Question

If $f(x)=\cot ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$ and $g(x)=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$, then $\displaystyle\lim _{x \rightarrow a} \frac{f(x)-f(a)}{g(x)-g(a)}, 0 < a < \frac{1}{2}$, is

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

Answer 1

Answer: (D)

Given that, $f(x)=\cot ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$
and $g(x)=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
Put $x=\tan \theta$ in $f(x)$, we get
$f(x)=\cot ^{-1}\left(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right)$
$\Rightarrow f(x)=\frac{\pi}{2}-3 \tan ^{-1} x$
Again put $x=\tan \theta$ in $g(x)$, we get
$g(x)=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)$
Again put $x=\tan \theta$ in $g(x)$, we get
$g(x)=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)$
$g(x)=2 \tan ^{-1} x$
$\therefore \displaystyle\lim _{x \rightarrow a} \frac{f(x)-f(a)}{g(x)-g(a)}$
$=\displaystyle\lim _{x \rightarrow a} \frac{\left(\frac{\pi}{2}-3 \tan ^{-1} x\right)-\frac{\pi}{2}+3 \tan ^{-1} a}{2 \tan ^{-1} x-2 \tan ^{-1} a}$
$=\displaystyle\lim _{x \rightarrow a} \frac{-3\left(\tan ^{-1} a-\tan ^{-1} x\right)}{2\left(\tan ^{-1} a-\tan ^{-1} x\right)}=-\frac{3}{2}$