Let $f: R \rightarrow R$ be defined by $f(x)=2x+3$. If $\alpha$ and $\beta$ are the roots of the equation $f(x^2)-2f(\frac{x}{2})-1=0$,then $\alpha^2+\beta^2=$

Match the items of List-$I$ with those of the items of List-$II$:

List-$I$	List-$II$
$A$. Range of $\sec ^{-1}\left[1+\cos ^2 x\right]$, where $[.]$ denotes the greatest integer function	$I$. Odd function
$B$. Domain of $f(x)$ where $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$	$II$. $\left\{0, \frac{1}{2}\right\}$
$C$. $f(x+y)=f(x)+f(y) ; f(1)=5$	$III$. $\left\{\sec ^{-1} 5, \sec ^{-1} 4\right\}$
$D$. $\sin ^{-1} x-\cos ^{-1} x+\sin ^{-1}(1-x)=0 \Rightarrow x \in$	$IV$. $R$
	$V$. $\left\{\sec ^{-1} 1, \sec ^{-1} 2\right\}$

Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases}$,where $[x]$ denotes the greatest integer less than or equal to $x$. Then for all $x$,$f(g(x)) = $

If $f(x) = \frac{2^{2x}}{2^{2x} + 2}$,$x \in R$,then $f\left(\frac{1}{2023}\right) + f\left(\frac{2}{2023}\right) + \dots + f\left(\frac{2022}{2023}\right)$ is equal to

If $f : R \rightarrow R$ is defined by $f(x) = \begin{cases} x + 4, & x < -4 \\ 3x + 2, & -4 \leq x < 4 \\ x - 4, & x \geq 4 \end{cases}$ then the correct matching of List-$I$ from List-$II$ is :
List-$I$
$(A) f(-5) + f(-4)$
$(B) f(|f(-8)|)$
$(C) f(f(-7) + f(3))$
$(D) f(f(f(f(0)))) + 1$
List-$II$
$(i) 14$
$(ii) 4$
$(iii) -11$
$(iv) -1$
$(v) 1$
$(vi) 0$

Let $f: R \rightarrow (0,1)$ be a continuous function. Then,which of the following function$(s)$ has(have) the value zero at some point in the interval $(0,1)$?