Question

Let `f:R - {-4/3} -> R` be a function defined as `f(x) = (4x)/(3x + 4)`. The inverse of f is map g Range `f -> R -{- 4/3}`

(A) `g(y) = (3y)/(3-4y)`

(B) `g(y) = (4y)/(4 - 3y)`

(C) `g(y) = (4y)/(3 - 4y)` 

(D) `g(y) = (3y)/(4 - 3y)`

Answer 1

t is given that `f:R -{-4/3} -> R` is defined as `f(x) = (4x)/(3x - 4)`

Let y be an arbitrary element of Range f.

Then, there exists x ∈ `R - {-4/3}` such that `y = f(x)`

`=> y = (4x)/(3x +  4)`

=>3xy + 4y = 4x  Let us define g: Range `f -> R -{-4/3} " as " g(y) = (4y)/(4 -3y)`

=> x(4 - 3y) = 4y

Now `(gof)(x) = g(f(x)) = g((4x)/(3x + 4))`

`=> x = (4y)/(4 -3y)`

`= (4(4x/(3x + 4)))/(4 - 3("4x"/(3x + 4)))  = "16x"/(12x + 16 - 12x)  = "16x"/16 = x`

And `(fog)(y) = f(g(y)) = f((4y)/(4 - 3y))`

`= (4("4y"/(4-3y)))/(3("4y"/(4 - 3y)) + 4) = (16y)/(12y + 16 -  12y) = (16y)/16 = y`

:. `gof = I_(R - (-4/3))` and fog = I_"Range f"

Thus, g is the inverse of f i.e., f⁻¹ = g.

Hence, the inverse of f is the map `g: Range f -> R - {-4/3}` which is given by

`g(y) = "4y"/(4 - 3y)`

The correct answer is B.