TMJC Term 4 Revision Chapter 3 Functions Solutions

Term 4 Revision Topical Quick Check: Functions TMJC 2024 JC1 H2 Mathematics (9758) Term 4 Revision Topical Quick Check Chapter 3 Functions Revision Guide Chapter 3 Page 3-4 6. One-one functions (b) Horizontal Line Test (MUST SKETCH GRAPH) The function f is one-one if any horizontal line yb= ( b ) cuts the graph of f at most once. • If not one -one: find a counter -example (sketch a horizontal line yk= that cuts the graph twice (providing the value of k) or find f,Dab  such that ( ) ( )ff ab= ). INVERSE FUNCTIONS 7. The inverse of a function f is denoted by 1f− . (Do not confuse inverse function ( ) 1f− with reciprocal graph ( ) 1 f x   ) 8. 1f− exists  f is one-one. 9. -1 ffDR = and -1 ffRD = 10. The graph of 1f− can be obta ined by reflecting the graph of f about the line yx= (provided that f has an inverse). 11. ( ) ( ) 11ff f f x x x−− == ; but 1ff− and 1ff− have different domains. (a) -1 -1 fff fD D R== (b) -1 fffDD = 12. If  occurs in the process of finding inverse, use the domain of f to determine whether to choose (positive square root) or − (negative square root). COMPOSITE FUNCTIONS 13. If f and g are two functions such that fgR D, then the composite function g of f (i.e. gf) exists and is defined by ( ) ( )( )gf g fxx= for all fD.x 14. For gf to exist, fgR D . 15. Domain of composite function gf, gf fD. D= 16. Range of composite function gf, gf gR. R (a) Method 1: Sketch both functions and use f f gf fgD R R⎯⎯ → ⎯⎯ → [**Using fR as restricted domain of g] (b) Method 2: Find the composite function and sketch the graph, keeping in mind of the domain of the first function, i.e., fD since gf fD D .=

Term 4 Revision Topical Quick Check: Functions TMJC 2024 Let’s Try Now 1 YIJC Promo 9758/2022/Q9 (modified) The function f is defined by 2f : 4 5xx x −− , for x , 2x . (a) Find 1f ( )x− and state the domain of 1f− . [3] (b) On the same diagram, sketch the graphs of f and 1f− . [3] (c) Find the exact solution of the equation 1f ( ) f ( )xx −= . [3] The function g is defined by 2g: 1x x− , for , 1x x . (d) Explain why the composite function fg exists and find the range of fg. [3]

Term 4 Revision Topical Quick Check: Functions TMJC 2024 1 YIJC Promo 9758/2022/Q9 (modified) The function f is defined by 2f : 4 5xx x −− , for x , 2x . (a) Find 1f ( )x− and state the domain of 1f− . [3] Q1 Solution (a) Let 2 45y x x= − − . Then 2 2 ( 2) 9 9 ( 2) 2 9 29 yx yx x y xy = − − + = − −= + =  + Since ( fD ,2=−  , 29xy= − + . Hence, 1f ( ) 2 9xx− = − + )1fD 9,− = −  (b) On the same diagram, sketch the graphs of f and 1f− . [3] Q1 Solution (b) yx=

Term 4 Revision Topical Quick Check: Functions TMJC 2024 (c) Find the exact solution of the equation 1f ( ) f ( )xx −= . [3] Q1 Solution (c) 2 2 f ( ) 45 5 5 0 xx x x x xx = − − = − − = ( ) ( )( ) ( ) 2( 5) 4 1 5 21 5 2 55 3 x x −−− =  −−= Since ( fD ,2=−  , then we reject 5 3 5 2x += . Therefore, 3 55 2x= − . The function g is defined by 2g: 1x x− , for , 1x x . (d) Explain why the composite function fg exists and find the range of fg. [3] Q1 Solution (d)   g f 2D R ( ,1 (, = − = − Since gfR D , then the composite function fg exists. fgR [ 8, )= − 

Term 4 Revision Topical Quick Check: Functions TMJC 2024 Let’s Try Now 2 HCI Promo 9758/2022/Q6 The function h is defined as follows. ( ) 321h : 6 32 ,8x x x −+ .x (i) Explain why h does not have an inverse. [1] (ii) If the domain of h is restricted to 0, xk state the largest value of k for which the function 1h− exists. [1] Use the domain of h in part (ii) for the rest of this question. (iii) Sketch the graphs of h and 1h− on the same diagram, showing clearly the relationship between the two graphs, and the coordinates of the end points for both graphs. [3] (iv) Deduce the solution(s) of the equation ( ) ( ) 1hhxx −= . [2] (v) The function g is defined as follows. ( )( ) 2 g : ln 3 1 ,xx −+ .x Given that the composite function gh exists, find the exact range of gh. [2]

Term 4 Revision Topical Quick Check: Functions TMJC 2024 2 HCI Promo 9758/2022/Q6 The function h is defined as follows. ( ) 321h : 6 32 ,8x x x −+ .x (i) Explain why h does not have an inverse. [1] 2 Solution (i) Since 2y= cuts the graph of h more than once, h is not one to one function, hence 1h− doesn’t exist. (ii) If the domain of h is restricted to 0, xk state the largest value of k for which the function 1h− exists. [1] 2 Solution (ii) Largest value of k is 4. Use the domain of h in part (ii) for the rest of this question. (iii) Sketch the graphs of h and 1h− on the same diagram, showing clearly the relationship between the two graphs, and the coordinates of the end points for both graphs. [3] 2 Solution (iii) (iv) Deduce the solution(s) of the equation ( ) ( ) 1hhxx −= . [2] 2 Solution (iv) From GC, ( )h2x x x=  = 2 or 0 or 4.x x x = = = y x y = 2

Term 4 Revision Topical Quick Check: Functions TMJC 2024 (v) The function g is defined as follows. ( )( ) 2 g : ln 3 1 ,xx −+ .x Given that the composite function gh exists, find the exact range of gh. [2] 2 Solution (v)     g hR 0, 4 0,ln10= ⎯⎯ →  ghR 0,ln10= OR 232 6 32gh( ) ln 3 1 8 xxx  −+= − +  ghR 0,gh(4)=  ghR 0,ln10=

Term 4 Revision Topical Quick Check: Functions TMJC 2024 Let’s Try Now 3 CJC Promo 9758/2022/Q10(b) Functions g and h are defined by 2 1g : , , 1,1 h : 1 2 , . x x x x x x x − − (i) Explain why the composite function gh does not exist. [2] (ii) Find ( )hg x . [1] (iii) Find the range of ( )hg x . [2] (iv) By using the result in part (ii), or otherwise find ( ) ( ) 1 hg 4 − . [3]

Term 4 Revision Topical Quick Check: Functions TMJC 2024 3 CJC Promo 9758/2022/Q10(b) Functions g and h are defined by 2 1g : , , 1,1 h : 1 2 , . x x x x x x x − − (i) Explain why the composite function gh does not exist. [2] Q3 Solution (i) For gh to exist, hgRD . ( ) ( ) h g R, D 1, = −  = Since hgRD , gh does not exist. (ii) Find ( )hg x . [1] Q3 Solution (ii) ( ) 22 2 1 1 2hg h 1 2 1 1 1 1x x x x    = = − = −   − − −    (iii) Find the range of ( )hg x . [2] Q3 Solution (iii) Method : Composite Function )(hg gD D 1,= =  )(hgR 1,= Method : Mapping Method x y x = 1 y = 1 y = hg(x)

Term 4 Revision Topical Quick Check: Functions TMJC 2024 )( )( )( gh g g hgD 1, R ,0 R 1,=  ⎯⎯ → = − ⎯⎯ → =  (iv) By using the result in part (ii), or otherwise find ( ) ( ) 1 hg 4 − . [3] Q3 Solution (iv) Hence Let ( ) ( ) 1 hg 4 a − = ( ) 2 2 2 2 2 4 hg 241 1 2 31 2 3 3 35 5 3 55 or33 a a a a a a a = =− − =−− =− + = = =− Since ( )hg gD D 1,= =  , 5 3a= x y x = 1 y = 0 y = g(x) x y ½ 1 y = h(x)