Let f : $R - \left\{ { - \cfrac{4}{3}} \right\} \to R$ be a function defined as $f(x) = \cfrac{{4x}}{{3x + 4}}.$ The inverse of f is the map g : Range $f \to R - \left\{ { - \cfrac{4}{3}} \right\}$ given by
(A) $g(y) = \cfrac{{3y}}{{3 - 4y}}$
(B) $g(y) = \cfrac{{4y}}{{4 - 3y}}$
(C) $g(y) = \cfrac{{4y}}{{3 - 4y}}$
(D) $g(y) = \cfrac{{3y}}{{4 - 3y}}$
Option B is correct
Given $:f:R - \left\{ { - \cfrac{4}{3}} \right\} \to R$ .
We have, $f(x) = \cfrac{{4x}}{{3x + 4}}$
Now, range of f is $R - \left\{ { - \cfrac{4}{3}} \right\}.$ Let $y = f(x). \to y = \cfrac{{4x}}{{3x + 4}}$
$\Rightarrow$ 3xy + 4y $=$ 4x $\Rightarrow$ x(4$-$3y)$=$4y
$\Rightarrow$ $x = \cfrac{{4y}}{{4 - 3y}}$
Therefore,${f^{ - 1}}(y) = g(y) = \cfrac{{4y}}{{4 - 3y}}$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Relations and Functions. Curated by Sachin Sharma. Free for all students.