Question

Solution set of $\left[{\sin }^{-1}x\right]>\left[{\cos }^{-1}x\right]$, where [ , ] denotes greatest integer function

• A. $\left[\sin 1,1\right]$
• B. $[\frac{1}{\sqrt{2}},1]$
• C. $(\cos 1,\sin 1)$
• D. $Noneofthese$

Answer 1

The correct option is B $[\frac{1}{\sqrt{2}},1]$

According to the question:

Given that:

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

here

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

Therefore

function is defined when

$-1\le x\le 1eq\left(1\right)$

Using the formula

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

Implies that

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

Put into

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

Hence

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

Therefore,

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

Implies that

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

Implies that

$x>\frac{1}{2}eq\left(2\right)$

From eq 1 and 2 we get

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

Implies that

$x\in \left(\frac{1}{\sqrt{2}},1\right)$