If domain of the function log e 6 x 2 + 5 x + 1 2 x - 1 + cos - 1 2 x 2 - 3 x + 4 3 x - 5 is α , β ∪ γ , δ , then 18 α 2 + β 2 + γ 2 + δ 2 is equal to

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We need to find the domain of the function $\log_{e} (\frac{6 x^{2} + 5 x + 1}{2 x - 1}) + \cos^{- 1} (\frac{2 x^{2} - 3 x + 4}{3 x - 5})$ .
$\Rightarrow \frac{6 x^{2} + 5 x + 1}{2 x - 1} > 0 . . . . . (1)$ and
$\Rightarrow - 1 \leq \frac{2 x^{2} - 3 x + 4}{3 x - 5} \leq 1 . . . . . . (2)$
Now From $(1)$ ,
$\Rightarrow \frac{6 x^{2} + 5 x + 1}{2 x - 1} > 0$
$\Rightarrow \frac{(3 x + 1) (2 x + 1)}{(2 x - 1)} > 0$
Now we get the common region here as
$\Rightarrow x \in (- \frac{1}{2}, - \frac{1}{3}) \cup (\frac{1}{2}, \infty) . . . . . . . (a)$
From $(2)$ we get that
$\Rightarrow - 1 \leq \frac{2 x^{2} - 3 x + 4}{3 x - 5} \leq 1$
$\Rightarrow \frac{2 x^{2} - 3 x + 4}{3 x - 5} \geq - 1 . . . . . . (3)$
and $\frac{2 x^{2} - 3 x + 4}{3 x - 5} \leq 1 . . . . . (4)$
From $(3)$ we get
$\Rightarrow \frac{2 x^{2} - 3 x + 4}{3 x - 5} + 1 \geq 0$
$\Rightarrow x \in [\frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}}] \cup (\frac{5}{3}, \infty) . . . . . (b)$
From $(4)$ we get
$\Rightarrow \frac{2 x^{2} - 3 x + 4}{3 x - 5} - 1 \leq 0$
$\Rightarrow \frac{2 x^{2} - 6 x + 9}{3 x - 5} \leq 0$
As $D = {(- 6)}^{2} - 4 (2) (9) < 0 and 2 > 0$ , we observe that $2 x^{2} - 6 x + 9 > 0 \forall x \in ℝ$
$\Rightarrow \frac{1}{3 x - 5} \leq 0 (∵ 2 x^{2} - 6 x + 9 > 0 \forall x \in R)$
$\Rightarrow x \in (- \infty, \frac{5}{3}) . . . . . . (c)$
$\Rightarrow$ Intersection of $(a), (b)$ and $(c)$ gives us
$x \in (- \frac{1}{2}, - \frac{1}{3}) \cup (\frac{1}{2}, \frac{1}{\sqrt{2}})$ .
On comparing this with $(α, β) \cup (γ, δ)$ ,
$\Rightarrow α^{2} + β^{2} + γ^{2} + δ^{2} = \frac{10}{9}$
$\Rightarrow 18 (α^{2} + β^{2} + γ^{2} + δ^{2}) = 20$ .
Hence this is the required answer.

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