Question

If $\left\{\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(e{x}^2\right)}\right\}+{\tan }^{-1}\left(\frac{3+2\log x}{1-6\log x}\right)$, then $\frac{d^{n}y}{d{x}^n}$ is

• A. ${\tan }^{-1}\left\{\right(\log x{)}^n\}$
• B. $0$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

Answer: (B)

$0$

Let $f\left(x\right)={\tan }^{-1}\left(\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(e{x}^2\right)}\right)+{\tan }^{-1}\left(\frac{3+2\log x}{1-6\log x}\right)$
$={\tan }^{-1}\left(\frac{1-\log x^2}{1+\log x^2}\right)+{\tan }^{-1}\left(3\right)+{\tan }^{-1}\left(2\log x\right)$
Thus $\frac{\pi }{4}=2{\tan }^{-1}\left(2\log x\right)+{\tan }^{-1}\left(3\right)+{\tan }^{-1}\left(2\log x\right)$
$\frac{dy}{dx}=0$
Therefore, $\frac{d^{n}y}{d{x}^n}=0$