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If $f(x)=\cot^{-1}\big[\large\frac{3x-x^3}{1-3x^2}\big]$ and $g(x)=\cos^{-1}\big[\large\frac{1-x^2}{1+x^2}\big]$ then $\lim\limits_{x\to a}\large\frac{f(x)-f(a)}{g(x)-g(a)}$$\quad0 < a < \large\frac{1}{2}$ is

$\begin{array}{1 1}(a)\;\large\frac{3}{2}&(b)\;\large\frac{3}{2(1+x^2)}\\(c)\;-\large\frac{3}{2}&(d)\;\large\frac{3}{2(1+a^2)}\end{array}$

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Let $x=\tan\theta$
$f(x)=\cot^{-1}\big(\large\frac{3x-x^3}{1-3x^2}\big)$
$\qquad=\large\frac{\pi}{2}$$-\tan^{-1}\big(\large\frac{3x-x^3}{1-3x^2}\big)$
$\qquad=\large\frac{\pi}{2}$$-3\tan^{-1}x$
$g(x)=\cos^{-1}\big(\large\frac{1-x^2}{1+x^2}\big)$
$\qquad=\cos^{-1}\big(\large\frac{1-\tan^2\theta}{1+\tan^2\theta}\big)$
$\qquad=\cos^{-1}(\cos 2\theta)$
$\qquad=2\theta$
$\qquad=2\tan^{-1}x$
$\lim\limits_{x\to a}\large\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(a)}{g'(a)}$
$\Rightarrow \large\frac{-3/1+a^2}{2/1+a^2}=-\frac{3}{2}$
Hence (c) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
 

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