Animated Solution for Mathematics - Functions: If domain of the function ${\log }_e\left(\frac{6x^2+5x+1}{2x-1}\right)+{\cos }^{-1}\left(\frac{2x^2-3x+4}{3x-5}\right)$ is $(\alpha ,\beta )\cup (\gamma ,\delta )$, then $18({\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2)$ is equal to

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a problem; we are dissecting the anatomy of a function. Domain problems are the gatekeepers of calculus, testing your ability to see the hidden constraints of the mathematical universe.

We are looking at the function:

At first glance, this looks like a chaotic mess of quadratics and fractions. However, mathematics is the art of imposing order on chaos. For the function to exist, both components must be valid simultaneously; this is our guiding star: the intersection of conditions.

The Logarithm's Gatekeeper

The natural logarithm, ${\log }_e(u)$, is a picky function. It demands that its argument $u$ be strictly positive. Therefore, our first constraint is:

We begin by factorizing the numerator. The quadratic $6x^2+5x+1$ splits beautifully into $(3x+1)(2x+1)$. Now our inequality looks much friendlier:

We identify our critical points by setting each factor to zero: $x=-\frac{1}{2}$, $x=-\frac{1}{3}$, and $x=\frac{1}{2}$. Using the Wavy Curve Method, we test the intervals. We find that the expression is positive in the intervals $(-\frac{1}{2},-\frac{1}{3})\cup (\frac{1}{2},\infty )$.

The Inverse Cosine's Boundary

Now, we turn to the inverse cosine, ${\cos }^{-1}(u)$. This function is only defined when its argument $u$ is trapped within the closed interval $[-1,1]$. This gives us two inequalities to solve:

1. $\frac{2x^2-3x+4}{3x-5}\ge -1$ 2. $\frac{2x^2-3x+4}{3x-5}\le 1$

Let's tackle the first one. We bring the $-1$ to the left, add it to the fraction, and simplify to obtain:

The roots are $\pm \frac{1}{\sqrt{2}}$ and $\frac{5}{3}$. Applying the Wavy Curve Method again, we find the solution set: $[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]\cup (\frac{5}{3},\infty )$.

The Hidden Trap

Now, we solve the second inequality: $\frac{2x^2-3x+4}{3x-5}\le 1$. Subtracting $1$ from both sides yields:

Here is where many students stumble. Look at the numerator: $2x^2-6x+9$. If you calculate the discriminant $D=(-6{)}^2-4(2)(9)=36-72=-36$.

Because $D<0$ and the leading coefficient is positive, this quadratic is always positive. It never touches the $x$-axis. Consequently, the only way the entire fraction can be less than or equal to zero is if the denominator is negative:

The Final Convergence

We now intersect our findings. For the ${\cos }^{-1}$ function, we intersect $[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]\cup (\frac{5}{3},\infty )$ with $(-\infty ,\frac{5}{3})$, which simplifies to $[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$.

Finally, we intersect this with our logarithm's domain: $(-\frac{1}{2},-\frac{1}{3})\cup (\frac{1}{2},\infty )$. The overlap gives us the final domain:

Comparing this to the form $(\alpha ,\beta )\cup (\gamma ,\delta )$, we identify our parameters: $\alpha =-\frac{1}{2}$, $\beta =-\frac{1}{3}$, $\gamma =\frac{1}{2}$, and $\delta =\frac{1}{\sqrt{2}}$.

We are asked to calculate $18({\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2)$. Substituting our values:

The final answer is 20. You have navigated the constraints, avoided the traps, and arrived at the truth. Keep this rigor in your toolkit—it will serve you well in every challenge to come.