Let f: R → R be given by f(x) = \ cases x ^ 5 + 5 x ^ 4 + 10 x ^ 3...

As given $f: R \to R$
f\left(x\right)= \left\{\begin{cases}{x}^{5} + 5{x}^{4} + 10{x}^{3} + 10{x}^{2} + 3x + 1 & x < 0 \\ {x}^{2} - x + 1 & 0 \le x < 1 \\ \dfrac{2}{3}{x}^{3} - 4{x}^{2} + 7x - \dfrac{8}{3} & 1 \le x < 3 \\ ( x - 2 ) \ln ( x - 2 ) - x + \dfrac{10}{3} & x \ge 3\end{cases}\right)<br >f ( {0}^{} ) = f ( {0}^{+} ) = f ( 0 ) = 1$<br >$f$\left({1}^{}\right) = f\left({1}^{+}\right) = f\left(1\right) = 1$
$f\left({3}^{}\right) = f\left({3}^{+}\right) = f\left(3\right) = \dfrac{1}{3}$
Hence $f\left(x\right)$ is continuous everywhere.
Now, $\lim_{x \to \infty}f\left(x\right) = - \infty$ and $\lim_{x \to \infty}f\left(x\right) = + \infty$
Hence, range of $f ( x )$ is \left(- \infty, \infty\right)$<br >Consider$f ' \left(x\right) = \left\{$\begin{cases}5{x}^{4} + 20{x}^{3} + 30{x}^{2} + 20x + 3 = 5{( x + 1 )}^{4} - 2 & x 3\end{cases}$\right)
Option $( B ) :$
For $x < 0{f}^{′}\left(x\right) = 5{\left(x + 1\right)}^{4} - 2,$ which takes both positive and negative values.
Hence, $f ( x )$ is non-monotonic for $x < 0$
Option $( A ) :$
{f}^{″} \left(x\right) = \left\{\begin{cases}20{( x + 1 )}^{3} & x 3\end{cases}\right)<br >Now,f”\left({1}^{′}\right) = 4 < 0$and$f”\left({1}^{-}\right) = 2 > 0$<br >Given of f′ ( x ) in$x \in \left(0,3\right)$<br ><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/ztefblhrtscbegodiotqbwcrmzzsaxequgujmexuteasspllul.png"/><br >$x = 1$is maxima for${f}^{′}\left(x\right) -----Option \left(D\right)
and ${f}^{′}\left(x\right)$ is not differential at $x = 1$
Option $( C ) :$
$\lim_{x \to \infty}f\left(x\right) = \lim_{x \to \infty}\left(x - 2\right)\ln \left(x - 2\right) - x + \dfrac{10}{3}$
$\lim_{x \to \infty}f\left(x\right) = \lim_{x \to \infty}\left(x - 2\right)\left(\ln \left(x - 2\right) - 1\right) + \dfrac{4}{3} = \infty$
$\lim_{x \to - \infty}f\left(x\right) = - \infty$
Hence, range of f $\in \left(- \infty, \infty\right)$