Let $f(x) = 2^{10} \cdot x + 1$ and $g(x) = 3^{10} \cdot x - 1$. If $(f \circ g)(x) = x$,then $x$ is equal to

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 $f(x) = e^x$ and $g(x) = x^2$. Then,the number of solutions of $f(g(x)) = g(f(x))$ is equal to:

Let $N$ be the set of natural numbers and two functions $f$ and $g$ be defined as $f, g : N \to N$ such that $f(n) = \begin{cases} \frac{n+1}{2} & \text{if } n \text{ is odd} \\ \frac{n}{2} & \text{if } n \text{ is even} \end{cases}$ and $g(n) = n - (-1)^n$. Then $fog$ is

If $g(x) = x^2 + x - 2$ and $\frac{1}{2} (g \circ f)(x) = 2x^2 - 5x + 2$,then $f(x)$ is equal to:

Consider functions $f: A \rightarrow B$ and $g: B \rightarrow C$ $(A, B, C \subseteq \mathbb{R})$ such that $(g \circ f)^{-1}$ exists. Then: