Animated Solution for Mathematics - Inverse Trigonometric Functions: Considering only the principal values of the inverse trigonometric functions, the domain of the function $f(x)={\cos }^{-1}\left(\frac{x^2-4x+2}{x^2+3}\right)$ is :

Welcome, fellow learners! Today, we are going to embark on a journey to solve a classic JEE Advanced problem. We are tasked with finding the domain of the function $f(x)={\cos }^{-1}\left(\frac{x^2-4x+2}{x^2+3}\right)$.

At first glance, this might look like a standard algebra problem, but it is actually a test of your conceptual clarity regarding inverse trigonometric functions and inequality handling. Let us break this down step by step.

The Domain Constraint

Whenever you see an inverse trigonometric function, your first instinct should be to recall its domain. For the function ${\cos }^{-1}(u)$, the input $u$ is strictly constrained. It must lie within the closed interval $[-1,1]$.

This is the golden rule. If the input falls outside this range, the function is undefined. Therefore, our first step is to set up the inequality:

This double inequality is the heart of the problem. We need to find all values of $x$ that satisfy this condition.

The Denominator Trap

Now, here is where many students stumble. We have a rational expression. The urge to cross-multiply is strong, but we must pause.

In the world of inequalities, cross-multiplication is dangerous unless you know the sign of the denominator. If the denominator were negative, the inequality signs would flip, and your entire solution would collapse.

Let us analyze $x^2+3$. Since $x^2$ is always greater than or equal to zero for any real number $x$, it follows that $x^2+3$ is always greater than or equal to $3$.

This is fantastic news! The denominator is strictly positive. We can safely multiply the entire inequality by $(x^2+3)$ without worrying about the signs changing.

Solving the Inequality

After multiplying, we get:

To solve this, we split it into two manageable parts. We need to satisfy both:

1) $x^2-4x+2\le x^2+3$

2) $x^2-4x+2\ge -(x^2+3)$

Let us tackle the first one. Subtracting $x^2$ from both sides, we get $-4x+2\le 3$. This simplifies to $-4x\le 1$.

Dividing by $-4$ requires us to flip the inequality sign, giving us $x\ge -\frac{1}{4}$.

Now, let us look at the second inequality: $x^2-4x+2\ge -x^2-3$. Rearranging this, we bring all terms to the left:

This is a quadratic inequality. To solve it, we check the discriminant $D=b^2-4ac$. Here, $D=(-4{)}^2-4(2)(5)=16-40=-24$.

Since $D<0$, the quadratic has no real roots. Because the leading coefficient is positive, the parabola opens upwards and never touches the $x$-axis. Thus, $2x^2-4x+5$ is always positive for all real $x$. This condition is satisfied for all $x\in \mathbb{R}$.

The Final Intersection

We have two conditions: $x\ge -\frac{1}{4}$ and $x\in \mathbb{R}$. To find the domain of the original function, we take the intersection of these two sets.

The intersection of "all real numbers" and $x\ge -\frac{1}{4}$ is simply $x\ge -\frac{1}{4}$.

Thus, the domain of our function is $