Question

If $f\left(x\right)={\tan }^{-1}\left[\begin{array}{c}\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(e{x}^2\right)}\end{array}\right]+{\tan }^{-1}\left[\begin{array}{c}\frac{3+2\log x}{1-6\log x}\end{array}\right]$ , then the value of $f^{\prime \prime }\left(x\right)$ is

• A. $x^2$
• B. $x$
• C. $1$
• D. $0$

Answer 1

Answer: (D)

$0$

$f\left(x\right)={\tan }^{-1}\left[\frac{\log \left(e\right)-\log \left(x^2\right)}{\log e+\log \left(x^2\right)}\right]+{\tan }^{-1}\left(\frac{3+2\log x}{1-6\log x}\right)$

$={\tan }^{-1}\left[\frac{1-\log \left(x^2\right)}{1+\log \left(x^2\right)}\right]+{\tan }^{-1}\left(\frac{3+2\log x}{1-6\log x}\right)$

$={\tan }^{-1}\left[\frac{\tan {45}^0-\tan \left({\tan }^{-1}\left(\log \left(x^2\right)\right)\right)}{1+\tan {45}^0\tan \left({\tan }^{-1}\left(\log \left(x^2\right)\right)\right)}\right]+{\tan }^{-1}\left[\frac{{\tan }^{-1}\left(\tan 3\right)+{\tan }^{-1}\left(\tan \left(2\log x\right)\right)}{1-\left({\tan }^{-1}\left(\tan 3\right)\right)\left({\tan }^{-1}\left(\tan \left(2\log x\right)\right)\right)}\right]$

$={\tan }^{-1}\left(\tan (\frac{\pi }{4}-{\tan }^{-1}\left(\log \left(x^2\right)\right))\right)+{\tan }^{-1}\left(\tan ({\tan }^{-1}3+{\tan }^{-1}\left(2\log \left(x\right)\right))\right)$

$=[\frac{\pi }{4}-{\tan }^{-1}\left(\log \left(x^2\right)\right)]+[{\tan }^{-1}3+{\tan }^{-1}\left(2\log x\right)]$

$=[\frac{\pi }{4}-{\tan }^{-1}\left(2\log x\right)]+[{\tan }^{-1}3+{\tan }^{-1}\left(2\log x\right)]$

$f\left(x\right)=\frac{\pi }{4}+{\tan }^{-1}3$

$f^{\prime }\left(x\right)=0$

$f"\left(x\right)=0$.