If the domain of the function

$f(x)=\log _{e}\left(4 x^{2}+11 x+6\right)+\sin ^{-1}(4 x+3)+\cos ^{-1}\left(\frac{10 x+6}{3}\right)$ is $(\alpha, \beta]$, then

$36|\alpha+\beta|$ is equal to :

<p>To find the domain of the function, we need to consider the individual functions and their respective domains. We have:</p> <p><ol> <li>$f_1(x) = \ln(4x^2 + 11x + 6)$</li> <li>$f_2(x) = \sin^{-1}(4x + 3)$</li> <li>$f_3(x) = \cos^{-1}\left(\frac{10x + 6}{3}\right)$</li> </ol></p> <p><ol> <li>For $f_1(x)$:</li> </ol></p> <p>$4x^2 + 11x + 6 &gt; 0$</p> <p>Factoring the quadratic expression:</p> <p>$(4x + 3)(x + 2) &gt; 0$</p> <p>From this inequality, we have:</p> <p>$x \in (-\infty, -2) \cup \left(-\frac{3}{4}, \infty\right)$</p> <ol> <li>For $f_2(x)$:</li> </ol> <p>$-1 \le 4x + 3 \le 1$</p> <p>From these inequalities, we get:</p> <p>$x \in \left[-1, -\frac{1}{2}\right]$</p> <ol> <li>For $f_3(x)$:</li> </ol> <p>$-1 \le \frac{10x + 6}{3} \le 1$</p> <p>From these inequalities, we get:</p> <p>$x \in \left[-\frac{9}{10}, -\frac{3}{10}\right]$</p> <p>Now, we need to find the intersection of the domains of the three functions:</p> <p>$$ \left(-\infty, -2\right) \cup \left(-\frac{3}{4}, \infty\right) \cap \left[-1, -\frac{1}{2}\right] \cap \left[-\frac{9}{10}, -\frac{3}{10}\right] $$</p> <p>To find the intersection, let&#39;s analyze the intervals:</p> <ul> <li>The interval $(-\infty, -2) \cup \left(-\frac{3}{4}, \infty\right)$ contains all $x$ values less than $-2$ and greater than $-\frac{3}{4}$.</li> <li>The interval $\left[-1, -\frac{1}{2}\right]$ contains all $x$ values between $-1$ and $-\frac{1}{2}$.</li> <li>The interval $\left[-\frac{9}{10}, -\frac{3}{10}\right]$ contains all $x$ values between $-\frac{9}{10}$ and $-\frac{3}{10}$.</li> </ul> <p>Looking at the intervals, we can see that the intersection is:</p> <p>$x \in \left(-\frac{3}{4}, -\frac{1}{2}\right]$</p> <p>Thus, the domain of the function is $(\alpha, \beta] = \left(-\frac{3}{4}, -\frac{1}{2}\right]$. Now, we need to find the value of $36|\alpha + \beta|$:</p> <p>$36\left|-\frac{3}{4} - \frac{1}{2}\right| = 36\left|-\frac{5}{4}\right| =45$</p>