1. (a) B, D A1A1 N2 2. A1A1 N2 Note: Award A1 for. 2xe. e and A1 for 2x.

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1 1. (a) B, D N (b) (i) f () = e N Note: Award for e and for. (ii) finding the derivative of, i.e. () evidence of choosing the product rule e.g. e e e 4 e f () = (4 ) e AG N0 5 (c) valid reasoning R1 e.g. f () = 0 attempting to solve the equation e.g. (4 ) = 0, sketch of f () p = , q N3 4 (d) evidence of using second derivative to test values on either side of POI M1 e.g. finding values, reference to graph of f, sign table correct working e.g. finding any two correct values either side of POI, checking sign of f on either side of POI reference to sign change of f () R1 N0 4 [15]. (a) finding A Bˆ C = 110 (= 1.9 radians) () evidence of choosing cosine rule e.g. AC = AB + BC (AB)(BC) cos A Bˆ C correct substitution e.g. AC = (5)(40) cos 110 AC = 53.9 (km) N3 IB Questionbank Maths SL 1

2 (b) METHOD 1 correct substitution into the sine rule sin BÂC sin110 e.g B ÂC = 44. bearing = 074 N1 METHOD correct substitution into the cosine rule e.g. cos BÂC (5)(53.9) B ÂC = 44.3 bearing = 074 N1 [7] 3. (a) Note: Award for approimately correct shape, for right endpoint at (5, 0) and for maimum point in circle. N3 (b) (i) recognizing that d is the area under the curve e.g. v (t) correct epression in terms of t, with correct limits A N3 e.g. d = vd 0 ( 15 t 3t), d t (ii) d = (m) (accept 149 to 3 sf) N1 IB Questionbank Maths SL

3 [7] 4. evidence of integrating the acceleration function 1 e.g. 3sin t t correct epression ln t 3 cos t + c evidence of substituting (1, 0) e.g. 0 = ln 1 3 cos + c c = cos ln1or cos () v = ln t cos t 0.64 lnt cos t cos or lnt cos t cos ln1 () v(5) =.4 (accept the eact answer ln cos cos ) N3 [7] 5. (a) t = h = 50 5( ) = 50 0 = 30 () OR h = 90 40() + 5( ) = 30 () 1 (b) h t (A4) 4 Note: Award () for marked scales on each ais, () for each section of the curve. IB Questionbank Maths SL 3

4 (c) (i) dh d (50 5t ) = 0 10t = 10t () (ii) dh d (90 40t + 5t ) = t = t () (d) When t = (i) dh dh = 10() or = = 0 = 0 () (e) dh = 0 10t = 0(0 t ) or t = 0( t 5) t = 0 or t = 4 ()() 3 (f) When t = 4 h = 90 40(4) + 5(4 ) = = 10 () 3 [15] 6. (a) N3 3 Note: Award for labelling 4 with horizontal, for labelling [AU] 5 metres, for drawing [TU]. IB Questionbank Maths SL 4

5 (b) TÂU = 86º () evidence of choosing cosine rule correct substitution e.g. = (5)(36) cos 86º = 4.4 N3 4 [7] 7. (a) attempt to substitute points into the function e.g. 8 = p( ) 3 + q( ) + r( ), one correct equation 8 = 8p + 4q r, = p + q + r, 0 = 8p + 4q + r N4 (b) attempt to solve system e.g. inverse of a matri, substitution p = 1, q = 1, r = A N3 Notes: Award for two correct values. If no working shown, award N0 for two correct values. [7] 8. (a) METHOD 1 choosing cosine rule substituting correctly e.g. AB = (3.9)(3.9) cos1. 8 AB = 6.11(cm) N METHOD evidence of approach involving right-angled triangles substituting correctly 1 e.g. sin 0.9 =, AB = 3.9 sin AB = 6.11 (cm) N METHOD 3 choosing the sine rule substituting correctly sin sin1.8 e.g. 3.9 AB AB = 6.11 (cm) N IB Questionbank Maths SL 5

6 (b) METHOD 1 refle A ÔB = π 1.8 (= 4.483) (A) correct substitution A = 1 (3.9) ( ) area = 34.1 (cm ) N METHOD finding area of circle A = π(3.9) (= ) finding area of (minor) sector A = 1 (3.9) (1.8) (= ) () () subtracting M1 e.g. π(3.9) 0.5(3.9) (1.8), area = 34.1 (cm ) N METHOD 3 finding refle A ÔB = π 1.8 (= 4.483) (A) finding proportion of total area of circle π 1.8 e.g. π(3.9), πr π π area = 34.1 (cm) N [7] 9. (a) combining terms () e.g. log 3 8 log 3 4, log log3 4 epression which clearly leads to answer given 8 4 e.g. log3, log3 3 f() = log 3 AG N0 (b) attempt to substitute either value into f e.g. log 3 1, log 3 9 f(0.5) = 0, f(4.5) = N3 3 (c) (i) a =, b = 3 N1N1 IB Questionbank Maths SL 6

7 (ii) Note: Award for sketch approimately through (0.5 ± 0.1, 0 ± 0.1) for approimately correct shape, for sketch asymptotic to the y-ais. N3 (iii) = 0 (must be an equation) N1 [6] (d) f 1 (0) = 0.5 N1 1 IB Questionbank Maths SL 7

8 (e) N4 4 Note: Award for sketch approimately through (0 ± 0.1, 0.5 ± 0.1), for approimately correct shape of the graph reflected over y =, for sketch asymptotic to -ais, for point ( ± 0.1, 4.5 ± 0.1) clearly marked and on curve. [16] 10. (a) = $59 (G) OR = 134 = $59 () IB Questionbank Maths SL 8

9 (b) Money ($) <0 <40 <60 <80 <100 <10 <140 Customers () 140 number of customers money ($) (A4) 5 Note: Award () for the correct scale, () for the points, and (A) for the curve. (c) (i) t = d /3 + 3 Mean d = 59 Mean t (59) / min. (3 sf) (accept 33.) () (ii) t > 37 d /3 + 3 > 37 d /3 > 34 d /3 > 17 () d > (17) 3/ d > 70.1 From the graph, when d = 70.1, n = 8 () number of shoppers = () = 5 () 8 [15] IB Questionbank Maths SL 9

10 11. (a) substituting (0, 13) into function M1 e.g. 13 = Ae = A + 3 A = 10 AG N0 (b) substituting into f(15) = 3.49 e.g = 10e 15k + 3, = e 15k evidence of solving equation e.g. sketch, using ln k = 0.01 ln0.049 accept N 15 (c) (i) f() = 10e f () = 10e (=.01e 0.01 ) N3 Note: Award for 10e 0.01, for 0.01, for the derivative of 3 is zero. (ii) valid reason with reference to derivative R1 N1 e.g. f () < 0, derivative always negative (iii) y = 3 N1 (d) finding limits , (seen anywhere) evidence of integrating and subtracting functions correct epression e.g g( ) f ( )d, [( 1 4) (10e 3)]d area = 19.5 A N4 [16] 1. (a) Since the verte is at (3, 1) h = 3 () k = 1 () (b) (5, 9) is on the graph 9 = a(5 3) + 1 = 4a + 1 () = > 9 1 = 4 a = 8 () = > a = (AG) 3 Note: Award ()(A0) for using a reverse proof, ie substituting for a, h, k and showing that (5, 9) is on the graph. (c) y = ( 3) + 1 IB Questionbank Maths SL 10

11 = (AG) 1 (d) (i) Graph has equation y = dy = 4 1 () d (ii) At point (5, 9), gradient = 4(5) 1 = 8 () (iii) Equation: y 9 = 8( 5) () 8 y 31 = 0 OR 9 = 8(5) + c c = 31 y = 8 31 () 4 [10] 13. (a) METHOD Finding gradient m = ( 5) 10 () y 13 = 5( ) y = METHOD u 3 = 13 and u 11 = 53 u 1 = 3 and d = 5 y = Note: Award no marks for showing that (, 13) and (10, 53) satisfy y = (AG) (N0) () (AG) (N0) (b) 3 kg () (N1) (c) Increase is 5 kg (per week) () (N1) (d) 98 = = 95 = 19 () (N) [6] IB Questionbank Maths SL 11

12 14. (a) Correct mid interval values 14, 3, 3, 41, 50 () Substituting into f w f M1 eg w = 7(14) 1(3) 13(3) 50 10(41) 8(50) w = w = 3 (kg) AG N0 (b) METHOD 1 Total weight of other boes = () Total number of other boes = 50 () Setting up their equation M1 eg = 30, = = 5 N3 METHOD Let z be the number of other boes in Class E (accept any symbol in the working, even including ). Total weight of other boes = z Total number of other boes = 4 + z Setting up their equation () () M1 eg z z = 30, z = z z = 3 = 5 N3 IB Questionbank Maths SL 1

13 (c) Setting up their inequality M1 Correct substitution eg (10 50 y y) , y y y y () 8y 50 (y 6.5) 6 N1 Note: If candidates don t use the mid-interval values, but assume that all the new boes weigh the minimum amount for Class D, award marks as follows: Setting up their inequality Correct substitution M1 eg y y y y () 3.5y 50 (y ) 14 N1 [1] IB Questionbank Maths SL 13