Find the domain of the function f(x) = 1 √(|x| - x) — JEE Mathematics

For the function to be real and defined:

1. The expression inside the square root must be non-negative: $|x| - x \ge 0$.
2. The denominator cannot be zero: $|x| - x \ne 0$.

Combining these, we get:

$$|x| - x > 0 \implies |x| > x$$

Let's evaluate this inequality:

• If $x \ge 0$, then $|x| = x$, so $x > x$ is false.
• If $x < 0$, then $|x| = -x$. Since $x$ is negative, $-x$ is positive, so $-x > x$ is true.

Thus, the condition holds true for all negative real numbers.
Domain $= (-\infty, 0)$.

Answer: $(-\infty, 0)$