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

We need to find the domain of $$f(x) = \log_e\left(\frac{6x^2 + 5x + 1}{2x - 1}\right) + \cos^{-1}\left(\frac{2x^2 - 3x + 4}{3x - 5}\right)$$, expressed as $$(\alpha, \beta) \cup (\gamma, \delta)$$, and compute $$18(\alpha^2 + \beta^2 + \gamma^2 + \delta^2)$$.

Domain of the logarithm.

We need $$\frac{6x^2 + 5x + 1}{2x - 1} > 0$$.

Factoring the numerator: $$6x^2 + 5x + 1 = (2x + 1)(3x + 1)$$.

Critical points: $$x = -\frac{1}{2},\; -\frac{1}{3},\; \frac{1}{2}$$.

Sign analysis of $$\frac{(2x+1)(3x+1)}{2x-1}$$:

$$x < -\frac{1}{2}$$: $$\frac{(-)(-)}{(-)} = \frac{(+)}{(-)} < 0$$ ✘

$$-\frac{1}{2} < x < -\frac{1}{3}$$: $$\frac{(+)(-)}{(-)} = \frac{(-)}{(-)} > 0$$ ✔

$$-\frac{1}{3} < x < \frac{1}{2}$$: $$\frac{(+)(+)}{(-)} < 0$$ ✘

$$x > \frac{1}{2}$$: $$\frac{(+)(+)}{(+)} > 0$$ ✔

Domain from log: $$\left(-\frac{1}{2}, -\frac{1}{3}\right) \cup \left(\frac{1}{2}, \infty\right)$$

Domain of $$\cos^{-1}$$.

We need $$-1 \leq \frac{2x^2 - 3x + 4}{3x - 5} \leq 1$$ and $$x \neq \frac{5}{3}$$.

Let $$g(x) = \frac{2x^2 - 3x + 4}{3x - 5}$$.

Condition (i): $$g(x) \leq 1$$

$$\frac{2x^2 - 3x + 4 - (3x - 5)}{3x - 5} \leq 0 \implies \frac{2x^2 - 6x + 9}{3x - 5} \leq 0$$

The discriminant of $$2x^2 - 6x + 9$$ is $$36 - 72 = -36 < 0$$, and the leading coefficient is positive, so $$2x^2 - 6x + 9 > 0$$ for all $$x$$.

Therefore we need: $$3x - 5 < 0 \implies x < \frac{5}{3}$$.

Condition (ii): $$g(x) \geq -1$$

$$\frac{2x^2 - 3x + 4 + (3x - 5)}{3x - 5} \geq 0 \implies \frac{2x^2 - 1}{3x - 5} \geq 0$$

$$2x^2 - 1 = 0 \implies x = \pm\frac{1}{\sqrt{2}}$$.

Critical points: $$-\frac{1}{\sqrt{2}},\; \frac{1}{\sqrt{2}},\; \frac{5}{3}$$.

Sign analysis of $$\frac{2x^2 - 1}{3x - 5}$$:

$$x < -\frac{1}{\sqrt{2}}$$: $$\frac{(+)}{(-)} < 0$$ ✘

$$-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$$: $$\frac{(-)}{(-)} > 0$$ ✔

$$\frac{1}{\sqrt{2}} < x < \frac{5}{3}$$: $$\frac{(+)}{(-)} < 0$$ ✘

$$x > \frac{5}{3}$$: $$\frac{(+)}{(+)} > 0$$ ✔

Including zeros: $$\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] \cup \left(\frac{5}{3}, \infty\right)$$.

Combining conditions (i) and (ii): the $$\cos^{-1}$$ domain is $$\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$$.

Intersect with the log domain.

Log domain: $$\left(-\frac{1}{2}, -\frac{1}{3}\right) \cup \left(\frac{1}{2}, \infty\right)$$

$$\cos^{-1}$$ domain: $$\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$$

Since $$-\frac{1}{\sqrt{2}} \approx -0.707 < -\frac{1}{2}$$ and $$\frac{1}{\sqrt{2}} \approx 0.707$$:

$$\left(-\frac{1}{2}, -\frac{1}{3}\right) \cap \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] = \left(-\frac{1}{2}, -\frac{1}{3}\right)$$

$$\left(\frac{1}{2}, \infty\right) \cap \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] = \left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right)$$

At $$x = \frac{1}{\sqrt{2}}$$: $$g\left(\frac{1}{\sqrt{2}}\right) = \frac{1 - \frac{3}{\sqrt{2}} + 4}{\frac{3}{\sqrt{2}} - 5} = \frac{5 - \frac{3}{\sqrt{2}}}{\frac{3}{\sqrt{2}} - 5} = -1$$, and $$\cos^{-1}(-1) = \pi$$ is defined. However, the problem specifies open intervals $$(\alpha, \beta) \cup (\gamma, \delta)$$.

So the domain is: $$\left(-\frac{1}{2}, -\frac{1}{3}\right) \cup \left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right)$$

Here $$\alpha = -\frac{1}{2},\; \beta = -\frac{1}{3},\; \gamma = \frac{1}{2},\; \delta = \frac{1}{\sqrt{2}}$$.

Compute the answer.

$$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = \frac{1}{4} + \frac{1}{9} + \frac{1}{4} + \frac{1}{2} = \frac{9 + 4 + 9 + 18}{36} = \frac{40}{36} = \frac{10}{9}$$

$$18\left(\alpha^2 + \beta^2 + \gamma^2 + \delta^2\right) = 18 \times \frac{10}{9} = 20$$

The answer is 20.