If domain of the function _ e ( 6 x ^ 2 + 5x + 1 2x - 1 ) + ^ - 1 (...

We need to find the domain of the function $\log_{e}\left(\dfrac{6{x}^{2} + 5x + 1}{2x - 1}\right) + \cos^{- 1}\left(\dfrac{2{x}^{2} - 3x + 4}{3x - 5}\right)$.

$⇒ \dfrac{6{x}^{2} + 5x + 1}{2x - 1} > 0 . . . . . ( 1 )$ and

$⇒ - 1 \le \dfrac{2{x}^{2} - 3x + 4}{3x - 5} \le 1 . . . . . . ( 2 )$
Now From \left(1\right)$,<br />$⇒ \dfrac{6{x}^{2} + 5x + 1}{2x - 1} > 0$</p><p>$⇒ \dfrac{( 3x + 1 ) ( 2x + 1 )}{( 2x - 1 )} > 0$</p><p>Now we get the common region here as<br />$⇒ x \in \left(- \dfrac{1}{2}, - \dfrac{1}{3}\right) \cup \left(\dfrac{1}{2}, \infty\right) . . . . . . . ( a )$</p><p>From \left(2\right)$ we get that
$⇒ - 1 \le \dfrac{2{x}^{2} - 3x + 4}{3x - 5} \le 1$

$⇒ \dfrac{2{x}^{2} - 3x + 4}{3x - 5} \ge - 1 . . . . . . \left(3\right)$

and $\dfrac{2{x}^{2} - 3x + 4}{3x - 5} \le 1 . . . . . \left(4\right)$

From \left(3\right)$we get<br />$⇒ \dfrac{2{x}^{2} - 3x + 4}{3x - 5} + 1 \ge 0$</p><p>$⇒ x \in \left[\dfrac{- 1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right] \cup \left(\dfrac{5}{3}, \infty\right) . . . . . \left(b\right)$</p><p>From \left(4\right)$ we get
$⇒ \dfrac{2{x}^{2} - 3x + 4}{3x - 5} - 1 \le 0$

$⇒ \dfrac{2{x}^{2} - 6x + 9}{3x - 5} \le 0$

As $D = {\left(- 6\right)}^{2} - 4\left(2\right)\left(9\right) 0$, we observe that $2{x}^{2} - 6x + 9 > 0 \forall x \in \mathbb{R}$
$⇒ \dfrac{1}{3x - 5} \le 0\left(∵ 2{x}^{2} - 6x + 9 > 0 \forall x \in R\right)$

$⇒ x \in \left(- \infty , \dfrac{5}{3}\right) . . . . . . \left(c\right)$

$⇒$ Intersection of $( a ) , ( b )$ and $( c )$ gives us
$x \in \left(- \dfrac{1}{2}, - \dfrac{1}{3}\right) \cup \left(\dfrac{1}{2}, \dfrac{1}{\sqrt{2}}\right)$.

On comparing this with \left(\alpha, \beta\right) \cup \left(\gamma, \delta\right)$,<br />$⇒ {\alpha}^{2} + {\beta}^{2} + {\gamma}^{2} + {\delta}^{2} = \dfrac{10}{9}$<br />$⇒ 18\left({\alpha}^{2} + {\beta}^{2} + {\gamma}^{2} + {\delta}^{2}\right) = 20.

Hence this is the required answer.