Cambridge IGCSE Mathematics - Additional 0606 — 2017 May/June Paper 2 · Variant 3

1 (a) Solve the equation 7 2x + 5 = 2.5 , giving your answer correct to 2 decimal places. [3] 5 q 3 b (b) Express ^ h 1 in the form 5a p q c, where a, b and c are constants. [3] 4 625p 12 q ^ h

Mark scheme: Question Answer Marks Guidance 1(a)  2.5 M1 correct first anti-logging step 2 x = log 7 2.5 = 2 x + 5 or log 7  5  7  or (2 x + 5)log7 = log2.5 log 7 2.5 − 5 M1 isolates x [ x = ] 2 1  2.5 x = or log 7  5 2  7  1  log2.5  or x =  − 5  2  log7  −2.26(4...) A1 1(b) 5 B3 B1 for each term 5 2 p −3 q 4 oe If B0 then allow M1 for numerator of 3 1 125q 2 or denominator of 5p 3 q 4

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2 y y A B O x O x C y D y O x O x The four graphs above are labelled A, B, C and D. (i) Write down the letter of each graph that represents a function, giving a reason for your choice. [2] (ii) Write down the letter of each graph that represents a function which has an inverse, giving a reason for your choice. [2]

Mark scheme: 2(i) B and C with valid reason B2 B1 for one graph and valid reason or both graphs and no reason 2(ii) B only with valid reason B2 B1 for graph B or valid reason

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3 3 y (1, 13) (0.2, 5) O 1 x Variables x and y are such that when 3 y is plotted against 1, a straight line graph passing through the x points (0.2, 5) and (1, 13) is obtained. Express y in terms of x. [4]

Mark scheme: 3 13 − 5 M1 or 13 = m + c and 5 = 0.2 m + c and [ m = ] or 10 soi subtracting/substituting to solve for m or 1 − 0.2 c, condone one error Y − 13 = their 10( X − 1) M1 or using their m or their c to find their c or their m, without further error or Y − 5 = their 10( X − 0.2) or 13 = their 10 + c or 5 = their 10 × 0.2 + c 1 M1 their m and c must be validly obtained 3 y = (their m ) + (their c ) or x 3  1  y = (their m )  − 1  + 13 or  x  3  1  y = (their m )  − 0.2  + 5  x  3 A1  10  y =  + 3   x    1   3 or y =  10  − 1  + 13    x     1   3 or y =  10  − 0.2  + 5  cao, isw   x  

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4 (a) Vectors a, b and c are such that a = KK OO, b = KK OO and 3a + c = b. - 6 - 15 L P L P (i) Find c. [1] (ii) Find the unit vector in the direction of b. [2] (b) P p R O Q q In the diagram, OP = p and OQ = q . The point R lies on PQ such that PR = 3RQ. Find OR in terms of p and q, simplifying your answer. [3]

Mark scheme: 4(a)(i)  −4  B1    3  4(a)(ii) 2 2 M1 11 + ( −15) or better 1  11  A1   346  −15  4(b) uuur uuur 3 uuur M1 uuur uuur 1 uuur OR = OP + PQ soi or OR = OQ − PQ soi 4 4 uuur 3 M1 uuur 1  OR =  + ( q − p ) or  OR =  − ( q − p )  p  q 4 4 uuur 1 3 A1  OR =  p + q oe   4 4

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5 (a) How many 5-digit numbers are there that have 5 different digits and are divisible by 5? [3] (b) A committee of 8 people is to be selected from 9 men and 5 women. Find the number of different committees that can be selected if the committee must have at least 4 women. [3]

Mark scheme: 5(a) ( 9 × 8 × 7 × 6 × 1) + ( 8 × 8 × 7 × 6 × 1 ) soi M2 M1 for one correct product of the sum 5712 A1 5(b) 9C4 × 5C4 + 9C3 × 5C5 oe M2 M1 for one correct product of the sum [630 + 84 = ] 714 A1

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6 The first three terms of the binomial expansion of 2 - ax n are 64 - 16 bx + 100 bx 2 . Find the value of each of the integers n, a and b. [7]

Mark scheme: 6 64 = 2 n M1 n = 6 A1 their (6 1) M1 their 6 ( 2 ) −× ( − a ) = − 16b oe 6 × (6 − 1) their (6 − 2) 2 M1 their ( 2 ) × ( − a ) = 100b oe 2 attempts to solve DM1 dep on both M1 marks being awarded; must have correctly or correct FT eliminated one unknown a = 5 A1 b = 60 A1

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7 Differentiate with respect to x, (i) 1 + 4x 10 cos x , [4] ^ h e 4x - 5 (ii) . [4] tanx

Mark scheme: 7(i) 9 M1 k (1 + 4 x ) 9 A1 4 × 10 (1 + 4 x ) or better (1 + 4 x )10 ( their − sin x ) + M1 clearly applies product rule cos x their 4 × 10 × (1 + 4 x ) 9 ( ( ) ) 10 9 A1 all correct (1 + 4 x ) ( − sin x ) + cos x 4 × 10 × (1 + 4 x ) ( ) 7(ii) d 4 x − 5 4 x − 5 B1 e = 4e soi ( ) dx d 2 B1 ( tan x ) = sec x soi dx clearly applies correct form of quotient rule M1 or correct form of product rule to 4 x − 5 4 x − 5 e 4 x − 5 (tan x ) −1 tan x ( their 4e their sec 2 x ) − e ( ) ( tan x ) 2 4e 4 x − 5 (tan x ) −1 + e 4 x − 5 (tan x ) −2 × sec 2 x 4 x − 5 4 x − 5 A1 all correct tan x ( 4e sec 2 x ) − e ( ) isw ( tan x ) 2

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8 A rr cmcm 2› cm θrad O B The diagram shows a circle, centre O of radius r cm, and a chord AB. Angle AOB = θ radians. The length of the major arc AB is 5 times the length of the minor arc AB. The minor arc AB has length 2r cm. (i) Find the value of θ and of r. [2] (ii) Calculate the exact perimeter of the shaded segment. [2] (iii) Calculate the exact area of the shaded segment. [4]

Mark scheme: 8(i) π B1 3 6 [cm] B1 8(ii)  π  M1 [major arc = ]  2π − their  their r  3  10π + 6 cao A1 8(iii) 1 2  π  M1 1 2  π  (their 6)  2π − their  (their 6)  their  2  3  2  3  1 2  π  M1 1 2  π  ( their 6) sin  their  ( their 6) sin  their  2  3  2  3  Sector + triangle M1 π ×their 6 2 − (Sector − triangle) 30π + 9 3 A1

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9 The functions f and g are defined, for x 2 1 , by f ( x) = 9 x - 1 , g ( )x = x 2 + 2 . (i) Find an expression for f -1 ( )x , stating its domain. [3] (ii) Find the exact value of fg(7). [2] (iii) Solve gf ( )x = 5 x 2 + 83x - 95 . [4]

Mark scheme: 9(i) y M1 attempt to swop; may be in later work that = x − 1 with attempt to swop x and y at contains an error 9 some point x or = y − 1 9 2 A1 condone y = ... etc; must be a function of x −1  x   f ( x ) =    + 1 oe   9  x > 0 B1 9(ii) f(51) M1 2 or fg( x ) = 9 x + 1 9 50 oe A1 9 x − 1 + 29(iii) [ gf ( x ) = ]( 2 M1 ) [ gf( x ) = ] 81( x − 1) + 2 or better A1 their ( 81x − 79 ) = 5 x 2 + 83 x − 95 → M1 provided their (81x − 79) of the form 2 ax + b for non-zero a and b their 5 x + 2 x − 16[ = 0] ( ) 1.6 oe only A1 must disregard other solution

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10 Solve the equation (a) 2 sinx = 1 for - r G x G r radians, [3] (b) 3 tan ( 2y + 15c) = 1 for 0c G y G 180c, [4] (c) 3 cot 2 z = cosec 2 z - 7 cosec z + 1 for 0c G z G 360c. [5]

Mark scheme: 10(a) sin x = 0.5 , sin x = − 0.5 M1 π π 5π 5π A2 A1 for any correct pair of angles , − , , − oe 6 6 6 6 if M0 then SC1 for a correct pair of angles 10(b) −1 1  M1 2 y + 15 = tan   soi  3  18.43(49...) and 198.43(49...) M1 1.7, 91.7 A2 A1 for each 10(c) Uses cot 2 z = cosec 2 z − 1 oe M1 for using correct identity or identities to obtain an equation in terms of a single trigonometric ratio 2cosec 2 z + 7cosec z − 4 = 0 ⇒ DM1 for dealing with quadratic ( 2cosec z − 1)( cosec z + 4 ) 1 M1 [sin z = 2 ] sin z = − 4 194.5, 345.5 A2 A1 for each

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11 y 4y = 5x + 20 Q y = 5+ 10x P O R x The diagram shows part of the curve y = 5 + 10 x and the line 4y = 5x + 20 . The line and curve intersect at the points P(0, 5) and Q. The line QR is parallel to the y-axis. (i) Find the coordinates of Q. [4] (ii) Find the area of the shaded region. You must show all your working. [6]

Mark scheme: 11(i) 5 x + 20 M1 or better; equates and solves as far as 5 + 10 x = → 20 + 4 10 x = 5 x + 20 clearing the fraction 4  x  4 10 M1 Simplifies as far as x =L = x = oe    x  5 x = 6.4 cao A1 squares and simplifies to 6.4 [ y = ]13 B1 11(ii) (area of trapezium = ) their 57.6 B1 FT x = their 6.4, y = their 13 using any valid method 6.4 M1 ∫0 ( 5 + 10 x ) d x 1 3 M1 1 3 2 10 x ) 2 dx = k (10 x ) 10 x 2 d x = k 10 ( x ) ∫ ( 2 or or ∫  3  A1  1 3  2 10 x ) 10 ) ( x ) 2   2 ( 2   2 ( 5 x + or 5 x +  3 × 10   3       3  M1 limits used correctly or correct FT and 10 × 6.4 )  2 ( 2  subtraction of trapezium; their 5(6.4) + − their 57.6 oe  3 × 10  992   their − their 57.6 15 128 A1 allow 8.5333333.... rot to 4 or more sf or 8.53 oe 15

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