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 $|a|\le \frac{1}{2\sqrt{2}}$

$-\pi /2\le {\sin }^{-1}x\le \pi /2\Rightarrow -\pi /2\le 2{\sin }^{-1}2a\le \pi /2$
$\Rightarrow -\pi /4\le {\sin }^{-1}2a\le \pi /4\Rightarrow -\frac{1}{\sqrt{2}}\le 2a\le \frac{1}{\sqrt{2}}$
$\Rightarrow -\frac{1}{2\sqrt{2}}\le a\le \frac{1}{2\sqrt{2}}\Rightarrow |a|\le \frac{1}{2\sqrt{2}}$