Question

If ${\cos }^{-1}x>{\sin }^{-1}x$, then $x$ belongs to the interval

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

Answer 1

Answer: (D)

$[-1,\frac{1}{\sqrt{2}}]$

Given, ${\cos }^{-1}\left(x\right)>{\sin }^{-1}\left(x\right)$

${\cos }^{-1}\left(x\right)>\frac{\pi }{2}-{\cos }^{-1}\left(x\right)$ ....... $[\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}]$

$2{\cos }^{-1}\left(x\right)>\frac{\pi }{2}$

${\cos }^{-1}\left(x\right)>\frac{\pi }{4}$

$x<\cos \left(\frac{\pi }{4}\right)$ ...($\cos \theta$ is a decreasing function).

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

Now, we know that the domain of $co{s}^{-1}$ is $[-1,1]$

Hence, $xϵ[-1,\frac{1}{\sqrt{2}}]$