The range of the function $f(x) = \left[ \frac{1}{\ln(x^2 + e)} \right] + \frac{1}{\sqrt{1 + x^2}}$ is,where $[*]$ denotes the greatest integer function and $e = \lim_{\alpha \to 0} (1 + \alpha)^{1/\alpha}$.

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

The function $f(x) = x^{\frac{1}{\ln x}}$ is:

Suppose $f:[-2, 2] \to R$ is defined by $f(x) = \begin{cases} -1 & \text{for } -2 \le x \le 0 \\ x - 1 & \text{for } 0 < x \le 2 \end{cases}$,then find the set $\{ x \in (-2, 2) : x \le 0 \text{ and } f(|x|) = x \}$.

Consider a function $f: R \to R$ such that $f(x + a) = \frac{1}{2} + \sqrt{f(x) - f^2(x)}$,where $a$ is a real constant. Then $f(x)$ must be

Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be functions defined by
$f(x)=\begin{cases} x|x| \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{cases}$ and $g(x)=\begin{cases} 1-2x, & 0 \leq x \leq \frac{1}{2} \\ 0, & \text{otherwise} \end{cases}$
Let $a, b, c, d \in R$. Define the function $h: R \rightarrow R$ by
$h(x)=a f(x)+b\left(g(x)+g\left(\frac{1}{2}-x\right)\right)+c(x-g(x))+d g(x), x \in R$
Match each entry in $List-I$ to the correct entry in $List-II$.

$List-I$	$List-II$
$(P)$ If $a=0, b=1, c=0$ and $d=0$,then	$(1)$ $h$ is one-one
$(Q)$ If $a=1, b=0, c=0$ and $d=0$,then	$(2)$ $h$ is onto
$(R)$ If $a=0, b=0, c=1$ and $d=0$,then	$(3)$ $h$ is differentiable on $R$
$(S)$ If $a=0, b=0, c=0$ and $d=1$,then	$(4)$ the range of $h$ is $[0,1]$
	$(5)$ the range of $h$ is $\{0,1\}$

The correct option is