Question

The trigonometric equation $\sin^{-1}x = 2 \sin^{-1}2a$ has a real solution if

• A. $\left|a\right| > \frac{1}{\sqrt{2}}$
• B. $ \frac{1}{2\sqrt{2}} < \left|a\right| > \frac{1}{\sqrt{2}}$
• C. $ \left|a\right| > \frac{1}{2\sqrt{2}}$
• D. $ \left|a\right| \le \frac{1}{2\sqrt{2}}$

Answer 1

Answer: (D)

We know that, $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$
$\Rightarrow \frac{-\pi}{2} \leq 2 \sin ^{-1} 2 a \leq \frac{\pi}{2}$
$\Rightarrow \frac{-\pi}{4} \leq \sin ^{-1} 2 a \leq \frac{\pi}{4}$
$\Rightarrow \sin \left(\frac{-\pi}{4}\right) \leq 2 a \leq \sin \left(\frac{\pi}{4}\right)$
$\Rightarrow \frac{-1}{\sqrt{2}} \leq 2 a \leq \frac{1}{\sqrt{2}}$
$\Rightarrow \frac{-1}{2 \sqrt{2}} \leq a \leq \frac{1}{2 \sqrt{2}}$
$\Rightarrow |a| \leq \frac{1}{2 \sqrt{2}}$