$f: R \rightarrow R$ is a function defined by $f(x) = \frac{1}{e^x + 2e^{-x}}$. Assertion $(A): f(c) = \frac{1}{3}$ for some values of $c \in R$. Reason $(R): 0 < f(x) \leq \frac{1}{2\sqrt{2}}$ for all $x \in R$. Then which of the following options is correct?

$f(x) = \frac{x}{\ln x}$ and $g(x) = \frac{\ln x}{x}$. Identify the $CORRECT$ statement.

Let $R$ denote the set of all real numbers and $R^{+}$ denote the set of all positive real numbers. For the subsets $A$ and $B$ of $R$, define $f: A \rightarrow B$ by $f(x) = x^2$ for $x \in A$. Match the following lists:
| Column $I$ | Column $II$ |
| :--- | :--- |
| $A$. $f$ is one-one and onto, if | $1$. $A = R^{+}, B = R$ |
| $B$. $f$ is one-one but not onto, if | $2$. $A = B = R$ |
| $C$. $f$ is onto but not one-one, if | $3$. $A = R, B = R^{+}$ |
| $D$. $f$ is neither one-one nor onto, if | $4$. $A = B = R^{+}$ |

For equality of functions $f$ and $g$,which of the following conditions must be satisfied?
$(i)$ $\text{domain of } f = \text{domain of } g$
(ii) $f(x) = g(x)$ for all $x$ in the domain
(iii) $x \in \text{domain of } f$

Consider a function $f : N \rightarrow R$,satisfying $f(1)+2 f(2)+3 f(3)+\ldots+x f(x)=x(x+1) f(x)$ for $x \geq 2$,with $f(1)=1$. Then $\frac{1}{f(2022)}+\frac{1}{f(2028)}$ is equal to

The number of non-constant functions $f: X \to Y$ where $X = \{0, 1, 2\}$ and $Y = \{1, 2, 3, 4, 5, 6, 7, 8\}$ such that $f(i) \leq f(j)$ whenever $i < j$ is: