MCQ
Function $y = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$ is not differentiable for
  • A
    $|x|\, < 1$
  • $x = 1, - 1$
  • C
    $|x|\, > 1$
  • D
    None of these

Answer

Correct option: B.
$x = 1, - 1$
b
(b) $y' = \frac{1}{{\sqrt {1 - {{\left( {\frac{{2x}}{{1 + {x^2}}}} \right)}^2}} }}.\frac{{2(1 + {x^2}) - 4{x^2}}}{{{{(1 + {x^2})}^2}}} = \frac{{2(1 - {x^2})}}{{\sqrt {{{(1 - {x^2})}^2}.(1 + {x^2})} }}$

==> $y' = \left\{ \begin{array}{l}\frac{2}{{1 + {x^2}}}\,\,\,\,\,\,{\rm{for}}\,\,\,\,|x| < 1\\\frac{{ - 2}}{{1 + {x^2}}}\,\,\,\,\,\,{\rm{for}}\,\,\,\,|x| > 1\end{array} \right.$

Hence for $|x| = 1$, the derivative does not exist.

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