JEE Main 2023 — Inverse Trigonometric Functions Question with Solution

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 $36|\alpha +\beta |$:

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