Practicing Success
Let f(x) = Degree of $\left(u^{x^2}+u^2+2 u+3\right)$. Then, at $x=\sqrt{2}, f(x)$ is |
continuous but not differentiable differentiable discontinuous none of these |
continuous but not differentiable |
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 $x=\sqrt{2}$. |