Let $A = \{(x, y) : y = e^x, x \in R\}$ and $B = \{(x, y) : y = e^{-x}, x \in R\}$. Then:

Let $S = \mathbb{N} \cup \{0\}$. Define a relation $R$ from $S$ to $\mathbb{R}$ by: $R = \{(x, y) : \log_e y = x \log_e \left(\frac{2}{5}\right), x \in S, y \in \mathbb{R}\}$. Then,the sum of all the elements in the range of $R$ is equal to

If $R \subset A \times B$ and $S \subset B \times C$,then the relation $(SoR)^{-1} = $

The function $f(x) = \sec \left[ \log \left( x + \sqrt{1 + x^2} \right) \right]$ is . . . . . . function.

Let $N$ be the set of positive integers. For all $n \in N$,let $f_n = (n+1)^{1/3} - n^{1/3}$ and $A = \{n \in N : f_{n+1} < \frac{1}{3(n+1)^{2/3}} < f_n\}$. Then,

Let $S = \{1, 2, 3, 4\}$. Then the number of elements in the set $\{f: S \times S \rightarrow S : f \text{ is onto and } f(a, b) = f(b, a) \geq a; \forall (a, b) \in S \times S\}$ is