TMJC H2 Chapter 3 Functions Discussion Questions Solutions 2024

Chapter 3 Functions TMJC 2024 Page 1 of 20 H2 Mathematics (9758) Chapter 3 Functions Discussion Questions (Suggested Solutions) Level 1 1 Sketch the graphs of each of the following functions. State its domain and give its corresponding range. (a) 23f : , , 1 1 xx x x x − − (b) ( ) 2 g : 2 4, , 2x x x x− − +   (c) ( )h : ln 1 , ,1 3x x x x −    1 Solution (a) ( )fD 1,= ( )fR , 2= − (b) gD ( ,2]= − gR ( ,4]= − y x O y = f (x) y x y = g(x) 23 1 xy x −= − Vertical asymptote: 1x= Horizontal asymptote: 2y= Find by referring to the graph of f to determine the range of possible y values. (i.e. minimum and maximum y values) Take the domain of f into consideration when sketching the graph of f ( )2, 4 Use a closed circle at the point since is included in the domain of g.

Chapter 3 Functions TMJC 2024 Page 2 of 20 (c) ( )hD 1,3= ( )hR ,ln 2= − y x O y = h(x) Use an open circle at the point since is excluded from the domain of h. ( )3,ln 2 Ensure the graph approaches the vertical asymptote

Chapter 3 Functions TMJC 2024 Page 3 of 20 2 The function f is defined by ( ) 2 f : 2 4,xx − − + , 0xx . (i) Sketch the graph of f. (ii) State the domain and range of f. (iii) Explain why 1f− does not exist. 2 Solution (i) (ii) ( )fD 0,= ( fR , 4= − (iii) Method 1: Horizontal Line Test Since the horizontal line 3y= cuts the graph of ( )fyx= more than once, f is not a one-one function. Hence 1f− does not exist. Method 2: Algebraic Method Since ( ) ( )f 1 f 3 3== , f is not a one-one function. Hence 1f− does not exist. y x 3y= Check both the rule and domain when sketching the graph. Open circle since 0x= is not included. Show that f is not a one-one function by giving a specific example of a horizontal line that cuts the graph the graph more than once. Remark: Any horizontal line yk= with 04 k will be accepted. To show that inverse does not exist, show that f is not a one-one function Show that f is not a one-one function by giving a specific example of two inputs that result in the same output.

Chapter 3 Functions TMJC 2024 Page 4 of 20 3 The function f is defined as 23f : , , 1 1 xx x x x − − . (i) Explain why 1f− exists. (ii) Find 1f− in a similar form. 3 Solution (i) Since any line ya= , where a , cuts the graph of f at most once, f is one-one. Hence 1f− exists. (ii) Let 23 1 xy x −= − ( ) ( ) 1 2 3 23 3 2 y x x x y y yx y − = − − = − −= − ( ) ( ) 1 1 3f 2 3f 2 yxy y xx x − − −== − −= − ( )1 ffD R , 2− = = − 1 3f : , 2,2 xxx x x− − −  y x O y = f (x) ya= To explain that 1f− exists  show f is one-one (horizontal line test) Note: you must draw the graph if it is not given by the question. To write the expression of in similar form, state the rule and domain of

Chapter 3 Functions TMJC 2024 Page 5 of 20 4 Functions g and h are defined by 2g 1, 0: x x x ,x+   , h 2 3 2: x x , x ,x .+   (i) Show that gh exists. (ii) Find gh in a similar form. (iii) Find the range of gh. 4 Solution (i) ( )hR 7,=  and  )gD 0,= Since hgR D, gh exists. (ii) ( )( ) ( ) 2 g h 2 3 1xx = + + ( ) 2 gh : 2 3 1, 2x x x ,x + +   (iii) Method 1: Using rule and domain of gh ( )ghR 50,=  Method 2: Using mapping method ( ) ( ) ( )h h g hg hD 2, R 7, R 50,=  ⎯⎯ → =  ⎯⎯ → =  ( )ghR 50,=  y x y = g (x) with restricted domain O y x y = h (x) O y x y = g (x) O

Chapter 3 Functions TMJC 2024 Page 6 of 20 Level 2 5 The function g is defined as ( ) 2g : ln , , 0x x x x  . (i) State the domain and range of g. (ii) Give a reason why 1g− exists. (iii) Find the rule, domain and range of 1g− . 5 Solution (i) gD ( ,0)= − gR ( , )= −  (ii) Since any line ya= , where a , cuts the graph of g at most once, g is one- one. Hence 1g− exists. (iii) Let ( ) 2lnyx= 2e e y y x x = = Since 0x , e yx=− ( ) 1ge xx− =− ( )1 ggD R ,− = = −  = ( )1 ggR D ,0− = = − y x O y = g (x) y x O y = g (x) Ensure the graph approaches the vertical asymptote Take the domain of g into consideration when sketching the graph of g Find by referring to graph of g to determine the range of possible y values. (i.e. minimum and maximum y values) To explain that 1g− exists  show g is one-one (horizontal line test) ya= Use the domain of g to decide which expression to pick.

Chapter 3 Functions TMJC 2024 Page 7 of 20 6 2013/CJC Prelim/II/3 (modified) Functions f and g are defined by 2f : ( 2) 1, , 2x x x x − −   2g : ln( 1), x x x + Only one of the composite functions fg and gf exists. Give the rule and domain of the composite function that exists, and explain why the other composite does not exist. 6 Solution  )gR 0,= ( )fD , 2= − Since gfRD , fg does not exist ( )fR 1,= −  gD = Since fgRD , gf exists. ( ) ( )( ) ( )( ) 2 22 gf g 2 1 ln 2 1 1 xx x = − − = − − + ( )gf fD D , 2= = − ( )( ) 22 gf : ln 2 1 1 , , 2x x x x  − − +   x y O y

Chapter 3 Functions TMJC 2024 Page 8 of 20 7 2018/ACJC Promo/Q9(part) Functions f and g are defined by 3f : , , 4,4 1g : , , 0. xx x x x x x x x + −  (i) Show that the composite function fg exists. [1] (ii) Find the range of fg. [2] (iii) Find an expression for ( )fg x and hence, or otherwise, find ( ) 1 1fg 2 −   . [3] 7 Solution (i) ( )gR ,0= −  fD \ 4= gfRD Therefore, fg exists.  (ii) Method 1: Using rule and domain of fg ( ) 1 3 13fg 1 414 xxx x x + +== −− ( )fg gD D ,0= = − fg 3R 1, 4 =− Method 2: Using mapping method ( ) ( ) gf g g fg 3D ,0 R ,0 R 1, 4 = − ⎯⎯ → = − ⎯⎯ → = −  y x y = −1 x = 4 (−3,0) ൬0, 3 4൰ y = f(x) with domain (−∞, 0) y x y = 0 x = 0 y = g(x) with domain (−∞, 0) O y x y = 0 x = 0 y = g(x) with domain (−∞, 0) O O y x y = y = fg(x) O Find fgR by referring to the graph of fg to determine the range of possible y values. (i.e. minimum and maximum y values)

Chapter 3 Functions TMJC 2024 Page 9 of 20 (iii) ( ) 1 3 13fg 1 414 xxx x x + +== −− ( ) ( ) ( ) 1 11Let fg fg 22 1 3 1 4 1 2 2 1 3 4 1 23 3 2 aa a a aa a a −  =  = + =− + = − =− =− ( ) 1 13fg 22 −  =−  ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1fg 2 1fg fg fg 2 1 fg2 [since fg fg ] a a a xx − − −  = =  = =

Chapter 3 Functions TMJC 2024 Page 10 of 20 8 It is given that ( ) ( ) 2 f 1 2, , 0 2x x x x= − +    and that ( ) ( )f f 2xx=+ for all real values of x. (i) State the period of f. (ii) Evaluate ( )f1 and ( )f2− . (iii) Sketch the graph of ( )fyx= for 23 x−   . 8 Solution (i) 2 (ii) ( ) ( ) 2 f 1 1 1 2 2= − + = ( ) ( ) ( ) ( ) 2 f 2 f 2 2 f 0 0 1 2 3− = − + = = − + = (iii) ( ) ( )f 3 =f 1 2 = 3 2 y x –1 1 2