If f(x) = ^ - 1 ( e x^2 ) (e x^2 ) + ^ - 1 ( 3 + 2 x 1 - 6 x ), t… - Vidyadip

MCQ

If $f(x) = {\tan ^{ - 1}}\left\{ {{{\log \left( {{e \over {{x^2}}}} \right)} \over {\log (e{x^2})}}} \right\} + {\tan ^{ - 1}}\left( {{{3 + 2\log x} \over {1 - 6\log x}}} \right)$, then ${{{d^n}y} \over {d{x^n}}}$ is $(n \ge 1)$

A
${\tan ^{ - 1}}\{ {(\log x)^n}\} $
✓
$0$
C
${1 \over 2}$
D
None of these

✓

Answer

Correct option: B.

$0$

b
(b) We have $y = {\tan ^{ - 1}}\left( {\frac{{\log e - \log {x^2}}}{{\log e + \log {x^2}}}} \right) + {\tan ^{ - 1}}\left( {\frac{{3 + 2\log x}}{{1 - 6\log x}}} \right)$

$ = {\tan ^{ - 1}}\left( {\frac{{1 - 2\log x}}{{1 + 2\log x}}} \right) + {\tan ^{ - 1}}\left( {\frac{{3 + 2\log x}}{{1 - 6\log x}}} \right)$

$ = {\tan ^{ - 1}}1 - {\tan ^{ - 1}}(2\log x) + {\tan ^{ - 1}}3 + {\tan ^{ - 1}}(2\log x)$

$ \Rightarrow y = {\tan ^{ - 1}}1 + {\tan ^{ - 1}}3 $

$\Rightarrow \frac{{dy}}{{dx}} = 0 \Rightarrow \frac{{{d^n}y}}{{d{x^n}}} = 0.$

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