Cambridge IGCSE Mathematics - Additional 0606 — 2017 Oct/Nov Paper 2 · Variant 3

1 (a) On each of the diagrams below, shade the region which represents the given set. A B A B C C A B C' A B' C [2] (b) P Q 2 7 4 8 1 3 5 6 R The Venn diagram shows the number of elements in each of its subsets. Complete the following. n(P') = ........................................................................ n((Q  R)  P) = ...................................................... n(Q'  P) = ................................................................ [3]

Mark scheme: Question Answer Marks Guidance 1(a) B2 B1 for each 1(b) n ( P ' ) = 18 B1 n ( ( Q ∪ R ) ∩ P ) = 11 B1 n ( Q ′∪ P ) = 29 B1

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2 Solve the equation 3x - 1 = 5 + x . [3]

Mark scheme: 2 3 x −=1 5 + x x = 3 B1 3 x −=1 −−5 x oe M1 M1 not earned if incorrect equation(s) present x = –1 A1

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4 Solve the simultaneous equations log 3 x + 1 = 1 + log 3 y , ^ h log 3 x - y = 2 . [5] ^ h

Mark scheme: 4 log 3 3 = 1 or log 3 9 = 2 B1 implied by one correct equation x + 1 = 3 y B1 x − y = 9 B1 solve correct equations for x or y M1 x = 14 and y = 5 A1

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5 D A X O B C The diagram shows points O, A, B, C, D and X. The position vectors of A, B and C relative to O are 3 OA = a , OB = b and OC = b . The vector CD = 3a . 2 (i) If OX = m OD express OX in terms of m, a and b. [1] (ii) If AX = n AB express OX in terms of n, a and b. [2] (iii) Use your two expressions for OX to find the value of m and of n. [3] AX(iv) Find the ratio . [1] XB OX (v) Find the ratio . [1] XD

Mark scheme: JJJG 5(i) OX = λ(1.5b + 3a ) B1 JJJG JJJG 5(ii) AB = b − a or BA = a – b B1 JJJG OX = a + µ( b − a ) B1 JJJG JJJG 5(iii) 1.5λ= µ or 3λ= 1 − µ M1 OX = OX and equate for a or b 1 2 A1 for each µ= λ= A2 3 9 5(iv) AX 1 B1 1 = Accept 1 : 2 but not :1 XB 2 2 5(v) OX 2 B1 2 = Accept 2 : 7 but not :1 XD 7 7

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6 The functions f and g are defined for real values of x by f x = x + 2 2 + 1 , ^ h ^ h x - 2 1 g x = , x ! . ^ h 2x - 1 2 (i) Find f 2 - 3 [2] ^ h. (ii) Show that g -1 x = g x [3] ^ h ^ h. 8 (iii) Solve gf x = . [4] ^ h 19

Mark scheme: 6(i) f 2 = f(f) used M1 numerical or algebraic algebraic ([(x + 2)2 + 1] + 2)2 + 1 17 A1 6(ii) y − 2 M1 change x and y x = 2 y − 1 2 xy − x = y − 2 → y ( 2 x − 1) = x − 2 M1 M1dep multiply, collect y terms, factorise x − 2 A1 correct completion y = =  g ( x )  2 x − 1 6(iii)  ( x + 2 ) 2 + 1 − 2 B1   oe gf ( x ) = 2 2  ( x + 2 ) + 1 − 1   ( x + 2 ) 2 − 1 8 M1 their gf = 8 and simplify to = 2 19 2 ( x + 2 ) + 1 19 2 quadratic equation 3 ( x + 2 ) = 27 oe 3x2 + 12x – 15 = 0 solve quadratic M1 M1dep Must be of equivalent form x = 1 x = −5 A1

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7 A particle moving in a straight line passes through a fixed point O. Its velocity, v ms -1 , t s after passing through O, is given by v = 3 cos 2t - 1 for t H 0 . (i) Find the value of t when the particle is first at rest. [2] r (ii) Find the displacement from O of the particle when t = . [3] 4 (iii) Find the acceleration of the particle when it is first at rest. [3]

Mark scheme: 7(i) 1 M1 set v = 0 and solve for cos2t v = 0 → cos2t = 3 →=t 0.615 or 0.616 A1 7(ii) 3 M1A1 M1 for sin2t and ± t s = sin2t − t ( + c ) 2 π π A1 t = → s = 1.5 − ( = 0.715 ) 4 4 7(iii) a = −6sin2t M1A1 M1 for −sin2t t = 0.615 → a = –5.66 or –5.65 or −2 8 A1 condone substitution of degrees

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8 B C 1 ms–1 50 m 3 ms–1 a A A man, who can row a boat at 3 ms−1 in still water, wants to cross a river from A to B as shown in the diagram. AB is perpendicular to both banks of the river. The river, which is 50 m wide, is flowing at 1 ms−1 in the direction shown. The man points his boat at an angle a° to the bank. Find (i) the angle a, [2] (ii) the resultant speed of the boat from A to B, [2] (iii) the time taken for the boat to travel from A to B. [2] On another occasion the man points the boat in the same direction but the river speed has increased to 1.8 ms−1 and as a result he lands at the point C. (iv) State the time taken for the boat to travel from A to C and hence find the distance BC. [2]

Mark scheme: 8(i) 1 M1 cosα= oe 3 α = 70.5° A1 8(ii) 2 2 M1 Pythagoras/trig ratio/cosine rule speed = 3 − 1 8 or 2 2 or 2.83 m s–1 A1 8(iii) 50 M1 time = their 8 25 2 A1 or 17.7s 2 8(iv) their 8 ( iii ) seen B1 BC = 10 2 or 14.1 m or 14.2 m B1

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9 (i) Show that 3 = 4 . [3] dx x x ln x (ii) Find the exact coordinates of the stationary point of the curve y = 3 . [3] x J ln x N(iii) Use the result from part (i) to find dx . [4] y KK 4 OO x L P

Mark scheme: 9(i) d 1 B1 seen (ln x ) = and dx x d 3 2 d −3 −4 x = 3 x or x = −3 x d x dx Substitution of their derivatives into quotient rule M1 3 1 2 A1 correct completion x × − 3 x ln x d  ln x x oe  3 = 6 d x  x  x 9(ii) d y 1 M1 dy = 0 →−1 3ln x = 0 lnx = equate given to zero and solve d x 3 dx for lnx or x 1 A1 seen x = e 3 1 A1 seen y = 3e 9(iii) lnx 1 − 3lnx M1 use given statement in (i) dx oe 4 x 3 =∫ x 1 −1 B1 seen anywhere ∫ x dx = 4 3 x 3 ln x 1 ln x A2 A1 for each term ∫ x d x = − − (+C) oe 4 9 x 3 3 x 3

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10 (a) Show that + = 2 cosec x . [3] 1 + cos x sin x (b) Solve the following equations. (i) cot 2 y + cosec y - 5 = 0 for 0° G y G 360° [5] r 3(ii) cos 2z + =- for 0 G z G r radians [4] ` 4 j 2 Question 11 is printed on the next page.

Mark scheme: 10(a) 2 2 B1 correct addition of fractions sin x + (1 + cos x ) LHS = sin x (1 + cos x ) 1 + 2cos x + 1 B1 expansion and use of identity = sin x (1 + cosx ) 2 (1 + cos x ) B1 factorisation and completion = = 2cosecx sinx (1 + cosx ) 10(b)(i) cosec 2 y −+1 cosecy − 5 = 0 M1 use of identity for cot2y to obtain quadratic in cosecy cosec2y + cosecy – 6 = 0 ( cosecy − 2 )( cosecy + 3 ) = 0 M1 solve 3 term quadratic for cosecy 1 1 M1 obtain values for siny sin y = , sin y = − 2 3 y = 30°, 150°, 199.5 °, 340.5 ° A2 A1 for 2 values 10(b)(ii) π 5π 7π M2 5π 2 z + = or (2.6…, 3.6…) M1 equate to 4 6 6 6 7π M1 equate to 6 7π 11π A2 A1 for 1 value z = or (0.916, 1.44) 24 24

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11 The cubic equation x3 + ax2 + bx − 36 = 0 has a repeated positive integer root. (i) If the repeated root is x = 3 find the other positive root and the value of a and of b. [4] (ii) There are other possible values of a and b for which the cubic equation has a repeated positive integer root. In each case state all three integer roots of the equation. [4]

Mark scheme: 11(i) Other root = 4 B1 f ( x ) = ( x − 3 )( x − 3 )( x − 4 ) M1 multiply out (x – 3)(x – 3)(x ± p) = x 3 − 10 x 2 + 33 x − 36 a = –10 b = 33 A2 A1 for each Can be implied by correct cubic 11(ii) x = 6, x = 6, x = 1 B4 B1 for each of first two sets x = 2, x = 2, x = 9 B2 for third set x = 1, x = 1, x = 36

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