Cambridge IGCSE Mathematics - Additional 0606 — 2023 Oct/Nov Paper 2 · Variant 2

1 (a) A straight line passes through the points (4, 23) and (-8, 29). Find the point of intersection, P, of this line with the line y = 2x + 5 . [5] (b) Find the distance of P from the origin. [2]

Mark scheme: Question Answer Marks Guidance 1(a) 1 3 29 − 23 1 y = − x + 25 isw M1 for m = oe or − 2 −−8 4 2 and y − 23  1  M1 FT for = their  −  oe x − 4  2  or 1 1 y = their − x + c and 23 = −  4 + c oe ( 2 ) 2 OR M1 for solving 23 = 4m + c 29 = –8m + c 1 for m = − or c = 25 2 and M1 FT for correctly using their m or their c to find c or m Solves their linear equation simultaneously M1 1 FT their y = − x + 25 oe with y = 2x + 5 to find x or y 2 (8, 21) A1 1(b) 2 2 M1 FT their (8, 21) 8 + 21 oe 505 isw or 22.5 A1 or 22.47[22…] rot to 2 or more dp

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2 Find the non-zero value of k for which the line y =- 2x - 6k - 1 is a tangent to the curve y = x ( x + 2k). [5]

Mark scheme: 2 x 2 + 2 kx = −2 x − 6 k − 1 M1 x 2 + ( 2k + 2 ) x + 6k + 1 = 0 A1 Correctly uses b2 – 4ac [*0] for their M1 where * is any inequality sign or =; equation FT their 3-term quadratic in x and k (2k + 2)2 − 4(6k + 1) [*0] 4k2 − 16k [*0] nfww A1 k = 4 A1 dep on all previous marks awarded 2 Alternative method 2x + 2k (M1) k = –x – 1 or x = –k – 1 oe (A1) –2(–k – 1) – 6k – 1 = (–k – 1)2 + 2k(–k – 1) (M1) FT their k of the form ax + b where a and b oe are non-zero constants or –2x – 6(–1 – x) – 1 = x (x + 2(–1 – x)) oe or their x of the form ck + d where c and d are non-zero constants k2 – 4k [= 0] (A1) or x2 + 6x + 5 [= 0] and x = –5 [ x = –1] nfww k = 4 (A1) dep on all previous marks awarded

3 DO NOT USE A CALCULATOR IN THIS QUESTION. A cylinder has base radius ( 2 + 3) m and volume r ( 16 + 9 3) m3 . Find the exact value of its height, giving your answer in its simplest form. [4]

Mark scheme: 3 16 + 9 3 B1 (2 + 3) 2 M1 16 + 9 3 7 − 4 3 c 16 + 9 3 ( )( ) ( ) FT where a, b and c are non- 7 + 4 3 7 − 4 3 a + b 3 ( )( ) zero constants 16 + 9 3 7 − 4 3 or  7 + 4 3 7 − 4 3 112 − 64 3 + 63 3 − 108 A1 −112 + 64 3 − 63 3 + 108 or −1 4 − 3 or − 3 + 4 cao, nfww A1 Alternative method 16 + 9 3 (B1) (2 + 3) 2 2 16 + 9 3 2 − 3 ) ( )( 16 + 9 3 ( 2 − 3 ) 2 (M1) 2 2 or 2  2 ( 2 + 3 ) ( 2 − 3 ) ( 2 + 3 ) ( 2 − 3 ) 112 − 64 3 + 63 3 − 108 (A1) or 64 − 32 3 − 32 3 + 48 + 36 3 − 54 − 54 + 27 3 4 − 3 or − 3 + 4 cao, nfww (A1)

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4 Solve the following equations. x + 1 2 ( e ) (a) = 10 [4] x e 1 (b) 2 log 9 y - log 9 ( 4y - 9) = [5] 2

Mark scheme: 4(a) e 2 x + 2 B1 = 10 oe, soi x e 2 e1.5 x + 2 = 10 oe M1 e 2 x + k e kx + 2 FT = 10 oe or = 10 oe x x e 2 e 2 where k is an integer and k > 0 e 2 x + 2 or = 10 oe x e n where n is an integer and n > 1 or n = –2 1.5 x + 2 = ln10 oe M1 FT an expression of, or equivalent to, the forme ax + b = 10 oe where a and b are non-zero constants 2 A1 x = ( ln10 − 2 ) oe, isw or 0.202 3 or 0.2017[23…] rot to 4 or more dp isw 4(b) 2 1 M2 M1 for at least one correct log law used in a y 2 = 9 nfww correct equation e.g. 4 y − 9 y 2 1 2 2 1 or log 9 = log 9 9 oe log 9 y − log 9 (4 y − 9) = 4 y − 9 2 y 2 1 or log 9 = 4 y − 9 2 1 or 2log 9 y − log 9 (4 y − 9) = log 9 9 2 y 2 − 12 y + 27[ = 0] nfww A1 ( y − 3 )( y − 9 ) = 0 DM1 dep on at least M1 previously awarded y = 3, y = 9 nfww A1

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5 (a) Find the equation of the normal to the curve y = x 3 - 7x 2 + 12x - 5 at the point (1, 1). [5] (b) Find the x-coordinates of the two points where the normal cuts the curve again. Give your answers in the form x = a ! b where a and b are integers. [5]

Mark scheme: 5(a) Correct first derivative: M2 M1 for two terms of x 3 − 7 x 2 + 12 x − 5 3 x 2 − 14 x + 12 differentiated correctly [At x = 1] gradient of tangent: 1 A1 y – 1 = their(–1)(x – 1) oe M1 −1 FT or y = –x + c and 1 = –1 + c soi dy their dx x =1 y – 1 = –1(x – 1) or y = −+x 2 oe, isw A1 5(b) x 3 − 7 x 2 + 12 x − 5 = their ( −+x 2) B1 FT their y = ax + b where a is a non-zero constant Uses the correct linear factor x – 1 and the M1 correct cubic x3 − 7 x 2 + 13x − 7  = 0 to find a quadratic factor with at least two terms correct x 2 − 6 x + 7 A1 Correct use of formula or completing the M1 FT their 3-term quadratic providing it is square on their 3-term quadratic, e.g., from an attempt at finding a quadratic factor 2 and the discriminant is not negative −−( 6 )  ( −6) − 4[1](7) x = 2[1] 6  36 − 4[1](7) or x = 2 x = 3  2 A1

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7 A particle is travelling in a straight line. Its displacement, s metres, from the origin at time t seconds is given by s = 1.5e 2 t + 2e -2 t - t . (a) Find expressions for the velocity, v ms -1 , and acceleration, a ms -2 , of the particle. [3] (b) Find the time, T seconds, when the particle is at rest. [4] (c) Find the acceleration of the particle at time T seconds. [2]

Mark scheme: 7(a) Velocity: 3e 2 t − 4e −2 t − 1 isw B2 B1 for 3e 2 t or −4e−2 t Acceleration: 6e 2 t + 8e−2 t isw B1 FT me 2 t + ne−2 t + k where m, n and k are constants 7(b) 3e 4 t − e 2 t − 4 = 0 B1 2 or 3 e 2 t − e 2 t − 4 = 0 ( ) 2 t 2 t M1 FT their 3-term quadratic in e2t oe 3e − 4 e + 1 = 0 ( )( ) 2 t 4 A1 e = nfww 3 1 4 A1 ln oe, isw or 0.144 2 3 or 0.1438[41…] rot to 4 or more dp and no other solutions 7(c) 2  1 ln 4  −2  1 ln 4  M1 FT p e 2 t + q e−2 t where p and q are non-zero 6  e  2 3  + 8  e  2 3  constants and their positive 1 ln 4 from part 2 3 (b) 14 nfww A1

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8 A curve has equation y = x sin 2x . dy (a) Find . [2] dx (b) Find the equation of the tangent to the curve at x = r . [3] 4 r (c) Use your answer to part (a) to find the exact value of 2x cos 2xdx . [5] 6y0

Mark scheme: 8(a) Derivative of sin2x: 2cos2x soi B1 Product rule: x  2cos2x + [1]sin 2x isw B1 FT their 2cos2x 8(b) π B1 y = soi, isw 4 gradient of tangent: 1 soi B1 dep on correct derivative y = x or y – x = 0 or x – y = 0 B1 dep on correct derivative 8(c) π M3 M2 for x sin2 x + k cos2 x  1  6 x sin 2 x + cos2 x nfww 1    2  0 where k > 0 or k = − ; nfww 2 or M1 for x2 cos2 x dx = x sin2 x − sin2 x dx − cos2 x or + x2 cos2 x dx = x sin2 x 2 π π 1 π 1 A1 sin + cos − cos0 6 3 2 3 2 A1 π 3 1 π 3 − 3 − or 12 4 12

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9 (a) An arithmetic progression has twelve terms. The sum of the first three terms is -36 and the sum of the last three terms is 72. Find the first term and the common difference. [5] (b) The first three terms of a geometric progression are 1, 1.2 and 1.44. Find the smallest value of n such that the sum of the first n terms is greater than 500. [5]

Mark scheme: 9(a) Correct pair of simplified linear equations B3 B2 for one correct simplified equation in a and d with terms collected, e.g., or B1 for 3a + 3d = −36 isw or a + d = –12 isw a + a + d + a + 2d = −36 3a + 30d = 72 isw or a + 10d = 24 isw 3 or 2 a + (3 − 1) d  = −36 2 or a + 9d + a + 10d + a + 11d = 72 12 9 or 2a + (12 − 1) d  − 2a + (9 − 1) d  = 72 2 2 or 12a + 66d –9a –36d = 72 3 or 2( a + 9 d ) + (3 − 1) d  = 72 2 Solves two linear equations for d or a e.g. M1 FT their linear equations in a and d 27d = 108 → d = … providing at least B1 earned and the or 9d = 36 → d = … equations have a solution or a + 10(–12 – a) = 24 → a = … 27a = –432 → a = … d = 4 and a = −16 nfww A1 9(b) 1.2n *101 B3 where * is any inequality sign or =; [1](1.2 n −1) B2 for *500 (1.2 −1) or B1 for r = 1.2 soi nlog1.2*log101 or log1.2 101soi M1 FT 1.2n * their 101 providing B2 has been awarded and (their 101) > 0 n = 26 A1 dep on all previous marks awarded

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10 (a) By writing cotx and tanx in terms of cosx and sinx, show that sin x cos x + = sin x + cos x. [5] 1 - cot x 1 - tan x (b) Solve the equation 9 cot x + 3 cosec x = tan x , for 0° 1 x 1 360° . [5]

Mark scheme: 10(a) Writes cotx and tanx in terms of sinx and M1 OR cosx:  sin x   cos x  sin x  1 −  + cos x  1 −  sin x cos x  cos x   sin x  + cos x sin x  cos x  sin x  1 − 1 −  1 −  1 −  sin x cos x  sin x  cos x  Simplifies denominator: A1 OR sin x cos x  cos x − sin x   sin x − cos x  + sin x   + cos x   sin x − cos x cos x − sin x  cos x   sin x  sin x cos x  sin x − cos x  cos x − sin x      sin x  cos x  Writes as two simple algebraic fractions: A1 OR writes as a single simple algebraic sin 2 x cos 2 x fraction: + 2 2 sin x (cos x − sin x ) + cos x (sin x − cos x ) sin x − cos x cos x − sin x (sin x − cos x )(cos x − sin x ) Writes as a difference with a common A1 sin 2 x (cos x − sin x ) − cos 2 x (cos x − sin x ) OR denominator: (sin x − cos x )(cos x − sin x ) sin 2 x cos 2 x − sin x − cos x sin x − cos x Correct simplification to given answer, e.g., A1 All steps correct and final step fully justified (sin x − cos x )(sin x + cos x ) by factorising = sin x + cos x (sin x − cos x ) or (sin x − cos x ) (sin x + cos x )  = sin x + cos x  (sin x − cos x ) 10(b) 10cos 2 x + 3cos x − 1[ = 0] B2 9cos x 3 sin x B1 for + = or better or sec 2 x − 3sec x − 10[ = 0] sin x sin x cos x 3tan x 2 or 9 + = tan x or better sin x OR M1 for one sign error in 10cos 2 x + 3cos x − 1[ = 0] or sec 2 x − 3sec x − 10[ = 0] ( 5cos x − 1)( 2cos x + 1) = 0  M1 FT their 3-term quadratic in cosx or secx or ( sec x − 5 )( sec x + 2 ) = 0  1 1 A2 A1 for any two correct angles [cosx = and cosx = − 5 2 1 1 [found using cosx = and cosx = − 5 2 OR OR secx = 5 and secx = –2 leading to] secx = 5 and secx = –2]; 78.5 or 78.46[30…] rot to 2 or more dp ignore extras 281.5 or 281.53[69…] rot to 2 or more dp 120 240 and no extras in range 0  x  360

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