Cambridge IGCSE Mathematics - Additional 0606 — 2018 Oct/Nov Paper 2 · Variant 1

1 Solve the inequality (2 - x)(x + 9) 1 10 . [4]

Mark scheme: Question Answer Marks Partial Marks 1 x 2 + 7 x − 8 ( > 0) 2 M1for expanding and collecting terms x < –8 or x > 1 2 M1 for factorising ( x + 8 )( x − 1) > 0

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2 (a) Solve 3 - 1 2 = 10 . [3] (b) Solve 2e 1 - 2y = 3e 3y + 2 . [4]

Mark scheme: 2(a)  x  M1 Take logs: − 1 log3 = log10    2   log10  M1 Make x the subject: x = 2  + 1   log3  6.19 A1 2(b) 5 y+1 2 2 M1 for attempt to combine e = exponential terms 3 –0.281 2 M1 for taking natural logs:  2  5 y + 1 = ln    3 

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3 Do not use a calculator in this question. (a) Simplify ( 2 + 2 5 )(4 2 - 3 5 ) , giving your answer in the form a + b c , where a, b and c are integers. [3] 4 - 3 6 (b) Simplify , giving your answer in the form p 3 + q 2 , where p and q are integers. [4] 3 + 2

Mark scheme: 3(a) Expand 4 terms: M1 8 + 8 10 − 3 10 − 30 –22 A1 5 10 A1 3(b) M1 Multiply numerator and 4 − 3 6 3 − 2 ( ) ( ) × denominator by 3 − 2 ( ) 3 + 2 3 − 2 ( ) ( ) M1 Expand 4 3 − 3 18 − 4 2 + 3 12 3 − 2 10 3 −13 2 A2 A1for each term

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5 A = . - 1 1 (i) Find A2. [2] (ii) Find constants p and q such that pA 2 + qA = I . [4]

Mark scheme: 5(i) 2  7 8  2 Minus 1 each error. A =    −4 −1  5(ii) 7 p + 3q = 1 2 M1 forms two equations in p and q 8 p + 2 q = 0 A1 Both correct −4 p − q = 0 , − p + q = 1 1 4 2 M1 solves equations to find p and q p = − , q = 5 5

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6 A 5-digit code is to be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9. Each digit can be used once only in any code. Find how many codes can be formed if (i) the first digit of the code is 6 and the other four digits are odd, [2] (ii) each of the first three digits is even, [2] (iii) the first and last digits are prime. [2]

Mark scheme: 6(i) 120 2 B2 5 × 4 × 3 × 2 or B1for pattern n ( n − 1)( n − 2)( n − 3) 6(ii) 720 2 B1 4 × 3 × 2 B1 dep × 6 × 5 = 720 6(iii) 2520 2 B1 4 ×…×…×…× 3 B1 Dep × 7 × 6 × 5 = 2520

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7 (i) Show that - = 2 cosec x cot x . [4] 1 - cos x 1 + cos x 1 1 (ii) Hence solve the equation - = sec x for 0 G x G 2 r radians. [4] 1 - cos x 1 + cos x

Mark scheme: 7(i) (1 + cos x ) − (1 − cos x ) M1 Taking common denominator (1 − cos x )(1 + cos x ) 2cosx A1 = 2 1 − cos x 2cosx M1 Using 1 − cos 2 x = sin 2 x = 2 sin x 2cosx 1 A1 Fully correct completion = × AG sinx sinx = 2cosecxcot x 7(ii) 2cosecxcotx = secx M1 2 1 A1 cot x = 2 0.955, 2.19, 4.10, 5.33 A2 A1 for 2 correct values A1 for further 2 correct values

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8 y y = x + e 5 - 2x A B 0 5 x The diagram shows part of the curve y = x + e 5 - 2x , the normal to the curve at the point A and the line x = 5 . The normal to the curve at A meets the y-axis at the point B. The x-coordinate of A is 2.5. (i) Find the equation of the normal AB. [4] (ii) Showing all your working, find the area of the shaded region. [6]

Mark scheme: 8(i) dy 2 − 5 x B1 = 1 − 2e dx dy B1 x = 2.5 → = −1 and y = 3.5 dx −1 M1 Grad of normal = dy dx y = x + 1 A1 Equation of normal 8(ii) 1 M1 Area of trapezium = × 2.5 × 4.5 2 5.625 sq units A1 5 M1 Area under curve x + e dx ∫ ( 5 − 2 x ) 2.5 5 A1 2  x 1 ( 5 − 2 x )  = − e    2 2  2.5 M1 insert limits and subtract (= 9.87) Shaded area = 15.5 A1 5.625 + 9.87

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9 In this question, all lengths are in metres. y 2r The diagram shows a window formed by a semi-circle of radius r on top of a rectangle with dimensions 2r by y. The total perimeter of the window is 5. (i) Find y in terms of r. [2] 2 r r 2 (ii) Show that the total area of the window is A = 5r - - 2r . [2] 2 (iii) Given that r can vary, find the value of r which gives a maximum area of the window and find this area. (You are not required to show that this area is a maximum.) [5]

Mark scheme: 9(i) 2 y + 2r + πr = 5 B1 5 − 2 r −πr B1 Dep y = 2 9(ii) πr 2 M1 A = 2 yr + 2 πr 2 A1 = r ( 5 − 2 r −πr ) + 2 2 πr 2 = 5 r − 2 r − 2 9(iii) M1 differentiate dA A1 = 5 −πr − 4 r dr dA M1 set to zero and attempt to solve = 0 dr 5 A1 r = = 0.7 π+ 4 A = 1.75 A1

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10 The line y = 12 - 2x is a tangent to two curves. Each curve has an equation of the form y = k + 6 + kx - x 2 , where k is a constant. (i) Find the two values of k. [5] The line y = 12 - 2x is a tangent to one curve at the point A and the other curve at the point B. (ii) Find the coordinates of A and of B. [3] (iii) Find the equation of the perpendicular bisector of AB. [3]

Mark scheme: 10(i) 12 − 2 x = k + 6 + kx − x 2 M1 * Equate and collect terms → x 2 − ( 2 + k ) x + 6 − k = 0 b 2 − 4ac = 0 M1 Dep* → ( 2 + k ) 2 = 4 ( 6 − k ) k 2 + 8k − 20 = 0 A1 ( k + 10 )( k − 2 ) = 0 M1 k = −10 or 2 A1 M1 Insert values of k in equations10(ii) ( − 4, 20 ) and ( 2, 8 ) 3 and solve for x A1 x 2 + 8 x + 16 = 0 → x = −4 → y = 20 A1 x 2 − 4 x + 4 = 0 → x = 2 → y = 8 10(iii) 1 B1 Grad of perpendicular = 2 Midpoint ( − 1 ,1 4 ) B1 FT y − 14 1 1 B1 FT Eqn = → y = x + 14.5 x + 1 2 2

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11 R H x 15 9 x x – 2 N There are 70 girls in a year group at a school. The Venn diagram gives some information about the numbers of these girls who play rounders (R), hockey (H) and netball (N). n(R) = 28 n(H) = 38 n(N) = 35. Find the value of x and hence the number of girls who play netball only. [6]

Mark scheme: B1 ′11 n ( ( R ∩ H ) ∩ N ) = 14 − x B1 ′ n ( ( R ∩ N ) ∩ H ) = 5 n( N ∩ (R ∪ H )') = 21 − x B1 x + 9 + x + 15 + 14 − x + 5 + 21 − x + x − 2 M1 correctly form equation in x and = 70 attempt to solve x = 8 A1 n( N ∩ (R ∪ H )') = 13 A1

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