If $f: R-\left\{\frac{3}{7}\right\} \rightarrow R-\left\{\frac{3}{7}\right\}$ is given by $f(x)=\frac{3…
If $f: R-\left\{\frac{3}{7}\right\} \rightarrow R-\left\{\frac{3}{7}\right\}$ is given by $f(x)=\frac{3 x+5}{7 x-3}$, then the statement which is not true, is
$f^{-1}(x)=f(x)$
$(f \circ f)(x)=x$
$(f \circ f \circ f)(x)=x$
$(f \circ f \circ f \circ f)(x)=x$
Solution
Given, function $f: R-\left\{\frac{3}{7}\right\} \rightarrow R-\left\{\frac{3}{7}\right\}$ is define by $f(x)=\frac{3 x+5}{7 x-3}$.
Let $\quad f(x)=y \Rightarrow \frac{3 x+5}{7 x-3}=y$
$\Rightarrow \quad x=\frac{3 y+5}{7 y-3}$, so $f(x)$ is a bijective function and $f^{-1}(x)=f(x)$.
$(f \circ f)(x)=x$ and $(f \circ f \circ f \circ f)(x)=x$
But $(f \circ f \circ f)(x) \neq x$.
Hence, option (c) is correct.