Question

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

$\lim_{x \rightarrow a}\frac{f(x) - f(a)}{g(x) - g(a)},0 < a < \frac{1}{2}$ is

• A. $\frac{3}{2(1 + a^{2})}$
• B. $\frac{3}{2(1 + x^{2})}$
• C. $\frac{3}{2}$
• D. $- \frac{3}{2}$

Answer 1

Answer: (D)

$- \frac{3}{2}$

Answer 2

$f(x) = \cot^{- 1}\left\{ \frac{3x - x^{3}}{1 - 3x^{2}} \right\}$ and $g(x) = \cos^{- 1}\left\{ \frac{1 - x^{2}}{1 + x^{2}} \right\}$

Put $x = \tan\theta$ in both equation

$f(\theta) = \cot^{- 1}\left\{ \frac{3\tan\theta - \tan^{3}\theta}{1 - 3\tan^{2}\theta} \right\} = \cot^{- 1}\left\{ \tan 3\theta \right\}$

$f(\theta) = \cot^{- 1}{\cot\left( \frac{\pi}{2} - 3\theta \right)} = \frac{\pi}{2} - 3\theta \Rightarrow f^{'}(\theta) = - 3$ ..….(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$\lim_{x \rightarrow a}\left( \frac{f(x) - f(a)}{g(x) - g(a)} \right) = \lim_{x \rightarrow a}\left( \frac{f(x) - f(a)}{x - a} \right)\frac{1}{\lim_{x \rightarrow a}\left( \frac{g(x) - g(a)}{x - a} \right)} = f^{'}(x).\frac{1}{g^{'}(x)} = - 3 \times \frac{1}{2} = - \frac{3}{2}$.