MCQ
જો $sin^{-1}x\leq cos^{-1}x$ હોય તો, $x\in ...............$
- A$\left( -1,\frac{1}{\sqrt{2}} \right)$
- ✓$\left[ -1,\frac{1}{\sqrt{2}} \right]$
- C$\left[ \frac{1}{\sqrt{2}},1 \right]$
- Dઆમાંથી એકપણ નહિં.
$\sin^{-1} x \leq \cos^{-1} x$
$\therefore \sin^{-1} x \leq \frac{\pi}{2} - \sin^{-1} x $
$\therefore 2\sin^{-1} x \leq \frac{\pi}{2}$
$\therefore \sin^{-1} x \leq \frac{\pi}{4}$
વળી, $\therefore - \frac{\pi}{2} \leq \sin^{-1}x \leq \frac {\pi}{4}$
$\sin \left( { - \frac{\pi }{2}} \right) \leqslant x \leqslant \sin \frac{\pi }{4}$
$- 1 \leq x \leq \frac {1}{\sqrt{2} }$
અહી $x \in \left[-1, \frac{1}{\sqrt{2}}\right]$
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