Q101 / 180 · COMEDK 2025 (10 May Shift 2) · Mathematics
Domain of the function $f(x)=\sqrt{\sin ^{-1}(2 x)+\dfrac{\pi}{6}}$ for real valued of $x$ is
- A$\left[-\dfrac{1}{2}, \quad \dfrac{1}{2}\right]$
- B$\left[-\dfrac{1}{4}, \quad \dfrac{1}{4}\right]$
- Correct answerC$\left[\begin{array}{ll}-\dfrac{1}{4}, & \dfrac{1}{2}\end{array}\right]$
- D$\left(-\dfrac{1}{2}, \quad \dfrac{1}{9}\right)$
Solution
For the function $f(x) = \sqrt{\sin^{-1}(2x) + \dfrac{\pi}{6}}$ to be defined, the expression inside the square root must be non-negative:
$\sin^{-1}(2x) + \dfrac{\pi}{6} \ge 0$
$\sin^{-1}(2x) \ge -\dfrac{\pi}{6}$
Taking the sine of both sides, since $\sin(x)$ is an increasing function on its domain $[-\pi/2, \pi/2]$:
$2x \ge \sin\left(-\dfrac{\pi}{6}\right)$
$2x \ge -\dfrac{1}{2}$
$x \ge -\dfrac{1}{4}$
Additionally, the domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$. Therefore, for $\sin^{-1}(2x)$ to be defined:
$-1 \le 2x \le 1$
$-\dfrac{1}{2} \le x \le \dfrac{1}{2}$
Combining the two conditions $x \ge -\dfrac{1}{4}$ and $-\dfrac{1}{2} \le x \le \dfrac{1}{2}$, we get:
$-\dfrac{1}{4} \le x \le \dfrac{1}{2}$
Answer: $\left[\begin{array}{ll}-\dfrac{1}{4}, & \dfrac{1}{2}\end{array}\right]$
$\sin^{-1}(2x) + \dfrac{\pi}{6} \ge 0$
$\sin^{-1}(2x) \ge -\dfrac{\pi}{6}$
Taking the sine of both sides, since $\sin(x)$ is an increasing function on its domain $[-\pi/2, \pi/2]$:
$2x \ge \sin\left(-\dfrac{\pi}{6}\right)$
$2x \ge -\dfrac{1}{2}$
$x \ge -\dfrac{1}{4}$
Additionally, the domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$. Therefore, for $\sin^{-1}(2x)$ to be defined:
$-1 \le 2x \le 1$
$-\dfrac{1}{2} \le x \le \dfrac{1}{2}$
Combining the two conditions $x \ge -\dfrac{1}{4}$ and $-\dfrac{1}{2} \le x \le \dfrac{1}{2}$, we get:
$-\dfrac{1}{4} \le x \le \dfrac{1}{2}$
Answer: $\left[\begin{array}{ll}-\dfrac{1}{4}, & \dfrac{1}{2}\end{array}\right]$
Hence, the correct answer is option: 'C'.