If domain of the function loge(6x2+5x+12x−1)+cos−1(2x2−3x+43x−5) is (α,β)∪(γ,δ), then 18(α2+β2+γ2+δ2)2 is equal to ___________

We need to find the domain of the function loge(6x2+5x+12x−1)+cos−1(2x2−3x+43x−5) ⇒6x2+5x+12x−1>0.....(1) and ⇒−1≤2x2−3x+43x−5≤1.....(2)Now from (1) , ⇒6x2+5x+12x−1>0⇒(3x+1)(2x+1)(2x−1)>0Now we get the common region here as ⇒x∈(−12,−13)∪(12,∞)....(a)Now from (2) we get that ⇒−1≤2x2−3x+43x−5≤1 ⇒2x2−3x+43x−5≥−1.....(3)And 2x2−3x+43x−5≤1.....(4)From (3) we get ⇒2x2−3x+43x−5+1≥0 ⇒x∈[−12,12]∪(53,∞)......(b)From (4) we get ⇒2x2−3x+43x−5−1≤0⇒2x2−6x+93x−5≥0 As D=(−6)2−4(2)(9) 0We observe that 2x2−6x+9>0 ∀x∈R ⇒13x−5≤0(∵ 2x2−6x+9)>0 ∀x∈R ⇒x∈(−∞,53)....(c)⇒ Intersection of (a),(b) and (c) gives us ⇒x∈(−12,−13)∪(12,12)On comparing this with ⇒(α,β)∪(γ,δ), ⇒α2+β2+γ2+δ2=109 ⇒18(α2+β2+γ2+δ2)=20Hence this is the required answer.

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If domain of the function $\log_{e} (\frac{6 x^{2} + 5 x + 1}{2 x - 1}) + \cos^{- 1} (\frac{2 x^{2} - 3 x + 4}{3 x - 5})$ is $(α, β) \cup (γ, δ),$ then $\frac{18 (α^{2} + β^{2} + γ^{2} + δ^{2})}{2}$ is equal to ___________

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Detailed Solution

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
$\Rightarrow x \in (- \frac{1}{2}, - \frac{1}{3}) \cup (\frac{1}{2}, \infty) .... (a)$
Now 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
$\Rightarrow \frac{2 x^{2} - 3 x + 4}{3 x - 5} - 1 \leq 0 \Rightarrow \frac{2 x^{2} - 6 x + 9}{3 x - 5} \geq 0$
$As D = {(- 6)}^{2} - 4 (2) (9) < 0 and 2 > 0$
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 $\Rightarrow x \in (- \frac{1}{2}, - \frac{1}{3}) \cup (\frac{1}{2}, \frac{1}{\sqrt{2}})$
On comparing this with $\Rightarrow (α, β) \cup (γ, δ),$
$\Rightarrow α^{2} + β^{2} + γ^{2} + δ^{2} = \frac{10}{9}$
$\Rightarrow 18 (α^{2} + β^{2} + γ^{2} + δ^{2}) = 20$
Hence this is the required answer.