Question

If $f\left(x\right)=ta{n}^{-1}\left(\frac{ln\left(e^2/x^3\right)}{ln\left(e^2{x}^3\right)}\right)+$ ${\left(tan\right)}^{-1}\left(\frac{ln\left(e^2{x}^3\right)}{ln\left(e/x^6\right)}\right)$ for all $x\ge e$ , then the incorrect statement is

• A. $f\left(x\right)$ is a constant function
• B. $f\left(x\right)\ge 0$
• C. $f\left(x\right)$ is an even function
• D. $f\left(x\right)\ge \pi$

Answer 1

Answer: (C)

We have,
$\begin{array}{l}f(x)={\tan }^{-1}\left(\frac{2-3\ln x}{1+3\ln x}\right)+{\tan }^{-1}\left(\frac{2+3\ln x}{1-6\ln x}\right)\\ ={\tan }^{-1}2-{\tan }^{-1}(3\ln x)+{\tan }^{-1}2+{\tan }^{-1}(3\ln x)+\pi \\ \text{ Using, }{\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\left(\frac{A-B}{1+AB}\right)\text{ and }\\ {\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)+\pi (\because AB>1),\text{ we get, }\\ f^{\prime }(x)={\tan }^{-1}2+{\tan }^{-1}(2)+\pi =\text{ constant }\end{array}$