Let f(x) = Degree of ≤ft(u^x^2+u^2+2 u+3). Then, at x=√2, f(x) is

We have,

f(x) = Degree of $\left(u^{x^2}+u^2+2 u+3\right)$

$\Rightarrow f(x)= \begin{cases}x^2, & x>\sqrt{2} \\ 2, & x \leq \sqrt{2}\end{cases} $

∴  $\lim\limits_{2 \rightarrow \sqrt{2}^{-}} f(x)=2=f(2)$

and,

$\lim\limits_{x \rightarrow \sqrt{2}^{+}} f(x)=\lim\limits_{x \rightarrow \sqrt{2}} x^2=(\sqrt{2})^2=2$

∴  $\lim\limits_{x \rightarrow \sqrt{2}^{-}} f(x)=\lim\limits_{x \rightarrow \sqrt{2}^{+}} f(x)=f(\sqrt{2})$

So, f(x) is continuous at $x=\sqrt{2}$

Now, (LHD at $x=\sqrt{2})=\left(\frac{d}{d x}(2)\right)_{x=\sqrt{2}}=0$

and, (RHD at $x=\sqrt{2})=\left(\frac{d}{d x}\left(x^2\right)\right)_{x=\sqrt{2}}=(2 x)_{x=\sqrt{2}}=2 \sqrt{2}$

Clearly, (LHD at $x=\sqrt{2}$) ≠ (RHD at $x=\sqrt{2}$)

So, f(x) is not differentiable at $x=\sqrt{2}$.