MCQ
${d \over {dx}}{\tan ^{ - 1}}\left[ {{{3{a^2}x - {x^3}} \over {a({a^2} - 3{x^2})}}} \right]$ at $x = 0$ is
- A${1 \over a}$
- ✓${3 \over a}$
- C$3a$
- D$3$
✓
Answer
Correct option: B.
${3 \over a}$
b
(b) $\frac{d}{{dx}}{\tan ^{ - 1}}\left[ {\frac{{3{a^2}x - {x^3}}}{{a({a^2} - 3{x^2})}}} \right]$
(b) $\frac{d}{{dx}}{\tan ^{ - 1}}\left[ {\frac{{3{a^2}x - {x^3}}}{{a({a^2} - 3{x^2})}}} \right]$
Put $x = a\tan \theta $
$ \Rightarrow \frac{d}{{dx}}{\tan ^{ - 1}}\left[ {\frac{{3{a^3}\tan \theta - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3{a^3}{{\tan }^2}\theta }}} \right]$
$ = \frac{d}{{dx}}{\tan ^{ - 1}}(\tan 3\theta ) = \frac{d}{{dx}}(3\theta ) = \frac{{3a}}{{{x^2} + {a^2}}}$
If $x = 0,$ then $\frac{d}{{dx}}{\tan ^{ - 1}}\left[ {\frac{{3{a^2}x - {x^3}}}{{a({a^2} - 3{x^2})}}} \right] = \frac{3}{a}$.
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