Question

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

• A. $f\left(x\right)$ is a constant function
• B. $f\left(x\right)\geq 0$
• C. $f\left(x\right)$ is an even function
• D. $f\left(x\right)\geq \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+A B}\right) \text { and } \\ \tan ^{-1} A+\tan ^{-1} B=\tan ^{-1}\left(\frac{A+B}{1-A B}\right)+\pi \quad(\because A B>1), \text { we get, } \\ f^{\prime}(x)=\tan ^{-1} 2+\tan ^{-1}(2)+\pi=\text { constant } \end{array} $