Exponential and Log Functions Quiz (non-calculator) [46 marks] Let f(x) = log p (x + ) for x >. Part of the graph of f is shown below. The graph passes through A(6, 2), has an x-intercept at ( 2, 0) and has an asymptote at x =. 1a. Find p. evidence of substituting the point A e.g. 2 = log p (6 + ) manipulating logs e.g. p 2 = 9 p = A2 In part (a), many candidates successfully substituted the point A to find the base of the logarithm, although some candidates lost a mark for not showing their manipulation of the logarithm equation into the exponential equation. 1b. The graph of f is reflected in the line y = x to give the graph of g. (i) Write down the y-intercept of the graph of g. [ marks] (ii) Sketch the graph of g, noting clearly any asymptotes and the image of A.
(i) y = 2 (accept (0, 2)) N1 (ii) N4 Note: Award for asymptote at y =, for an increasing function that is concave up, for a positive x-intercept and a native y-intercept, for passing through the point (2, 6). [ marks] A number of candidates who correctly stated the y-intercept was 2 had difficulty sketching the graph of the reflection in the line y = x. A number of candidates graphed directly on the question paper rather than sketching their own graph; candidates should be reminded to show all working for Section B on separate paper. Some correct sketches did not have the position of A indicated. Many candidates had difficulty reflecting the asymptote. The graph of f is reflected in the line y = x to give the graph of g. 1c. Find g(x).
METHOD 1 recognizing that g = f 1 evidence of valid approach (R1) e.g. switching x and y (seen anywhere), solving for x correct manipulation e.g. x METHOD 2 recognizing that g(x) = () N identifying vertical translation (R1) () e.g. graph shifted down units, f(x) evidence of valid approach e.g. substituting point to identify the base = y + g(x) = x g(x) = x a x N + b Part (c) was often well done, with candidates showing clear and correct working. The most successful candidates clearly appreciated the linkage between the question parts. Let f(x) = x, for x. Find f. 2a. 1 (2) METHOD 1 attempt to set up equation 2 = y, 2 = x correct working () 4 = y, x = 2 2 + f 1 (2) = 9 METHOD 2 interchanging x and y (seen anywhere) x = y correct working () x 2 = y, y = x 2 + f 1 (2) = 9 Candidates often found an inverse function in which to substitute the value of 2. Some astute candidates set the function equal to 2 1 and solved for x. Occasionally a candidate misunderstood the notation as asking for a derivative, or used. f(x)
Let g be a function such that g exists for all real numbers. Given that, find. 2b. 1 g(0) = (f g 1 )() recognizing g 1 () = 0 f(0) correct working () (f g 1 )() = 0, 2 (f g 1 )() = Note: Award A0 for multiple values, ±. For part (b), many candidates recognized that if g(0) = then g () = 0, and typically completed the question successfully. 1 Occasionally, however, a candidate incorrectly answered 2 = ±. Find the value of log 40. a. 2 log 2 evidence of correct formula loga logb = log a, log( ), log8 + log log b Note: Ignore missing or incorrect base. 40 correct working log 2 8, 2 = 8 40 = log 2 log 2 () 40 Many candidates readily earned marks in part (a). Some interpreted log 2 40 log 2 to mean, an error which led to no further marks. Others left the answer as log 2 where an inter answer is expected. log 2 log 2 b. Find the value of 8 log 2.
attempt to write 8 as a power of 2 (seen anywhere) (2 ) log 2, 2 = 8, 2 a multiplying powers 2 log 2, alog 2 correct working () log 2 2 12, log 2, ( 2 ) log 2 log 8 2 = 12 N Part (b) proved challenging for most candidates, with few recognizing that changing 8 to base 2 is a helpful move. Some made it as far as 2 log 2 yet could not make that final leap to an inter. The following diagram shows the graph of y = f(x), for 4 x. 4a. Write down the value of f 1 (1). f 1 (1) = 0 (accept y = 0) N1 4b. Find the domain of f 1.
domain of f 1 is range of \(f\) (R1) Rf = Df 1 correct answer x, x [, ] (accept < x <, y ) 4c. On the grid above, sketch the graph of f 1. Note: Graph must be approximately correct reflection in y = x. Only if the shape is approximately correct, award the following: for x-intercept at 1, and for endpoints within circles. Let f(x) = x 2 and g(x) =, for x 0. x a. Find f 1 (x).
interchanging x and y x = y 2 f 1 (x) = (accept y =, ) b. Show that (g )(x) =. f 1 attempt to form composite (in any order) correct substitution g ( ), ( ) +2 x (g f 1 )(x) = AG N0 Let h(x) =, for x 0. The graph of h has a horizontal asymptote at y = 0. c. Find the y-intercept of the graph of h. valid approach 2 h(0), 0+2 y = (accept (0, 2.)) d. Hence, sketch the graph of h.
A2 N Notes: Award for approximately correct shape (reciprocal, decreasing, concave up). Only if this is awarded, award A2 for all the following approximately correct features: y-intercept at (0, 2.), asymptotic to x- axis, correct domain x 0. If only two of these features are correct, award. e. For the graph of h 1, write down the x-intercept; x = (accept (2., 0)) 2 N1 f. For the graph of h 1, write down the equation of the vertical asymptote. x = 0 (must be an equation) N1 g. Given that h 1 (a) =, find the value of a.
METHOD 1 attempt to substitute into h (seen anywhere) correct equation a = 1 METHOD 2 () attempt to find inverse (may be seen in (d)) correct equation, a = 1 () h(), a = +2 +2 y+2, h() = a x =, = 2, + 2 h 1 x x 2 = x International Baccalaureate Organization 2016 International Baccalaureate - Baccalauréat International - Bachillerato Internacional Printed for North Hills Preparatory