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

Give examples of two functions $f: N \rightarrow N$ and $g: N \rightarrow N$ such that $g \circ f$ is onto but $f$ is not onto.

Let $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. Define $f: S \rightarrow S$ as $f(n) = \begin{cases} 2n, & \text{if } n = 1, 2, 3, 4, 5 \\ 2n - 11, & \text{if } n = 6, 7, 8, 9, 10 \end{cases}$. Let $g: S \rightarrow S$ be a function such that $f \circ g(n) = \begin{cases} n + 1, & \text{if } n \text{ is odd} \\ n - 1, & \text{if } n \text{ is even} \end{cases}$. Then $g(10) \cdot (g(1) + g(2) + g(3) + g(4) + g(5))$ is equal to

If $f(x) = \frac{3x - 4}{2x - 3}$,then $f(f(f(x)))$ will be

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=2x+3$ and $g(x)=x^2+7$,then the values of $x$ such that $g(f(x))=8$ are

If $f(x)=2^{100} x+1$ and $g(x)=3^{100} x+1$,then the set of real numbers $x$ such that $f(g(x))=x$ is