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 $\underset{x\to a}{\lim }\frac{f(x)-f(a)}{g(x)-g(a)},0<a<\frac{1}{2}$, 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, $f(x)={\cot }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$
and $g(x)={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
Put $x=\tan \theta$ in $f(x)$, 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
$g(x)={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$
$g(x)=2{\tan }^{-1}x$
$\therefore \underset{x\to a}{\lim }\frac{f(x)-f(a)}{g(x)-g(a)}$
$=\underset{x\to a}{\lim }\frac{\left(\frac{\pi }{2}-3{\tan }^{-1}x\right)-\frac{\pi }{2}+3{\tan }^{-1}a}{2{\tan }^{-1}x-2{\tan }^{-1}a}$
$=\underset{x\to a}{\lim }\frac{-3\left({\tan }^{-1}a-{\tan }^{-1}x\right)}{2\left({\tan }^{-1}a-{\tan }^{-1}x\right)}=-\frac{3}{2}$