Continuity and Differentiability
Differentiability of Piecewise Functions
IIT-JEE 2012 Paper 1
Grade 12
Question:
Let $f(x) = \begin{cases} x^2 \left|\cos \frac{\pi}{x}\right|, & x \neq 0 \\ 0, & x = 0 \end{cases}$, $x \in \mathbb{R}$. Then $f$ is:
(1) differentiable at $x = 0$ and $f'(0) = 0$
(2) not differentiable at $x = 0$
(3) differentiable at $x = 0$ and $f'(0) = 1$
(4) differentiable everywhere with $f' \equiv 0$
Step-by-Step Solution
Key Concept: $f'(0) = \lim_{x \to 0} \frac{x^2 |\cos(\pi/x)|}{x} = \lim_{x \to 0} x|\cos(\pi/x)|$. Since $|\cos(\pi/x)| \leq 1$ and $x \to 0$, the limit is 0. So $f$ is differentiable at $x = 0$ with $f'(0) = 0$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)