Animated Solution for Mathematics - Functions: Let the domains of the functions $f(x)={\log }_4{\log }_3{\log }_7(8-{\log }_2(x^2+4x+5))$ and $g(x)={\sin }^{-1}(\frac{7x+10}{x-2})$ be $(\alpha ,\beta )$ and $[\gamma ,\delta ]$, respectively. Then ${\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2$ is equal to :-

Unlocking the Domain

Imagine you are an architect designing a structure that must hold up under extreme pressure. In mathematics, the 'domain' of a function is exactly that—the structural integrity of the expression. If you step outside the domain, the function doesn't just break; it ceases to exist.

Today, we are going to explore two such structures: $f(x)={\log }_4{\log }_3{\log }_7(8-{\log }_2(x^2+4x+5))$ and $g(x)={\sin }^{-1}(\frac{7x+10}{x-2})$. Our goal is to find the boundaries of their existence and uncover the hidden numerical beauty within.

When we look at $f(x)$, we see a nested sequence of logarithms. Think of this as a series of security checkpoints. To pass through, the argument of each logarithm must be strictly positive.

We start at the outermost gate: ${\log }_4(\dots )>0$. Since the base $4>1$, we can safely exponentiate without flipping the inequality. This gives us ${\log }_3{\log }_7(8-{\log }_2(x^2+4x+5))>4^0$, which simplifies to ${\log }_3{\log }_7(8-{\log }_2(x^2+4x+5))>1$.

We continue this process, peeling away the layers of the onion. As we strip away each layer, we are essentially solving:

Finally, we reach the core: $x^2+4x+5<2^1$, which simplifies to $x^2+4x+3<0$. Factoring this quadratic gives us $(x+3)(x+1)<0$.

By the Wavy Curve method, we find the domain of $f(x)$ is $x\in (-3,-1)$. Thus, $\alpha =-3$ and $\beta =-1$. We have successfully mapped the first territory!

Now, let us turn our attention to $g(x)={\sin }^{-1}(\frac{7x+10}{x-2})$. The inverse sine function is a strict guardian; it only accepts inputs in the range $[-1,1]$. This forces us to solve the double inequality:

This is a classic JEE trap. Many students try to cross-multiply, but that is a dangerous path. Instead, we treat this as two separate inequalities.

First, $\frac{7x+10}{x-2}\ge -1$, which rearranges to $\frac{8x+8}{x-2}\ge 0$. Using the Wavy Curve, we find $x\in (-\infty ,-1]\cup (2,\infty )$.

Second, $\frac{7x+10}{x-2}\le 1$, which rearranges to $\frac{6x+12}{x-2}\le 0$. This simplifies to $\frac{x+2}{x-2}\le 0$, yielding $x\in [-2,2)$.

To find the domain, we must find the intersection of these two sets. The overlap between $(-\infty ,-1]\cup (2,\infty )$ and $[-2,2)$ is precisely $[-2,-1]$. Therefore, $\gamma =-2$ and $\delta =-1$.

We have navigated the constraints and found our boundaries: $\alpha =-3,\beta =-1,\gamma =-2,\delta =-1$. The problem asks for the sum of their squares: ${\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2$.

Substituting our values:

It is truly elegant how these complex, nested functions collapse into such a clean, integer result. You have successfully navigated the layers of logarithmic constraints and the rigid boundaries of inverse trigonometry.

Remember, in JEE Advanced, the math is not just about calculation; it is about maintaining clarity under pressure. You have done exactly that. Keep this momentum, and keep questioning the 'why' behind every step!