TMJC H2 Chapter 3 Functions Assignment Solutions 2024
Uploaded by KSKS · 28 September 2024
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Chapter 3 Functions TMJC 2024 Page 1 of 5 H2 Mathematics (9758) Chapter 3 Functions Assignment Solutions 1 2011/9740/II/3 The function f is defined by 1f : ln 2 1 3, , . 2x x x x (i) Find 1f x and write down the domain and range of 1f . [4] (ii) Sketch on the same diagram the graphs of fy x and 1fy x , giving the equations of any asymptotes and the exact coordinates of any points where the curves cross the x- and y-axes. [4] (iii) Explain why the x-coordinates of the points of intersection of the curves in part (ii) satisfy the equation ln 2 1 3,x x and find the values of these x-coordinates, correct to 4 significant figures. [3] Q1 Solution (i) Let ln 2 1 3y x . 3 3 2 1 e e 1 2 y y x x 3 1 e 1f 2 x x 1 1 f ff f 1, , 2D R R D (ii) x y fy x 1fy x (0,3) (3,0) 3e 10, 2 3e 1 , 02 1 2y 1 2 x The functions are still increasing, so do not ‘bend back on itself’. Question asks for ‘exact coordinates’. y x Note: all 3 graphs, ݕ= f(ݔ) , ݕ= f ିଵ(ݔ) and ݕ= ݔ all intersect at the same points (this includes the asymptotes) Reminder to use the same scale for x- and y-axis
Chapter 3 Functions TMJC 2024 Page 2 of 5 (iii) The points of intersection also lies on the line y = x, thus the x-coordinates also satisfy the equation of f ln 2 1 3 ln 2 1 3x x x x x x . By GC, x = –0.4847 or x = 5.482 (4 s.f.) There are 2 values of x so make sure the graphs in part (ii) intersect at 2 points. Note that GC does not show the intersection but you can still find the intersection.
Chapter 3 Functions TMJC 2024 Page 3 of 5 2 2017/PJC Promo/Q7 (modified) Functions f and g are defined by 2 f : 1 +2 for x x x , 1g : for , 5 . 5x x x x (i) Only one of the composite functions fg and gf exists. Give a definition (including the domain) of the composite that exists, and explain why the other composite does not exist. For the composite function that exist, find its range. [5] (ii) If the domain of f is restricted to x k , k , state the greatest value of k for which the function 1f exists. For the rest of the question, use the value of k found in part (ii). (iii) Sketch on the same diagram the graphs of fy x for x k and 1fy x , showing clearly the relationship between the two graphs. [4] (iv) Find 1f x and state the domain of 1 f . [3] Q2 Solution (i) gR \ 0 , fD Since g fR D , fg exists 1fg = f 5x x 2 1 1 +25x 2 1fg : 1 +2, , 5 5x x x x fgD = \ 5 fR [2, ) , gD = \ 5 Since f gR D , gf does not exist. g f fg g g fgD
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