Cambridge IGCSE Mathematics - Additional 0606 — 2023 Oct/Nov Paper 2 · Variant 3

1 The functions f and g are defined as follows, for all real values of x. f ( x) = 2 sin x + 3 cos x g ( x) = e 3 x - 1 (a) Find fg(0). [2] (b) Find gg(x). [1] -1 1 (c) Solve the equation g ( x) = ln 5 . [3] 3

Mark scheme: Question Answer Marks Guidance 1(a) 3 B2 B1 for g(0) = 0 or [fg(x) =] 2sin(e3x – 1) + 3cos(e3x – 1) soi 1(b) 3( e 3 x −1) B1 gg ( x ) = e − 1 oe, isw 1(c) 3 y = ln( x + 1) M1 condone one error or 3x = ln( y + 1) and swops the variables at some point g−1(x) = 13 ln( x + 1) soi A1 [x =] 4 A1 Alternative method x = g( 13 ln5 ) soi (B1) e3(13ln 5) −1 oe (M1) [x =] 4 (A1)

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3 (a) Solve the following simultaneous equations. 3 log 2 x + 2 log 2 y = 24 5 log 2 x - 3 log 2 y = 2 [5] 2 t + 4 (b) Solve the equation = 512 . [4] l - 2 t 2

Mark scheme: 3(a) Correctly eliminates log 2 x or log 2 y M1 A correct equation in log 2 x only or log 2 y only x = 16 or y = 64 A2 A1 for log 2 x = 4 or log 2 y = 6 y = 64 or x = 16 A2 A1 for log 2 y = 6 or log 2 x = 4 Alternative method 3 2 24 x 5 2 (M1) x y = 2 and = 2 oe y 3 y = 64 or x = 16 (A2) A1 for y19 = 2114 oe or x19 = 276 oe x = 16 or y = 64 (A2) A1 for x3 642 = 224 oe x 5 2 or 3 = 2 oe 64 OR 163  y2 = 224 oe 16 5 2 or 3 = 2 oe y 3(b) t + 4 −−(1 2 t ) 9 B2 t + 4 −−(1 2 t ) 2 = 2 B1 for 2 = 512 or 2t + 4 = 29 +1− 2 t oe, soi 2 t + 4 9 or 1− 2 t = 2 soi 2 OR OR t + 4 − (1 − 2t ) = log 2 512 oe, soi ( t + 4 ) log a 2 − (1 − 2t )log a 2 = log a 512 log a 512 or t + 4 − (1 − 2t ) = oe log a 2 3t + 3 = 9 or better M1 t = 2 A1

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5 The curved surface area of a cylinder with radius r and height h is 2rrh . A closed cylinder has radius r cm and volume 1000 cm 3. 2 2000 2 (a) Show that the total surface area of the cylinder is 2rr + cm . [3] r (b) Find the value of r which makes this area a minimum. You should show that your value of r gives a minimum for this area. [5]

Mark scheme: 5(a) Correct use of r 2 h = 1000 to find an expression that M2 M1 for r 2 h = 1000 soi can be used to eliminate h 1000 1000 e.g. h = or πrh = πr 2 r Correct substitution and completion to given answer A1 2 1000 2 2000 e.g. 2πr + 2πr  = 2πr + πr 2 r 5(b) 2000 B2 B1 for one correct term Correct derivative: 4πr − oe, isw r 2 2000 M1 FT their derivative providing that at 4πr − = 0 and solves for r r 2 least one term is correct 2000 A1 r = 3 oe, isw 4π or 5.42 or 5.419[26…] rot to 3 or more dp d 2 S –3 A1 Dep on previous mark Second derivative = 4π + 4000r and d r 2 d 2 S When r = 5.42, > 0 oe [hence minimum] d r 2 or 4π + 4000(5.42) –3 > 0 oe [hence minimum] d 2 S or = 12 or 37 to 38 [hence minimum] d r 2 d 2 S or as r  0 , > 0 [hence minimum] d r 2 OR correctly finds the values of the first derivative at 5.42 oe  h, where h is small [hence minimum]

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6 A particle travels in a straight line. Its displacement, s metres, from the origin, at time t seconds, where t 2 2 , is given by s = ln ( 4t 2 - 5 ) - t . (a) Find expressions for the velocity, v ms -1 , and acceleration, a ms -2 , of the particle. [4] (b) Find the time when the particle is at rest. [3] (c) Find the acceleration at this time. [2]

Mark scheme: 6(a) 8t B2 f ( t ) 8t Velocity: − 1 B1 for 2 − 1 or for 2 + g(t) 2 4t − 5 4t − 5 4t − 5 Correct structure of quotient rule or equivalent product M1 FT their v if possible; must be of rule equivalent difficulty (4t 2 − 5)(8) − (8t )(8t ) A1 8t Acceleration: oe, isw FT 2 + k where k is a constant 2 2 (4t − 5) 4t − 5 6(b) 4t2 − 8t – 5 = 0 oe B1 ( 2t + 1)( 2t − 5 ) = 0 M1 FT their 3-term quadratic in t t = 2.5 and no other values A1 dep on correct quadratic seen 6(c) (4  2.52 − 5)(8) − (8  2.5)(8  2.5) M1 Substitutes a value of t  2 in an oe, soi 2 2 expression for a which has at least one (4  2.5 − 5) 1 term with a factor of or 2 4t 2 − 5 ( ) 1 16t 4 − 40t 2 + 25 3 A1 a = − oe only 5

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7 DO NOT USE A CALCULATOR IN THIS QUESTION. A You may use the following 60° trigonometrical ratios. 3 2 sin 60 ° = , sin 45 ° = 2 2 2 1 2 cos60 ° = , cos45° = 2 2 75° tan60° = 3, tan45° = 1 45° B C 3 + 3 6 + 2 (a) Given that the area of triangle ABC is , show that sin75° = . [5] 4 4 (b) Hence find the exact length of AC. [2]

Mark scheme: 7(a) BC 2 M1 = oe sin60 sin 45 BC = 3 oe A1 1 3 + 3 M1 FT their BC providing it has not been  2  3  sin75 = found using the given result for sin75 2 4 OR  3 + 3  2 height =    oe    4  3 2(3 + 3) A1 dep on all previous marks being sin75 = oe awarded 4 6 Isolates sin 75 correctly or deals with surds on LHS of correct OR equation  3 + 3  2  1 + 3  1  3 + 3      6   e.g. 6 sin75 =      4  3 2 1 + 3 2 4 sin 75 = oe  = =    2  2 2 2      6 + 2 A1 must be convincing with an correct completion to given answer intermediate step if needed 4 Alternative methods (finding AC first) 1 3 + 3 (M1)  2  AC  sin60 = oe 2 4 2 2 3 + 3 (A1) Isolates AC correctly AC =   oe 2 3 4 2 + 6 (A1) Must be convinced no calculator is AC = being used 2 sin75 sin45 (M1) FT their AC = oe AC 2 + 6 2 May simplify to sin75 = before 2 2 inserting their AC 6 + 2 (A1) dep on all previous marks being correct completion to given answer awarded 4 must be convincing with an intermediate step if needed 7(b) AC 2 AC 3 M1 = or = or better 6 + 2 2 6 + 2 3 4 2 4 2 6 + 2 A1 nfww 2

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sin x cos x(b) Hence solve the equation - = 1 for 0° 1 x 1 360° . [5] tan x - 1 tan x + 1

Mark scheme: 8(a) sin x cos x M1 sin x (tan x + 1) − cos x (tan x − 1) − OR sin x sin x (tan x − 1)(tan x + 1) − 1 + 1 cos x cos x sin x cos x A1 OR − sin x − cos x sin x + cos x sin x tan x + sin x − cos x tan x + cos x cos x cos x (tan x − 1)(tan x + 1)  sin x   sin x  sin x  + 1  − cos x  − 1   cos x   cos x   sin x   sin x  OR sin x  + 1  − cos x  − 1  2 sin x  cos x   cos x  − 1 OR cos 2 x  sin x  sin x   − 1  + 1   cos x  cos x  sin x cos x cos 2 x A1 sin 2 x − + sin x − sin x + cos x sin x − cos x sin x + cos x OR cos x  sin x + cos x   sin x − cos x  sin 2 x sin x   − cos x   2 − 1  cos x   cos x  cos x OR sin 2 x − cos 2 x cos 2 x sin 2 x cos x + sin x cos 2 x − cos 2 x sin x + cos 3 x A1 sin 2 x + cos 2 x sin 2 x + cos 2 x sin 2 x + cos xsinx − cosxsinx − cos 2 x cos x cos x OR or OR sin 2 x − cos 2 x 1 − 2cos 2 x sin 2 x + sin x cos x − cos x sin x + cos 2 x cos 2 x cos 2 x cos 2 x  cos x sin 2 x − cos 2 x Fully correct justification of given answer e.g. A1 All steps correct and final step justified 1 cos 2 x cos x cos x sin 2 x + cos 2 x ( ) cos x  = = oe 2 2 2 2 2 2 2 2 cos x sin x − cos x sin x − cos x sin x − cos x sin x − cos x oe 2 x − cos 2 x or8(b) 2cos 2 x + cos x − 1  = 0 B2 B1 for cos x = 1 − cos better ( 2cos x − 1)( cos x + 1) = 0  M1 FT their 3-term quadratic in cosx [ x =] 60,300,180 A2 A1 for any two correct , ignoring and no extras in range extras

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9 A curve has equation y = xe 2 x . d y. [2] (a) Find d x (b) Find the equation of the normal to the curve at x = 1. [4] (c) Use your answer to part (a) to find the exact value of 2 xe 2 x d x . [5] 2y0

Mark scheme: 9(a) Derivative of e 2 x : 2e 2 x soi B1 2 x 2 x B1 FT their 2e2x x  2e + e isw 9(b)  When x = 1 y = e2 B1 dy B1 FT their derivative which must include  gradient tangent =  their x dx x =1 at least one term in 2e −1 B1 −1 Gradient of normal = 2 FT their (3e ) d y their d x x =1 −1 B1 dep on 2 marks awarded in part (a) and y – e2 = (x – 1) oe, isw 3e 2 all previous marks awarded in this part 9(c)  2 x 1 2 x  2 M3 M2 for xe 2 x + ke 2 x where k < 0 or xe − e    2  0 k = 1 2 e 2 x dx or M1 for  2 xe 2 x dx = xe 2 x −  4 1 4 1 A1 2e − e −− ( 2 ) ( 2 ) 1.5e4 + 0.5 or exact equivalent A1

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10 (a) In an arithmetic progression the 5th term is 11. The 7th term is three times the 2nd term. Find the 1st term and the common difference. [4] (b) A different arithmetic progression (AP) and a geometric progression (GP) have the following properties. • The 1st terms of the AP and GP are both 3. • The 2nd term of the AP is the same as the 3rd term of the GP. • The 6th term of the AP is the same as the 5th term of the GP. • The common ratio of the GP is greater than 1. Find the common difference of the AP and the common ratio of the GP. [6]

Mark scheme: 10(a) a + 4d = 11 oe B1 a + 6d = 3(a + d) oe B1 Correctly eliminates one unknown and solves for a or d M1 FT their linear equations in a and d providing B1 earned. d = 2, a = 3 A1 10(b) 3 + d = 3r 2 B1 3 + 5d = 3r 4 B1 2 M1  3 + d  3 + 5d = 3 oe    3  or 3 + 5 3r 2 − 3 = 3r 4 oe ( ) d 2 − 9d  = 0 or 3r 4 − 15r 2 + 12  = 0 A1 d = 9 and r = 2 and no other values A2 A1 for d = 9 and no other value of d or for r = 2 and no other value of r

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