Question

If $f\left(x\right)=\cos \left({\cot }^{-1}\left(\sin \left({\tan }^{-1}x\right)\right)\right),$ then which of the following is(are) TRUE?

• A. $\underset{x\to \infty }{lim}f\left(x\right)=\frac{1}{\sqrt{2}}$
• B. $\underset{x\to -\infty }{lim}f\left(x\right)=\frac{-1}{\sqrt{2}}$
• C. Range of $f$ is $(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}})$
• D. $f\left(1\right)=\frac{1}{\sqrt{3}}$

Answer 1

Answer: (A), (B), (C), (D)

$f\left(1\right)=\frac{1}{\sqrt{3}}$

$\underset{x\to \infty }{lim}f\left(x\right)=\cos \left({\cot }^{-1}\left(\sin \left(\frac{\pi }{2}\right)\right)\right)=\frac{1}{\sqrt{2}}$
$\underset{x\to -\infty }{lim}f\left(x\right)=\cos \left({\cot }^{-1}\left(\sin (-\frac{\pi }{2})\right)\right)=\frac{-1}{\sqrt{2}}$

Also, $\sin \left({\tan }^{-1}x\right)\in (-1,1)$
$\Rightarrow {\cot }^{-1}\left(\sin \right({\tan }^{-1}x\left)\right)\in (\frac{\pi }{4},\frac{3\pi }{4})$
$\Rightarrow \cos \left({\cot }^{-1}\right(\sin \left({\tan }^{-1}x\right))\in (\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}})$
$\therefore$ Range of $f$ is $(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

$f\left(1\right)=\cos \left({\cot }^{-1}\left(\sin \frac{\pi }{4}\right)\right)$
$=\cos \left({\cot }^{-1}\frac{1}{\sqrt{2}}\right)$
$=\cos \left({\tan }^{-1}\sqrt{2}\right)$ $(\because {\cot }^{-1}x={\tan }^{-1}\frac{1}{x},x>0)$
$=\cos \left({\cos }^{-1}\frac{1}{\sqrt{3}}\right)$
$=\frac{1}{\sqrt{3}}$