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 the sets $A$ and $B$ are defined as $A = \{(x, y) : y = e^x, x \in R\}$ and $B = \{(x, y) : y = x, x \in R\}$,then:

If $f(a) = \log \left| \frac{1-a}{1+a} \right|$ for $a \neq \{-1, 1\}$,then the set of values of all $a$,for which $f\left( \frac{2a}{1+a^2} \right) > 0$ is

If a function $g(x)$ is defined in $[-1, 1]$ and two vertices of an equilateral triangle are $(0, 0)$ and $(x, g(x))$ and its area is $\frac{\sqrt{3}}{4}$,then $g(x)$ equals :-

Let $f, g: R \rightarrow R$ be functions defined by $f(x) = \begin{cases} [x] & x < 0 \\ |1-x| & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} e^x - x & x < 0 \\ (x-1)^2 - 1 & x \geq 0 \end{cases}$ where $[x]$ denotes the greatest integer less than or equal to $x$. Then,the function $(f \circ g)(x)$ is discontinuous at exactly

If $f: R \rightarrow R$ is defined as $f(x)=\frac{2020^x}{2020^x+\sqrt{2020}}$,$\forall x \in R$,then $\sum_{r=1}^{4039} 2 f\left(\frac{r}{4040}\right)=$