A) \[\frac{1}{a}\]
B) \[\frac{3}{a}\]
C) \[3a\]
D) 3
Correct Answer: B
Solution :
\[\frac{d}{dx}{{\tan }^{-1}}\left[ \frac{3{{a}^{2}}x-{{x}^{3}}}{a({{a}^{2}}-3{{x}^{2}})} \right]\] 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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