Function Practice. 1. (a) attempt to form composite (M1) (c) METHOD 1 valid approach. e.g. g 1 (5), 2, f (5) f (2) = 3 A1 N2 2

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1 1. (a) attempt to form composite e.g. ( ) 3 g 7 x, 7 x + (g f)(x) = 10 x N (b) g 1 (x) = x 3 N1 1 (c) METHOD 1 valid approach e.g. g 1 (5),, f (5) f () = 3 N METHOD attempt to form composite of f and g 1 e.g. (f g 1 )(x) = 7 (x 3), 13 x (f g 1 )(5) = 3 N [5]. (a) valid approach e.g. b 4ac, = 0, ( 4k) 4( k)( 1) correct equation e.g. ( 4k) 4(k)(1) = 0, 16k = 8k, k k = 0 correct manipulation e.g. 8k ( k 1 ), 8 ± k = A N3 5 (b) recognizing vertex is on the x-axis M1 e.g. (1, 0), sketch of parabola opening upward from the x-axis P 0 N1 [7] 3. (a) f (x) = 10(x + 4)(x 6) N IB Questionbank Maths SL 1

2 (b) METHOD 1 attempting to find the x-coordinate of maximum point e.g. averaging the x-intercepts, sketch, y = 0, axis of symmetry attempting to find the y-coordinate of maximum point e.g. k = 10(1+ 4)(1 6) f (x) = 10(x 1) + 50 N4 4 METHOD attempt to expand f (x) e.g. 10(x x 4) attempt to complete the square e.g. 10((x 1) 1 4) f (x) = 10(x 1) + 50 N4 4 (c) attempt to simplify e.g. distributive property, 10(x 1)(x 1) + 50 correct simplification e.g. 10(x 6x + 4x 4), 10(x x +1) + 50 f (x) = x 10x AG N0 (d) (i) valid approach e.g. vertex of parabola, v (t) = 0 t =1 N (ii) recognizing a(t) = v (t) a(t) = 0 0t speed is zero t = 6 () a(6) = 100 (m s ) N3 7 [15] 4. (a) (1, ) N (b) g (x) = 3(x 1) (accept p =1, q = ) N IB Questionbank Maths SL

3 (c) (1, ) N [6] 5. (a) combining terms () e.g. log 3 8x log 3 4, log 1 3 x + log3 4 expression which clearly leads to answer given 8x 4x e.g. log3, log3 3 f(x) = log 3 x 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 3

4 (ii) Note: Award for sketch approximately through (0.5 ± 0.1, 0 ± 0.1) for approximately correct shape, for sketch asymptotic to the y-axis. N3 (iii) x = 0 (must be an equation) N1 [6] (d) f 1 (0) = 0.5 N1 1 IB Questionbank Maths SL 4

5 (e) N4 4 Note: Award for sketch approximately through (0 ± 0.1, 0.5 ± 0.1), for approximately correct shape of the graph reflected over y = x, for sketch asymptotic to x-axis, for point ( ± 0.1, 4.5 ± 0.1) clearly marked and on curve. [16] 6. (a) attempt to form composite e.g. f(x 5) h(x) = 6x 15 N IB Questionbank Maths SL 5

6 (b) interchanging x and y evidence of correct manipulation () e.g. x y x, = y x h ( x) = +15 N3 3 6 [5] 7. (a) evidence of setting function to zero e.g. f(x) = 0, 8x = x evidence of correct working 8 ± 64 e.g. 0 = x(4 x), 4 x-intercepts are at 4 and 0 (accept (4, 0) and (0, 0), or x = 4, x = 0) N1N1 (b) (i) x = (must be equation) N1 (ii) substituting x = into f(x) y = 8 N [7] 8. (a) q =, r = 4 or q = 4, r = N (b) x = 1 (must be an equation) N1 (c) substituting (0, 4) into the equation e.g. 4 = p(0 ( ))(0 4), 4 = p( 4)() correct working towards solution () e.g. 4 = 8p 4 1 p = = 8 N [6] IB Questionbank Maths SL 6

7 9. (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(x) = 10e 0.01x + 3 f (x) = 10e 0.01x 0.01 (=.01e 0.01x ) N3 Note: Award for 10e 0.01x, for 0.01, for the derivative of 3 is zero. (ii) valid reason with reference to derivative R1 N1 e.g. f (x) < 0, derivative always negative (iii) y = 3 N1 (d) finding limits , (seen anywhere) evidence of integrating and subtracting functions correct expression x e.g. g( x) f ( x)dx, [( x + 1x 4) (10e + 3)]dx area = 19.5 A N4 [16] IB Questionbank Maths SL 7

8 10. (a) N3 (b) x = 1.3, x = 1.68 (accept x = 1.41, x = 1.39 if working in degrees) N (c) 1.3 < x < 1.68 (accept 1.41 < x < 1.39 if working in degrees) A N [7] 11. (a) (i) 1.15, 1.15 N (ii) recognizing that it occurs at P and Q e.g. x = 1.15, x = 1.15 k = 1.13, k = 1.13 N3 (b) evidence of choosing the product rule e.g. uv + vu derivative of x 3 is 3x () derivative of ln (4 x x ) is 4 x () correct substitution 3 x e.g. x + ln(4 x ) 3x 4 x 4 x g (x) = + 3x ln(4 x ) 4 x AG N0 IB Questionbank Maths SL 8

9 (c) N (d) w =.69, w < 0 A N [14] 1. (a) (i) interchanging x and y (seen anywhere) M1 e.g. x = e y+3 correct manipulation e.g. ln x = y + 3, ln y = x + 3 f 1 (x) = ln x 3 AG N0 (ii) x > 0 N1 (b) collecting like terms; using laws of logs ()() 1 e.g. ln x ln 3, ln ln 3; ln x = x + x = = 3, ln x = 3 x 1 x simplify () 3 e.g. ln x =, x = e x = e ( = e ) N [7] 13. (a) for interchanging x and y (may be done later) e.g. x = y 3 g 1 x + (x) = x accept y =, N IB Questionbank Maths SL 9

10 (b) METHOD 1 g(4) = 5 () evidence of composition of functions f(5) = 5 N3 METHOD f g(x) = (x 3) f g(4) = ( 4 3) () = 5 N3 [5] 14. (a) A N (b) evidence of appropriate approach 1 e.g. reference to any horizontal shift and/or stretch factor, x = 3 + 1, y = P is (4, 1) (accept x = 4, y = 1) N3 [5] 15. (a) (i) coordinates of A are (0, ) N IB Questionbank Maths SL 10

11 (ii) derivative of x 4 = x (seen anywhere) () evidence of correct approach e.g. quotient rule, chain rule finding f (x) A e.g. f (x) = 0 ( 1) (x 4) ( x 4)(0) (0)(x) (x), ( x 4) substituting x = 0 into f (x) (do not accept solving f (x) = 0) M1 at A f (x) = 0 AG N0 (b) (i) reference to f (x) = 0 (seen anywhere) (R1) reference to f (0) is negative (seen anywhere) R1 evidence of substituting x = 0 into f (x) M finding f (0) = = 3 ( 4) then the graph must have a local maximum AG (ii) reference to f (x) = 0 at point of inflexion, (R1) recognizing that the second derivative is never 0 N e.g. 40(3x + 4) 0, 3x + 4 0, x 4, the numerator is 3 always positive Note: Do not accept the use of the first derivative in part (b). (c) correct (informal) statement, including reference to approaching y = 3 N1 e.g. getting closer to the line y = 3, horizontal asymptote at y = 3 (d) correct inequalities, y, y > 3, FT from (a)(i) and (c) N [16] IB Questionbank Maths SL 11