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

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving an inequality; we are embarking on a journey to find the 'common ground' of a function.

When we look at a complex function like $f(x)={\cos }^{-1}(\frac{2-|x|}{4})+({\log }_e(3-x){)}^{-1}$, it is easy to feel overwhelmed. But remember, a function is like a delicate ecosystem; it only thrives where all its components are allowed to exist.

Our goal is to find the set of all $x$ values where this ecosystem remains stable. Let us break this down into two distinct acts.

The Inverse Cosine's Boundary

The first part of our function is $f_1(x)={\cos }^{-1}(\frac{2-|x|}{4})$. The inverse cosine function, ${\cos }^{-1}(u)$, is a picky eater; it refuses to accept any input $u$ that lies outside the interval $[-1,1]$.

Thus, we must enforce the constraint:

To solve this, we multiply the entire inequality by $4$, giving us $-4\le 2-|x|\le 4$. Now, we subtract $2$ from every part:

Here is where the magic happens. When we multiply by $-1$, we must flip the inequality signs—a classic JEE trap! This yields:

The condition $|x|\ge -2$ is always true for any real number, because the absolute value is never negative. The real restriction is $|x|\le 6$, which tells us that $x$ must live in the interval $[-6,6]$.

The Logarithmic Gatekeeper

Now, let us turn our attention to the second part: $f_2(x)=\frac{1}{{\log }_e(3-x)}$. This function has two strict demands.

First, the argument of the logarithm, $3-x$, must be strictly positive. If it were zero or negative, the logarithm would be undefined:

Second, because this logarithm sits in the denominator, the entire expression ${\log }_e(3-x)$ cannot be zero. We know that ${\log }_e(u)=0$ only when $u=1$.

Therefore, $3 - x eq 1$, which means $x eq 2$. We have now identified the second playground: $x$ must be less than $3$, but it cannot be $2$.

The Intersection of Worlds

We have two playgrounds. The first, from the inverse cosine, is $[-6,6]$. The second, from the logarithm, is $(-\infty ,3)\setminus \{2\}$.

To find the domain of the total function $f(x)$, we must find the intersection of these two sets. Imagine placing these two intervals on a number line.

The first interval covers everything from $-6$ to $6$. The second interval starts from the left and stops abruptly at $3$, with a tiny hole at $2$.

Where do they overlap? They overlap from $-6$ to $3$. But we must remember to keep that hole at $2$. Thus, our final domain is $[-6,3)\setminus \{2\}$.

The problem states that the domain is in the form $[-\alpha ,\beta )-\{\gamma \}$. By comparing our result $[-6,3)-\{2\}$ with this form, we can identify our variables:

$-\alpha =-6\Rightarrow \alpha =6$ $\beta =3$ $\gamma =2$

The final step is a simple sum:

We have navigated the traps, respected the constraints, and arrived at the solution. Remember, in JEE Advanced, the math is not just about calculation; it is about understanding the boundaries of existence for every function you encounter.