Animated Solution for Mathematics - Functions: If the domain of the function $f(x)={\log }_e(\frac{2x-3}{5+4x})+{\sin }^{-1}(\frac{4+3x}{2-x})$ is $[\alpha ,\beta )$, then ${\alpha }^2+4\beta$ is equal to

Welcome, my dear students. Today, we are embarking on a journey to decode the DNA of a function. When we look at a function like

we are not just looking at a collection of symbols; we are looking at a set of rules. A function is only as strong as its domain—the set of all possible inputs that make the function 'happy' and defined.

Our goal is to find the exact interval $[\alpha ,\beta )$ where this function exists.

The Logarithmic Constraint

Let us first examine the natural logarithm, ${\log }_e\left(\frac{2x-3}{5+4x}\right)$. The rule for a logarithm is absolute: the argument must be strictly greater than zero.

We set up our first inequality:

To solve this, we identify the critical points where the numerator or denominator becomes zero: $x=\frac{3}{2}$ and $x=-\frac{5}{4}$.

Using the Wavy Curve Method, we test the intervals. We find that the expression is positive when $x<-\frac{5}{4}$ or $x>\frac{3}{2}$. This gives us our first 'blue' region on the number line: $(-\infty ,-\frac{5}{4})\cup (\frac{3}{2},\infty )$.

The Inverse Sine Constraint

Now, we turn to the inverse sine function, ${\sin }^{-1}\left(\frac{4+3x}{2-x}\right)$. The inverse sine function is even more restrictive, as it demands that its argument must live within the closed interval $[-1,1]$.

This gives us the double inequality:

To conquer this, we split it into two manageable battles.

Case 1: $\frac{4+3x}{2-x}\ge -1$. By moving the $-1$ to the left and simplifying, we get $\frac{2x+6}{2-x}\ge 0$. Multiplying by $-1$ to normalize the denominator, we get $\frac{2(x+3)}{x-2}\le 0$, which yields the interval $[-3,2)$.

Case 2: $\frac{4+3x}{2-x}\le 1$. Similarly, moving the $1$ to the left and simplifying, we get $\frac{4x+2}{2-x}\le 0$. Normalizing gives $\frac{2(2x+1)}{x-2}\ge 0$, which yields $(-\infty ,-\frac{1}{2}]\cup (2,\infty )$.

The intersection of these two cases gives us the 'green' region: $[-3,-\frac{1}{2}]$.

The Grand Intersection

We are now at the climax of our problem. We have the 'blue' region from the logarithm and the 'green' region from the inverse sine.

The domain of the entire function $f(x)$ is the intersection of these two sets:

The overlap is the interval $[-3,-\frac{5}{4})$. This is our final domain!

We are given that the domain is $[\alpha ,\beta )$. By comparing this to our result $[-3,-\frac{5}{4})$, we identify $\alpha =-3$ and $\beta =-\frac{5}{4}$.

The final step is a simple calculation: ${\alpha }^2+4\beta$. Substituting our values, we get:

And there you have it! Through careful analysis and systematic inequality solving, we have arrived at the final answer of 4. Remember, in JEE Advanced, the beauty lies not just in the final number, but in the logical path you take to get there.