MCQ
સમીકરણ ${{\sin }^{-1}}x=2{{\sin }^{-1}}a$ ને ............. માટે ઉકેલ મળે.
- Aબધી જ વાસ્તવિક સંખ્યાઓ
- B$|a|<\frac{1}{2}$
- ✓$|a|\le \frac{1}{\sqrt{2}}$
- D$|a|<\frac{1}{\sqrt{2}}$
$-\frac{\pi}{2}\leq sin^{-1}x\leq\frac{\pi}{2}$
$\therefore-\frac{\pi}{2}\leq2sin^{-1}a\leq\frac{\pi}{2}$
$\therefore-\frac{\pi}{4}\leq sin^{-1}a\leq\frac{\pi}{4}$
$\therefore -\frac{1}{\sqrt{2}}\leq\frac{1}{\sqrt{2}}\Rightarrow|a|\leq\frac{1}{\sqrt{2}}$
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