Q. If $f(x) = x(\sqrt{x} + \sqrt{x + 1})$, then
A) $f(x)$ is continuous but not differentiable at $x = 0$
B) $f(x)$ is differentiable at $x = 0$
C) $f(x)$ is not differentiable at $x = 0$
D) None of the above
Explanation
(c)
Given,$f(x) = x\left( \sqrt{x} + \sqrt{x + 1} \right)$. At $x = 0$ LHL of $\sqrt{x}$ is not defined, therefore it is not continuous at $x = 0$
Hence, it is not differentiable at $x = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.