The graph of $y = f(x)$ is shown. The number of solutions of the equation $f(f(x)) = 2$ is:

If $f:[-6,6] \rightarrow R$ is defined by $f(x)=x^2-3$ for $x \in R$,then $(f \circ f \circ f)(-1)+(f \circ f \circ f)(0)+(f \circ f \circ f)(1)$ is equal to

If $g(x)=x^2+x-1$ and $(g \circ f)(x)=4 x^2-10 x+5$,then $f(2)$ is equal to

If $f(x) = x^2 - 1$ and $g(x) = 3x + 1$,then $(gof)(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:

Define $f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$. If $f \circ f(x)$ is discontinuous at $a$ and $b$ in $[0, 3]$ and $a < b$,then $2 a + 3 b = $