Chp 4 Tutorial 4 APQ solutions (RVHS)

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 FUNCTIONS - Additional Practice Questions’ Solutions 1. [EJC/2019/JC1 Promo/Q7] The function f is defined by f : , for , ,bx bx x xax b a  where a and b are non-zero constants. (i) Find   1f x and state the domain of 1f . [3] (ii) Hence show that   2f xx , and write down  f n x where n is an odd number. [2] The function g is defined by g : 2 e , for , 0.xx x x    If 2a and 1b for function f, (iii) Explain why the composite function gf does not exist. [1] (iv) Find an expression for  fg x and state the domain and exact range of fg . [3] EJC JC1 Promo 9758/2019/01/Q7 (Solutions) Suggested solution (i) f : , for , ,bx bx x xax b a  Let  bxy y ax b bxax b    axy bx by byx ay b      1f : , for , , bx bx x xax b a   (ii) Since 1f ( ) f ( )xx  ,     21f f fx x x        32f f f fx x x  , therefore for    ffn bxxx ax b  for n odd, (iii) If 2a and 1b for function f, 1f : , for , ,2 1 2 xx x x x  f 1R\ 2    D 0,g  As fRD g , gf does not exist. (iv)    f2f eg xx    2e 2 2 e 1 2e ,03 2e x x x x x         fg gD D [0, )   Range of fg 32R = , 53  

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 2. [MI/2019/J1 Promo/01/Q5] The function f is defined by 2 1f : for , 2 4x x x x   . (i) Find   1f x and write down the domain of 1f . [3] (ii)On the same diagram, sketch the graphs of  fyx ,   1fyx  and   1ffyx  , stating the equations of any asymptotes and showing the relationships between the graphs clearly. [4] M1 JC1 Promo 9758/2019/01/Q5 (Solutions) 5(i)       2 2 2 1 1 1 = 4 14 14 114 rejected 2 or 4 1f 4 Domain of f Range of f 0, y x x y x y x x x yy x x                 5(ii) 3. [TMJC/2019/JC2 Prelim/01/Q7] The function f is defined by x y O x y O

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022   for 1, where is any positive odd integer, f for 1, where is any positive even integer .2 n x n x n n x nx n x n n         (i) Show that  f 1.5 0.5 and find  f 2.5 . [2] (ii) Sketch the graph of  fyx for 15 x . [2] (iii) Does f have an inverse for 15 x ? Justify your answer. [2] (iv) The function g is defined by 21g : , , 1 1 xx x x x    . For 23 x , find an expression for  gf x and hence, or otherwise, find   1 2gf 3    . [4] TMJC JC2 Prelim 9758/2019/01/Q7 (Solutions) For 12 x ,  f1 xx  f 1.5 1 1.5 0.5 (shown)    For 23 x ,  f1 xx  f 2.5 2.5 1 1.5    For 12 x ,  f1 xx For 23 x ,  f1 xx For 34 x ,  f3 xx For 45 x ,  f2 xx Since f (1) 0 f (3) , f is not a one-to-one function. Hence, f does not have an inverse. Alternative Since 0y cuts the graph of f at 2 distinct points, f is not a one-to-one function. Hence, f does not have an inverse. For 23 x ,  f1 xx 1 2 3 4 5 1 2 3 1 1

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022       2 1 1gf 11 xx x   = 23x x  Let   1 2gf 3x      2gf 3x  2 3 2 3 x x   9 4x Alternative: For 23 x ,  f1 xx ,       2 1 1gf 11 xx x   = 23x x    23Let 23 23 23 3 2 xy x xy x x xy xy x y           1 3gf 2x x      1 2 3 9gf 234 2 3     

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 4(i)    fgR ,16 , 25 D      gf exists. Method 1 (Using graph of gf ( )yx ): From graph,  gfR 3, Method 2 (Using graph of g( )yx :       fg, 1 ,16 3,          gfR 3, (ii) Let   35y x x      2 2 35 15 2 16 ( 1) y x x xx x          2( 1) 16 1 16 xy xy       Since 1x , 1 16xy    1f ( ) 1 16xx     1 ffD R ,16    (iii) 1f ( ) f ( ) f ( ) xx xx   2 2 2 15 2 3 15 0 3 3 4(1)( 15) 3 69 2(1) 2 3 69 ( 1)2 x x x xx x xx               y x O (-1, 3) gfR

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 5(i) Greatest k =  . 5(ii)    22 22 2 1 2 2 Let . ( 0) f ( ) , 0, yx xy x y x x x x                 5(iii) 1 1 2 2 fg f g( ) 1 125 1 1 24 11 D , 526 x x x xx                 OR 1 1 2 2 fg f g( ) 1 g( ) f (1) 1 24 1 24 11 D , 526 x x x x xx                x y   0 2 22yx 

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 6(i) Least 6a . Let 2 8 12y x x      2 2 8 12 0 8 8 4 1 12 2 44 x x y yx xy           As 6x , 44xy   Hence -1f : 4 4xx  , 0x Alternatively, Let 2 8 12y x x       2 2 2 2 8 12 0 4 4 12 0 44 44 44 x x y xy xy xy xy                   As 6x , 44xy   . Hence -1f : 4 4xx  , 0x (ii) Reflect  fyx about the line yx to get   1fyx      -1ff xx  f xx    2 2 2 8 12 9 12 0 9 9 4 1 12 2 9 33 2 x x x xx x x       9 33 2x  as 6x

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 7(a)   2 2 2 2 2 g( ) 2 33 28 18 3 3 2 8 18 08 18 2 19 39 0.8 18 x x xx x x x xx xx xx            Since   22 8 18 4 2 0x x x      for all ,x the inequality reduces to    2 2 19 39 0 2 13 3 0 133. 2 xx xx x           7(b)(i) Given that the range of f is [ 50,50] , we have   2 50 2( ) 4 f( ) 50kk      4 10 2k   3k (since 0k  ) 7(b)(ii) Consider the inequality g( ) 0 3 3 0 1. x x x      So to solve gf( ) 0,x  we replace x with f( )x to obtain     2 2 f( ) 1 50 2 4 1 2 4 49 7 2 4 7 3 11 .22 x x x x x                Since gf f [ 3,3],DD   3 3.2 x   8(i) x O y

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022   2 2 f ( ) 2 11 x x x a xa        From the graph, range of f [ 1 , )a    1 ,22 ggRD      1 fgRD  Therefore, composite function -1fg exists Finding range of -1fg : 1 ,22gR    1 1 1 2 , [ 1, )2 2 4g g fgD R R a a               Therefore, range of -1 2 fg 1, 4R a a       (ii) when 3a , 2 2 23 ( 1) 4 4 1 4 1( ) y x x yx x y or y rej             1f : 4 1 for , 4x x x x     

H2 Mathematics (9758) JC 1 River Valley High School, Mathematics Department, 2022 9. The functions f, g and h are defined by , , , . 2h : e , 0 xxx   . (i) Determine, stating your reasons, whether each of the following functions exist: (a) g-1, (b) fg. If the function exists, give its domain, rule and range. (ii) Given that f( ) h(ln ) 0 and 0,1  , show that 32 10      .   1 1 g 2g : , 2, , 3 xx x R        f : 2x x x   RIx 2g : 3 2xx   RIx