Cambridge IGCSE Mathematics - Additional 0606 — 2020 Oct/Nov Paper 2 · Variant 1

1 Solve the inequality 3x + 2 2 8 + x . [3]

Mark scheme: Question Answer Marks Partial Marks 1 3 x + 2 > 8 + x → x > 3 B1 −3 x − 2 > 8 + x M1 Correct inequality oe x < −2.5 A1

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2 Find the coordinates of the points of intersection of the curve x 2 + xy = 9 and the line y = x - 2 . 3 [5]

Mark scheme: 2 2  2  M1 Eliminate y x + x  x − 2  = 9  3  5 x 2 − 6 x − 27 = 0 A1 ( x − 3 )( 5 x + 9 ) = 0 M1 Factorise or formula (3, 0) A1 Or both x values  9 16  A1  − , −   5 5 

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3 Write 3 lg x + 2 - lg y as a single logarithm. [3]

Mark scheme: 3 Uses lg100 = 2 or 3lgx = lgx 3 . B1 a  B1 Uses lg a + lg b = lg ab or lg a − lg b = lg   b  lg 100 x 3  B1 Correct final answer  y 

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4 It is given that y = ln ( sin x + 3 cos x) for 0 1 x 1 r. 2 dy (a) Find . [3] dx dy 1 (b) Find the value of x for which = - . [3] dx 2

Mark scheme: 4(a) d y cos x − 3sin x 3 M1 for attempt at chain rule must have = function in numerator and denominator d x sin x + 3cos x A1 for denominator A1 for numerator (b) –2 cos x – 3 cos x = sin x – 6 sin x M1 Expand and collect terms in sin x and cos x 1 = tan x M1 sinx Use = tanx cos x π A1 Must be radians x = 4

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5 The first three terms in the expansion of a + bx ( 1 + x) are 32 - 208x + cx 2 . Find the value of each of the integers a, b and c. [7]

Mark scheme: 5 a 5 + 5 a 4 bx + 10 a 3 b 2 x 2 2 B1 for powers or for coefficients 5 5 4 3 2 4 2 2 M1 for multiplying to obtain 5 terms a + a + 5a b x + 10 a b + 5a b x ( ) ( ) A1 for all correct a 5 = 32 → a = 2 A1 32 + 80b = −208 → b = −3 A1 10 × 8 × 9 + 5 × 16 ×−=3 c → c = 480 A1

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6 DO NOT USE A CALCULATOR IN THIS QUESTION. In this question all lengths are in centimetres. A 3 - 1 3 + 1 C 15° B In the diagram above AC = 3 - 1, AB = 3 + 1, angle ABC = 15° and angle CAB = 90° . (a) Show that tan15° = 2 - 3 . [3] (b) Find the exact length of BC. [2]

Mark scheme: 6(a) 3 − 1 M1 Correct use of tan tan15 ° = 3 + 1 M1 3 − 1 3 − 1 Multiply by 3 − 1 ( ) ( ) ( ) tan15 ° = × 3 + 1 3 − 1 ( ) ( ) tan15 ° = 2 − 3 A1 AG So all working must be seen 6(b) 2 2 2 M1 Correct use of Pythagoras ( BC ) = 3 − 1 + 3 + 1 ( ) ( ) BC = 8 or 2 2 A1

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7 DO NOT USE A CALCULATOR IN THIS QUESTION. p ( x) = 2x 3 - 3x 2 - 23 x + 12 1 (a) Find the value of p [1] b 2 l. (b) Write p ( x) as the product of three linear factors and hence solve p ( x) = 0 . [5]

Mark scheme: 7(a)  1   1   1   1  B1 Working must be seen p   = 2   − 3   − 23   + 12 = 0  2   8   4   2  7(b) − 12 p ( x ) = ( 2 x − 1)( x 2 − x − 12 ) 2 M1 for terms x 2 and A1 for −x p ( x ) = ( 2 x − 1)( x − 4 )( x + 3 ) 2 M1 for solving quadratic A1 for all three correct factors 1 A1 f ( x ) = 0 → x = , 4, − 3 2

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8 The population P, in millions, of a country is given by P = A # bt , where t is the number of years after January 2000 and A and b are constants. In January 2010 the population was 40 million and had increased to 45 million by January 2013. (a) Show that b = 1.04 to 2 decimal places and find A to the nearest integer. [4] (b) Find the population in January 2020, giving your answer to the nearest million. [1] (c) In January of which year will the population be over 100 million for the first time? [3]

Mark scheme: 8(a) 40 = A × b10 and 45 = A × b13 B1 3 45 M1 Divide to find b 3. b = 40 b = 1.04 A1 A = 27 A1 8(b) 59 B1 P = 27 × 1.04 20 8(c) 100 = 27 × 1.04 t M1 Insert P = 100 in their expression  100  M1 Rearrange to make t the subject log    27  t = oe log1.04 t = 33.4 → Year 2034 A1

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9 A particle moves in a straight line such that, t seconds after passing a fixed point O, its displacement from O is s m, where s = e 2 t - 10 e t - 12 t + 9 . (a) Find expressions for the velocity and acceleration at time t. [3] (b) Find the time when the particle is instantaneously at rest. [3] (c) Find the acceleration at this time. [2]

Mark scheme: 9(a) v = 2e 2 t − 10e t − 12 3 M1 for correctly differentiating 2e t . a = 4e 2 t − 10e t A1 for v correct A1 for a correct 9(b) v = 0 → e 2 t − 5e t − 6 = 0 M1 Factorise quadratic Solve and discard et = –1 → = 0 ( e t + 1)( e t − 6 ) e t = 6 A1 t = ln6 = 1.79 A1 9(c) t = ln6 → a = 4 × 36 − 10 × 6 = 84 2 M1 for inserting their value of t into a

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10 The gradient of the normal to a curve at the point (x, y) is given by . x + 1 (a) Given that the curve passes through the point (1, 4), show that its equation is y = 5 - ln x - x . [5] (b) Find, in the form y = mx + c , the equation of the tangent to the curve at the point where x = 3 . [3]

Mark scheme: 10(a) d y (1 + x )  1  2 M1 for using m1 × m2 = −1 = − = −  + 1  dx x  x  y = − lnx − x + C 2 1 M1 for integrating x A1 for all correct including C 4 = −ln1 −+1 C A1 Insert (1, 4) and arrive at correct C = 5 → y = 5 − lnx − x answer. AG 10(b) x = 3 → y = 2 − ln3 B1 d y 1 4 and = − −=1 − d x 3 3 y − ( 2 − ln3 ) 4 M1 = − x − 3 3 4 A1 y = − x + 6 − ln3 3 or y = −1.33 x + 4.90

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11 The equation of a curve is y = x 16 - x 2 for 0 G x G 4 . (a) Find the exact coordinates of the stationary point of the curve. [6] d 2 23 2(b) Find 16 - x and hence evaluate the area enclosed by the curve y = x 16 - x and the d x ` j lines y = 0, x = 1 and x = 3 . [5]

Mark scheme: 11(a) d y 1 2 − 12 2 12 3 d 2 12 B1 for = x × (16 − x ) × ( − 2 x ) + (16 − x ) (16 − x ) d x 2 d x − 1 2 1 = 2 × ( − 2 x ) (16 − x ) 2 M1 for product rule A1 for all correct 1 2 3 dy d y 2 2 x M1 for setting = 0 and attempt to = 0 → 16 − x = ( ) 1 dx dx 2 2 16 − x ( ) solve 2 2 M1 for obtaining x = k x = 8 2 2, 8 ( ) A1 11(b) 1 2 M1 for attempt at chain rule 3 2 2 × ( − 2 x ) A1 for all correct unsimplified (16 − x ) 2 3 1 3 3 3 3   1 2 2 M1 for obtaining k − 2 dx = ( 16 − x 2 ) Area =   ( 16 − x 2 ) ( 16 − x 2 ) x 3 1  1 1  32 32  A1 for obtaining k = − 1 = −  7 − 15  = 13.2 3 3   A1 for 13.2

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12 C 3 cm 4 cm A B 5 cm D The diagram shows a shape consisting of two circles of radius 3 cm and 4 cm with centres A and B which are 5 cm apart. The circles intersect at C and D as shown. The lines AC and BC are tangents to the circles, centres B and A respectively. Find (a) the angle CAB in radians, [2] (b) the perimeter of the whole shape, [4] (c) the area of the whole shape. [4]

Mark scheme: 12(a) 4 M1 Correct use of tan oe tan CAB = 3 CAB = 0.927 A1 isw 12(b)  π  B1 Angle CBD = 2  − 0.927  = 1.287  2  Perimeter 3 = 3 ( 2 π − 2 × 0.927 ) + 4 ( 2 π − 1.287 ) M1 for correct plan of two arcs A1 for either arc = 13.287 + 19.985 A1 = 33.3 12(c) Area of two right-angled triangles B1 1 = × 3 × 4 × 2 = 12 2 Area of Sectors 3 32 4 2 M1 for correct plan of two sectors plus = ( 2 π − 2 × 0.927 ) + ( 2 π − 1.287 ) triangles 2 2 A1 for either sector = 19.93 + 39.97 A1 Total = 71.9

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