Animated Solution for Mathematics - Functions: If the functions $f(x)$ and $g(x)$ are defined on $R\to R$ such that $f(x)=\{\begin{array}{ll}0,& x\in \text{rational}\\ x,& x\in \text{irrational}\end{array}$; $g(x)=\{\begin{array}{ll}0,& x\in \text{irrational}\\ x,& x\in \text{rational}\end{array}$ then $(f-g)(x)$ is

A Journey into Piecewise Functions

Welcome, fellow explorers of mathematics! Today, we are going to dissect a problem that often trips up even the most seasoned JEE aspirants. It involves two functions, $f(x)$ and $g(x)$, that seem to play a game of musical chairs depending on whether the input $x$ is a rational or an irrational number.

This is a classic JEE Advanced setup—a problem that tests your ability to look past the intimidating notation and see the underlying structure.

We are given two functions defined on $\mathbb{R}\to \mathbb{R}$:

We are asked to analyze the nature of the function $h(x)=(f-g)(x)$. To understand $h(x)$, we must evaluate it in two distinct scenarios. Let us define $h(x)=f(x)-g(x)$ by looking at these two cases.

The Rational Realm

When $x$ is a rational number, we look at our definitions: $f(x)=0$ and $g(x)=x$.

Substituting these into our expression for $h(x)$, we get:

So, for every rational number $x$, our function acts like the line $y=-x$. It is simple, elegant, and predictable.

The Irrational Realm

Now, let us step into the world of irrational numbers. Here, the roles are reversed: $f(x)=x$ and $g(x)=0$.

Substituting these into $h(x)$, we get:

Thus, for every irrational number $x$, our function acts like the line $y=x$.

Combining these, we have a complete definition for our function:

Note that $0$ is a rational number, so $h(0)=-0=0$, meaning the origin $(0,0)$ is a solid, valid point on our graph.

Avoiding the Trap

Is this function one-one? To be one-one, every $y$ value must correspond to exactly one $x$ value. Let us test this by setting $h(x)=c$ for some constant $c$.

If $c$ is rational, we have two possibilities: either $x$ is rational (so $-x=c\Rightarrow x=-c$) or $x$ is irrational (so $x=c$). However, if $c$ is rational, then $x=c$ would be a rational number, which contradicts our assumption that $x$ is irrational. Thus, the only valid solution is $x=-c$.

If $c$ is irrational, we again have two possibilities: either $x$ is rational (so $-x=c\Rightarrow x=-c$) or $x$ is irrational (so $x=c$). If $c$ is irrational, then $x=-c$ would be an irrational number, which contradicts our assumption that $x$ is rational. Thus, the only valid solution is $x=c$.

In both scenarios, for any given $y$, there is exactly one valid $x$. The function is strictly one-one.

Covering the Real Line

Finally, is the function onto? We need to see if the range of $h(x)$ is the entire set of real numbers $\mathbb{R}$.

For any real number $y$, can we find an $x$ such that $h(x)=y$?

If $y$ is rational, we choose $x=-y$. Since $y$ is rational, $-y$ is rational, and $h(-y)=-(-y)=y$.

If $y$ is irrational, we choose $x=y$. Since $y$ is irrational, $x$ is irrational, and $h(y)=y$. In both cases, we have successfully mapped an $x$ to every possible $y$. The range is indeed $\mathbb{R}$.

We have navigated the piecewise landscape and found that our function $h(x)$ is both one-one and onto. It is a beautiful example of how even the most fragmented functions can possess perfect symmetry and order. Keep practicing, keep questioning, and remember: the math is always more elegant than it first appears!