RI Functions C4 Tut Sect A (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ___________________________________________________________________________ _________________ Tutorial 4: Functions Page 1 of 3 Tutorial 4: Functions Section A (Basic Questions) 1 Find the range of the following functions: (a) 2f: 2 3 , , 1 5xx x x x     , (b) g: ( 5 ) ( 8 ) , , 3 5xx x x x x      , (c) 2h: t a n , , , 63 2xx x xx     [(a) [4,12] (b) [151,66.5] (c) 1,3 , 3        ] (a) 2f: 2 3 , , 1 5xx x x x     2 23 (3 ) (1 )yx x x x   Min turning point is at (1,-4) When 5x , f (5) 12 fR [ 4,12] 1(b) g: ( 5 ) ( 8 ) , , 3 5xx x x x x      Find the maximum and minimum point using GC and Rg = [151,66.5] ( 2.7859,66.530) (4.7859, 150.530) O x y (-1,0) 3

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _________________ Tutorial 4: Functions Page 2 of 3 (c) 2h: t a n , , , 63 2xx x xx     h 1R, 3 , 3       2 For each of the following functions, sket ch the graph and determine if the inverse function exists. If it does, find it in a similar form. (a) 2f( ) 4 3 , , 1x xx x x    , (b) 2 3g( ) ,xx x  , (c)  1h( ) ln 2 1 , , 2xx x x    . (a) 2f( ) 4 3 ( 1 ) ( 3 ) , , 1x xx xx x x      Since every horizontal line  ,0 ,yk k  cuts the graph of f at one and only one point, f is a one-one function and hence the inverse function of f exists. Let  22 43 2 1yx x x   Then  2 21x y  21x y  Since 1x , 21x y  1f( ) 2 1 , 0xx x    . 1(, )6 3  2(,3 )3   2x  x y O

Raffles Institution H2 Mathematics 2025 Year 5 __________________________________________________________________________________________ _________________ Tutorial 4: Functions Page 3 of 3 (b) 2 3g( ) ,xx x  Note : g is an even function. Gradient at 0x is undefined. Since g( 8) g(8) 4  , g is not a one-one function. Therefore, inverse function of g does not exist. OR: Since the horizontal line 1y cuts the graph of g at more than one point, g is not a one-one function and therefore the inverse function of g does not exist. (c)  1h( ) ln 2 1 , , 2xx x x    3 The function f is defined by 2f : 2 ( 1 ) , , x xx x k   . Determine the largest value of k for which the function 1f exists. [largest value of k = - 1] In order for 1f to exist, f needs to be a 1-1 function. The function f has a turning point at (1 , 2 )- and therefore 1.k  Therefore the largest value of is 1k  . Let  ln 2 1 .yx Then e2 1y x 1 e12 yx  1 1h ( ) e 1 , 2 xxx    Since every horizontal line y = k, k  cuts the graph of h at one and only one point, h is a one-one function and hence the inverse function of h exists. O x y O x y y = k