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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
${d \over {dx}}\left\{ {{{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)} \right\} = $
  • A
    ${1 \over {1 + {x^2}}}$
  • B
    $ - {1 \over {1 + {x^2}}}$
  • C
    $ - {2 \over {1 + {x^2}}}$
  • ✓
    ${2 \over {1 + {x^2}}}$
✓

Answer

Correct option: D.
${2 \over {1 + {x^2}}}$
d
(d) $\frac{d}{{dx}}\left\{ {{{\cos }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)} \right\}$

Let $\frac{{1 - {x^2}}}{{1 + {x^2}}} = \cos \theta $

==> $1 - {x^2} = (1 + {x^2})\cos \theta $

==> $ - {x^2}(1 + \cos \theta ) = \cos \theta - 1$

==> ${x^2} = \frac{{1 - \cos \theta }}{{1 + \cos \theta }} = \frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2{{\cos }^2}\frac{\theta }{2}}} = {\tan ^2}\frac{\theta }{2}$

or $x = \tan \frac{\theta }{2}$ or $\theta = 2{\tan ^{ - 1}}x$

So, $\frac{d}{{dx}}[\theta ] = \frac{d}{{dx}}[2{\tan ^{ - 1}}x] = \frac{2}{{1 + {x^2}}}$.

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