If the domain of the function f (x) = _ e (4 x ^ 2 + 11x + 6) + ^ -...

Given that $f\left(x\right) = \log \left(4{x}^{2} + 11x + 6\right) + \sin^{- 1}\left(4x + 3\right) + \cos^{- 1}\left(\dfrac{10x + 6}{3}\right)$

We know that the domain of $A\left(x\right) + B\left(x\right) + C\left(x\right)$ is $D\left(A\right) \cap D\left(B\right) \cap D\left(C\right)$.

Let us find the domain of $\log \left(4{x}^{2} + 11x + 6\right)$.

We know that domain of $\log \left(f\left(x\right)\right)$ is $f\left(x\right) > 0$.

$⇒ 4{x}^{2} + 11x + 6 > 0$

$⇒ 4{x}^{2} + 8x + 3x + 6 > 0$

$⇒ \left(4x + 3\right)\left(x + 2\right) > 0$

$⇒ x \in \left(- \infty , - 2\right) \cup \left(\dfrac{- 3}{4}, \infty\right)$

Domain of $\log \left(4{x}^{2} + 11x + 9\right)$ is $x \in \left(- \infty , - 2\right) \cup \left(\dfrac{- 3}{4}, \infty\right)$.

Let us find the domain of $\sin^{- 1}\left(4x + 3\right)$

$⇒ - 1 \le 4x + 3 \le 1$

$⇒ - 1 \le x \le \dfrac{- 1}{2}$

$⇒ x \in \left[- 1, \dfrac{- 1}{2}\right]$

Let us find the domain of $\cos^{- 1}\left(\dfrac{10x + 6}{3}\right)$.

$⇒ - 1 \le \dfrac{10x + 6}{3} \le 1$

$⇒ - 3 \le 10x + 6 \le 3$

$⇒ \dfrac{- 9}{10} \le x \le \dfrac{- 3}{10}$.

Now the common domain is $\dfrac{- 3}{4} \le x \le \dfrac{- 1}{2}$.

$⇒ x \in \left[\dfrac{- 3}{4}, \dfrac{- 1}{2}\right]$

But given that the domain of $f\left(x\right)$ is $x \in \left[\alpha, \beta\right]$.

$⇒ \alpha = \dfrac{- 3}{4}, \beta = \dfrac{- 1}{2}$.

$⇒ \left|36\left(\alpha + \beta\right)\right| = \left|36\left(\dfrac{- 3}{4} + \left(\dfrac{- 1}{2}\right)\right)\right|$

$⇒ \left|36\left(\alpha + \beta\right)\right| = \left|36 \times \dfrac{- 5}{4}\right|$

$⇒ \left|36\left(\alpha + \beta\right)\right| = 45$

Therefore, the required answer is $45$.