4.Functions Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Functions T1 1 x y Solutions Functions 1 2 2 2xxye              2 2 2 2 ln 2 1 1 ln 2 1 1 ln 2 1 1 ln 2 x x y xy xy xy                Since x < 0,  1 1 ln 2xy    Df–1 = fR =  1,   1f : 1 1 ln 2 , 1x x x       1 1,gg fR D R            -1gf2, 1,1 1 1 ln 3, 0        2 (i) Since any horizontal line y = k, k  R will cut the graph of f at most once, hence f is one-one. Thus, f -1 exists. (ii) Let xxy 1 42 1 2 2  yyx But x > 0, 42 1 2 2  yyx When g( ) 1x  23(2) 1 21g    gD 2, Least 2k 

River Valley High School, Mathematics Department 2023 Functions T2 Hence, f -1 : x  42 1 2 2  xx , x  (- , ) (iii) [B1] – correct graph of f -1 with y = x [B1] - intercepts (iv) Df : (0,) and Rg : [-1 , 1]. Since Rg  Df , therefore fg does not exist. (v) Largest Dg : (0 ,) fg(x) = f(sin x) = xx sin 1sin  Hence, fg : x  xx sin 1sin  , x  (0 ,). 3a  1f ( ) 2 tan 2x x x  ' 2 4f ( ) 1 14 x x   2 2 2 2 30 4 1 1 4 4 11 14 14 414 14 x x x x         Since 2 414 14 x   , 2 410 14 x     , 'f ( ) 0x  So f ( )x decreases when x increases. 1 x y 1

River Valley High School, Mathematics Department 2023 Functions T3 f 3 2 3 2,2 3 2 3R     Or f ( 1.23,1.23)R  For ff to exist, ffRD 3 2 3 2 ,2 3 2 3 fR     or f ( 1.23,1.23)R  33,22fD  or g ( 0.866,0.866)R  ffRD ff does not exist. Let  12 tan 2y x x  . Then  12 tan 2x y y  b) 2h( ) 2xx     24 2 2gh 2 5 1 4x x x x          22g 2 2 3 4xx          2 g 3 4xx   Alternatively, 2h( ) 2xx         42 22 2 2 222 gh 2 5 2 4 4 2 5 2 6 2 13 x x x x x x xx                 2g 6 13x x x    4 (i)    D , and R 2,fg     Since    R 2, D ,gf       function fg exists.         2 41 3, fg x f g x x fg x x x         (ii) Since    R 3, D 0,fg g      function gfg exists        3 4 7, 0h x gfg x g fg x x x x       

River Valley High School, Mathematics Department 2023 Functions T4 (iii)    1h x g g x , x > 0    1g fg x g g x           2 2 7, 7 70 1 1 4 1 7 1 1 292 1 2 1 1 29 sin2 x x x xx xx x x ce x                   5 (i) g( ) f ( 1)xx The transformation is a translation of +1 unit in the direction of the x-axis Alternatively, 1 11g( ) 2 (2 ) f ( )22 xxxx    The transformation is a scaling parallel to the y-axis by factor 1 2 (ii) A horizontal line y = k (k > 9 4 ) cuts the graph at 2 points, h is not one- one , so 1h does not exist (iii) 1 2a 2h : 2x x x  , 1, 2xx  Let 2 1h( ) 2, 2y x x x x     2 19 24yx    2 1 9 4 9 2 4 4 yxy      491 22 yx   491 22 yx   (since 1 2x )

River Valley High School, Mathematics Department 2023 Functions T5 1 1 4 9 9h : , , 24 xx x x      (iv) f (0, )R  1h 9( , )4D     1f hRD  . Hence 1hf exists. 11h f( ) h (2 ) xx  = 1 4(2 ) 9 2 x   1 1 4(2 ) 9h f : , 2 x xx     1hfR (1, )  6 2Let ( 1) 1 11 yx xy        Since x ≤ 1, 1 g ( ) 1 1 , 1x x x     2( 1) 2fg( ) 3 2e xx  ; x , x ≤ 1 Rfg = 2( , 3 2e ]  1 1xy    g f gf R [ 1, ) D R D , fg exists    

River Valley High School, Mathematics Department 2023 Functions T6 7(a) fg(x) = gf(x)  f( )2ln( x ) = g( )2ln( x )  ln [ )2ln( x + 2 ] = ln [ )2ln( x + 2 ]  )2ln( x = )2ln( x y = )2ln( x y = )2ln( x Consider )2ln()2ln(  xx  2 1)2(  xx  1)2)(2(  xx  14 2  x  32 x  3x (reject 3x ) Hence, for fg(x) = gf(x), 0x or 3x (b)(i) g(x) = )2ln( x . Note that both 1x and 1x are in the domain of g. Clearly 1  1, But g(1) = g(1) = ln3. Therefore, function g is not 1-1 and inverse of function of g cannot be formed. (ii) largest a = 0. (iii) Let y = g(x) = )2ln( x . Since 0x , y = )2ln( x  yex  2  yex  2  yey  2)(g 1 Hence xex  2:g 1 , ),2[ln x

River Valley High School, Mathematics Department 2023 Functions T7 8(i)   fg fg , ( ,1 Since , largest =1. RD RD        (ii) (iii)       1 2 2 ff 12 5 3 0 5 13 2 5 13 5 13 is rejected as 222 xx xx xx x x               9(i) (ii) The horizontal line y = 1 cuts the graph of f at two points.  f is not one-one. Thus f does not have an inverse. Alt Since  32f f 1 5 17    , f is not one-one. Thus f does not have an inverse. (iii) Least value of k = 1/5 -1ff ( )yx (−2, 1) x O -1f ( )yx −3 −3 y = f(x) (1, −2) (1, 1) −1 −1 y  f 0,2R 

River Valley High School, Mathematics Department 2023 Functions T8 Let     2 21f , . 51 5 1 y x x x       2 21 5 1 x y   25 1 1x y   Since 1 , 5 1 0.5xx    12115x y     1 12f 1 1 5x x     (iv)    1ff xx    f xx    2 2 1 5 1 x x      21 25 10 1 2x x x     3225 10 2 2 0x x x    10 Since any horizontal line y = k, 05 k cuts the graph of f ( )yx once and only once, f is a one-one function. Hence 1f  exists. Let 2 55 , 0y ax x a    22 5y ax 2 2 5 yx a  5 a x O y

River Valley High School, Mathematics Department 2023 Functions T9 (i) (ii) 255 , since 0yxx aa    2 1 5f : , 0 5 xxx a     2 5f ( ) = for all , 0x x x x a   Given: 2f ( )xx 1f ( ) f ( )xx  2 2 55 xax aa    515, aaa   1a (shown) Since gfR (1,2] D [0, 5]   ,  fg exists. Method 1 2fg( ) 5 (1 ) , 0 xx e x     fgR [1, 2) Method 2 [0, ) (1,2] [1,2)   fgR [1, 2) 11 (i)   2f 4 4x x x          22 2 2 2 2 4 2 4 4 x x           Since horizontal line y = 0 cuts the graph  fyx twice, f is not a one-one function.  f does not have an inverse. g f x O y y =1 2

River Valley High School, Mathematics Department 2023 Functions T10 (ii) Largest value of k = 2 Let   2 22 4 4yx     22 4 4xy     Since 2x  , 22 4 4xy     1 2 2f : 2 4 4 , 4 4x x x          (iii)  1f ,2R      g ,1D   Given 1 2  ,    1 gf , 2 ,1RD        1gf  exists   1gf ln 1 2 ,R     12 (i)   3f2 2x x   (ii) 12let 2 xy x     1 2 1 2 2 1 2 12 2 12f : , 2 2 xy y x x y y yx y xxx x            iii)  ,gR     ,2fD   Since    , ,2gfRD      , fgdoes not exist  2,fR     3,gD    Since    2, 3,fgRD       , gf exists