If the domain of the function log _eleft(frac | vedclass

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(A) For the function to be defined,both parts must be valid.$1$. For $\log_e\left(\frac{6x^2+5x+1}{2x-1}\right)$,we need $\frac{6x^2+5x+1}{2x-1} > 0$.

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$20$

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(A) For the function to be defined,both parts must be valid.$1$. For $\log_e\left(\frac{6x^2+5x+1}{2x-1}ight)$,we need $\frac{6x^2+5x+1}{2x-1} > 0$.$\frac{(3x+1)(2x+1)}{2x-1} > 0$.Using the wavy curve method,the critical points are $x = -1/2, -1/3, 1/2$.The inequality holds for $x \in (-1/2, -1/3) \cup (1/2, \infty) \dots (A)$.$2$. For $\cos^{-1}\left(\frac{2x^2-3x+4}{3x-5}ight)$,we need $-1 \le \frac{2x^2-3x+4}{3x-5} \le 1$ and $3x-5 
eq 0$.Solving $\frac{2x^2-3x+4}{3x-5} \le 1 \implies \frac{2x^2-6x+9}{3x-5} \le 0$. Since $2x^2-6x+9$ has $D Solving $\frac{2x^2-3x+4}{3x-5} \ge -1 \implies \frac{2x^2-1}{3x-5} \ge 0$.Critical points are $x = -1/\sqrt{2}, 1/\sqrt{2}, 5/3$.The inequality holds for $x \in [-1/\sqrt{2}, 1/\sqrt{2}] \cup (5/3, \infty) \dots (C)$.Intersection $A \cap B \cap C = (-1/2, -1/3) \cup (1/2, 1/\sqrt{2}]$.Here $\alpha = -1/2, \beta = -1/3, \gamma = 1/2, \delta = 1/\sqrt{2}$.$18(\alpha^2+\beta^2+\gamma^2+\delta^2) = 18(1/4 + 1/9 + 1/4 + 1/2) = 18(1/2 + 1/9 + 1/2) = 18(1 + 1/9) = 18 + 2 = 20$.

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

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