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 $f: R \rightarrow R$ be defined as $f(x)=x-1$ and $g: R -\{1,-1\} \rightarrow R$ be defined as $g(x)=\frac{x^{2}}{x^{2}-1}$. Then the function $f \circ g$ is

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 $f(x) = ax + b$ and $g(x) = cx + d$,then $f(g(x)) = g(f(x))$ is equivalent to

Let $f(x)=3+2x$ and $g_n(x)=(f \circ f \circ f \circ \dots \text{n times})(x)$. For all $n \in N$,if all the lines $y=g_n(x)$ pass through a fixed point $(\alpha, \beta)$,then $\alpha+\beta=$

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = x^{2} - 3x + 4$ and $g(x) = 2x + 1$,then the value of $x$ for which $f(x) = (f \circ g)(x)$ is