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