Question

If ${\cos }^{-1}x>{\sin }^{-1}x$, then

• A. $\frac{1}{\sqrt{2}}<x⩽1$
• B. $0<x<\frac{1}{\sqrt{2}}$
• C. $-1\le x<\frac{1}{\sqrt{2}}$
• D. $x>0$

Answer 1

The correct option is C

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

Explanation for the correct options:

Step 1. Find the intervals in which $x$ lies:

We have given ${\cos }^{-1}x>{\sin }^{-1}x$,

and we know that,

$\begin{array}{rcl}{\sin }^{-1}x+{\cos }^{-1}x& =& \frac{\mathrm{\pi }}{2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{\cos }^{-1}x& =& \frac{\mathrm{\pi }}{2}-{\sin }^{-1}x\end{array}$

But $\begin{array}{rcl}\frac{\mathrm{\pi }}{2}-{\sin }^{-1}x& >& {\sin }^{-1}x\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{\mathrm{\pi }}{2}& >& 2{\sin }^{-1}x\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{\mathrm{\pi }}{4}& >& {\sin }^{-1}x...........\left(1\right)\end{array}$

Also $-\frac{\mathrm{\pi }}{2}\le {\sin }^{-1}x\le \frac{\mathrm{\pi }}{2}........\left(2\right)$

Step 2. From equation $\left(1\right)\&\left(2\right)$, we get

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

taking$\text{'}\sin \text{'}$ all sides,

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

$\mathbf{\therefore }\mathbf{-}\mathbf{1}\mathbf{\le }\mathit{x}\mathbf{<}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}$

Hence, option $\left(C\right)$ is the correct.