Animated Solution for Mathematics - Inverse Trigonometric Functions: The number of real roots of the equation ${\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{4}$ is :

Welcome, fellow traveler, to the fascinating world of JEE mathematics. Today, we are dissecting a problem that looks like a standard equation but is, in reality, a masterclass in domain analysis.

We are given the equation:

At first glance, you might be tempted to start manipulating the terms, perhaps by taking the tangent of both sides. But stop! Before you dive into the algebra, you must respect the domain. In the realm of inverse trigonometric functions, the domain is not just a suggestion; it is the law.

The Square Root

Let us look at the first term: ${\tan }^{-1}\sqrt{x(x+1)}$. For this to exist in the set of real numbers, the expression inside the square root must be non-negative.

Thus, we have the condition:

This simplifies to $x^2+x\ge 0$. This is our first boundary, which tells us that $x$ cannot lie in the open interval $(-1,0)$.

The Sine Inverse Trap

Now, consider the second term: ${\sin }^{-1}\sqrt{x^2+x+1}$. The domain of ${\sin }^{-1}(u)$ is strictly $u\in [-1,1]$.

Therefore, we must have:

Squaring this inequality gives us $0\le x^2+x+1\le 1$. Focus on the right side: $x^2+x+1\le 1$. If we subtract $1$ from both sides, we get:

Now, look at what we have achieved. We have two conflicting requirements for the expression $x^2+x$.

The first constraint from the tangent term demanded $x^2+x\ge 0$. The second constraint from the sine inverse term demands $x^2+x\le 0$.

The only way both can be true is if:

This is the beautiful moment where the problem collapses into a simple quadratic equation: $x(x+1)=0$. This gives us two candidates: $x=0$ and $x=-1$.

Many students stop here, assuming they have found the roots. But in the JEE, you must always verify.

Let us test $x=0$:

This is clearly not $\frac{\pi }{4}$. Now, let us test $x=-1$:

Again, this is not $\frac{\pi }{4}$. Both candidates fail. We have exhausted all possibilities, and none of them satisfy the equation.

Thus, the number of real roots is 0. This problem teaches us that sometimes, the most powerful tool in your arsenal is not a complex formula, but the simple, rigorous check of the domain.