Question

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

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

Answer 1

The correct option is D $|a|\le \frac{1}{2\sqrt{2}}$

We know
$-\frac{\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}$
So, the equation holds true if
$-\frac{\pi }{2}\le 2{\sin }^{-1}2a\le \frac{\pi }{2}$
$\Rightarrow -\frac{\pi }{4}\le {\sin }^{-1}2a\le \frac{\pi }{4}$
$\Rightarrow \sin (-\frac{\pi }{4})\le 2a\le \sin \left(\frac{\pi }{4}\right)$
$\Rightarrow -\frac{1}{2\sqrt{2}}\le a\le \frac{1}{2\sqrt{2}}$
$\Rightarrow |a|\le \frac{1}{2\sqrt{2}}$