Animated Solution for Mathematics - Differentiation: Let $f(x)=\{\begin{array}{ll}|x|,& \text{for }0<|x|\le 2\\ 1,& \text{for }x=0\end{array}$ then at $x=0,f$ has

Imagine you are standing on a graph. You are walking along a path, and suddenly, you encounter a strange, broken landscape. This is exactly what we face with the function $f(x)=|x|$ for $0<|x|\le 2$, with a special, isolated point at $f(0)=1$.

Many students, when they see this, immediately jump to conclusions. They see the V-shape and think, "Oh, that's a minimum at the origin!" But in the world of JEE Advanced, intuition without rigor is a dangerous path. Let's dissect this piece by piece.

First, let's draw the function. For all $x$ except zero, $f(x)=|x|$. This is our classic V-shape. It starts at $(-2,2)$, descends toward the origin, and then ascends to $(2,2)$.

But wait—look at the condition $0<|x|\le 2$. The origin $(0,0)$ is strictly excluded. There is a "hole" in our V-shape right at the origin.

Now, look at the second part of the definition: $f(0)=1$. We have taken that point, which was missing from the V-shape, and lifted it up to the coordinate $(0,1)$. Our graph now consists of a V-shaped valley with a single, isolated point hovering right above the hole at the origin.

Now, we ask: does this function have a local maximum at $x=0$? To answer this, we must abandon our reliance on derivatives and continuity. We must return to the fundamental definition of a local maximum.

A function $f(x)$ has a local maximum at $x=c$ if there exists a small neighborhood $(c-\delta ,c+\delta )$ such that for all $x$ in this neighborhood:

Notice what is not in that definition. It does not say the function must be continuous. It does not say the derivative must be zero. It only asks for a comparison of values.

Let's choose a tiny neighborhood around $x=0$. Let's pick a $\delta$ such that $0<\delta <1$. In this interval $(-\delta ,\delta )$, what is happening?

For any $x$ that is not zero, the function is defined by $f(x)=|x|$. Since we chose $\delta <1$, the maximum value of $|x|$ in this interval is strictly less than $1$.

Now, compare this to our isolated point. We know $f(0)=1$. For any $x$ in our neighborhood (excluding $x=0$), $f(x)=|x|<1$. Therefore, $f(0)>f(x)$ for all $x$ in the neighborhood.

The condition $f(0)\ge f(x)$ is satisfied for all $x$ in our neighborhood. The point at $(0,1)$ is strictly higher than all the points on the V-shape immediately surrounding it. It is a peak. It is a local maximum.

This problem is a beautiful reminder that in mathematics, definitions are the ultimate truth. When you encounter a problem that seems to defy your visual intuition, don't panic. Go back to the definition.

The definition of a local extremum is a powerful tool, and it doesn't care about holes or jumps. It only cares about the relative height of the point. You have successfully navigated the trap. Keep this rigor in your toolkit, and you will be ready for any challenge the JEE throws your way.