Q78 / 150 · MHT CET 2024 (09 May Shift 1) · Mathematics

Domain of definition of the real valued function $f(x)=\sqrt{\sin ^{-1}(2 x)+\frac{\pi}{6}}$ is

  • Correct answer
    A
    $\left[-\frac{1}{4}, \frac{1}{2}\right]$
  • B
    $\left[\frac{-3}{2}, \frac{1}{2}\right]$
  • C
    $\left[\frac{-3}{2}, \frac{1}{9}\right]$
  • D
    $\left[-\frac{1}{4}, \frac{3}{4}\right]$

Solution

$\begin{aligned} & \\ & f(x)=\sqrt{\sin ^{-1}(2 x)+\frac{\pi}{6}} \text { to find domain } \\ & \\ & \sin ^{-1}(2 x)+\frac{\pi}{6} \geq 0 \\ & \\ & \text { But } \frac{-\pi}{2} \leq \sin ^{-1} \theta \leq \frac{\pi}{2} \\ & \\ & \frac{-\pi}{6} \leq \sin ^{-1}(2 x) \leq \frac{\pi}{2} \\ & \\ & \Rightarrow-\sin \frac{\pi}{6} \leq 2 x \leq \sin \frac{\pi}{2} \\ & \\ & \Rightarrow \frac{-1}{2} \leq 2 x \leq 1 \\ & \\ & \Rightarrow \frac{-1}{4} \leq x \leq \frac{1}{2} \\ & \therefore \quad \text { Domain of } \sqrt{\sin ^{-1}(2 x)+\frac{\pi}{6}} \text { is }\left[-\frac{1}{4}, \frac{1}{2}\right]\end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions