Cambridge IGCSE Mathematics - Additional 0606 — 2023 Oct/Nov Paper 2 · Variant 1

1 (a) Write 19 - 12x - 3x 2 in the form a ( x + b) 2 + c where a, b and c are integers. [4] (b) Hence find the maximum value of 19 - 12x - 3 x 2 and the value of x at which this maximum occurs. [2] (c) Use your answer to part (a) to solve the equation 19 - 12 u - 3 u = 0 . [3]

Mark scheme: Question Answer Marks Partial marks 1(a) −3(x + 2)2 + 31 B4 B2 for −3(x + 2)2 or B1 for (x + 2)2 or a = –3 and b = 2 B2 for c = 31 or B1 for –4  –3 + 19 soi 1(b) Maximum value 31 when x = –2 B2 Strict FT their c from part (a) and −their b from part (a) B1 for either without contradiction 1(c) 2 M1 FT an expression of correct form from part (a) −3 u + 2 = −31 oe ( ) 31 A1 Rearranges as far as u = −2 3 1.48 cao or 1.475[13…] rot to 3 or more dp A1 43 − 4 93 or isw 3

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2 Solve the following simultaneous equations. 5x - 3 ln y = 2 [4] x + ln y = 1

Mark scheme: 2 Correct method to eliminate y: M1 5 x − 3 (1 − x ) = 2 oe or adds 3x + 3lny = 3 5x – 3lny = 2 to obtain 3x + 5x = 3 + 2 or better 5 A1 x = oe 8 3 M1 ln y = oe 8 3 A1 y = e 8 oe or 1.45 2 Alternative method Correct method to eliminate x: (M1) 5(1 – ln y) – 3ln y = 2 oe or subtracts 5x – 3lny = 2 from 5x + 5lny = 5 to obtain 5lny – (–3lny) = 5 – 2 or better 3 (M1) ln y = oe 8 3 (A1) y = e 8 oe or 1.45 5 (A1) x = oe 8

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3 (a) Find 4x + 5 - dx . [3] 2x + 3 3 1 (b) Hence find the exact value of 4x + 5 - dx , simplifying your answer. [3] + 3 o y1 e 2x

Mark scheme: 3(a) 2 1 B3 2 1 2 x + 5 x − ln ( 2 x + 3 ) + c oe B2 for 2 x + 5 x − ln ( 2 x + 3 ) 2 2 2 1 or 2 x + 5 x − ln2 x + 3 + c 2 or 2 x 2 + 5 x + k ln ( 2 x + 3 ) + c with k ≠ 0 or B1 for 2 x 2 + 5 x +…+c 1 or ... − ln2 x + 3 2 or ... + k ln ( 2 x + 3 ) with k ≠ 0 3(b) Substitutes limits and subtracts in correct M1 FT their part (a) providing it includes a term order k ln ( 2 x + 3 ) with k ≠ 0  1   1  A1 18 + 15 − ln9 − 2 + 5 − ln5      2   2  1 9 1 5 A1 26 − ln or 26 + ln oe 2 5 2 9

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4 In this question a and b are integers. Three terms in the expansion of ( 2 + ax) 5 ( 1 + bx) are 32 + 112x - 240 x 2 . Find the values of a and b. [7]

Mark scheme: 4 ( 2 + ax )5 = 25 + 5  2 4 ax + 10  23 a 2 x 2 + ... B1 ( 2 + ax )5 (1 + bx ) = M1 32 + 80ax + 80a2x2 + 32bx + 80abx2... 80a + 32b = 112 oe, isw A1 80 a 2 + 80 ab = −240 oe, isw A1 3a 2 − 7 a − 6 = 0 oe M1 (3a + 2)(a – 3) = 0 M1 a = 3 and b = –4 and no other values A1

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5 In this question p and q are constants. p The normal to the curve y = 2 + 5x - 2 , at the point where x = 1, has equation y =- x + q . x Find the values of p and q. [6]

Mark scheme: 5 [ mtangent =] − 2 px−3 + 5 oe B1 −1 B1 −1 [When x = 1, mnormal = ] FT if appropriate −2 p + 5 dy their or gradient of tangent = 1 nfww dx x =1 −1 = −1 oe M1 −1 their ( −2 p + 5) FT dy or their( –2p + 5) = 1 their dx x =1 d y − 1 or their and their evaluation of d x x =1 −1 p = 2 nfww A1 [When x = 1] their5 = −+1 q or y = –x + 6 M1 FT y = (their p) + 3 providing at least 2 of the first 3 marks awarded q = 6 nfww A1

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6 Find the value of the constant a for which the line y = ( 2a + 1 ) x - 10 is a tangent to the curve y = ax 2 - 5 x + 2 . [6]

Mark scheme: 6 ax 2 − 5 x + 2 = 2 ax + x − 10 M1 ax 2 − ( 2a + 6 ) x + 12  = 0 oe A1 Correct use of b2 – 4ac [*0]: M1 where * is any inequality sign or =; (– (2a + 6))2 – 4(a)(12) [*0] oe FT their 3-term quadratic in x and a 4 a 2 − 24 a + 36[*0] A1 Factorises or solves their 3-term quadratic M1 FT their 3-term quadratic in a in a a = 3 A1 6 Alternative method 3 a + 3 (2) M1 for 2ax – 5 a = oe or x = x − 1 a  6   3  2 (M1) k  + 1  x − 10 =   x − 5 x + 2 oe FT their a of the form where k, b, c are  x − 1   x − 1  bx + c or non-zero constants 2 da + e  a + 3   a + 3   a + 3  or their x of the form where d, e, f are (2 a + 1)   − 10 = a   − 5   + 2 fa  a   a   a  oe non-zero constants 3x2 – 12x + 12 [= 0] oe or a 2 − 6 a + 9[ = 0] (A1) Solves their 3-term quadratic in x as far as (M1) x = ... or factorises or solves their 3-term quadratic in a a = 3 (A1)

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7 A particle moves in a straight line. At time t seconds after passing through a fixed point O, its velocity, v ms -1 , is given by v = 10 sin 2 t - 6 cos 2 t . (a) Find an expression for the acceleration of the particle. [2] (b) Find the acceleration when t = r . [1] 4 (c) Find the first time at which the acceleration is zero. [3] (d) Find the displacement of the particle between t = r and t = r . [4] 4 2

Mark scheme: 7(a) 20cos2t + 12sin2t 2 B1 for 20cos2t or 12sin2t 7(b) 12 B1 7(c) 20  M1 FT acos2t + bsin2t where a and b are non-zero tan 2t = their −  oe integers  12  t = 1.06 or 1.055[60…] rot to 3 or more dp A2 A1 for 2t = –1.030[3…] or 2t = 2.111[2…] 7(d) s = −5cos2t −3sin2t (+ c) B2 B1 for −5cos2t or −3sin2t π π M1 FT providing at least B1 previously awarded −5cosπ − 3sinπ − (−5cos − 3sin ) 2 2 or s = −5cos2t −3sin2t + 5 and s π − s π = 10 − 2 2 4 8 A1

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8 DO NOT USE A CALCULATOR IN THIS QUESTION. Solve the equation ( 2 - 10) x 2 + x + ( 2 + 10) = 0 , giving your answers in the form a + b 10 , where a and b are rational. [7]

Mark scheme: 8 B1 2 − 10 2 + 10 = −6 ( )( ) [2] M1 −1 1 − 4(2 − 10)(2 + 10) x = 2 2 − 10 ( ) −1 1 − 4( − 6) A1 x = 2 2 − 10 ( ) −6 4 A1 oe , oe 2 2 − 10 2 2 − 10 ( ) ( ) M1 k 2 + 10 −6 ( ) FT where k is a non-zero constant  2 2 − 10 ( ) 2 2 − 10 2 + 10 ( ) ( ) 2 + 10 4 ( ) or  2 2 − 10 2 + 10 ( ) ( ) −6(2 + 10) 1 A1 Must have sufficient detail shown oe = 1 + 10 2 ( −6 ) 2 4(2 + 10) 2 1 A1 Must have sufficient detail shown oe = − − 10 2 ( −6 ) 3 3

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9 The functions f and g are defined as follows, for all real values of x. f (x) = 2 x 2 - 1 g (x) = e x + 1 (a) Solve the equation fg ( x) = 8 . [3] (b) For each of the functions f and g, either explain why the inverse function does not exist or find the inverse function, stating its domain. [4]

Mark scheme: 9(a) 2(ex + 1)2 − 1 [= 8] M1 9 A1 ex = –1+ oe 2  3  A1 x = ln − 1   isw or 0.115  2  or 0.1145[06…] rot to 4 or more dp 9(b) f is not one-one, hence f–1 does not exist oe B1 g−1(x) = ln(x − 1) 2 M1 for x = ln(y − 1) and a swop of variables at some point or y = ln(x + 1) or ex = y – 1 and y = lnx – 1 x > 1 B1

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10 In this question all lengths are in centimetres. C 6 O A 8 B The diagram shows a circle centre O with radius 6. The line AB is a tangent to the circle at the point B. The point C lies on the circle such that AOC is a straight line. AB = 8 . (a) Find the perimeter of the shaded region. [6]

Mark scheme: 10(a) 2 2 B1 OA= 6 + 8 oe or 10 [Angle AOB = ] M1 0.9272[95…] rot to 4 or more dp or [Angle OAB = ] 0.6435[01…] rot to 4 or more dp [Angle COB =] 2.214[297…] rot to 3 or A1 more dp [Arc CB =] 6(their 2.214) M1 FT their COB [Perimeter =] 8 + (their 10 + 6) + 6(their M1 FT their arc CB and OA 2.214) 37.3 or 37.28[578…] rot to 2 or more dp A1 10(b) 1 1 2 M2 FT their 2.21 +8 6  6  2.214 oe, soi 2 2 1 2 M1 for  6  2.214 soi 2 63.9 or 63.85[735…] rot to 2 or more dp A1

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(b) Solve the equation 3 tan 2 ( y + r ) = 1 for - 2r 1 y 1 0 . [4] 4

Mark scheme: 11(a) 1 1 M1 + 1 1 1 1 − + cos x sin x cos x sin x Simplifies denominator A1 1 1 + sin x − cos x sin x + cos x sin x cos x sin x cos x Writes as two simple algebraic fractions: A1 OR writes as a single simple algebraic fraction: sin x cos x sin x cos x sin x cos x (sin x + cos x ) + sin x cos x (sin x − cos x ) + sin x − cos x sin x + cos x (sin x − cos x )(sin x + cos x ) Combines and simplifies: A1 2sin 2 x cos x sin 2 x − cos 2 x Correct simplification to given answer e.g. A1 All steps correct nd fully justified 2sin 2 x cos x 2cos x = 2 cos 2 x 1 − cot 2 x sin x − sin 2 x sin 2 x (2cos x )  2cos x  or = 2 2  2  sin x (1 − cot x )  1 − cot x  11(a) Alternative method Common denominator: (M1) sec x + cosec x + sec x − cosec x (sec x − cosec x )(sec x + cosec x ) 2sec x (A1) Simplifies: sec 2 x − cosec 2 x Rewrites in terms of sinx and cosx: (A1) OR multiplies numerator and denominator by 2 cos2x: cos x 2sec x cos 2 x  1 1 2 2 2 − sec x − cosec x cos x cos 2 x sin 2 x 2 (A1) cos x cos 2 x  1 1 cos 2 x − cos 2 x sin 2 x Correct simplification to given answer e.g. (A1) 2cos 2 x cos x 2cos x = cos 2 x cos 2 x 1 − cot 2 x − cos 2 x sin 2 x 11(b)    B1 tan  y +  = 1  4  3  π     7  M1 y + = , or − , or − , or − oe    4  6 6 6 6  5 13 17  A2 No extras within range  y =  − , − , − , − oe 12 12 12 12 A1 for two correct, ignoring extras

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