If $f(x) = \frac{4x + 3}{6x - 4}$, $x \neq \frac{2}{3}$ and $(f \circ f)(x) = g(x)$ where $g : R - \{\frac{2}{3}\} \to R - \{\frac{2}{3}\}$, then $(g \circ g \circ g \circ g \circ g)(3) = $

Let $R$ be the set of real numbers and the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x) = x^{2} + 2x - 3$ and $g(x) = x + 1$. Then, the value of $x$ for which $f(g(x)) = g(f(x))$ is

Two functions $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined as follows: $f(x) = \begin{cases} 0, & x \text{ is rational} \\ 1, & x \text{ is irrational} \end{cases}$ and $g(x) = \begin{cases} -1, & x \text{ is rational} \\ 0, & x \text{ is irrational} \end{cases}$. Then, $(f \circ g)(\pi) + (g \circ f)(e)$ is equal to:

Let $f^1(x) = \frac{3x + 2}{2x + 3}$,$x \in R - \left\{-\frac{3}{2}\right\}$. For $n \geq 2$,define $f^n(x) = f^1 \circ f^{n-1}(x)$. If $f^5(x) = \frac{ax + b}{bx + a}$ and $\gcd(a, b) = 1$,then $a + b$ is equal to $............$.

If $f(x) = \sqrt{x} - 1$ and $g\{f(x)\} = x + 2\sqrt{x} + 1$,then $g(x) = $

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=x^3-x$ and $g(x)=\sin 2x$,then the values of $x \in (0, 2\pi)$ that satisfy $f(g(x)) > 0$ lie in the interval