Functions 1 Question 3
3.
Domain of definition of the function
$f(x)=\sqrt{\sin ^{-1}(2 x)+\frac{\pi}{6}}$ for real valued $x$, is
(2003, 2M)
(a) $-\frac{1}{4}, \frac{1}{2}$
(b) $-\frac{1}{2}, \frac{1}{2}$
(c) $-\frac{1}{2}, \frac{1}{9}$
(d) $-\frac{1}{4}, \frac{1}{4}$
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Answer:
Correct Answer: 3. (a)
Solution:
- Here, $f(x)=\sqrt{\sin ^{-1}(2 x)+\frac{\pi}{6}}$, to find domain we must have,
$ \begin{aligned} & \sin ^{-1}(2 x)+\frac{\pi}{6} \geq 0 \quad \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} \\ & \sin \frac{-\pi}{6} \leq 2 x \leq \sin \frac{\pi}{2} \Rightarrow \frac{-1}{2} \leq 2 x \leq \frac{1}{2} \\ & \frac{-1}{4} \leq x \leq \frac{1}{2} \\ & \because \quad x \in \frac{-1}{4}, \frac{1}{2} \end{aligned} $
