RI Functions C4 Add Prac (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ________________________________________ Additional Practice Questions for Chapter 4: Functions Page 1 of 17 Additional Practice Questions for Chapter 4: Functions (Solutions) 1 9758/2017/02/3b The function g is defined by 1g: 1 , w h e r e , .1xx x a x   (i) State the value of a and explain why this value has to be excluded from the domain of g. [2] (ii) Find  2g x and  1g, x giving your answers in simplified form. [4] (iii) Find the values of b such that   21gg .bb  [2] Solution: (i) a = 1. This value has to be excluded from domain of g as 1x will make the denominator 1 x zero. (ii)    2gg g 1g1 1 11 111 1 11 xx x x x x              Let 11 1y x  1 11 11 1 11 1 yx x y x y         1 1g1 1x x    (iii)     21gg 11 1 11 1 20 0 or 2 bb b b bb b b bb bb            

Raffles Institution H2 Mathematics 2025 Year 5 ______________________________________________________________________________________________ ________________________________________ Additional Practice Questions for Chapter 4: Functions Page 2 of 17 2 RVHSPromo9740/2009/03 The function f is defined by  3 f: l n 8 ,x x 8x . (a) Solve the inequality f( )x x . [3] (b) Find, in a similar form, the inverse function 1f . [3] Solution (a)  3 f( ) ln 8 x x x x   From GC, [4.09, 8)x (b)   3 Let ln 8 , then 3ln 8 .yx y x    3 3 ln 83 e8 8e y y y x x x    fR   Therefore, f-1: 38e , x xx  .

Raffles Institution H2 Mathematics 2025 Year 5 ______________________________________________________________________________________________ ________________________________________ Additional Practice Questions for Chapter 4: Functions Page 3 of 17 3 9758/2021/02/03 (a) The function h is defined by 21h: 3 , 2xx for x  . The function g is defined by 1g: , 51 xx x   for x  ,0 . 2 .x (i) Find gh(2). [2] (ii) Find the value of x for which g( ) 1.4.x = [1] (b) The function f is defined by f: , 2 x ax x b   for x  ,.x k (i) Give an expression for k and explain why this value of x has to be excluded from the domain of f. [2] The function f is such that 1f( ) f ( )x x for all x in the domain of f. (ii) Determine the possible values of a and of b. [3] (iii) Find an expression for 1f( 4 ) .  [1] Solution (a)(i)   211gh(2) g 2 3 g 524    . (a)(ii)  1g1 . 4 51 xx x   From GC, 0.4.x (b)(i) When 20 , . 2 bxb x   This value of x has to be excluded from the domain of f as 2 bx will make the denominator 2x b zero. 2 bk  (b)(ii) 1 2f( ) 22 2 baxax x bx b   by long division and fD\ 2 b     , f 1R\ 2     . Since 1f( ) f ( ) ,x x and 11 ffff 1DR \, RD \ 22 b         ,

Raffles Institution H2 Mathematics 2025 Year 5 ______________________________________________________________________________________________ ________________________________________ Additional Practice Questions for Chapter 4: Functions Page 4 of 17 So we have 1 1 2f( ) , 22 ba x x b     x  1,. 2x Hence from the denominator, we have 120 ,2 b  so 1.b Also since f is a rational function and its graph is not a horizontal line, else 1f does not exist, so 0.2 ba Hence 1 .2a 1\ and 1.2ab       (b)(iii)    1 4 4f( 4 ) 241 9 ab a     since 1.b

Raffles Institution H2 Mathematics 2025 Year 5 ______________________________________________________________________________________________ ________________________________________ Additional Practice Questions for Chapter 4: Functions Page 5 of 17 4 HCICT8864/2006/07 The function f is defined by f: 2 1 4 ,xx x    . (i) Show that the composite function ff exists and give the corresponding definition of ff . State the range of ff . (ii) Solve the equation f( ) 1 0 0x  . (iii) The function g is defined by g: 2 1 4 , .x xx k  Find the greatest value of k for 1g to exist. Using this value of k , find 1g . Solution (i)  fR4 , and fD   Since ffRD , so ff exists.  ff f 2 1 4 22 1 7 4 22 1 1 1 xx xx        ff : 2 2 1 11, xx x    ffR1 1 , (ii)  f( ) 1 0 0 21 4 1 0 0 21 9 6 97 952 1 96 or 2 1 96 or 22 x x x xxx x            (iii) Greatest value of k is 1 2 . For 1 2x  ,    g2 1 4 2 5xx x     . Let 25yx  , 1 2x  1 52x y   1 1g: 5 , 4 2xx x  .

Raffles Institution H2 Mathematics 2025 Year 5 ______________________________________________________________________________________________ ________________________________________ Additional Practice Questions for Chapter 4: Functions Page 6 of 17 5 MIPromo9740/2012/06 Functions g and h are defined by 2g: 2 ,xx m x x  where m is a constant,  h: l n 1 2 , 1x xx  . (i) Sketch the graph of h( )yx , stating the exact coordinates of the point of intersection with the axis and the equation of the asymptote. [2] (ii) Find 1h( ) x and write down the domain of 1h . [3] (iii) On the same diagram as in part (i), sketch the graph of 1h( ) .yx  Find the solution of the equation 1h( ) h ( )x x . [3] State the condition for the composite function hg to exist. Find the maximum range of values of m such that hg exists. [3] Solution (i) (ii) 1 2 12 h Let h( ) ln( 1) 2 1 e h ( ) 1 e with domain, D y x yx x x x            (iii) Using GC, 2x  is the solution to the equation.  2 2 g2 24 mmxx    For hg to exist, 2 gh2,R D ( 1 , ) .4 m      2 21 4 22 m m      x 1y h( )yx O 1x  2(1 e , 0) yx 1h( )yx  y

Raffles Institution H2 Mathematics 2025 Year 5 ______________________________________________________________________________________________ ________________________________________ Additional Practice Questions for Chapter 4: Functions Page 7 of 17 6 NJCCommonTest8864/2007/Q2 The functions f , g and h are defined as follows: 2 2 f: 4 1 , 1g : 1 , \{0} h : e , x xx x x k xx x xx        (i) State the smallest value of k such that 1f exists and express 1f in a similar form. [3] (ii) Express each of the following functions in terms of f, g and h as appropriate. (a) ln , 0xx x  (b) 2 11, e xxx  (c) , 3xx x  [4] Solution (i)  22f( ) 4 1 2 3xx x x   Least value of 2k    2 2 Let f ( ). 23 32 3 2 or 3 2 (rejected, since 2) yx yx yx xy x y x              1f: 32 , 3xx x    (ii) (a) 1h( ) l n , 0xx x  (b) 2 1gh( ) 1 , e xxx   (c) 3 ,)(ff 1  xxx

Raffles Institution H2 Mathematics 2025 Year 5 ______________________________________________________________________________________________ ________________________________________ Additional Practice Questions for Chapter 4: Functions Page 8 of 17 7 SAJCPromo9740/2012/Q6 Functions f and g are defined by 2f : 2 for , where is an unknown constant g: t a n f o r , . 22 xx x a x a xx x x         (i) Find the range of f in terms of a. [2] (ii) Show that the composite function 1fg exists. Find the range of 1fg in terms of a. [3] (iii) It is given that 3a . The domain of f is now restricted to  :1xx . For the new domain, give a definition (including the domain) of -1f. [3] Solution (i)  2 2 f( ) 2 11 x xx a x a    From the graph, range of f [1 , )a  (ii) -1 ggRD , 22     and fD   Since -1 fgRD  , the comp