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:

Show that if $f: A \rightarrow B$ and $g: B \rightarrow C$ are one-one,then $g \circ f: A \rightarrow C$ is also one-one.

If $f(x) = \frac{2x+3}{3x-2}$,$x \neq \frac{2}{3}$,then $(f \circ f)(x)$ is:

Let $R = \{(1, 3), (2, 2), (3, 2)\}$ and $S = \{(2, 1), (3, 2), (2, 3)\}$ be two relations on the set $A = \{1, 2, 3\}$. Then $R \circ S = $

Let the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x) = \begin{cases} x+2, & x < 0 \\ x^2, & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x^3, & x < 1 \\ 3x-2, & x \geq 1 \end{cases}$. Then,the number of points in $R$ where $(f \circ g)(x)$ is $NOT$ differentiable is equal to

If $f(x) = \frac{3x - 4}{2x - 3}$, then $f(f(f(x)))$ will be