MCQ
વિધેય $y = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$ એ . . . .બિંદુએ વિકલનીય નથી.
- A$|x|\, < 1$
- ✓$x = 1, - 1$
- C$|x|\, > 1$
- Dએકપણ નહી.
==> $y' = \left\{ \begin{array}{l}\frac{2}{{1 + {x^2}}}\,\,\,\,\,\,{\rm{for}}\,\,\,\,|x| < 1\\\frac{{ - 2}}{{1 + {x^2}}}\,\,\,\,\,\,{\rm{for}}\,\,\,\,|x| > 1\end{array} \right.$
Hence for $|x| = 1$, the derivative does not exist.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.