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 6 6 1ln 6 accept ln 6 a a a a e e e e a − − − −= −= = = =−  Q5 2009(9740)/II/3 (i) Let ( )f, xy= 1f : , , axy bx a bxy ay ax bxy ax ay ayx by a ax ax x x bx a b − = − − =  − = = − − ( ) ( ) ( ) ( ) ( ) 1 2 1Since f f , f ff f fx x x x x x−−= = = = . 2ff fD \ R \ R \a a a b b b      = → = → =           2 f : , , ax x x x b 2fR\ a b =  (ii) Since ( ) 2f xx= , 2023 2022f ( ) ff ( ) f ( ) axx x x bx a= = = − 2023f (1) a ba= − (iii) ( ) 1g , , 0x x xx=    gR \ 0= , fD\ a b =  Observe that gRa b but fDa b . Since gfR D , the composite function fg does not exist. Note: ffR \ D a b ==  , that’s why 2f exists.

Chapter 3 Functions TMJC 2024 Page 5 of 8 (iv) ( ) 1f xx− = ( ) 2 2 20 20 20 or ax xbx a ax bx ax bx ax x bx a axx b =− =− −= −= == Q 6 Suggested Solutions Things to note: • Proper scale for x-axis (e.g. 1cm denote 1 unit) • Open circles for excluded endpoints and closed circles for included endpoints • Draw dotted reference lines for critical y-values to ensure you always end at the same height Check for continuous/disjoint at 1:x= When 1,x= 2f ( ) 1 1x == When 1,x= ( )f ( ) 3 2 1 1x = − = Thus, the graph is continuous at 1.x= Do NOT “cancel” x, always do factorisation and conclude from there. Calculating the coordinates of end-points for 47 x−   : ( ) ( ) ( ) ( ) ( ) ( ) 2f 7 f 4 f 1 1 1 closed circle f 4 ... f 2 3 2 2 1 closed circle = = = =  − = = = − =−  ( ) ( )f 3 fxx+=  The period of the graph is 3 units. The graph repeat itself every 3 units. Axial intercept: 3 2 0 3 2 x x −= =

Chapter 3 Functions TMJC 2024 Page 6 of 8 ( ) ( )( ) ( ) ( )f 26 f 8 3 2 f 2 3 2 2 1= + = = − =− Q7 2009/NYJC/I/2 (i) ( ) ( )fgD , and R 2,= −  =  Since gfRD function fg exists. ( ) ( ) ( ) 2 fg f 4 4 1 fg : 3, x x x x x x + = + = + − + (ii) ( )fgR 3,= ( )gD 0,= Since fg gRD function gfg exists ( ) ( ) ( ) ( )h gfg g fg 3 4 7x x x x x= = = + + = + (iii) ( ) ( ) -1h g gxx= , x > 0 ( ) ( ) 1g fg g gxx −= ( )( ) ( ) ( ) ( ) 2 2 7, 7 70 1 1 4 1 7 1 1 292 1 2 1 1 29 since2 x x x xx xx x xx + + + =  += − − =  − −= =  = + 

Chapter 3 Functions TMJC 2024 Page 7 of 8 Q8 IJC/Prelim2013/P1/Q13 [modified] (i) (ii) 3f ( 3) f (4.5) 3 4.5 2 − + =− + − =−  (iii) Greatest value of a 3 2=− (iv) (v) Since the curves intersect at the line y = x, solving 1h( ) h ( )xx −= is equivalent to solving h( )xx= . 2 3 (2 3) 2 xx =+− ( ) 23 4 12 9 2x x x= + + − 324 12 7 3 0x x x+ + − = Using GC, 1.781x=− or 1.5x=− (rej, since 3 2x ) or 0.2808x= (rej, since 3 2x ) Q9 2016/H2 Specimen Paper/II/1 (i) ( )Let 3cos 2sin cos cos cos sin sin x x R x R x R x   − = + =− 2 2 2 1 cos 3 ---(1), sin 2 ---(2) (1) (2) : 13 13 (2) 2: tan(1) 3 2tan 3 RR R R    − == += = = = y x 1 2 ൬5 2 , − 3 2൰ ൬− 5 2 , −3൰ −1 −2 − 3 2 −3 y x ൬− 3 2 , − 3 2൰ (−2, −2) (−3, −2) (−2, −3) 𝑦 = h(𝑥) 𝑦 = h−1(𝑥) 𝑦 = h−1h(𝑥)

Chapter 3 Functions TMJC 2024 Page 8 of 8 1 2 3cos 2sin 13 cos tan 3x x x − − = +  fR 13, 13 = − ( )cos 0 or 22 or 22 x xx xx     += + = + =− = − =− − (ii) 1 1 11 11 1 1 1 2Largest value of tan 3 213 cos tan 3 2tan cos3 13 2cos tan 313 2f ( ) cos tan 313 b yx yx yx xx  − − −− −− − − − =− =+  += =−  = − 1 1 1 2f : cos tan , 13 13 313 xxx− − − − −   x O y ( )0,3 1 2tan ,023  −−− 1 2tan ,023  − −