Cambridge IGCSE Mathematics - Additional 0606 — 2021 May/June Paper 2 · Variant 2

1 Using the binomial theorem, expand ( 1 + e 2 x) 4 , simplifying each term. [2]

Mark scheme: Question Answer Marks Partial Marks 1 1 + 4e 2 x + 6e 4 x + 4e 6 x + e 8 x B2 mark final answer for B2 B1 for any 3 correct simplified terms in a sum or all 5 simplified terms listed but not summed or for a correct, simplified expansion that is not their final answer or M1 for a correct unsimplified expansion e.g. 2 x 4 × 3 2 x 2 4 × 3 × 2 2 x 3 2 x 4 1 + 4e + (e ) + (e ) + (e ) 2 6 If 0 scored, SC1 for a complete, correct, simplified expansion as final answer found by multiplying out the brackets

More questions on Use the binomial theorem for expansion of

2 On the axes, sketch the graph of y = 3 ( x - 3)( x - 1)( x + 2) stating the intercepts with the coordinate axes. [3] y x O

Mark scheme: 2 Correct graph and intercepts B3 B1 for correct shape; the ends must extend above and below the x-axis y B1 for correct roots indicated; must have 18 attempted a cubic shape B1 for correct y-intercept indicated; must have attempted a cubic shape -2 O 1 3 x

More questions on Sketch the graphs of cubic polynomials and

4 The polynomial p ( x) = mx 3 - 29 x 2 + 39x + n , where m and n are constants, has a factor 3x - 1, and remainder 6 when divided by x - 1. Show that x - 2 is a factor of p ( x) . [6]

Mark scheme: 4 m 29 39 B1 − + + n = 0 oe 27 9 3 m – 29 + 39 + n = 6 oe B1 Eliminates one unknown correctly for a M1 pair of linear equations in m and n and solves for one unknown m = 6, n = −10 A2 A1 for either [p(2) =] 48 – 116 + 78 – 10 = 0 oe, A1 nfww

More questions on Know and use the remainder and factor

5 The function f is defined, for 0° G x G 810° , by f ( x) =- 2 + cos . 3 (a) Write down the amplitude of f. [1] (b) Find the period of f. [2] (c) On the axes, sketch the graph of y = f ( x) . [2] y 5 x 00° 90° 180° 270° 360° 450° 540° 630° 720° 810° – 5

Mark scheme: 5(a) 1 B1 5(b) 2 M1 360 ÷ oe 3 540 A1 If 0 scored, SC1 for 3π 5(c) Correct sketch for domain B2 B1 for correct cosine shape from (0, −1) with 0° ⩽ x ⩽ 810° amplitude 1 for 0° ⩽ x ⩽ 810° 5 B1 for attempt at correct cosine shape with period 540° for 0° ⩽ x ⩽ 810° If 0 scored, SC1 for a fully correct graph for 0° ⩽ x ⩽ 540° 90 180 270 360 450 540 630 720 810 Maximum of 1 mark if not fully correct. -5

More questions on Understand and use the amplitude and

6 The points A(5, - 4 ) and C(11, 6) are such that AC is the diagonal of a square, ABCD. (a) Find the length of the line AC. [2] (b) (i) The coordinates of the centre, E, of the square are (8, y). Find the value of y. [1] (ii) Find the equation of the diagonal BD. [3] (iii) Given that the x‑coordinate of B is less than the x‑coordinate of D, write EB and ED as column vectors. [2]

Mark scheme: 6(a) 2 2 M1 (11 − 5) + (6 −−4) oe 11.7 or 11.66[19...] rot to 4 or more A1 figs 6(b)(i) [y = ] 1 B1 6(b)(ii) 6 −−4 10 B1 m AC = or nfww oe 11 − 5 6 −1 M1 m BD = oe 10 their 6 A1 FT their 1 from (b)(i) and their perpendicular y – their 1 = − 3( x − 8) oe isw gradient 5 6(b)(iii)  − 5   5  B2 B1 for either   and    3   − 3  If 0 scored, SC1 for −5i + 3j and 5i −3j

More questions on Solve problems involving midpoint and length

7 B 18 cm A 7rrad C 9 D DAB is a sector of a circle, centre A, radius 18 cm. The lines CB and CD are tangents to the circle. 7 r Angle DAB is radians. 9 (a) Find the perimeter of the shaded region. [3] (b) Find the area of the shaded region. [3]

Mark scheme: 7(a) [Arc length + 2 × tangent length] M2 M1 for 7π 7π 7π 18 × + 2 × 18 × tan oe [Arc length] 18 × oe 9 18 9 7π or [Tangent length] 18 × tan oe 18 18 or [Tangent length] oe π tan 9 18 7π or [Tangent length] × sin oe π 18 sin 9 143 or 142.9 or awrt 142.9 (cm) A1 7(b) [Area of kite – area of sector] M2 FT their BC or CD from (a) providing it is not  7π  1 2 7π 18 18 × their  18 × tan  − × 18 ×  18  2 9 1 2 7π oe M1 for [area of sector] × 18 × oe 2 9  7π  or [area of kite] 18 × their  18 × tan  oe  18  or [area of kite] 18 × their(18×tan70) oe 494 or 494.3 or awrt 494.3 (cm2) A1

More questions on Solve problems involving the arc length and

8 A particle moves in a straight line so that, t seconds after passing through a fixed point O, its velocity, v ms -1 , is given by v = 3t 2 - 30t + 72 . (a) Find the distance between the particle’s two positions of instantaneous rest. [6] (b) Find the acceleration of the particle when t = 2 . [2]

Mark scheme: 8(a) Factorises or solves M1 3t 2 − 30t + 72 = 0 or t 2 − 10t + 24 = 0 t = 4, t = 6 A1 Integrates v to find F(t): M2 M1 for any two terms correct 3t 3 30t 2 − + 72t 3 2 Correct substitution for M1 dep on at least M1 for integration F(6) – F(4) or F(4) – F(6) FT their 4 and their 6 provided they are both positive 4 (m) A1 dep on all previous marks being awarded 8(b) [ a = ]6t − 30 B1 [When t = 2: a = 6(2) – 30 =] B1 −18 (ms−2) cao

More questions on Apply differentiation and integration

9 Solve the following simultaneous equations. 4 x 2 + 3 xy + y 2 = 8 xy + 4 = 0 [6]

Mark scheme: 9 Correctly eliminates x or y e.g. M1 2  4   4  2 4 x + 3 x  −  +  −  = 8 oe  x   x   4  2  4  2 or 4  −  + 3  −  y + y = 8 oe  y   y  Rearranges to a 3-term quadratic in x2 A1 or y2 soi e.g. 4 x 4 − 20 x 2 + 16 = 0 or y 4 − 20 y 2 + 64 = 0 Factorises or solves their 3-term M1 quadratic in x2 or y2 soi : (x2 – 1)(x2 − 4) or (y2 – 16)(y2 − 4) x 2 = 1 , x 2 = 4 oe, nfww A1 or y 2 = 16 , y 2 = 4 oe, nfww x = ±1 x = ± 2 A2 A1 for all 4 x values or all 4 y values y = ∓4 y = ∓2 oe, nfww

More questions on Solve simultaneous equations in two

10 (a) Find ( e x + 1 ) 3 d x . [2] (b) (i) Differentiate, with respect to x, y = x sin 4x . [2] r 3 1 r 3 (ii) Hence show that 4x cos 4xdx = - . [4] yr 8 6 4

Mark scheme: 10(a) 1 3 x + 3 1 3 3 x B2 3 x + 3 3 3 x 1 e + c or e × e + c nfww B1 for k e or ke × e where k ≠ or 0 3 3 3 10(b)(i) d(sin4x ) B1 = 4cos4 x soi dx Applies correct form of product rule: B1 FT their 4 cos 4x if possible 4x cos 4x + [1] sin 4x isw M1 FT use of their mx cos 4x + n sin 4x where m10(b)(ii)  (4 x cos4 x )d x =  x sin 4 x − sin 4 xd x     and n are constants 1 A1 x sin4 x + cos4 x [ + c ] soi 4 π  π  1  π  A1 sin  4 ×  + cos  4 ×  − 3  3  4  3   π  π  1  π    sin  4 ×  + cos  4 ×    4  4  4  4   Correct completion to given answer A1 1 π 3 − 8 6

More questions on Differentiate products and quotients of

11 In this question all lengths are in centimetres. 4 3 2 The volume and surface area of a sphere of radius r are rr and 4rr respectively. 3 x y The diagram shows a solid object made from a hemisphere of radius x and a cylinder of radius x and height y. The volume of the object is 500 cm3. (a) Find an expression for y in terms of x and show that the surface area, S, of the object is given by 5 2 1000 S = r x + . [4] 3 x (b) Given that x can vary and that S has a minimum value, find the value of x for which S is a minimum. [4]

Mark scheme: 11(a) 4 3 2 M1 500 = πx + πx y oe 6 1  4 3  A1 if first M0, SC1 for oe, isw 500 − y =  πx  1  4 3  6   πx 2 y = 2  500 − πx  oe seen πx  3  2 2  500 2  M1 dep on first M1 S = 2πx + πx + 2πx  2 − x   πx 3  Correct completion to given answer: A1 5 2 1000 S = πx + 3 x 11(b) 10 1000 B2 B1 for each term Differentiates S: πx − oe 3 x 2 10 1000 M1 d S πx − = 0 and attempt to solve FT their providing at least B1 awarded 2 3 x d x 300 A1 x = 3 isw or 4.57[07...] nfww π

More questions on Apply differentiation to practical problems

12 DO NOT USE A CALCULATOR IN THIS QUESTION. y A 1 y = 2x + 1 5y = x - 1 C O B x 1 The diagram shows part of the curve y = and part of the line 5y = x - 1. 2x + 1 The curve meets the y‑axis at point A. The line meets the x‑axis at point B. The line and curve intersect at point C. (a) (i) Find the coordinates of A and B. [1] (ii) Verify that the x‑coordinate of C is 2. [2] (b) Find the exact area of the shaded region. [5] Question 13 is printed on the next page.

Mark scheme: 12(a)(i) A(0,1) and B(1, 0) B1 12(a)(ii) 1 2 − 1 B2 1 [ y = ] and [ y = ] B1 for [ y = ] and 5 y = 2 − 1 oe 2(2) + 1 5 2(2) + 1 and 1 evaluates both expressions as 5 Alternative 1 (B2) 1 1 2 − 1 1 1 1 1 [ y = ] = or[ y = ] = B1 for = and 5 × = x − 1 oe 2(2) + 1 5 5 5 2(2) + 1 5 5 and 2 1 1 1 1 or −= and = oe 1 5 5 5 2 x + 1 solves 5 × = x − 1 oe to get x = 2 5 1 1 or = oe to get x = 2 5 2 x + 1 Alternative 2 (B2) 2x2− x – 6 = 0 B1 for (2x + 1)(x – 1) = 5 or 2x2− x – 6 = 0 and solves or factorises to get (2x + 3)(x – 2) and states x = 2 OR shows 2(22) – 2 – 6 = 0 oe Alternative 3 (B2) (2x + 1)(x – 1) = 5 oe B1 for (2x + 1)(x – 1) = 5 and shows (2 × 2 + 1)(2 – 1) = 5 12(b) 1 B1 × 1 × 0.2 oe 2 2 2 2  12 1  or − −  −  oe 5 × 2 5  5 × 2 5  1 B2 [ F( x ) = ] ln(2 x + 1) [+c] oe 1 1 2 B1 for ln 2 x + 1 or ln x + 0.5 2 2 1 or ln( x + 0.5) [+c] oe or k ln(2 x + 1) or k ln( x + 0.5) , k ≠ 0.5 or 0 2 F(2) – F(0) – their 0.1 M1 FT their F(x) providing at least B1 for integration of curve awarded 0.5ln5 − 0.1 or exact equivalent A1

More questions on Evaluate definite integrals and apply

13 The functions f and g are defined, for x 2 0 , by 2x 2 - 1 f ( x) = , 3x 1 g ( x) = . x (a) Find and simplify an expression for fg ( x) . [2] (b) (i) Given that f -1 exists, write down the range of f -1 . [1] 2 -1 px + qx + r (ii) Show that f ( x) = , where p, q and r are integers. [4] 4

Mark scheme: 13(a) 2 M1  2  1 − 1  x  [ fg( x ) = ] oe  1  3    x  2 − x 2 2 −x A1 mark final answer or [ fg( x ) = ] 3 x 3 x 3 13(b)(i) f−1 > 0 B1 13(b)(ii) 2 x 2 − 3 xy −=1 0 B1 or 2 y 2 − 3 xy −=1 0 Correctly applies quadratic formula: M1 FT their 2 x 2 − 3 xy −=1 0 or 2 y 2 − 3 xy −=1 0 with at most one sign error −−( 3 y ) ± ( − 3 y ) 2 − 4(2)( − 1) [ x = ] oe in the equation 2(2) or −−( 3 x ) ± ( − 3 x ) 2 − 4(2)( − 1) [ y = ] oe 2(2) Justifies the positive square root at B1 some point 2 A1 must be a function of x −1 3 x + 9 x + 8 f ( x ) = cao 4

More questions on Find the domain and range of functions