If the domain of the function f(x)= _e(4 x^2+11 x+6)+ ^-1(4 x+3)+ ^-1((10 x+6)/(3)) is (, …

Question

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 :

Accepted Answer

To find the domain of the function, we need to consider the individual functions and their respective domains. We have:

$f_1(x) = \ln(4x^2 + 11x + 6)$
$f_2(x) = \sin^{-1}(4x + 3)$
$f_3(x) = \cos^{-1}\left(\frac{10x + 6}{3}\right)$

For $f_1(x)$:

$4x^2 + 11x + 6 > 0$

Factoring the quadratic expression:

$(4x + 3)(x + 2) > 0$

From this inequality, we have:

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

For $f_2(x)$:

$$
-1 \le 4x + 3 \le 1
$$

From these inequalities, we get:

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

For $f_3(x)$:

$$
-1 \le \frac{10x + 6}{3} \le 1
$$

From these inequalities, we get:

$$
x \in \left[-\frac{9}{10}, -\frac{3}{10}\right]
$$

Now, we need to find the intersection of the domains of the three functions:

$$
\left(-\infty, -2\right) \cup \left(-\frac{3}{4}, \infty\right) \cap \left[-1, -\frac{1}{2}\right] \cap \left[-\frac{9}{10}, -\frac{3}{10}\right]
$$

To find the intersection, let's analyze the intervals:

The interval $(-\infty, -2) \cup \left(-\frac{3}{4}, \infty\right)$ contains all $x$ values less than $-2$ and greater than $-\frac{3}{4}$.
The interval $\left[-1, -\frac{1}{2}\right]$ contains all $x$ values between $-1$ and $-\frac{1}{2}$.
The interval $\left[-\frac{9}{10}, -\frac{3}{10}\right]$ contains all $x$ values between $-\frac{9}{10}$ and $-\frac{3}{10}$.

Looking at the intervals, we can see that the intersection is:

$$
x \in \left(-\frac{3}{4}, -\frac{1}{2}\right]
$$

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 $36|\alpha + \beta|$:

$$
36\left|-\frac{3}{4} - \frac{1}{2}\right| = 36\left|-\frac{5}{4}\right| =45
$$

Solutions · 1