Animated Solution for Mathematics - Functions: Let $f(x)=\{\begin{array}{ll}-a& \text{if }-a\le x\le 0\\ x+a& \text{if }0<x\le a\end{array}$ where $a>0$ and $g(x)=(f(|x|)-|f(x)|)/2$. Then the function $g:[-a,a]\to [-a,a]$ is

A Journey into Piecewise Logic

Welcome, future engineers. Today, we are not just solving a problem; we are dissecting the anatomy of a function.

When you see a piecewise function like

do not be intimidated. Think of it as a machine with two different modes of operation.

Our goal is to understand the derived function

This is a classic JEE Advanced problem that tests your ability to handle absolute values and domain splitting with surgical precision.

The Domain Split

The first step in any piecewise analysis is to respect the boundaries. We are given the domain $[-a,a]$.

We must split this into two distinct territories: the negative side $x\in [-a,0]$ and the positive side $x\in (0,a]$.

Imagine you are standing at $x=0$. To the left, the world is defined by constant values. To the right, it is defined by linear growth.

When we introduce the absolute value $|x|$, we are essentially reflecting the negative side onto the positive side. This is why we must analyze $g(x)$ in these two separate cases.

The Algebra of the Left Side

Let us focus on the interval $x\in [-a,0]$. Here, $x$ is negative, so $|x|=-x$.

Now, look at the definition of $f(x)$. Since $-x$ is positive, we must use the second branch of $f(x)$, which is $f(x)=x+a$.

Therefore, $f(|x|)=f(-x)=(-x)+a=a-x$.

Next, we look at $|f(x)|$. In this interval, $f(x)=-a$. Thus, $|f(x)|=|-a|=a$.

Now, we combine these into our formula for $g(x)$:

See how the $a$ terms cancel out? It is elegant, isn't it? The complexity collapses into a simple linear equation $g(x)=-\frac{x}{2}$.

The Silence of the Right Side

Now, let us move to the interval $x\in (0,a]$. Here, $x$ is positive, so $|x|=x$.

Consequently, $f(|x|)=f(x)=x+a$. Since $x>0$ and $a>0$, the value $x+a$ is strictly positive.

This means $|f(x)|=|x+a|=x+a$.

When we substitute these into our expression for $g(x)$, something fascinating happens:

For the entire positive domain, the function is flat. It is a horizontal line at zero. This is the 'trap' that catches many students—they expect a complex result, but the math leads us to a beautiful, simple zero.

The Verdict

We have our function:

Now, let us test the properties. Is it one-one? A function is one-one if every output comes from a unique input.

But look at the interval $(0,a]$. Every single $x$ in this interval maps to $0$. Since multiple inputs map to the same output, the function is many-one, not one-one.

Is it onto? A function is onto if its range equals its codomain. The codomain is given as $[-a,a]$.

However, our range is only $[0,\frac{a}{2}]$. Since the range does not cover the entire codomain, the function is into, not onto.

We have successfully navigated the logic. The function is neither one-one nor onto. Keep this level of rigorous, step-by-step analysis in your toolkit, and no function will ever be able to hide its secrets from you.