Cambridge IGCSE Mathematics - Additional 0606 — 2020 Oct/Nov Paper 2 · Variant 2

1 Solve the inequality ( x - 8)( x - 10) 2 35 . [4] + 1

Mark scheme: Question Answer Marks Partial Marks 1 x2 – 18x + 45 (= 0) B1 Expand and simplify to three terms. (x – 15)(x – 3)(= 0) M1 Factorise or use formula on their 3 term 2 quadratic or complete the square 18 ± 18 −×4 45 or x = 2 or (x – 9)2 = –45 + 81 x = 15 and x = 3 A1 x < 3 or x > 15 A1 oe Do not accept ‘and’. or (–∞, 3) ∪ (15, ∞) Do not accept 3 > x > 15. Mark final answer.

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3 (a) Find the equation of the perpendicular bisector of the line joining the points (12, 1) and (4, 3), giving your answer in the form y = mx + c . [5] (b) The perpendicular bisector cuts the axes at points A and B. Find the length of AB. [3]

Mark scheme: 3(a) 3 − 1  1  B1 Gradient of line =  −  4 − 12  4  Gradient of perpendicular = 4 M1 − 1 their grad line Mid-point is (8, 2) B1 y − 2 M1 Using their perpendicular gradient and Equation: = 4 mid-point x − 8 y = 4x – 30 A1 3(b) x = 0 → (y) = –30 B1 FT equation must have 3 terms y = 0 → (x) = 7.5 B1 FT equation must have 3 terms B1 15 17 AB = 30 2 + 7.5 2 = 30.9 or better nfww Accept exact answer of 2

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4 Solve the simultaneous equations. log 3 ( x + y) = 2 2 log 3 ( x + 1) = log 3 ( y + 2) [6]

Mark scheme: 4 x + y = 9 B1 (x + 1)2 = y + 2 B1 x + (x + 1)2 – 2 = 9 M1 Replace y or x. Allow unsimplified using their three term expressions both or (10 – y)2 = y + 2 containing x and y terms. Condone one sign or arithmetic error. Result must be a quadratic function. x2 + 3x – 10 (= 0) A1 Correct 3 term quadratic or y2 – 21y + 98 (= 0) x = –5 and x = 2 M1 Dep on correct method to solve their or y = 7 and y = 14 quadratic or (x + 5)(x – 2) or (y – 7)(y – 14) x = 2 and y = 7 only A1 Reject x = –5, y = 14 as log –4 is not appropriate

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5 DO NOT USE A CALCULATOR IN THIS QUESTION. (a) Find the equation of the tangent to the curve y = x 3 - 6x 2 + 3x + 10 at the point where x = 1. [4] (b) Find the coordinates of the point where this tangent meets the curve again. [5]

Mark scheme: 5(a) x = 1 → y = 8 B1 dy M1 Attempt to differentiate. Powers reduced = 3x2 – 12x + 3 by 1 in all four terms. dx dy A1 x = 1 → = –6 dx y − 8 A1 Either form. = −→6 y = −6 x + 14 isw x − 1 5(b) x3 – 6x2 + 9x – 4 = 0 2 M1 for equating their tangent to curve and (x – 1)(x2 – 5x + 4) = 0 simplifying to 4 term cubic. or (x – 4)(x2 – 2x + 1) = 0 M1Dep for finding a factor or stating that (x – 1) is a factor or makes at least 3 attempts to find a factor. (x – 1)(x – 1)(x – 4) = 0 2 A1 for (x – 1) or x = 1 can be implied. nfww A1 for (x – 4) or x = 4 not repeated. nfww x = 4 → y = –10 only A1 nfww

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7 A geometric progression has a first term of 3 and a second term of 2.4. For this progression, find (a) the sum of the first 8 terms, [3] (b) the sum to infinity, [1] (c) the least number of terms for which the sum is greater than 95% of the sum to infinity. [4]

Mark scheme: 7 (a) 2.4 B1 a = 3 r = = 0.8 3 3(1 − 0.88 ) M1 Inserts their ܽ and ݎ into ଼ܵ S 8 = (1 − 0.8) = 12.48 awrt or 12.5 A1 7(b) 3 B1 S∞= = 15 (1 − 0.8) 7(c) Sn = 15(1 – 0.8n) > 0.95 × 15 M1 their correctly produced Sn > 0.95S∞ 0.8n < 0.05 A1 oe log0.05 M1 Dep takes logs correctly of their n < or n < log 0.8 0.05 expression with power of n. log0.8 n = 14 A1 nfww

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8 DO NOT USE A CALCULATOR IN THIS QUESTION. In this question lengths are in centimetres. B You may use the following trigonometric ratios. 2 3 + 1 1 sin30° = 2 3 30° cos30 ° = A 2 1 tan30° = 3 C (a) Given that the area of the triangle ABC is 5.5 cm2, find the exact length of AC. Write your answer in the form a + b 3 , where a and b are integers. [4] (b) Show that BC 2 = c + d 3 , where c and d are integers to be found. [4]

Mark scheme: 8(a) 1 11 M1 Correct use of area of a triangle (2 3 + 1) AC sin 30° = 2 2 (2 3 + 1) AC = 22 A1 oe 22 (2 3 − 1) M1 Multiply by their (2 3 + 1) AC = × (2 3 + 1) (2 3 − 1) AC = 4 3 –2 A1 8(b) 2 2 M1 Correct use of cosine rule with their AC. BC = 2 3 + 1 + 4 3 − 2 ( ) ( ) −2 2 3 + 1 4 3 − 2 cos30 ( )( ) BC 2 = 13 + 4 3  +  52 − 16 3  A2 A1 for one correct expanded bracket     A1 for the other two correct expanded +  − 22 3  brackets   BC 2 = 65 − 34 3 A1

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9 P a Q 2b R X b O S 3a In the diagram OP = 2b , O S = 3a , SR = b and PQ = a . The lines OR and QS intersect at X. (a) Find OQ in terms of a and b. [1] (b) Find Q S in terms of a and b. [1] (c) Given that QX = n Q S , find OX in terms of a, b and n. [1] (d) Given that OX = m OR , find OX in terms of a, b and m. [1] (e) Find the value of m and of n. [3] QX (f) Find the value of . [1] X S OR(g) Find the value of . [1] OX

Mark scheme: 9(a) 2 b + a B1 9(b) 2 a − 2 b B1   9(c) 2b + a + μ(2a − 2b ) B1 FT on their OQ and QR isw 9(d) λ( 3a + b ) B1 λ3a + b is B0 9(e) 3λ = 1 + 2μ 3 M1 for forming two simultaneous λ = 2 − 2μ equations equating correct terms. Each equation must have 3 terms. M1Dep for attempting to solve by 3 5 λ= , μ = removing µ or λto λ = or µ = 4 8 A1 for both 9(f) QX 5 B1 FT Must be positive from µ < 1 = XS 3 9(g) OR 4 B1 FT Must be positive from λ < 1 = OX 3

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10 The number, b, of bacteria in a sample is given by b = P + Q e t2 , where P and Q are constants and t is time in weeks. Initially there are 500 bacteria which increase to 600 after 1 week. (a) Find the value of P and of Q. [4] (b) Find the number of bacteria present after 2 weeks. [1] (c) Find the first week in which the number of bacteria is greater than 1 000 000. [3]

Mark scheme: 10(a) P + Q = 500 and P + Q2e = 600 B1 100 2 M1 for attempt to solve by removing P Q = = 15.7 or 15.6 2 from two equations both containing 3 (e − 1) terms A1 awrt P = 484 or 485 A1 awrt 10(b) B = 484.3 + 15.65e 4 = 1338 B1 Integer value rounded down from 1338… if seen. 10(c) 2 t 1 000 000 − 484.3 M1 Make e 2 t the subject e = 15.65  1 000 000 − 484.3  M1 Take logs correctly where e 2 t > 0 or 2t = ln   n  15.65  e > 0  =t 5.5 ( 3 ) or t = 5.5 ….→ 6th week. A1 nfww

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(b) Solve the equation 5 tan x - 3 cot x = 2 sec x for 0° G x G 360 ° . [6]

Mark scheme: 11(a) sinx M1 sinx sinx × Uses tanx = cosx cosx LHS = 1 − cosx 1 − cos 2 x M1 Dep Uses sin 2 x = 1 − cos 2 x to eliminate = cos x (1 − cosx ) sinx (1 − cos x )(1 + cosx ) 1 + cosx 2 M1Dep Factorise correctly and cancel = = secx + 1 correctly. cos x (1 − cosx ) cosx 1 A1 Uses = secx cosx 11(b) sinx cosx 2 B1 Change tan x, cot x and sec x into sin x and 5 − 3 = cos x correctly. cosx sinx cosx 2 2 M1 Multiply correctly by sin x cos x and use 5sin x − 3 1 − sin x = 2sin x ( ) 2 2 cos x + sin x = 1 8sin 2 x − 2sin x − 3 = 0 A1 Three term quadratic. ( 2sinx + 1)( 4sinx − 3 ) = 0 M1 Factorise or use formula on their quadratic 1 A1 sinx = − → x = 210° , 330° 2 3 A1 sinx = → x = 48.6°,131.4° 4

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