Cambridge IGCSE Mathematics - Additional 0606 — 2022 Oct/Nov Paper 2 · Variant 3

1 Solve the following inequality. ( 2x + 3)( x - 4) 2 ( 3x + 4)( x - 1) [5]

Mark scheme: Question Answer Marks Guidance 1 2 x 2 − 8 x + 3 x − 12 * 3 x 2 − 3 x + 4 x − 4 B1 Correctly expands all brackets * is any inequality or equals sign  0* x 2 + 6 x + 8 B1 Collects terms to correct 3-term quadratic in solvable form  0*( x + 2 )( x + 4 ) M1 Factorises or solves their 3-term quadratic −4 and −2 A1 Correct critical values −4 < x < −2 mark final answer A1

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2 The tangent to the curve y = ax 2 - 5x + 2 at the point where x = 2 has equation y = 7x + b . Find the values of the constants a and b. [5]

Mark scheme: 2 dy B1 = 2 ax − 5 dx 2a −=2 5 7 oe M1  dy FT their  = 7  dx x = 2  a = 3 A1 7 +2 b = their 4 M1 dep on previous M1 or b = 2 −4 theira where their 4 is an attempt to evaluate y = ax 2 − 5 x + 2 using x = 2 and their a b = −10 A1 Alternative ( −12 )2 − 4a ( 2 − b ) = 0 oe (B1) for use of discriminant on ax 2 − 12 x + 2 − b = 0 144 − 8 a + 4 a ( 4 a − 22 ) = 0 oe (M1) Condone one sign or arithmetic error or 144 − ( b + 22 )( 2 − b ) = 0 oe a 2 − 6a + 9  = 0 oe (A1) for correct 3-term quadratic in solvable form or b2 + 20b + 100  = 0 oe a = 3 (A2) A1 for a = 3 or b = −10 and b = −10

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4 The line y = kx + 6 intersects the curve y = x 3 - 4x 2 + 3kx + 2 at the point where x = 2 . (a) Find the value of k. [2] (b) Show that, for this value of k, the line cuts the curve only once. [4]

Mark scheme: 4(a) 2k + 6 = 8 − 16 + 6k + 2 oe M1 For equating line to curve and substituting x = 2, or vice versa k = 3 A1 4(b) x3 − 4 x 2 + ( 2  theirk ) x − 4  = 0 M1 FT their k in correct cubic or x3 − 4 x 2 + 6 x − 4  = 0 x 2 − 2 x + 2 A2 Correct quadratic factor from correct cubic A1 for a quadratic factor with two terms correct, from correct cubic ( −2) 2 − 4 (1)( 2 ) < 0 oe A1 Uses discriminant correctly on the correct quadratic factor or 4 – 8 < 0 oe [and so x = 2 is the only solution]

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5 (a) Show that + = 2 sec x . [4] 1 - sin x cos x i i cos 1 - sin 2 2 2 i (b) Hence solve the equation + = 8 cos for - 360° 1 i 1 360° . [4] i i 2 1 - sin cos 2 2

Mark scheme: 5(a) cos 2 x + (1 − sin x ) 2 M1 Correctly takes common denominator (1 − sin x ) cos x cos 2 x (1 − sin x ) 2 or + (1 − sin x ) cos x (1 − sin x ) cos x cos 2 x + 1 − 2sin x + sin 2 x A1 1 − sin 2 x + (1 − sin x ) 2 OR (1 − sin x ) cos x (1 − sin x ) cos x 1 + 1 − 2sin x A1 (1 − sin x )(1 + sin x ) + (1 − sin x ) 2 (1 − sin x ) cos x OR (1 − sin x ) cos x 1 − sin 2 x + 1 − 2sin x + sin 2 x or (1 − sin x ) cos x 2(1 − sin x ) A1 All steps correct and final step justified = 2sec x (1 − sin x ) cos x 2 − 2sin x 2 or = = 2sec x 1 + sin x + 1 − sin x (1 − sin x ) cos x cosx OR = 2sec x cos x or equivalent Alternative Must work with LHS only (cos x )(1 + sin x ) (1 − sin x )cos x (M1) Forms fractions with common + (1 − sin x )(1 + sin x ) ( cos x ) cos x denominator in different form (cos x )(1 + sin x ) (1 − sin x )cos x (A1) Uses difference of two squares and + 2 cos x cos 2 x sin 2 x + cos 2 x = 1 to write fractions with a common denominator in the same form 2cos x (A1) Combine as a single fraction and 2 collects terms cos x 2 (A1) All steps correct and final step justified = 2sec x cos x 5(b) 3 1 B1 cos = 2 4 1 M1  2  cos = 3 their soi dep on starting with 2sec = 8cos 2 4 2 2 101.9 awrt A2 and no extras in range A1 for either, ignoring extras in range If A0 then SC1 for 102 with no extras in range

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6 The first four terms in ascending powers of x in the expansion ( 3 + ax) 4 can be written as 2 3 3 81 + bx + cx + x . Find the values of the constants a, b and c. [6] 2

Mark scheme: 6 81 + 108ax + 54a 2 x 2 + 12a 3 x 3 soi M3 M2 for any 3 correct terms or 2 correct equations or M1 for any 2 correct terms, 1 3 3 2 or 12 a = b = 108a c = 54 a soi correct equation or for correct but 2 insufficiently simplified expansion e.g. 4 3 4  3 2 2 3 + 4  3  ax +  3  ( ax ) 2 4 3 2 3 + 3 ( ax ) 3  2 1 A1 a = oe 2 b = 54 A1 FT 108  their a, providing at least M1 awarded 27 A1 FT 54  (their a)2, providing at least c = oe 2 M1 awarded

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8 (a) Particle A starts from the point with position vector and travels with speed 26 ms -1 in the - 2 12 direction of the vector Find the position vector of A after t seconds. [3] e 5o. 67 (b) At the same time, particle B starts from the point with position vector It travels with speed e- 18o. -1 3 20 ms at an angle of a above the positive x‑axis, where tan a = . Find the position vector of B 4 after t seconds. [4]

Mark scheme: 8(a) 26  12  M2 M1 for 12 2 + 52 or 13 or 2 seen (Velocity vector =) oe    5  12 2 + 52  3   24  A1 (Position vector =)   + t   oe  −2   10  8(b) 4 B1 (Direction vector =)  soi 3 4 or x component: cos= 5 3 y component: sin = soi 5 20 4 M1 (Velocity vector =) oe  their  3 4 2 + 32 soi  their cos or 20   soi  their sin  67   16  A2  67   16  (Position vector =)   + t   oe A1 FT   + t  their    −18   12   −18   12  If zero scored, SC2 for one correct component, either 67 + 16t or −18 + 12t 8(c) 3 + 24t = 67 + 16t oe M1 FT Equates their x components, or or −+2 10t = −18 + 12t oe their y components from parts (a) and (b), providing of equivalent difficulty, e.g. a + bt = c + dt t = 8 A1 dep on full marks in (a) and (b)  195  A1 dep on full marks in (a) and (b) (Position of meeting =)    78 

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9 The equation of a curve is y = kxe - 2 x , where k is a constant. dy (a) Find . [2] dx (b) Find the coordinates of the stationary point on the curve y = 10xe - 2 x . [3] (c) Use your answer to part (a) to find 4xe - 2 x dx . [3] y 1 (d) Find the exact value of 4xe - 2 x dx . [2] y0

Mark scheme: 9(a) d −2 x −2 x B1 e = −2e soi ( ) dx dy −2 x −2 x B1 FT for use of product rule = ke − 2 kxe oe, isw dx −2 x  d −2 x  k .e + kx. their e  ( )   dx  Alternative d 2 x 2 x (B1) e = 2e soi ( ) dx d y k e 2 x − 2 kx e 2 x (B1) FT for use of quotient rule = oe, isw 2 x 2 x 2 k .e − kx. their 2e d x ( ) e 2 x ( ) 2 e 2 x ( ) 9(b) dy M1 FT their (a), provided of the form Equates = 0 and finds 10 – 20x = 0 oe −2 x −2 x 2 x 2 x dx me + nxe or me + nxe  1 5  A2 For both values:  ,  oe only 1  2 e  x = 0.5 and y = 5e− or 1.84 or 1.839[39...] rot to 4 or more sf 1 A1 for x = only 2 9(c) −2 xe −2 x − e −2 x + c B3 For fully correct answer or B2 for −2 xe −2 x − e −2 x or   4 xe −2 x dx  = −2 xe −2 x +  2e −2 x dx   or B1 for kxe −2 x =  ke −2 x − 2 kxe −2 x dx ( ) or better 9(d) −2e −2 − e −2 − 0 − e 0 oe M1 Correct substitution of limits into ( ) correct expression 3 2 A1 1 − or 1 − 3e− e 2

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10 (a) The third term of an arithmetic progression is 10 and the sum of the first 8 terms is 116. Find the first term and common difference. [5]

Mark scheme: 10(a) a + ( 3 − 1) d = 10 soi B1 8 2 a + ( 8 − 1) d  = 116 soi B1 2 Correct method to eliminate one unknown M1 dep on at least B1 awarded and attempt to solve to find a or d a = 4 and d = 3 A2 A1 for either 10(b) 30 B2 M1 FT their a and their d for S30 = 2 ( 4 ) + 29 ( 3 ) 2 30 S30 = 2 ( their 4 ) + 29 ( their 3 ) 11 2 and S11 = 2 ( 4 ) + 10 ( 3 ) 2 11 or S11 = 2 ( their 4 ) + 10 ( their 3 ) 2 Correct plan S30 − S11 attempted M1 FT their a and their d 1216 A1 10(b) Alternative 1 first term = 4 + 11×3 or 37 (M1) FT their a + 11×their d and an attempt at S19 19 2 ( 37 ) + (19 − 1)  3 oe (B2) M1 FT for their first term and their d 2 19 in 2 ( their 37 ) + (19 − 1)  their 3 19 2 or 37 + 91 oe or for their first term and their last 2 19 term in their 37 + their 91 2 1216 (A1) Alternative 2 Correct sum of terms: (M3) M2 FT their a and their d for sum 37 + 40 + 43 + 46 + 49 + 52 + 55 + 58 + 61 starting with their 37 and ending with + 64 + 67 + 70 + 73 + 76 + 79 + 82 + 85 + their 91, with at most one omission or 88 + 91 error or M1 FT their a and their d for sum starting with their 37 or ending with their 91, with at most two omissions or errors 1216 (A1)

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11 5a R S b Q 3b X O P 2a In the vector diagram, OP = 2a , SR = 5a , OS = 3b and QR = b . (a) Given that PX = mPS , write OX in terms of a, b and m. [3] (b) Given that O X = n OQ , write OX in terms of a, b and n. [2] (c) Find the values of m and n. [4] OX(d) Write down the value of . [1] OQ PX(e) Find the value of . [1] XS

Mark scheme: 11(a) 2a + ( 3b − 2a ) oe isw B3 B1 for PS = 3b − 2a soi or 3b − (1 − )( 3b − 2a ) oe isw and B1 for correct route using , either OX = OP + PS soi or OX = OS − (1 − ) PS soi 11(b) ( 5a + 2b ) isw B2 B1 for OQ = 3b + 5a − b oe soi 11(c) 2 − 2= 5 and 3= 2 oe M2 for correctly equating scalars for both components FT their (a) and (b) if possible M1 FT for equating scalars for either component 4 6 A1 Solves to find = or = 19 19 4 6 A1 = and = 19 19 11(d) 6 B1 isw 19 11(e) 4 B1 isw 15

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