x - 4)(b) Given that y = 3 show that y - - 2 can be written as 5 . [4] ( x - 3) dx 3 f d x p ( x - 3)

Mark scheme: 7(a) d 2 B1 (cos x ) = −2cos x sin x soi dx Attempts the quotient rule M1 d 2 2 2 FT their (cos x ) dy −2cos x sin x tan x − (1 + cos x )sec x dx = dx tan 2 x Fully correct isw A1 d 2 FT their (cos x ) only dx δy dy M1  their h dx x = π 4 δy − 4 h cao A1 7(a) Alternative method 1 d 3 2 (B1) (cos x ) = −3cos x sin x soi dx Attempts the quotient rule: (M1) d 3 2 3 FT their (cos x ) dy (sin x )( − sin x − 3cos x sin x ) − (cos x + cos x )cos x dx = dx sin 2 x Fully correct isw (A1) d 3 FT their (cos x ) only dx δy dy (M1)  their h dx x = π 4 δy − 4 h cao (A1) Alternative method 2 (B1) d  −2 sec 2 x  2 = −2(tan x ) dx  tan x  Attempts the product rule (M1) d  2  FT their   dy d x  tan x  = −2(tan x ) −2 sec 2 x − (sin x ( − sin x ) + cos x (cos x )) dx Fully correct isw (A1) d  2  FT their   only d x  tan x  δy dy (M1)  their h dx x = π 4 δy − 4 h cao (A1) 7(b) dy −4 B1 = −3( x − 3) oe, soi dx d 2 y −5 B1 = −−4 3( x − 3) oe, soi dx 2 ( x − 3) 2 + 3( x − 3) − 4 M1 dy −4 FT = k ( x − 3) ( x − 3) 5 dx d 2 y −5  x −+3 3 4  x ( x − 3) − 4 and = m ( x − 3) where k 2 or  4 − 5 =  5 dx  ( x − 3) ( x − 3)  ( x − 3) and m are constants Correct completion to given answer: A1 x 2 − 3 x − 4 ( x + 1)( x − 4) = ( x − 3)5 ( x − 3)5

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8 The function f is defined for x H 0 by f ( x) = 5 - 2e -x . (a) (i) Find the domain of f -1 . [2] (ii) Solve ff - 1 ( x) = 5 x - 4 . [3] (iii) On the axes, sketch the graph of y = f ( x) and hence sketch the graph of y = f -1 ( x) . Show clearly the positions of any points where your graphs meet the coordinate axes and the positions of any asymptotes. [4] y x O 3(b) The function g is defined for 0 G x G 0.2 by g ( x) = . 1 - x Find and simplify an expression for f - 1 g ( x) . [4]

Mark scheme: 8(a)(i) 3 ⩽ x < 5 B2 B1 for x ⩾ 3 or for x < 5 or for 3 and 5 in an incorrect inequality 8(a)(ii) x = 5 x − 4 and rearrangement to B1 x 2 − 5 x + 4  = 0 Factorises x 2 − 5 x + 4 or solves their x 2 − 5 x + 4 = 0 M1 x = 4 only, nfww A1 8(a)(iii) Correct pair of graphs. 4 B1 for correct shape for f; may not be over correct y y = f − 1 ( x ) domain but must have positive y-intercept and appear to tend to an asymptote in the 1st quadrant 5 y = f ( x ) B1 for (0, 3) and f in 1st 3 quadrant only; must have attempted correct shape x O 3 5 B1 for asymptote at y = 5; must have attempted correct y = x shape B1 for a correct reflection of their f in the line y = x Maximum of 3 marks if not fully correct 8(b) −1 5 − x −1 2 2 M1 for a complete attempt to f ( x ) = − ln or f ( x ) = ln oe find the inverse function with 2 5 − x at most one sign or arithmetic error: Putting y = f(x) and changing subject to x and swopping x and y or swopping x and y and changing subject to y −1 2 − 5 x 2 M1 FT for a correct Correct simplified form e.g.  f g( x ) =  − ln   unsimplified form of the 2(1 − x ) function; FT providing of −1 2 − 2 x or  f g( x ) =  ln equivalent difficulty   2 − 5 x

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9 In this question, all lengths are in centimetres and all angles are in radians. (a) C A 3r 8 D B O The diagram shows sectors AOB and COD of two circles with the same centre, O. Angle AOB is 3 r and the length of OC is 6.5. It is given that OAC and OBD are straight lines and 8 OA : OC is 4 : 5. Find the perimeter of the shaded region. [3] (b) Q a y P O z R The diagram shows a circle with centre O and radius a. Sector PQR is a sector of a different circle with centre R and radius y. Angle OPR is z. Find, in terms of a and z only, the total area of the three shaded regions. Simplify your answer. [4]

Mark scheme: 9(a)  3π   3π  M2  3π  6.5   + 5.2   + 2(6.5 − 5.2) M1 for 6.5    8   8   8   3π  or 5.2    8  16.38 to 16.4 A1 9(b) [Angle PRQ = ] 2 soi B1 a sin(π − 2) B1 y = 2a cosoe or y = oe sin y 2 = a 2 + a 2 − 2 a 2 cos(π − 2) oe or y 2 = a 2 + a 2 + 2 a 2 cos(2) oe Complete and correct plan soi: M1 FT their 2and their 2 1 2 expression for y or y2 in terms πa − (2 a cos) (2) oe 2 of a and  2 1  a sin(π − 2)  2 or πa −   (2) oe 2  sin  2 1 2 2 2 or πa − ( a + a − 2a cos(π − 2))(2) oe 2 2 1 2 2 2 or πa − ( a + a + 2a cos(2))(2) 2 2sin 2 (π − 2) A1 2 2 2 a π − 4cos  or πa −a ( ) sin 2  or πa 2 − 2( a 2 − a 2 cos(π − 2)) or πa 2 − 2( a 2 + a 2 cos2) oe

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10 A particle P moves in a straight line such that, t seconds after passing a fixed point O, its acceleration, a ms -2 , is given by a = 6t for 0 G t G 3, 18e 3 a = t for t H 3. e When t = 1, the velocity of P is 2 ms -1 and its displacement from O is – 4 m. (a) (i) Find the velocity of P when t = 3 . [3] (ii) Find the displacement of P from O when t = 3 . [3] (b) Find an expression in terms of t for the displacement of P from O when t H 3 . [4] Question 11 is printed on the next page.

Mark scheme: 10(a)(i) v = 3t 2 + c M1 v = 3t 2 − 1 A1 When t = 3 v = 26 A1 10(a)(ii) s = t 3 −+t c M1 FT kt 2 + c s = t 3 −−t 4 A1 When t = 3 s = 20 A1 10(b) −18e 3 M1 v = + c oe e t −18e 3 A1 FT (their 26) + 18 v = + 44 oe e t 18e 3 M1 dep on previous M1 s = + their 44t + d oe e t 18e 3 A1 s = + 44t − 130 oe, cao e t

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11 The normal to the curve y = sin ( 4 x - r ) at the point A(a, 0), where r 1 a 1 r , meets the y-axis at 2 the point B. Find the exact area of triangle OAB, where O is the origin. [9]

Mark scheme: 11 Solves sin(4 x − π) = 0 oe M1 3π A1 a = 4 dy B2 dy = 4cos(4 x − π) B1 for = k cos(4 x − π) , dx dx where k > 0 or k = −4  −1  1 B2 nπ = − FT their a = , n is a  3π  4 4  4cos(4  their − π)   4  positive integer B1 for −1 3π theirk cos(4  their − π) 4 1  3π  1  3π  M1 FT their perpendicular y − 0 = −  x −  or 0 = −   + coe gradient and their a 4  4  4  4   3π  B1 B  0,  soi  16  9π 2 B1 [Exact area =] 128

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