Choose the correct answer from the given four options. If $\cos^{-1} x > \sin^{-1} x$, then: 1.…
Choose the correct answer from the given four options.
If $\cos^{-1} x > \sin^{-1} x$, then:
1. $\frac{1}{\sqrt{2}} < x \leq 1$
2. $0 \leq x < \frac{1}{\sqrt{2}}$
3. $-1 \leq x < \frac{1}{\sqrt{2}}$
4. $x > 0$
Solution
Solution:
We have, $\cos^{-1}x > \sin^{-1}x$
$\Rightarrow \frac{\pi}{2} - \sin^{-1}x > \sin^{-1}x$
$\Rightarrow \frac{\pi}{2} > 2\sin^{-1}x$
$\Rightarrow \sin^{-1}x < \frac{\pi}{4}$ ....(i)
But $-\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2}$ ....(ii)
From (i) and (ii), $-\frac{\pi}{2} \leq \sin^{-1}x < \frac{\pi}{4}$
$\Rightarrow \sin\left(-\frac{\pi}{2}\right) \leq x < \sin\frac{\pi}{4}$
$\Rightarrow -1 \leq x < \frac{1}{\sqrt{2}}$