Question

The solution set of the equation ${\cos }^{-1}x<{\sin }^{-1}x$ is

• A. $[-1,1]$
• B. $\left[\frac{1}{\sqrt{2}},1\right]$
• C. $[0,1]$
• D. $(\frac{1}{\sqrt{2}},1]$

Answer 1

The correct option is D

$(\frac{1}{\sqrt{2}},1]$

Explanation for the correct option:

Find the solution set of the given inequality.

In the question, an inequality ${\cos }^{-1}x<{\sin }^{-1}x$ is given.

We know that, ${\cos }^{-1}x=\frac{\mathrm{\pi }}{2}-{\sin }^{-1}x$.

So,

$$\begin{gathered} \frac{\mathrm{\pi }}{2}-{\sin }^{-1}x<{\sin }^{-1}x \\ \Rightarrow 2{\sin }^{-1}x>\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\sin }^{-1}x>\frac{\mathrm{\pi }}{4} \end{gathered}$$

Since, ${\sin }^{-1}x\le \frac{\mathrm{\pi }}{2}$.

Therefore,

$\frac{\mathrm{\pi }}{4}<{\sin }^{-1}x\le \frac{\mathrm{\pi }}{2}$

Take $\text{sine}$ of each side.

$\frac{1}{\sqrt{2}}<x\le 1$

Therefore, the solution set of the given equation is $(\frac{1}{\sqrt{2}},1]$.

Hence, option $D$ is the correct answer.