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

A Journey Through Constraints

Welcome, fellow traveler of the mathematical landscape. Today, we are going to tackle a problem that, at first glance, looks like a battlefield of complex trigonometric identities.

You see the equation ${\tan }^{-1}\sqrt{x(x+1)}+{\sin }^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$ and your instinct might be to reach for your toolkit of identities—perhaps trying to convert the tangent inverse into a sine inverse or vice versa. But stop. Take a breath.

In the world of JEE Advanced, the most elegant solutions often come not from brute-force calculation, but from the quiet, observant analysis of constraints.

The Domain Detective

Whenever you see square roots, logarithms, or inverse trigonometric functions, you are looking at a function with a 'personality.' These functions have boundaries they cannot cross. Our equation is defined by two distinct terms.

Let us look at the first one: ${\tan }^{-1}\sqrt{x(x+1)}$. The tangent inverse function, ${\tan }^{-1}(y)$, is quite generous; it accepts any real number $y\in \mathbb{R}$. However, the square root function is far more demanding.

It refuses to accept negative numbers. Therefore, for the first term to exist, we must satisfy the condition:

This is a classic quadratic inequality. If we find the critical points, $x=0$ and $x=-1$, and apply the wavy curve method, we see that the expression is positive in the intervals $(-\infty ,-1]$ and $[0,\infty )$.

This is our first 'blue' region on the number line. It is the territory where the first term is allowed to breathe.

The Gatekeeper

Now, let us turn our attention to the second term: ${\sin }^{-1}\sqrt{x^2+x+1}$. Here, we encounter a strict gatekeeper. The function ${\sin }^{-1}(u)$ is only defined when $-1\le u\le 1$.

Furthermore, because the output of a square root is always non-negative, we are effectively constrained to the interval $[0,1]$. So, we must satisfy:

To solve this, we square the inequality. Since all terms are non-negative, the inequality signs remain unchanged:

Let us split this into two parts. The left side, $x^2+x+1\ge 0$, is always true for all real $x$ because the discriminant $D=1^2-4(1)(1)=-3$ is negative, and the leading coefficient is positive. It is a parabola that never touches the x-axis, always hovering above it.

But the right side, $x^2+x+1\le 1$, is where the magic happens. Subtracting $1$ from both sides, we get:

Factoring this, we arrive at $x(x+1)\le 0$. Using the wavy curve method again, we find that $x$ must lie in the closed interval $[-1,0]$. This is our 'green' region.

The Intersection

Now, we reach the climax of our story. For the entire equation to be valid, both terms must exist simultaneously. We need the intersection of our blue region, $(-\infty ,-1]\cup [0,\infty )$, and our green region, $[-1,0]$.

Look at the number line. Where do these regions overlap? The only way a number can be both $\ge 0$ and $\le 0$ is if it is exactly zero.

Similarly, the only way it can be $\le -1$ and $\ge -1$ is if it is exactly $-1$. The intersection is not a wide range; it is just two points: $x=0$ and $x=-1$.

We have narrowed the infinite possibilities of the real number line down to just two candidates. But we must verify them. Let us test $x=0$:

It works! Now, let us test $x=-1$:

It works again! Both candidates are valid. We have successfully navigated the problem by respecting the domain constraints, bypassing the need for complex identities entirely.

There are exactly two real solutions. Remember this lesson: in the heat of an exam, sometimes the most powerful move is to stop, look at the boundaries, and let the constraints guide you to the answer.