A) \[\frac{3}{2(1+{{a}^{2}})}\]
B) \[\frac{3}{2(1+{{x}^{2}})}\]
C) \[\frac{3}{2}\]
D) \[-\frac{3}{2}\]
Correct Answer: D
Solution :
\[f(x)={{\cot }^{-1}}\left\{ \frac{3x-{{x}^{3}}}{1-3{{x}^{2}}} \right\}\] and \[g(x)={{\cos }^{-1}}\left\{ \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right\}\] Put \[x=\tan \theta \] in both equations \[f(\theta )={{\cot }^{-1}}\left\{ \frac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } \right\}\]\[={{\cot }^{-1}}\left\{ \tan 3\theta \right\}\] \[f(\theta )={{\cot }^{-1}}\cot \left( \frac{\pi }{2}-3\theta \right)=\frac{\pi }{2}-3\theta \Rightarrow {f}'(\theta )=-3\] .?.(i) and \[g(\theta )={{\cos }^{-1}}\left\{ \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right\}\]\[={{\cos }^{-1}}(\cos 2\theta )=2\theta \] \[\Rightarrow {g}'(\theta )=2\] ?..(ii) Now \[\underset{x\to a}{\mathop{\lim }}\,\left( \frac{f(x)-f(a)}{g(x)-g(a)} \right)=\underset{x\to a}{\mathop{\lim }}\,\left( \frac{f(x)-f(a)}{x-a} \right)\frac{1}{\underset{x\to a}{\mathop{\lim }}\,\left( \frac{g(x)-g(a)}{x-a} \right)}\] \[={f}'(x).\frac{1}{{g}'(x)}=-3\times \frac{1}{2}=-\frac{3}{2}\].
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