Question:
If $\cos ^{-1} x>\sin ^{-1} x$, then
(a) $\frac{1}{\sqrt{2}} (b) $0 \leq x<\frac{1}{\sqrt{2}}$ (c) $-1 \leq x<\frac{1}{\sqrt{2}}$ (d) $x>0$
Solution:
$\cos ^{-1} x>\sin ^{-1} x$
$\Rightarrow \cos ^{-1} x>\frac{\pi}{2}-\cos ^{-1} x$
$\Rightarrow 2 \cos ^{-1} x>\frac{\pi}{2}$
$\Rightarrow \cos ^{-1} x>\frac{\pi}{4}$
$\Rightarrow x>\cos \frac{\pi}{4}$
$\Rightarrow x>\frac{1}{\sqrt{2}}$
We know that the maximum value of cosine fuction is 1 .
$\therefore \frac{1}{\sqrt{2}} Hence, the correct answer is option(a).
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.