Animated Solution for Mathematics - Functions: If the functions are defined as $f(x)=\sqrt{x}$ and $g(x)=\sqrt{1-x}$, then what is the common domain of the following functions: $f+g,f-g,f/g,g/f,g-f$ where $(f\pm g)(x)=f(x)\pm g(x),(f/g)(x)=\frac{f(x)}{g(x)}$

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to unravel a fundamental concept that often trips up even the brightest minds: the domain of combined functions.

Imagine you are an architect designing a bridge. You have two structural components, $f(x)$ and $g(x)$. If you want to combine them—add them, subtract them, or divide them—the bridge only stands where both components are physically capable of existing. This is the essence of finding a common domain.

The Individual Constraints

We begin with our two functions: $f(x)=\sqrt{x}$ and $g(x)=\sqrt{1-x}$. Before we even think about adding or dividing, we must respect the laws of mathematics. For any square root to yield a real number, the expression inside must be non-negative.

For $f(x)=\sqrt{x}$, the condition is simple: $x\ge 0$. This means our function $f$ is only 'alive' on the interval $[0,\infty )$.

Now, look at $g(x)=\sqrt{1-x}$. Here, the expression inside is $1-x$. We require $1-x\ge 0$, which rearranges beautifully to $x\le 1$. Thus, $g$ is only 'alive' on the interval $(-\infty ,1]$.

The Intersection (The Overlap)

Now, imagine these two intervals on a number line. $f$ covers everything from $0$ to the right, and $g$ covers everything from $1$ to the left.

For any operation like $f+g$ or $f-g$ to exist, we need a value of $x$ that satisfies both conditions simultaneously. This is the intersection:

This is our baseline. Any $x$ outside this interval makes at least one of the functions disappear, and the whole operation collapses.

The Trap of Division

Here is where the JEE examiners love to test your attention to detail. The problem asks for the domain of $f/g$ and $g/f$ as well. In the world of fractions, there is one cardinal sin: division by zero.

For $f/g=\frac{\sqrt{x}}{\sqrt{1-x}}$, the denominator is $\sqrt{1-x}$. We must ensure $\sqrt{1-x} eq 0$, which means $1-x eq 0$, or $x eq 1$. We must punch a hole in our interval at $x=1$.

For $g/f=\frac{\sqrt{1-x}}{\sqrt{x}}$, the denominator is $\sqrt{x}$. We must ensure $\sqrt{x} eq 0$, which means $x eq 0$. We must punch a hole in our interval at $x=0$.

We started with the closed interval $[0,1]$. We then applied the strict rules of division, which forced us to exclude the endpoints $0$ and $1$.

What remains is the open interval $(0,1)$. This is the only region where all our operations—addition, subtraction, and both divisions—are perfectly valid.

Remember, in your JEE journey, always look for these hidden constraints. They are not just rules; they are the boundaries of the mathematical universe you are exploring. Keep practicing, stay curious, and never stop questioning the 'why' behind the math!