IIT-JEE 2000 ScreeningQuestion 1
Let $f(x)=\begin{cases}|x|, & \text{for } 0<|x|\le2\\ 1, & \text{for } x=0\end{cases}$ then at $x=0$, $f$ has
Differentiation and its Application
Solution
Concept. Compare $f(0)$ with nearby values directly ($f$ is not continuous at $0$). Formulas used. (1) Local maximum at $c$: $f(c)\ge f(x)$ for all $x$ near $c$Step 1 - Nearby values. For $0<|x|<1$: $f(x)=|x|<1$. Step 2 - Conclude. $f(0)=1$ is larger, so by (1) $f$ has a local maximum at $0$. Answer: (A)