QJEEM23Ma5767: JEE Main 2023 Mathematics - Class 12 JEEM23Ma5767 Solution: If the domain of the function is , then is equal to :

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

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

1. For $f_1(x)$ :

4x^2 + 11x + 6 > 0

Factoring the quadratic expression:

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

From this inequality, we have:

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

1. For $f_2(x)$ :

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

From these inequalities, we get:

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

1. 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 $36|\alpha + \beta|$ :

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