1. Arithmetic sequence (M1) a = 200 d = 30 (A1) (a) Distance in final week = (M1) = 1730 m (A1) (C3) = 10 A1 3

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1 . Arithmetic sequence a = 00 d = 0 () (a) Distance in final week = = 70 m () (C) 5 (b) Total distance = [ ] = 5080 m () (C) Note: Penalize once for absence of units ie award A0 the first time units are omitted, the net time.. (a) a dv dt = 0 (b) s = vdt = 50t 5t + c 0 = 50(0) 5(0) + c c = 0 s = 50t 5t + 0 Note: Award and the first () in part (b) if c is missing, but do not award the final marks.. (a) r = 6 N (b) correct calculation or listing terms () 6 e.g.,8,,...,, u 6 = N (c) evidence of correct substitution in S e.g., S = 6 N [5] IB Questionbank Maths SL

2 . (a), 6, 9 N (b) (i) Evidence of using the sum of an AP M 0 eg 0 0 n n 60 N (ii) METHOD Correct calculation for 00 n n () 00 eg 99, 550 Evidence of subtraction eg n n 50 N METHOD Recognising that first term is 6, the number of terms is 80 ()() eg 6 00, n n 50 N 5. (a) Function displacement acceleration Graph A B AA N (b) t = A N IB Questionbank Maths SL

3 6. METHOD log log log 9 = + = log9 () = 9 = 7 () (C) METHOD log 8 + log 9 + log 9 = log (M) = log 9 7 () = 7 () (C) [] 7. (a) evidence of epanding M e.g. + ( ) + 6( ) + () +, ( + + )( + + ) ( + ) = A N (b) finding coefficients and ()() term is 5 N 8. (a) Mean = f f f = () (0) + () () + () (6) + () (k) + (5) (8) + (6) (6) + (7) (6) () f k + 0 () Using mean.6 = k k 0.6k + 8 = + k () 0.6k = 6 k = 0 () (C5) IB Questionbank Maths SL

4 (b) Mode = () (C) (accept 5, if k < 8 ) 9. evidence of anti-differentiation e.g. s = 6e d s = e t + t + C A substituting t = 0, 7 = + C C = 5 s = e t + t + 5 N [7] 0. (a) METHOD 5 + = 5 + = () = N METHOD Taking logs eg + = log 5 65, ( + )log 5 = log 65 log65 + = log5 () = N (b) METHOD Attempt to re-arrange equation + 5 = a a 5 N METHOD Change base to give log ( + 5) = log a + 5 = a a 5 N. Discriminant = b ac (= ( k) ) () IB Questionbank Maths SL

5 > 0 Note: Award (M0) for 0. (M) (k) > 0 k > 0 EITHER k > (k > ) () OR (k )(k + ) > 0 () OR (k )(k + ) > 0 () THEN k < or k > Note: Award () for < k <. ()() (C6). (a) (i) interchanging and y (seen anywhere) M e.g. = e y+ correct manipulation e.g. ln = y +, ln y = + f () = ln AG N0 (ii) > 0 N (b) collecting like terms; using laws of logs ()() e.g. ln ln, ln ln ; ln, ln simplify () e.g. ln =, = e = e ( e ) N [7] IB Questionbank Maths SL 5

6 . (a) METHOD recognizing that f(8) = e.g. = k log 8 recognizing that log 8 = e.g. = k k = () N METHOD attempt to find the inverse of f() = k log e.g. = k log y, y = substituting and 8 e.g. = k log 8, k k = 8 k = log 8 k N (b) METHOD recognizing that f() = e.g. log log = () f = (accept = ) A N METHOD attempt to find inverse of f() = log e.g. interchanging and y, substituting k = into y = k correct inverse () e.g. f () =, f = A N [7]. (a) d d ln, (seen anywhere) d d IB Questionbank Maths SL 6

7 attempt to substitute into the quotient rule (do not accept product rule) M e.g. ln correct manipulation that clearly leads to result ln ln e.g., ln, ln g AG N0 (b) evidence of setting the derivative equal to zero e.g. g () = 0, ln = 0 ln e N [7] 5. (a) evidence of using area of a triangle e.g. A sinθ A = sin N (b) METHOD P ÔA = () area OPA = sin θ (= sin ( )) since sin ( ) = sin R then both triangles have the same area AG N0 METHOD triangle OPA has the same height and the same base as triangle OPB R then both triangles have the same area AG N0 IB Questionbank Maths SL 7

8 (c) area semi-circle = area APB = sin + sin (= sin ) S = area of semicircle area APB (= sin ) M S = ( sin ) AG N0 (d) METHOD attempt to differentiate ds e.g. cos θ dθ setting derivative equal to 0 correct equation e.g. cos = 0, cos = 0, cos = 0 = N EITHER evidence of using second derivative S ( ) = sin S it is a minimum because S 0 OR evidence of using first derivative R N0 for <, S ( ) < 0 (may use diagram) for >, S ( ) > 0 (may use diagram) it is a minimum since the derivative goes from negative to positive R N0 METHOD sin is minimum when sin is a maimum R sin is a maimum when sin = (A) = A N IB Questionbank Maths SL 8

9 (e) S is greatest when sin is smallest (or equivalent) (R) = 0 (or ) N [8] 6. evidence of substituting for cos evidence of substituting into sin + cos = correct equation in terms of cos (seen anywhere) e.g. cos cos =, cos cos 5 = 0 evidence of appropriate approach to solve e.g. factorizing, quadratic formula appropriate working e.g. ( cos 5)(cos + ) = 0, ( 5)( + ), cos = correct solutions to the equation 9 e.g. cos = 5, cos =, = 5, = () = π N [7] d d 7. (a) sin cos, cos sin d d evidence of using the quotient rule correct substitution sin ( sin ) cos (cos ) e.g., sin (sin cos ) f () = sin f () = sin (seen anywhere) sin sin cos ()() M AG N0 (b) METHOD appropriate approach e.g. f () = (sin ) f () = (sin )(cos ) cos sin N Note: Award for sin, for cos. IB Questionbank Maths SL 9

10 METHOD derivative of sin = sin cos (seen anywhere) evidence of choosing quotient rule e.g. u =,v = sin sin 0 ( ) sin cos, f () = (sin ) f () = sin cos cos (sin ) sin N (c) evidence of substituting π π cos e.g., π π sin sin p =, q = 0 M NN (d) second derivative is zero, second derivative changes sign RR N [] 8. (a) tan θ = do not accept N (b) (i) sin θ = 5, cos θ = 5 correct substitution e.g. sin θ = 5 5 sin θ = 5 ()() N (ii) correct substitution e.g. cos θ = 7 cos θ = 5 5, 5 5 N [7] IB Questionbank Maths SL 0

11 9. (a) Lines on graph 00 students score 0 marks or fewer. N (b) Identifying 00 and 600 Lines on graph a = 55, b = 75 NN 0. (a) f () = + 5 N (b) evidence of attempting to solve f () = 0 evidence of correct working 6 0 e.g. ( + 5) ( ),, sketch = 5, = () so = 5 N IB Questionbank Maths SL

12 (c) METHOD f () = + (may be seen later) evidence of setting second derivative = 0 e.g. + = 0 = N METHOD evidence of use of symmetry e.g. midpoint of ma/min, reference to shape of cubic correct calculation 5 e.g., = N (d) attempting to find the value of the derivative when = f () = 6 valid approach to finding the equation of a line M e.g. y = 6( ), = 6 + b y = 6 6 N []. (a) substituting into the second derivative M e.g. f = 5 since the second derivative is negative, B is a maimum R N0 (b) setting f () equal to zero evidence of substituting = or e.g. f () correct substitution e.g. () p, p correct simplification 8 e.g. 6 + p = 0, + p = 0, + p = 0 p = AG N0 IB Questionbank Maths SL

13 (c) evidence of integration f() = c substituting (, ) or 58, 7 into their epression correct equation e.g. c, 8 c, 8 c f ( ) 0 N [] IB Questionbank Maths SL