If f(x) = cot^{-1} left( frac{3x - x^3}{1 - 3 | vedclass

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(D) Given $f(x) = \cot^{-1} \left( \frac{3x - x^3}{1 - 3x^2} ight)$ and $g(x) = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} ight)$.Let $x = 	an 	he

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$-\frac{3}{2}$

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(D) Given $f(x) = \cot^{-1} \left( \frac{3x - x^3}{1 - 3x^2} ight)$ and $g(x) = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} ight)$.Let $x = 	an 	heta$. For $0 $f(x) = \cot^{-1}(	an 3	heta) = \cot^{-1}(\cot(\frac{\pi}{2} - 3	heta)) = \frac{\pi}{2} - 3	heta = \frac{\pi}{2} - 3	an^{-1}x$.Thus,$f'(x) = -\frac{3}{1 + x^2}$.$g(x) = \cos^{-1}(\cos 2	heta) = 2	heta = 2	an^{-1}x$.Thus,$g'(x) = \frac{2}{1 + x^2}$.Using $L$'$H$ôpital's rule or the definition of the derivative:$\lim_{x 	o a} \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(a)}{g'(a)} = \frac{-\frac{3}{1 + a^2}}{\frac{2}{1 + a^2}} = -\frac{3}{2}$.

If $f(x) = \cot^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$ and $g(x) = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$,then for $0 < a < \frac{1}{\sqrt{3}}$,the value of $\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)}$ is

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