If \[f(x)\,={{\cot }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)\] and x\[g(x)={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\], then \[\underset{x\to a}{\mathop{\lim }}\,\frac{f(x)-f(a)}{g(x)\,-g(a)},\] \[0

If \[f(x)\,={{\cot }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \ri

1.	If \[f(x)\,={{\cot }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)\] and x\[g(x)={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\], then \[\underset{x\to a}{\mathop{\lim }}\,\frac{f(x)-f(a)}{g(x)\,-g(a)},\] \[0
A.	\[\frac{3}{2(1+{{a}^{2}})}\]
B.	\[\frac{3}{2(1+{{x}^{2}})}\]
C.	\[\frac{3}{2}\]
D.	\[-\frac{3}{2}\]
Answer» E.

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If \[f(x)\,={{\cot }^{-1}}\left( \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right)\] and x\[g(x)={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\], then \[\underset{x\to a}{\mathop{\lim }}\,\frac{f(x)-f(a)}{g(x)\,-g(a)},\] \[0

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