Cambridge IGCSE Mathematics - Additional 0606 — 2018 Oct/Nov Paper 2 · Variant 2

1 Solve the inequality (x - 3)(x + 4) 2 x + 13 . [3]

Mark scheme: Question Answer Marks Partial Marks 1 x 2 + x − 12 > x + 13 M1 expand and simplify → x 2 … 25 A1 x > 5 or x < − 5 A1 or x > 5 , x < −5 or x > 5 and x < − 5

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2 F C There are 105 boys in a year group at a school. Some boys play football (F) and some play cricket (C). • x boys play both football and cricket. • The number of boys that play neither game is the same as the number of boys that play both. • 40 boys play cricket. • The number of boys that only play football is twice the number of boys that only play cricket. Complete the Venn diagram and find the value of x. [5]

Mark scheme: 2 ′ B1 n ( F ∩ C ) = n ( F ∪ C ) = x n ( C ∩ F ′ ) = 40 − x B1 n ( F ∩ C ′ ) = 80 − 2 x or 2 ( 40 −x ) B1 x + x + 40 − x + 80 − 2 x = 105 M1 x = 15 A1 cao

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3 A curve has equation y = . Find sin 2x dy (i) , [3] dx (ii) the equation of the tangent to the curve at the point where x = r . [3] 4

Mark scheme: 3(i) 3 x 2 sin2 x − x 3 × 2cos2 x 3 M1 Quotient rule 2 A2/1/0 minus one each error ( sin2 x ) isw 3(ii) π3 B1 y = [ = 0.48…] 64 dy 3π 2 B1 = [=1.85] oe dx 16 3π 2 π 3 B1 cao y = x − 16 32 [ y = 1.85 x − 0.97 ]

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4 Solve (i) 2 3x - 1 = 6 , [3] 2 (ii) log 3 (y + 14) = 1 + . [5] log y 3

Mark scheme: 4(i) Take logs : ( 3 x − 1) log2 = log6 M1 log6 A1 + 1 log2 Make x the subject : x = oe 3 awrt 1.19 A1 or awrt 1.195 4(ii) 1 = log 3 3 B1 2 B1 = 2log 3 y log y 3 3 y 2 − y − 14 = 0 B1 ( 3 y − 7 )( y + 2 ) = 0 M1 Solve a three term quadratic 7 A1 y = only 3

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5 Solve the simultaneous equations 8 p + 1 11 q = 2 , 4 3 2p + 5 3q = 9 . [5] 1 27 3

Mark scheme: 5 3( p +1) 2 p + 5 M1 2 11 3 2 ( 3 q ) = 2 or = 3 2 2 q 3 1  3  3  x a a − b a b a + b M1 Use = x or x × x = x x b 3 p + 3 − 2 q = 11 and 2 p + 5 −=1 6 q A1 Allow unsimplified M1 solve p = 4 and q = 2 A1

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6 (a) A 5-character code is to be formed from the 13 characters shown below. Each character may be used once only in any code. Letters : A, B, C, D, E, F Numbers: 1, 2, 3, 4, 5, 6, 7 Find the number of different codes in which no two letters follow each other and no two numbers follow each other. [3] (b) A netball team of 7 players is to be chosen from 10 girls. 3 of these 10 girls are sisters. Find the number of different ways the team can be chosen if the team does not contain all 3 sisters. [3]

Mark scheme: 6(a) Number first B1 = 7 × 6 × 5 × 6 × 5 or 7 P3 × 6 P2 or 6300 Letter first B1 = 6 × 5 × 4 × 7 × 6 or 6 P3 × 7 P2 or 5040 6300 + 5040 = 11 340 B1 6(b) With 2 sisters = 7 C5 × 3C 2 = 63 3 B1 One combination evaluated 7 3 B1Another combination With 1 sister = C6 × C1 = 21 evaluated With no sister = 7 C7 = 1 and B1 Third combination and 85 Total 85 OR Total no of ways = 10 C7 = 120 B1 With 3 sisters = 7 C 4 = 35 B1 Without 3 sisters = 120 − 35 = 85 B1

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8 (i) Show that - = 2 tan x sec x . [4] 1 - sin x 1 + sin x 1 1 (ii) Hence solve the equation - = cosec x for 0° G x G 360° . [4] 1 - sin x 1 + sin x

Mark scheme: 8(i) (1 + sinx ) − (1 − sinx ) M1 (1 − sinx )(1 + sinx ) 2sin x A1 1 − sin 2 x 2sinx M1 cos 2 x 2sinx 1 A1 AG × = 2tan x secx cosx cosx 8(ii) M1 equate 2sec x tan x = cosecx 2 1 A1 tan x = 2 35.3°,144.7°, 215.3°, 324.7° 2 A1 two correct

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9 y y = 2√x A (4, 4) O B x The diagram shows part of the curve y = 2 x . The normal to the curve at the point A (4, 4) meets the x-axis at the point B. (i) Find the equation of the line AB. [4] (ii) Find the coordinates of B. [1]

Mark scheme: 9(i) dy − 12 B1 = x dx dy 1 B1 x = 4 → = dx 2 grad of normal = −2 M1 y − 4 A1 = −→2 [ y = −2 x + 12 ] x − 4 9(ii) (6, 0) B1 FT 9(iii) 1 B1 FT Area of triangle = × 2 × 4 = 4 2 1 M1 Area under curve 2 x 2 d x =∫ 3 A1 4 2 = x 3 2 A1 FT Total area = 14 [14.7 ] 3 OR Area of trapezium OBAP B1 FT 1 = ( 6 + 4 ) × 4 = 20 2 Area between curve and y- axis M1 y 2 = dy ∫ 4 y 3 A1 = 12 2 A1 FT Total area = 14 [14.7 ] 3

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10 Two lines are tangents to the curve y = 12 - 4x - x 2 . The equation of each tangent is of the form y = 2 k + 1 - kx , where k is a constant. (i) Find the two possible values of k. [5]

Mark scheme: 10(i) 2 k + 1 − kx = 12 − 4 x − x 2 M1 * x 2 + 4 x − kx + 2 k − 12 + 1 b 2 − 4ac M1 Dep* → ( 4 − k ) 2 − 4 ( 2 k − 11) k 2 − 16k + 60 A1 ( k − 6 )( k − 10 ) M1 k = 6 or1 0 A1 OR k = 4 + 2 x M1 * −4 x − 2 x 2 + 8 + 4 x + 1 = 12 − 4 x − x 2 M1 Dep*  k − 4   k − 4  2 or 2 k + 1 − k   = 12 − 2 ( k − 4 ) −    2   2  x 2 − 4 x + 3 A1 or k 2 − 16 k + 60 ( x − 1)( x − 3 ) M1 or ( k − 6 )( k − 10 ) x = 1 or x = 3 → k = 6 or 10 A1 10(ii) k = 6 → [ y ] = 13 − 6 x B1 FT k = 10 → [ y ] = 21 − 10 x B1 FT M1 solve x = 2 , y = 1. 2 cao

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11 The functions f and g are defined for real values of x H 1 by f ()x = 4x - 3 , 2x + 1 g ()x = . 3x - 1 (i) Find gf ()x . [2] (ii) Find g -1 ()x . [3] (iii) Solve fg ()x = x - 1. [4]

Mark scheme: 11(i) 2 ( 4 x − 3 ) + 1 M1 gf ( x ) = 3 ( 4 x − 3 ) − 1 8 x − 5 A1 = 12 x − 10 11(ii) y ( 3 x − 1) = 2 x + 1 B1 or x ( 3 y − 1) = 2 y + 1 ( 3 y − 2 ) x = y + 1 M1 or ( 3 x − 2 ) y = x + 1 -1 x + 1 A1 g ( x ) = 3 x − 2 11(iii)  2 x + 1 B1 − = x − 1] 4   3 [  3 x − 1  3 x 2 − 3 x − 6 oe B1 3 ( x + 1)( x − 2 ) M1 x = 2 only A1

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12 A plane that can travel at 260 km/h in still air heads due North. A wind with speed 40 km/h from a bearing of 310° blows the plane off course. Find the resultant speed of the plane and its direction as a bearing correct to 1 decimal place. [6]

Mark scheme: 12 Identifying angle with downward vertical of B1 wind as 50° Triangle drawn with sides 260,40 and included B1 angle of 50° . Cosine rule : M1 * ( rv ) 2 = 260 2 + 40 2 − 2 × 260 × 40cos50 ° rv = 236 A1 sinα sin50° M1 dep* Sine rule : = 40 rv or Cosine rule : 40 2 = 260 2 + 236 2 −×2 260 × 236cosα α = 7.5 ° A1 OR Using components Identifying angle with downward vertical of B1 wind as 50°  40cos40°  B1 v w =    −40cos50°  2 2 M1 rv = ( 40cos40° ) + ( 260 − 40cos50 ) v r = 236 A1 40cos40° M1 tanα= 260 − 40cos50° α= 7.5 ° A1

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