Let `f: R-{-4/3}->R`be a function as `f(x)=(4x)/(3x+4)`. The inverse of f is map, `g: R a ngef->R-{-4/3}`given by.(a) `g(y)=(3y)/(3-4y)` (b) `g(y)=(4y)/(4-3y)`(c) `g(y)=(4y)/(3-4y)` (d) `g(y)=(3y)/(4-3y)`A. `g(y) = ( 3y)/( 3-4y)`B. `g(y)= ( 4y )/( 4-3y)`C. `g(y) = ( 4y)/( 4-3y)`D. `g(y) = ( 4y)/( 3-4y) `

Let `f: R-{-4/3}->R`be a function as `f(x)=(4x)/(3x+4)`. The inverse o

1.	Let `f: R-{-4/3}->R`be a function as `f(x)=(4x)/(3x+4)`. The inverse of f is map, `g: R a ngef->R-{-4/3}`given by.(a) `g(y)=(3y)/(3-4y)` (b) `g(y)=(4y)/(4-3y)`(c) `g(y)=(4y)/(3-4y)` (d) `g(y)=(3y)/(4-3y)`A. `g(y) = ( 3y)/( 3-4y)`B. `g(y)= ( 4y )/( 4-3y)`C. `g(y) = ( 4y)/( 4-3y)`D. `g(y) = ( 4y)/( 3-4y) `
Answer» Correct Answer - b In `f: R - {- (4)/(3) } to R`, ` f(x) = ( 4x)/( 3x + 4) AA x in R - {-(4)/(3)}` Let for `y in R, x in R - {-(4)/(3)}` is such that `" " f(x)=y ` `rArr " " (4x)/( 3x + 4) = y rArr 4x = 3xy + 4y` `rArr x ( 4-3y) = 4y rArr x = (4y)/( 3-4y) ` Let, in `g: f ` range of `f to R - {- (4)/(3)}, g(y) = ( 4y )/( 3-4y )` Now `(gof) (x) = g{f(x)} = g((4x)/( 3x + 4))` `" " = (4((4x)/( 3x+ 4)))/( 4-3((4x )/( 3x + 4))) = ( 16x )/(12 x + 16 - 12x` `" " = ( 16x)/( 16) = x` and `(fog) (y) = f[g(y) ] = f[ ( 4y)/( 4-3y)]` `" " = ( 4(( 4y)/(3x + 4)))/( 3(( 4y )/( 4- 3y ))+4)` `" " = ( 16y )/( 12y + 16 - 12y ) = ( 16y)/(16) = y` `therefore " " gof = I_(R- {- (4)/(3)} ) and fog = I_R` ` therefore f^(-1) = g`.