ضابطهٔ وارون تابع $y=\left\{ \begin{matrix}
{{x}^{۲}}+۱\,\,\,\,\,\,\,\,x\le ۰  \\
۱-x\,\,\,\,\,\,\,\,\,\,\,\,\,x>۰  \\
\end{matrix} \right.$ کدام است؟

$y=\left\{ \begin{matrix}
-\sqrt{x-۱}\,\,\,\,\,\,\,\,\,\,x\ge ۰  \\
۱-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\lt ۰  \\
\end{matrix} \right.$

$y=\left\{ \begin{matrix}
\sqrt{x-۱}\,\,\,\,\,\,\,\,x\ge ۰\text{ }~\text{ }  \\
x+۱\,\,\,\,\,\,\,\,x\lt ۰  \\
\end{matrix} \right.$

$y=\left\{ \begin{matrix}
\sqrt{x-۱}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\ge ۱  \\
x+۱\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\lt ۱  \\
\end{matrix} \right.$

$y=\left\{ \begin{matrix}
-\sqrt{x-۱}\,\,\,\,\,x\ge ۱  \\
۱-x\,\,\,\,\,\,\,\,\,\,\,x\lt ۱  \\
\end{matrix} \right.$