Letbe a function defined as. The inverse of f is map g:


Let $f: \mathbf{R}-\left\{-\frac{4}{3}\right\} \rightarrow \mathbf{R}$ be a function defined as $f(x)=\frac{4 x}{3 x+4}$. The inverse of $f$ is map $g$ : Range $f \rightarrow \mathbf{R}-\left\{-\frac{4}{3}\right\}$ given by

(A) $g(y)=\frac{3 y}{3-4 y}$

(B) $g(y)=\frac{4 y}{4-3 y}$

(C) $g(y)=\frac{4 y}{3-4 v}$

(D) $g(y)=\frac{3 y}{4-3 y}$


Let $y$ be an arbitrary element of Range $f$.

Then, there exists $x \in \mathbf{R}-\left\{-\frac{4}{3}\right\}$ such that $y=f(x)$.

$\Rightarrow y=\frac{4 x}{3 x+4}$

$\Rightarrow 3 x y+4 y=4 x$    Let us define g: Rangeas

$\Rightarrow x(4-3 y)=4 y$

$\Rightarrow x=\frac{4 y}{4-3 y}$    Now,

$=\frac{4\left(\frac{4 x}{3 x+4}\right)}{4-3\left(\frac{4 x}{3 x+4}\right)}=\frac{16 x}{12 x+16-12 x}=\frac{16 x}{16}=x$

And, $(f \circ g)(y)=f(g(y))=f\left(\frac{4 y}{4-3 y}\right)$

$=\frac{4\left(\frac{4 y}{4-3 y}\right)}{3\left(\frac{4 y}{4-3 y}\right)+4}=\frac{16 y}{12 y+16-12 y}=\frac{16 y}{16}=y$

Thus, $g$ is the inverse of $f$ i.e., $f^{-1}=g$.

Hence, the inverse of $f$ is the map $g$ : Range $f \rightarrow \mathbf{R}-\left\{-\frac{4}{3}\right\}$, which is given by

$g(y)=\frac{4 y}{4-3 y}$

The correct answer is B.

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