Animated Solution for Mathematics - Inverse Trigonometric Functions: The domain of the function $f(x)={\sin }^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is $(-\infty ,-a]\cup [a,\infty )$. Then $a$ is equal to :

Welcome, future engineers! Today, we are going to peel back the layers of a function that might look intimidating at first glance, but is actually a beautiful exercise in logical deduction. We are tasked with finding the domain of $f(x)={\sin }^{-1}\left(\frac{|x|+5}{x^2+1}\right)$.

Think of the domain as the 'allowable territory' for our function. It is the set of all inputs $x$ for which the function produces a valid, real output.

Our journey begins with the gatekeeper of this function: the inverse sine. Recall that for any value $u$, the function ${\sin }^{-1}(u)$ is only defined when $-1\le u\le 1$. This is our non-negotiable constraint.

If our input $u$ wanders outside this interval, the function ceases to exist. So, our first mission is to ensure that the expression $\frac{|x|+5}{x^2+1}$ stays strictly within the bounds of $[-1,1]$. This gives us the double inequality:

Now, let's pause and analyze the left side of this inequality. We have $|x|+5$ in the numerator and $x^2+1$ in the denominator.

Since $|x|$ is always non-negative, $|x|+5$ is always at least $5$. Since $x^2$ is non-negative, $x^2+1$ is always at least $1$. A positive number divided by a positive number is always positive.

Therefore, the fraction is always greater than zero. Since it is always greater than zero, it is automatically greater than $-1$. The left side of our inequality is satisfied for all real numbers $x$!

This is where the algebra gets exciting. We want to know when this fraction is less than or equal to $1$.

Many students fall into the trap of cross-multiplying blindly. In inequalities, cross-multiplying is dangerous because if you multiply by a negative number, the inequality sign must flip.

However, in this problem, we are safe. Because $x^2+1$ is strictly positive for all real $x$, we can multiply both sides by $x^2+1$ without worrying about flipping the inequality sign. This leads us to:

Rearranging this, we get:

This looks like a quadratic, doesn't it? Let's make it even cleaner by using a substitution. Let $t=|x|$. Since $t$ is an absolute value, we must remember that $t\ge 0$.

Our inequality transforms into $t^2-t-4\ge 0$. To solve this, we first find the roots of the equation $t^2-t-4=0$. Using the quadratic formula:

We find the roots to be $t=\frac{1\pm \sqrt{17}}{2}$. We have two critical points: $t_1=\frac{1-\sqrt{17}}{2}$ and $t_2=\frac{1+\sqrt{17}}{2}$.

Since $\sqrt{17}$ is slightly larger than $4$, $t_1$ is negative, and $t_2$ is positive. We need the quadratic to be greater than or equal to zero, which occurs outside the roots.

However, we have the constraint $t\ge 0$. This means we must discard the region containing the negative root. We are left with:

Finally, we substitute back $t=|x|$. We have $|x|\ge \frac{1+\sqrt{17}}{2}$.

This inequality splits into two regions:

The domain of the function is:

You have successfully navigated the domain of this function! It was not just about solving an inequality; it was about understanding the physical constraints of the function and respecting the boundaries of the modulus. Keep this logical rigor in your toolkit, and no problem will ever be too complex for you.