Q64 / 170 · Manipal (MET) 2019 · Mathematics
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 $\lim _{x \rightarrow a} \frac{f(x)-f(a)}{g(x)-g(a)}, 0 \lt a \lt \frac{1}{2}$, i
- A$\frac{3}{2\left(1+a^2\right)}$
- B$\frac{3}{2\left(1+x^2\right)}$
- C$\frac{3}{2}$
- Correct answerD$-\frac{3}{2}$
Solution
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
$\begin{aligned}
f(x) & =\cot ^{-1}\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right) \\
\Rightarrow \quad f(x) & =\frac{\pi}{2}-3 \tan ^{-1} x
\end{aligned}$
Again put $x=\tan \theta$ in $g(x)$, we get $\begin{aligned} & \quad g(x)=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\ & g(x)=2 \tan ^{-1} x \\ & \therefore \lim _{x \rightarrow a} \frac{f(x)-f(a)}{g(x)-g(a)} \\ & =\lim _{x \rightarrow a} \frac{\left(\frac{\pi}{2}-3 \tan ^{-1} x\right)-\frac{\pi}{2}+3 \tan ^{-1} a}{2 \tan ^{-1} x-2 \tan ^{-1} a} \\ & =\lim _{x \rightarrow a} \frac{-3\left(\tan ^{-1} a-\tan ^{-1} x\right)}{2\left(\tan ^{-1} a-\tan ^{-1} x\right)}=-\frac{3}{2} \end{aligned}$
Again put $x=\tan \theta$ in $g(x)$, we get $\begin{aligned} & \quad g(x)=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\ & g(x)=2 \tan ^{-1} x \\ & \therefore \lim _{x \rightarrow a} \frac{f(x)-f(a)}{g(x)-g(a)} \\ & =\lim _{x \rightarrow a} \frac{\left(\frac{\pi}{2}-3 \tan ^{-1} x\right)-\frac{\pi}{2}+3 \tan ^{-1} a}{2 \tan ^{-1} x-2 \tan ^{-1} a} \\ & =\lim _{x \rightarrow a} \frac{-3\left(\tan ^{-1} a-\tan ^{-1} x\right)}{2\left(\tan ^{-1} a-\tan ^{-1} x\right)}=-\frac{3}{2} \end{aligned}$
Hence, the correct answer is option: 'D'.