Cambridge IGCSE Mathematics - Additional 0606 — 2021 Feb/March Paper 2 · Variant 2

1 Solve the equation 4x + 9 = 6 - 5 x . [3]

Mark scheme: Question Answer Marks Partial Marks 1 4x + 9 = 6 − 5x oe M1 or 4x + 9 = 5x – 6 oe 1 A2 not from wrong working; no extras x = − , x = 15 3 A1 for x = 15 ignoring extras implies M1 if no extras seen mark final answer If M0 then SC1 for any correct value with at most one extra value Alternative method: M1 for (4x + 9)2 = (6 – 5x)2 oe soi A1 for 9 x 2 − 132 x − 45 = 0 oe 1 A1 for x = − , x = 15 only; mark 3 final answer

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2 Find the values of the constant k for which the equation kx 2 - 3 ( k + 1) x + 25 = 0 has equal roots. [4]

Mark scheme: 2 Uses b 2 − 4 ac with at most one M1 error in substitution: ( −3( k + 1)) 2 − 4( k )(25)*0 9 k 2 − 82 k + 9*0 A1 Factorises or solves their 3-term M1 quadratic 1 A1 k = or 9; mark final answer 9

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3 y 2 1 – 2 – 1 0 1 2 x – 1 – 2 The diagram shows the graph of y = f ( x) , where f ( x) = a ( x + b) 2 ( x + c) and a, b and c are integers. (a) Find the value of each of a, b and c. [2] (b) Hence solve the inequality f ( )x G- 1. [3]

Mark scheme: 3(a) a = 2, b = 1, c = −1 B2 B1 for any two correct 3(b) Finds three correct critical values: B1 −1.5 to −1.4 inclusive −0.4 0.8 to 0.9 inclusive A correct pair of inequalities B2 B1 for either inequality correct

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4 The curve 2 + 2 = 1 and the line x + 2y = 0 intersect at two points. Find the exact distance x 4 y between these points. [6]

Mark scheme: 4 Correctly eliminates one unknown: M1 4 5 + = 1 ( −y2 ) 2 4 y 2 4 5 or 2 + 2 = 1 x  x  4  −   2  Simplifies and rearranges e.g. : M1 FT omitted brackets; condone one slip 4 5 2 + = 1 → 4 + 5 = 4 y 4 y 2 4 y 2 4 5 2 or + = 1 → 4 + 5 = x x 2 x 2 3 A2 3 y = ± and x = ±3 oe A1 for y = ± or x = ±3 2 2 2 2 M1 3 (3 −−3) + (1.5 −−1.5) FT their y = ± and x = ±3 provided that 2 no FT coordinate is 0 45 or 3 5 indicated as final A1 answer

5 A cube of side x cm has surface area S cm2. The volume, V cm3, of the cube is increasing at a rate of 480cm 3 s -1 . Find, at the instant when V = 512, (a) the rate of increase of x, [4] (b) the rate of increase of S. [2]

Mark scheme: 5(a)  d(x 3 )  2 3 B1  =  3 x and x = 512 soi  dx  OR  d( 3 V )  1 − 23  =  V  dV  3 dx dV dx B1 = × oe, soi dt dt dV 480 M1 dV 2 oe FT their = k (8) 2 3(8) dx d x − 23 or = k (512) k ≠ 0 d V 2.5oe A1 5(b) 12(8) ×their 2.5 soi M1 FT their 8 provided it is not 512 240 A1 FT provided at least M1 earned in (a)

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6 A B 16 cm 2 r rad C 7 7.5 cm O 2r AOB is a sector of a circle with centre O and radius 16 cm. Angle AOB is radians. The point C lies 7 on OB such that OC is of length 7.5 cm and AC is a straight line. (a) Find the perimeter of the shaded region. [3] (b) Find the area of the shaded region. [3]

Mark scheme: 6(a) M2 M1 for 2 2 2π 16 + 7.5 − 2(16)(7.5)cos 2π 7 16 2 + 7.5 2 − 2(16)(7.5)cos + (16 − 7.5) 7 2π + (16 − 7.5) + 16 × 2π 7 or for 16 × seen oe, soi 7 35.6 or 35.6 to 35.614 A1 6(b) 1 2 2π M2 1 2 2π × 16 × − M1 for either × 16 × or 2 7 2 7 1  2π  1  2π  × 16 × 7.5 × sin   oe × 16 × 7.5 × sin   2  7  2  7  68[.0] or 67.98 to 68.0 A1

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7 A curve has equation y = p ( x) , where p ( )x = x 3 - 4x 2 + 6x - 1. (a) Find the equation of the tangent to the curve at the point (3, 8). Give your answer in the form y = mx + c . [5] (b) (i) Given that p -1 exists, write down the gradient of the tangent to the curve y = p -1 ( x) at the point (8, 3). [1] (ii) Find the coordinates of the point of intersection of these two tangents. [2]

Mark scheme: 7(a) dy 2 B1 = 3 x − 8 x + 6 dx d y M1 condone one slip Finds their d x x = 3 mT1 = 9 A1 y – 8 = their 9(x – 3) M1 or y = 9x + c and 8 = 9(3) + c y = 9x – 19 cao A1 7(b)(i) 1 B1 FT their 9 mT2 = their 9 7(b)(ii) [Uses y = x in their ( y = 9x – 19) M1 to form] their ( x = 9x – 19) or their ( y = 9y – 19) oe and solves for x or y or solves e.g. x + 19 their (9x – 19) = their 9  19 19  A1 FT equal x and y coordinates providing at  ,  oe least 3 marks earned in (a)  8 8 

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8 A photographer takes 12 different photographs. There are 3 of sunsets, 4 of oceans, and 5 of mountains. (a) The photographs are arranged in a line on a wall. (i) How many possible arrangements are there if there are no restrictions? [1] (ii) How many possible arrangements are there if the first photograph is of a sunset and the last photograph is of an ocean? [2] (iii) How many possible arrangements are there if all the photographs of mountains are next to each other? [2] (b) Three of the photographs are to be selected for a competition. (i) Find the number of different possible selections if no photograph of a sunset is chosen. [2] (ii) Find the number of different possible selections if one photograph of each type (sunset, ocean, mountain) is chosen. [2]

Mark scheme: 8(a)(i) 479 001 600 oe B1 8(a)(ii) 3 × 10! × 4 oe M1 43 545 600 oe A1 8(a)(iii) 5! × 8 × 7! oe M1 4 838 400 oe A1 8(b)(i) 9 C 3 M1 84 A1 8(b)(ii) 3 C1 × 4 C1 × 5 C1 oe M1 60 A1

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9 (a) In the expansion of 2 k - , where k is a constant, the coefficient of x2 is 160. Find the value k of k. [3] 6 (b) (i) Find, in ascending powers of x, the first 3 terms in the expansion of 1 + 3x , simplifying ` j the coefficient of each term. [2] 6 2 (ii) When 1 + 3x a + x is written in ascending powers of x, the first three terms are ` j ` j 4 + 68x + bx 2 , where a and b are constants. Find the value of a and of b. [3]

Mark scheme: 9(a) Identifies the correct term: B1 5 3  1  2 2 [× x ] oe, soi C 2 × (2 k ) ×  −   k  8 k 3 M1 FT only for correct term with bracketing 10 × = 160 soi errors; condone one slip in simplification 2 k k = 2 nfww A1 9(b)(i) 1 + 18x + 135x2 B2 B1 for any 2 terms correct or for all 3 correct terms listed but not summed or M1 for a correct unsimplified expansion e.g. : 1 + 6(3x) + 15(3x)2 9(b)(ii) Uses constant/coefficient of x to B2 B1 for both a = 2 and −2 or find a = −2 only 17 for both a = and −2 9 b = 469 only B1 FT their calculated value of a

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10 The function f is defined by f ( )x = for 0.5 G x G 1 .5 . 2x The diagram shows a sketch of y = f ( x) . y 4x 2 - 1 y = 2x 0 x 0.5 1.5 (a) (i) It is given that f -1 exists. Find the domain and range of f -1 . [3] (ii) Find an expression for f -1 ( )x . [3] a 1 - 2 (b) The function g is defined by g ( )x = ex2 for all real x. Show that gf ( )x = e e bx o, where a and b are integers. [2]

Mark scheme: 10(a)(i) Range f−1: 0.5 ⩽ f−1 ⩽ 1.5 B1 2 2 2 2 Domain f−1: 0 ⩽ x ⩽ oe B2 B1 for 0 and in an incorrect inequality 3 3 2 2 or for x ⩾ 0 or x ⩽ 3 10(a)(ii) Correctly collects terms ready to M1 factorise e.g. 4 x 2 − 4 x 2 y 2 = 1 or 4 y 2 x 2 − 4 y 2 = −1 or simplifies to subject in one term 1 2 only e.g. = 1 −x or 4 y 2 1 2 − = y − 1 oe 4 x 2 Correctly factorises and/or M1 FT only if of equivalent difficulty rearranges at least as far as: 2 1 2 −1 x = or y = oe 4 − 4 y 2 4 x 2 − 4 −1 1 A1  f ( x ) =  or   2 4 − 4 x −1 [ y = ] 2 oe, isw 4 x − 4 10(b) Correct order of composition: M1 2 −1   4 x 2   gf(x) = e  2 x  1   A1  1−  2 gf ( x ) = e  4 x  isw

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11 (a) (i) Find 6 dx . [2] 10x - 1 e dd` j 2 c 2 x 3 + 5 (ii) Find ` j d x . [3] dd x e (b) (i) Differentiate y = tan ( 3 x + 1) with respect to x. [2] c 10r sec 2 ( 3x + 1) (ii) Hence find dd - sin x d x . [4] r e 2 o e12 Question 12 is printed on the next page.

Mark scheme: 11(a)(i) (10 x − 1) −5 B2 (10 x − 1) −5 1 ( + c ) isw B1 for k ( + c ) , where k ≠ −×5 10 −5 10 11(a)(ii)  5 2 25  B1  4 x + 20 x +  d x   x  4 6 20 3 B2 B1 for any 3 terms correct x + x + 25ln x + c 6 3 11(b)(i) 3sec 2 (3 x + 1) B2 B1 for k sec 2 (3 x + 1) where k ≠ 3 11(b)(ii) sec 2 (3 x + 1) tan(3 x + 1) B1 dx =  2 6 oe, soi B1 − sin x d x = cos x oe   π   π  M1 F   − F   where  10   12  F(x) = k1 tan(3 x + 1) + k 2 cos x oe 0.322 or 0.3222[32...] rot to 4 figs A1

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12 A particle P travels in a straight line so that, t seconds after passing through a fixed point O, its velocity, v ms -1 , is given by t v = for 0 G t G 2 , 2e t 2 v = e - for t 2 2 . Given that, after leaving O, particle P is never at rest, find the distance it travels between t = 1 and t = 3. [6]

Mark scheme: 12 t t 2 B1 For 0 ≤≤t 2 : dt =  2e 4e t t t − − − t 3 B2 − 2 2 2 2 For t > 2: e d t = −2e + oe B1 for e dt = −2e ( + c ) oe  e  3 M2 M1 for − 3 1 3 2 + − −2e e 4e 1 − 2 3 [s(1) =] and [ s (3) = ] − 2e + their 4e e 3 or at least one term correct in the difference: −   3 1 1 1   3 2 + − − −2e OR   +   −  3  1 e e e 4e     2e 2 + their  −  −  e  4e   or for one bracket correct in:  − 3   1 1  2 3 1  −2e + their −  +  −   e e   e 4e  0.565 or 0.5654 to 0.56541 nfww A1 12 Alternative method (using def int): 2   t 2 M1* for =    4e 1  4 1  M1 for  −  (dep*)  4e 4e oe 3  − t  M1** for  − 2e 2    2  − 3  2 2 M1 for  −2e +  (dep**)  e oe  − 3   4 1  2 2 M1 for  −2e +  +  −   e   4e 4e  oe A1 for 0.565 or 0.5654 to 0.56541 nfww

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