Let $f:R\to R$ be given by $f\left(x\right)$ $=\left\{\begin{array}{cc}{\mathrm{x}}^5+5{\mathrm{x}}^4+10{\mathrm{x}}^3+10{\mathrm{x}}^2+3\mathrm{x}+1,& \mathrm{x}<0;\\ {\mathrm{x}}^2-\mathrm{x}+1,& 0\le \mathrm{x}<1;\\ \frac{2}{3}{\mathrm{x}}^3-4{\mathrm{x}}^2+7\mathrm{x}-\frac{8}{3},& 1\le \mathrm{x}<3;\\ \left(\mathrm{x}-2\right){\mathrm{l}\mathrm{o}\mathrm{g}}_{\mathrm{e}}\left(\mathrm{x}-2\right)-\mathrm{x}+\frac{10}{3},& \mathrm{x}\ge 3\end{array}\right.$
Then which of the following options is/are correct?

As given $\mathrm{f}:\mathrm{R}\to \mathrm{R}$
$f\left(x\right)$$=\left\{\begin{array}{cc}{x}^5+5x^4+10x^3+10x^2+3x+1& x<0\\ x^2-x+1& 0\le x<1\\ \frac{2}{3}{x}^3-4x^2+7x-\frac{8}{3}& 1\le x<3\\ (x-2)\mathrm{l}\mathrm{n}(x-2)-x+\frac{10}{3}& x\ge 3\end{array}\right.$
$f\left(0^{}\right)=f\left(0^{+}\right)=f\left(0\right)=1$
$f\left(1^{}\right)=f\left(1^{+}\right)=f\left(1\right)=1$
$f\left(3^{}\right)=f\left(3^{+}\right)=f\left(3\right)=\frac{1}{3}$
Hence $f\left(x\right)$ is continuous everywhere.
Now, $\underset{x\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=-\infty$ and $\underset{x\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=+\infty$
Hence, range of $f\left(x\right)$ is $\left(-\infty ,\infty \right)$
Consider $f\text{'}\left(x\right)$ $=\left\{\begin{array}{cc}5x^4+20x^3+30x^2+20x+3=5{(x+1)}^4-2& x<0\\ 2x-1& 0<x<1\\ 2x^2-8x+7& 1<x<3\\ ln(x-2)& x>3\end{array}\right.$
Option $\left(B\right):$
For $x<0 f^{\prime }\left(x\right)=5{\left(x+1\right)}^4 -2,$ which takes both positive and negative values.
Hence, $f\left(x\right)$ is non-monotonic for $x<0$
Option $\left(A\right):$
$f^{″}\left(x\right)$ $=\left\{\begin{array}{cc}20{(x+1)}^3& x<0\\ 2& 0<x<1\\ 4x-3& 1<x<3\\ \frac{1}{x-2}& x>3\end{array}\right.$
Now, $f”\left(1^{\prime }\right)= 4<0$ and $f”\left(1^{-}\right)=2>0$
Given of $f\prime \left(x\right)$ in $x\in \left(\mathrm{0,3}\right)$

$x=1$ is maxima for $f^{\prime }\left(x\right)$ -----Option $\left(D\right)$
and $f^{\prime }\left(x\right)$ is not differential at $x=1$
Option $\left(\mathrm{C}\right):$
$\underset{x\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=\underset{x\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\left(x-2\right)\ln \left(x-2\right)-x+\frac{10}{3}$
$\underset{x\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=\underset{x\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\left(x-2\right)\left(\ln \left(x-2\right)-1\right)+\frac{4}{3}=\infty$
$\underset{x\to -\infty }{\mathrm{l}\mathrm{i}\mathrm{m}}f\left(x\right)=-\infty$
Hence, range of f  $\in \left( -\infty , \infty \right)$