Animated Solution for Mathematics - Inverse Trigonometric Functions: The domain of the function ${\cos }^{-1}\left(\frac{2{\sin }^{-1}\left(\frac{1}{4x^2-1}\right)}{\pi }\right)$ is:

Imagine you are standing before a complex, nested function. It looks intimidating, doesn't it?

Here is the secret of the JEE Advanced: every complex problem is just a series of simple, logical steps waiting to be unraveled. We treat this like an onion, peeling the layers one by one, starting from the outside.

A Trivial Constraint

The outermost function is ${\cos }^{-1}(u)$. We know from our fundamental trigonometry that for ${\cos }^{-1}(u)$ to be defined, the input $u$ must live in the interval $[-1,1]$.

So, we set our condition:

If we multiply this entire inequality by $\frac{\pi }{2}$, we get:

Now, pause and look at that inequality. The range of the ${\sin }^{-1}$ function is, by definition, $[-\frac{\pi }{2},\frac{\pi }{2}]$. This means the condition is always satisfied as long as the inner ${\sin }^{-1}$ function itself is defined.

The outer layer, which looked so imposing, was actually a gentle guardian, not a barrier!

The Inner Constraint

Now we reach the core: the ${\sin }^{-1}$ function. For ${\sin }^{-1}(y)$ to be defined, its input $y$ must also lie in the interval $[-1,1]$.

Our input here is $\frac{1}{4x^2-1}$. So, we must satisfy the condition:

This is equivalent to saying $\left|\frac{1}{4x^2-1}\right|\le 1$. If the reciprocal of a number has an absolute value less than or equal to $1$, then the absolute value of the number itself must be greater than or equal to $1$.

Thus, we require:

Splitting the Cases

An absolute value inequality $|A|\ge 1$ splits into two distinct paths: $A\ge 1$ or $A\le -1$. Let's walk these paths together.

Case 1: $4x^2-1\ge 1$. Adding $1$ to both sides, we get $4x^2\ge 2$, which simplifies to $x^2\ge \frac{1}{2}$. Taking the square root, we find $|x|\ge \frac{1}{\sqrt{2}}$.

This gives us the intervals:

Case 2: $4x^2-1\le -1$. Adding $1$ to both sides, we get $4x^2\le 0$. Now, think carefully: can the square of a real number be negative? Never!

It can only be zero or positive. So, the only way $x^2\le 0$ is if $x^2=0$, which means $x=0$. This is our hidden gem, the isolated point that many students overlook.

We have our intervals from Case 1 and our isolated point from Case 2. When we combine them, we get the complete domain:

By staying calm and peeling the layers, we didn't just solve the problem; we mastered it. Keep this logical rigor in your toolkit, and no function will ever be able to hide its domain from you again.