Animated Solution for Mathematics - Inverse Trigonometric Functions: If the domain of the function $f(x)={\sin }^{-1}\left(\frac{x-1}{2x+3}\right)$ is $\mathbb{R}-(\alpha ,\beta )$, then $12\alpha \beta$ is equal to :

Mastering the Domain of ${\sin }^{-1}(u)$

Welcome, future engineers! Today, we are going to peel back the layers of a classic JEE Advanced problem. We are looking at the domain of the function $f(x)={\sin }^{-1}\left(\frac{x-1}{2x+3}\right)$.

At first glance, it might look like just another algebraic headache, but I want you to see it as a beautiful puzzle of constraints.

The Fundamental Constraint

The sine inverse function, ${\sin }^{-1}(u)$, is a gatekeeper. It only accepts inputs that live in the world of $[-1,1]$.

If you try to feed it anything outside this range, the function simply ceases to exist in the real number system. So, our mission is to find all values of $x$ such that the argument $u=\frac{x-1}{2x+3}$ stays within this safe zone.

Mathematically, we must solve the double inequality:

This is not just one inequality; it is two conditions that must be satisfied simultaneously. We must solve $\frac{x-1}{2x+3}\ge -1$ AND $\frac{x-1}{2x+3}\le 1$.

The Rational Inequality Trap

Now, here is where many students stumble. You might be tempted to cross-multiply the denominator $(2x+3)$. Please, resist that urge!

Cross-multiplying is a dangerous game when the denominator contains a variable. We do not know if $(2x+3)$ is positive or negative. Instead, we use the 'bring to one side' strategy.

Let us tackle the first part: $\frac{x-1}{2x+3}\ge -1$. By adding $1$ to both sides, we get:

Combining these into a single fraction gives us:

This is the standard form we need for the wavy curve method. The critical points are $x=-\frac{2}{3}$ and $x=-\frac{3}{2}$. Plotting these on the number line, we find the solution set is $(-\infty ,-\frac{3}{2})\cup [-\frac{2}{3},\infty )$.

The Second Condition and the Sign Flip

Now, let us look at the second part: $\frac{x-1}{2x+3}\le 1$. Again, we bring everything to the left:

Combining terms, we get:

To make this easier to read, we multiply the entire inequality by $-1$. Remember the golden rule: multiplying by a negative number flips the inequality sign!

So, we get $\frac{x+4}{2x+3}\ge 0$. The critical points here are $x=-4$ and $x=-\frac{3}{2}$. Using the wavy curve method again, we find the solution set is $(-\infty ,-4]\cup (-\frac{3}{2},\infty )$.

The Intersection and the Final Victory

We have two conditions. The first gave us $(-\infty ,-\frac{3}{2})\cup [-\frac{2}{3},\infty )$, and the second gave us $(-\infty ,-4]\cup (-\frac{3}{2},\infty )$.

Since both must be true, we find their intersection. Looking at the number line, the overlapping region is $(-\infty ,-4]\cup [-\frac{2}{3},\infty )$.

The problem tells us the domain is $\mathbb{R}-(\alpha ,\beta )$. By comparing our result, we see that the gap in the real number line is the open interval $(-4,-\frac{2}{3})$.

Thus, $\alpha =-4$ and $\beta =-\frac{2}{3}$. Finally, we calculate $12\alpha \beta$:

And there you have it! A perfect, systematic victory over a complex domain problem. The final answer is 32.