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 = $

Let $f(x) = \frac{\alpha x}{x+1}$,$x \neq -1$. If $f(f(x)) = x$,then the value of $\alpha$ is . . . . . . .

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

For a suitably chosen real constant $a$,let the function $f: R-\{-a\} \rightarrow R$ be defined by $f(x)=\frac{a-x}{a+x}$. Further,suppose that for any real number $x \neq-a$ and $f(x) \neq-a$,$(f \circ f)(x)=x$. Then,$f\left(-\frac{1}{2}\right)$ is equal to

If $f(x)$ and $g(x)$ are two functions such that $g(x) = x - \frac{1}{x}$ and $(f \circ g)(x) = x^3 - \frac{1}{x^3}$,then $f'(1)$ is equal to:

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.