If domain of the function log _eleft(frac{6 x | vedclass

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Question

$\frac{6 x^2+5 x+1}{2 x-1} > 0$

$\frac{(3 x+1)(2 x+1)}{2 x-1} > 0$

$-\frac{1}{2}-\frac{1}{3} \quad \frac{1}{2}$

$x \in\left(\frac{-1}{2

Vedclass · Accepted Answer

$20$

Vedclass · Answer

$\frac{6 x^2+5 x+1}{2 x-1} > 0$

$\frac{(3 x+1)(2 x+1)}{2 x-1} > 0$

$-\frac{1}{2}-\frac{1}{3} \quad \frac{1}{2}$

$x \in\left(\frac{-1}{2}, \frac{-1}{3}ight) \cup\left(\frac{1}{2}, \inftyight).....(A)$

$x \in\left[\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}ight] \cup\left(\frac{5}{3}, \inftyight)..........(B)$

$x < \frac{5}{3}........(C)$

$A \cap B \cap C \equiv\left(\frac{-1}{2}, \frac{-1}{3}ight) \cup\left(\frac{1}{2}, \frac{1}{\sqrt{2}}ight]$

$	ext { So } 18\left(\alpha^2+\beta^2+\gamma^2+\delta^2ight)=18\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{4}+\frac{1}{2}ight)$

$=18+2=20$

If 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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