If f: R - {frac{3}{7}} rightarrow R - {frac{3 | vedclass

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(C) Given the function $f(x) = \frac{3x+5}{7x-3}$.To find $f^{-1}(x)$,let $y = \frac{3x+5}{7x-3}$.$y(7x-3) = 3x+5 \Rightarrow 7xy - 3y = 3x+5$.$x(7y-3

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$(f \circ f \circ f)(x) = x$

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(C) Given the function $f(x) = \frac{3x+5}{7x-3}$.To find $f^{-1}(x)$,let $y = \frac{3x+5}{7x-3}$.$y(7x-3) = 3x+5 \Rightarrow 7xy - 3y = 3x+5$.$x(7y-3) = 3y+5 \Rightarrow x = \frac{3y+5}{7y-3}$.Thus,$f^{-1}(x) = \frac{3x+5}{7x-3} = f(x)$. So,option $A$ is true.Now,$(f \circ f)(x) = f(f(x)) = \frac{3(\frac{3x+5}{7x-3})+5}{7(\frac{3x+5}{7x-3})-3} = \frac{9x+15+35x-15}{21x+35-21x+9} = \frac{44x}{44} = x$. So,option $B$ is true.Since $(f \circ f)(x) = x$,then $(f \circ f \circ f)(x) = f((f \circ f)(x)) = f(x) 
eq x$.Also,$(f \circ f \circ f \circ f)(x) = (f \circ f)(f \circ f)(x) = x$. So,option $D$ is true.Therefore,the statement that is not true is $(f \circ f \circ f)(x) = x$.

If $f: R - \{\frac{3}{7}\} \rightarrow R - \{\frac{3}{7}\}$ is given by $f(x) = \frac{3x+5}{7x-3}$,then the statement which is not true is:

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