Cambridge IGCSE Mathematics - Additional 0606 — 2024 Feb/March Paper 2 · Variant 2

1 (a) Solve the equation 2 8 - 4x + 5 = 25 . [3] 2 57 - 9x (b) Solve the inequality 16 x - 5x - 3 1 . [4] 6

Mark scheme: Question Answer Marks Partial Marks 1(a) 8 – 4x = 10 oe soi M1 and 8 – 4x = −10 oe soi OR 16 x 2 − 64 x − 36  = 0 oe 1 9 A2 mark final answer x = − , x = 2 2 1 9 A1 for x = − or x = 2 2 1(b) −30 x 2 + 105 x − 75 *0  oe M1 condone one sign or arithmetic error where * is any inequality sign or = Critical values 2.5 and 1 2 M1 for factorises or solves a 3-term quadratic to find critical values x < 1 x > 2.5 A1 mark final answer

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2 DO NOT USE A CALCULATOR IN THIS QUESTION. In this question all lengths are in centimetres. a + b 5 1 + 7 5 4 + 2 5 20 The diagram shows two similar triangles. The height of the smaller triangle is 1 + 7 5 and the height of the larger triangle is a + b 5 , where a and b are integers. Find the values of a and b. [4]

Mark scheme: 2 a + b 5 20 20(1 + 7 5) B1 = or oe, soi 1 + 7 5 4 + 2 5 4 + 2 5 + 7 5 4 − 2 5 M1 condone one slip providing it is not in  20  1  the rationalisation factor 4 + 2 5 4 − 2 5 + 7 5 2 − 5 or 10  1  oe 2 + 5 2 − 5 A1 20(28 5 − 70 + 4 − 2 5) 16 − 20 10(14 5 − 35 + 2 − 5) or oe 4 − 5 a = 330 and b = −130 oe, nfww A1 Alternative method a + b 5 20 (B1) = oe, soi 1 + 7 5 4 + 2 5 Cross multiplies and multiplies out: (M1) condone one sign or arithmetic error 20 + 140 5 = 4 a + 4b 5 + 2 a 5 + 10b Correct pair of simultaneous equations (A1) 4a + 10b = 20 oe 2a + 4b = 140 oe and solves for a = 330 or b = −130 a = 330 and b = −130 oe, nfww (A1)

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3 (a) B b P A O a The diagram shows a triangle OAB. The point P lies on AB. The ratio AP : PB is 1 : 3. Given that OA = a and OB = b , find an expression for OP in terms of a and b. Simplify your answer. [2] J 6N (b) Vector q has magnitude 12 5 and direction KK OO. - 3 L P J- 5N Vector r has magnitude 15 2 and direction KK OO. 5 L P Find the unit vector in the direction of q + r . [6]

Mark scheme: 3(a) 3 1 B2 1 3 a + b or equivalent simplified B1 for a + (b – a) or b + (a – b) 4 4 4 4 expression oe or for 3( OP – a) = b – OP oe 3(b)  24  2 1  6  oe, oe M1 for 12 5    q =    −3  6 2 + ( −3) 2  −12  soi  −15  2 1  −5  oe, oe M1 for 15 2    r =    5  ( −5) 2 + 5 2  15  soi If M0 M0, then SC1 for the unit 1  6  direction vectors   or better 45  −3  1  −5  and   or better 50  5  M1 FT their (q + r) providing at least M1 9 2 2 q + r =  = 9 + 3 previously awarded 3 1 9 A1 [unit vector in direction q + r =]  oe, 90 3 isw

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4 (a) (i) Given that y = 3 sin 2 x + cos x , show that y + cot x = k ( 1 + cos 2 x) , where k is an dx integer. [4] (ii) Using your value of k, solve the equation k ( 1 + cos 2 x) = 4 for - r G x G r . [4] (b) (i) Differentiate y = tan x - x with respect to x. [2] ` j 2 x - 1 (ii) Hence find dx . [2] y 2 x cos x - x ` j

Mark scheme: 4(a)(i) dy B2 B1 for an attempt to differentiate both = 6sin x cos x − sin x oe, isw terms with one term correct dx 2 cos x M1 d y 3sin x + cos x + ( 6sin x cos x − sin x ) FT their of the form sin x d x k sin x cos x  sin x Correct simplified step e.g. A1 3sin 2 x + cos x + 6cos 2 x − cos x or 3sin 2 x + 6cos 2 x or 3 + 3cos 2 x leading to 3(1 + cos 2 x ) nfww 4(a)(ii) 2 1 M1 FT their k providing 0 < k ≤ 4 cos x = 3 M1 dep on previous M1; FT their k cos x = 1 oe 3 0.955 or 0.9553[1...] rot to 4 or more sf A2 and no other angles in range 2.19 or 2.186[2...] rot to 4 or more sf A1 for any two correct angles, ignoring extras M1 for f( x)sec 2 ( x − x )4(b)(i)  1 − 12  2 2  1 − x  sec ( x − x ) oe, isw  2  4(b)(ii) Correctly writes M1 where k is a non-zero constant;  1 − 12  2 dependent on part (b)(i)  1 − x  sec ( x − x ) =  2  2 x − 1 2 x − 1sec 2 ( x − x ) or 2 x 2 x cos 2 ( x − x ) and states an answer k tan ( x − x ) or 1 2 x − 1 states dx = tan( x − x ) 2  x cos 2 ( x − x ) 2tan ( x − x ) + c nfww A1

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5 Variables x and y are related by the equation y = . Use differentiation to find the approximate ln 3x change in y when x increases from 1 to 1+ h , where h is small. [4]

Mark scheme: 5 Correct quotient rule: 2 M1 for  1   1  (ln3 x )[1] − x   3  (ln3 x )[1] − x  their   3  dy  3 x  dy  3 x  = oe = OR dx (ln3 x ) 2 dx (ln3 x ) 2 OR correct product rule using y = x(ln3x)–1: for dy  −2 3  −1 dy  −2 3  = x  − (ln3 x )   + 1 (ln3 x ) = x  their  −(ln3 x )   dx  3 x  dx  3 x  +[1](ln3 x ) −1 δ y ln3 − 1 M1 d y = oe, soi FT their providing quotient 2 h (ln3) d x x =1 rule or appropriate product rule attempted ln3 − 1 A1 must have evidence of correct δy = h or δy = 0.0817 h nfww 2 derivative (ln3)

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6 Find the exact area of the region enclosed by the curve y = e 2 - 4 x , the x-axis, the line x =- 0.25 and the line x = 0.5 . [4]

Mark scheme: 6  1 2 − 4 x  0.5 B2 B1 for ke 2 −x4 , k −4 − e oe    4  −0.25 Correct use of correct limits: M1 1 2 − 4 x FT their − e providing B1 4 1 0  1 3  awarded − e −− e  oe 4  4  1 1 3 A1 − + e or exact equivalent, isw 4 4

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7 (a) The curves 4x 2 - 3y 2 + xy = 24 and y = intersect at the points P and Q. Find the coordinates x of P and Q. [5] (b) Find the length of PQ. Give your answer in the form a b, where a is rational and b is the smallest possible integer. [2]

Mark scheme: 7(a) Correctly eliminates x or y e.g. M1 2  2  2  2  4 x − 3   + x   = 24 oe or  x   x   2  2 2  2  4 − 3 y + y = 24 oe      y   y  Rearranges to a 3-term quadratic in x2 or y2 A1 soi e.g. 4 x 4 − 22 x 2 − 12  = 0  or 2 x 4 − 11x 2 − 6  = 0  or 3 y 4 + 22 y 2 − 16  = 0  oe Factorises or solves their 3-term quadratic in M1 x2 or y2 soi e.g. (2x2 + 1)(x2 − 6) or (3y2 – 2)(y2 + 8) 2 2 2 A1 x = 6 oe, nfww or y = nfww 3  2   2  A1 and no other values;  6,    or   6,   oe, nfww dep on at least the first M1 A1  6   3  7(b) 2 2 M1 FT providing their xP, xQ and their yP, ( xP − xQ ) + ( y P − yQ ) oe, soi yQ are non-zero 4 A1 15 3

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8 Variables y and x are known to be connected by the relationship y = Abx where A and b are constants. The table shows values of y for certain values of x. x 1 3 5 10 12 y 38 150 600 20 500 82 000 (a) Draw the graph of lgy against x. [2] lg y 5 4 3 2 1 0 2 4 6 8 10 12 x (b) Use your graph to find values of A and b, giving each to 1 significant figure. [6] (c) Find an estimate of x when y = 1500 . [2]

Mark scheme: 8(a) Points plotted at B2 B1 for at least 4 correctly plotted points x 1 3 5 10 12 lg y 1.6 2.2 2.8 4.3 4.9 soi and ruled, single straight line of best fit 8(b) lgy = lgA + xlgb soi B1 lgA = their 1.3 soi M1 dep on using linear points 4.9 − 1.6 M1 dep on using linear points lgb = their oe or lgb = 0.3 oe soi 12 − 1 3 A2 3 A = 101.3 isw and b = 10 10 isw A1 for A = 101.3 isw or b = 10 10 isw A = 20 and b = 2 nfww A1 If zero scored, award SC1 for A = 20 and SC1 for b = 2 found without using the graph in any way 8(c) lg1500 = 3.2 or 3.17[60...] M1  1500  FT their A and b OR x = log theirb    theirA  lg1500 − their lg A FT their lgA and lgb OR x = their lg b awrt 6.2 to awrt 6.4 isw A1

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9 In this question all lengths are in centimetres and all angles are in radians. C B A 0.5 2 O 1 E D F The diagram shows a company logo. Each part of the logo is a sector of a circle with centre O. Sector AOB has radius x. Sector COD has radius x + 2 . Sector EOF has radius y. The shaded region has area A cm 2 and perimeter 24. It is given that x and y can vary. 91 2 (a) Show that A = x - 68 x + 132 . [4] 8

Mark scheme: 9(a) 1 2 1 2 1 2 B1  A =  x  0.5 + ( x + 2) +2 y [1] soi 2 2 2 [P = ] M1 Attempts to form an expression in x x + 0.5 x + 2 + 2( x + 2) + ( x + 2 − y ) + y + y and y for the perimeter using arc lengths and lengths of lines 9 A1 Equates P to 24 and rearranges: y = 16 − x 2 5 2 81 2 A1 A = x + 4 x + 4 + 128 − 72 x + x oe 4 8 leading to given answer 91 2 A = x − 68 x + 132 8 9(b) dA 91 M1 = x − 68 dx 4 dA M1 dA Solves = 0 for x FT their providing at least one dx dx term is correct 272 90 A1 x = or 2 or 91 91 2.99 or 2.989[01...] rot to 4 or more sf 2 M1 FT their x 91  272   272  A =   − 68   + 132 8  91   91  2764 34 A1 A = or 30 or 91 91 30.4 or 30.37[36...] rot to 4 or more sf

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10 The expansion of a + in ascending powers of x begins b 4 + 48b 3 x , where n, a and b are a positive integers. n 2 2 - 4 48 (a) Show that a = . [4] b n l (b) Given also that the third term is 1056 b 2 x 2, find the values of n, a and b. [6] Question 11 is printed on the next page.

Mark scheme: 10(a) n 4 n −1 1 3 M1 a = b and na  = 48b oe  a  Eliminates b from one equation using the M1 dep previous M1 other equation e.g. a n − 2 48 = 3 n n a 4 Simplifies a terms e.g. A1 n 3 n 3 2 48 4 − 6  48  4 a −= or a =   n  n  Uses an appropriate power and completes to A1 the given form e.g. 2 2 3 2 3 n n 3 − 6 − 2 3  48   48  4 4 ) 2 ) = or ( a ( a =      n   n  n 2 − 4  48  2 → a =    n  10(b) Correct equation in a, b, n M1 n ( n − 1) n − 2 1 2  a  = 1056b oe, soi 2 a 2 Correct equation in a, n A1 n ( n − 1) n − 2 1 n2  a  = 1056 a oe 2 2 a − 4  A1  n( n − 1) n2   a = 1056 oe →  2  Correct equation in n only n ( n − 1)  48  2    = 1056 oe 2  n  n2 – 12n = 0 or n – 12 = 0 oe A1 n = 12 only A1 a = 4 only and b = 64 only A1

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11 A cylinder, open at both ends, has base radius r cm and height 4r cm. Its curved surface area is S cm2. dS dr Given that r varies with time t, find S at the instant when = 6 . [5] dt dt

Mark scheme: 11 dS dS dr dS B1 =  or = 6 soi dt dr dt dr 2 B1 S = 2πr (4r ) or 8πr 16πr = 6 M1 FT their S = kr2 with k a positive integer to give 2kπr = 6 6 A1 r = oe, isw 16π 9 A1 S = oe, isw 8π

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