Animated Solution for Mathematics - Differentiation: If a function $f:[-2a,2a]\to R$ is an odd function such that $f(x)=f(2a-x)$ for $x\in [a,2a]$ and the left hand derivative at $x=a$ is 0 then find the left hand derivative at $x=-a$.

Welcome, student. Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of functional symmetry. In the JEE Advanced, you will often encounter problems that look like algebraic puzzles but are, in reality, geometric stories waiting to be told.

We are given a function $f:[-2a,2a]\to \mathbb{R}$ with two distinct, powerful symmetries. Our goal is to find the left-hand derivative at $x=-a$. Let us peel back the layers together.

The Odd Function and the Even Derivative

First, let us look at the definition of an odd function. We are told $f$ is odd, which means $f(-x)=-f(x)$. Imagine the graph of this function; it is symmetric about the origin.

If you have a point $(x,y)$ on the curve, there is a corresponding point $(-x,-y)$ on the other side. Now, let us apply the power of calculus to see how the slope behaves. We differentiate both sides of the odd function definition with respect to $x$:

On the left side, we apply the chain rule. The derivative of $f(-x)$ is $f^{\prime }(-x)$ multiplied by the derivative of the inner function, which is $-1$. On the right side, the derivative is simply $-f^{\prime }(x)$.

Canceling the negative signs, we arrive at a beautiful realization: $f^{\prime }(-x)=f^{\prime }(x)$. This means the derivative of an odd function is an even function. This is a fundamental property that will serve as our bridge later in the problem.

The Symmetry at $x=a$

Next, we are given a second condition: $f(x)=f(2a-x)$ for $x\in [a,2a]$. This tells us that the function is symmetric about the vertical line $x=a$. Imagine a mirror placed at $x=a$.

Let us differentiate this relation to see what it tells us about the slopes:

On the left, we have $f^{\prime }(x)$. On the right, we apply the chain rule again, where the derivative of $2a-x$ is $-1$. Thus, we get:

This equation is our second key. It tells us that the slope at $x$ is the negative of the slope at $2a-x$.

The Bridge to the Solution

Now, we connect the dots. We are given that the left-hand derivative at $x=a$ is zero, denoted as $f^{\prime }(a^{-})=0$.

Let us use our second key, $f^{\prime }(x)=-f^{\prime }(2a-x)$, and let $x$ approach $a$ from the left ($x\to a^{-}$). As $x$ approaches $a$ from the left, the term $(2a-x)$ approaches $a$ from the right ($a^{+}$). Therefore, our equation becomes:

Since we know $f^{\prime }(a^{-})=0$, it follows immediately that $f^{\prime }(a^{+})=0$. The slope is horizontal on both sides of $x=a$.

Finally, we need to find the left-hand derivative at $x=-a$, which is $f^{\prime }(-a^{-})$. Recall our first key: the derivative of an odd function is an even function, $f^{\prime }(-x)=f^{\prime }(x)$.

If we let $x$ approach $a$ from the right ($x\to a^{+}$), then $-x$ approaches $-a$ from the left ($-a^{-}$). Substituting this into our even function property:

Since we just calculated that $f^{\prime }(a^{+})=0$, we have our final result:

The Elegance of Symmetry

We started with abstract functional definitions and, through the rigorous application of the chain rule and symmetry properties, we arrived at a concrete value. The derivative at $x=-a$ is 0.

This is the beauty of JEE Advanced mathematics. It is not about memorizing formulas; it is about understanding the underlying structure of the problem. When you see symmetry, do not just look at the function—look at its derivative, its integral, and its behavior.