Q. If $f(x) = \cot^{- 1}{\lbrack(3x - x^{3})/(1 - 3x^{2})\rbrack}$ and $g(x) = \cos^{- 1}{\lbrack(1 - x^{2})/(1 + x^{2})\rbrack}$, then $\lim_{x \rightarrow a}{\frac{f(x) - f(a)}{g(x) - g(a)}\left( 0 < a < \frac{1}{2} \right)}$ is
A) $- \frac{3}{2}$
B) $\frac{1}{2}$
C) $\frac{3}{2}$
D) None of these
Explanation
(a)
$f(x) = \cot^{- 1}\left( \frac{3x - x^{3}}{1 - 3x^{2}} \right) = \frac{\pi}{2} - 3\tan^{- 1}x$
and $g(x) = \cos^{- 1}\left( \frac{1 - x^{2}}{1 + x^{2}} \right) = 2\tan^{- 1}x$
$\therefore\ \ \ \ \ \lim_{x \rightarrow a}\frac{f(x) - f(a)}{g(x) - g(a)}\$
$= \lim_{x \rightarrow a}\ \frac{\frac{\pi}{2} - 3\tan^{- 1}{x - \frac{\pi}{2} + 3\tan^{- 1}a}}{2\tan^{- 1}{x - 2\tan^{- 1}a}}$
$= - \frac{3}{2}\lim_{x \rightarrow a}\ \frac{\tan^{- 1}{x - \tan^{- 1}a}}{\tan^{- 1}{x - \tan^{- 1}a}} = - \frac{3}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.