TMJC H2 Chapter 3 Function Extra Practice Solutions

Chapter 3 Functions TMJC 2024 Page 1 of 8 H2 Mathematics (9758) Chapter 3 Functions Extra Practice Solutions Q1 2018/NYJC Promo/7 (i) 1g(0.5) g(1.5) 4== . Since g is not one-one, g does not have an inverse. Alternative: Since the horizontal line 0.5y= cuts the graph of g( )yx= at 2 different points, f is not one-one. Thus 1g− does not exist. (ii)  )  )gfR 0, D 0,=   =  Since gfRD , fg exists (iii) ( )  )  ) gf fg g g fgD D , 2 R 0, R 1,= = − ⎯⎯ → =  ⎯⎯ → =  Q2 2008(9740)/II/4 (i), (iii) y x y = g (x) (1,0) (0,1) (2,1) O y x y = f (x) with domain (0,1) O y x 𝑦 = f −1(𝑥) 𝑦 = f(𝑥) y = x (1, 4) (4, 1) gf fg f R D , RR = = O Sketch based on the domain given. Use open circle for exclusion of points. Graphs should be symmetrical about the line y = x (Use same scale for x- and y-axis) Graphs must pass vertical and horizontal line test. Be careful with the curvature, ensure that the graph of ( ) 1f x− does not curve downwards!

Chapter 3 Functions TMJC 2024 Page 2 of 8 (ii) ( ) ( ) ( ) 2 2 Let 4 1 14 41 41 yx yx xy xy = − + − = − − = − =  − Since 4x , 41xy = + − ( ) 1f 4 1xx− = + − ( )1 ffD R 1,− = =  1f : 4 1, , 1x x x x− + −   (iv) Reflect graph of ( )fyx= in the line yx= to obtain graph of ( ) 1fyx −= . At intersection, ( ) ( ) 1ffy x x x −= = = , Equating yx= and ( )fyx= we have: ( ) ( ) 2 2 2 41 9 17 0 9 9 4 17 9 13 22 xx xx x = − + − + = − == Since 9 134, 2xx +  = . Q3 2009/TPJC/I/9 (i) From graph, fR = (ii)  )gR 2,= , fD \{0}= Since gfRD , fg exists. (iii) Since the horizontal line 1y= cuts the graph of f ( )yx= at 2 different points, f is not one-one. Thus 1f− does not exist. (iv) Greatest value of 0a= . ( ) 2ln , 0y x x= 2 eyx = e yx= 2e y x=− since 0x -1fD = 1 2f : e , x xx− →−  y x = 0 x y = f (x) (1,0) (−1,0) Use the domain of f to decide which expression to pick. Find fR by sketching graph of f and find range of possible y values. (i.e. minimum and maximum y values)

Chapter 3 Functions TMJC 2024 Page 3 of 8 Q4 2018/SAJC Promo/6 (i) 2 2 1 1 1 1 1Let (1 )(3 ) 32 ( 1) 1 3 14 14 y y y y y xx xx x x x = +− + − = − − − =− − = − =  − 114 yx− =− − since 1x  114 yx= − − 1 1f ( ) 1 4x x − = − − (ii) ( )gR, = −  and ( )hD ln 2,= Since ghRD ,  hg does not exist. (iii) ( ) ( )( ) ( ) ( ) gh g h g4 ln 2 4 x x xx e e − − = = =− ( ) ( ) ( )h h ghD ln 2, R 0, 2 R ,ln 2=  → = → = − (iv) Let ( ) 1 4gh ln 3 a −  = ( ) 4gh ln 3a= y x y = h (x) with domain (ln2,2) O y x y = g (x) with domain (0,ln2) O y=g(x) x (1,0) x = 2 (0, ln 2) O

Chapter 3 Functions TMJC 2024 Page 4 of 8 ( ) 4ln 2 4 ln 3 424 3 1