Question

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

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

Answer: (C)

$0$

We have $y={\tan }^{-1}\left(\frac{\log e-\log x^2}{\log e+\log x^2}\right)+{\tan }^{-1}\left(\frac{3+2\log x}{1-6\log x}\right)$
$={\tan }^{-1}\left(\frac{1-2\log x}{1+2\log x}\right)+{\tan }^{-1}\left(\frac{3+2\log x}{1-6\log x}\right)$
$={\tan }^{-1}1-{\tan }^{-1}\left(2\log x\right)+{\tan }^{-1}3+{\tan }^{-1}\left(2\log x\right)$
$={\tan }^{-1}1+{\tan }^{-1}3$
or $\frac{dy}{dx}=0$ or $\frac{d^{2}y}{d{x}^2}=0$