Question

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

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

The correct option is C $0$

$y={\tan }^{-1}\left(\frac{\log e{x}^{-2}}{\log e{x}^2}\right)+{\tan }^{-1}\left(\frac{3+2\log x}{1-6\log x}\right)$
$={\tan }^{-1}\left(\frac{1-2\log x}{1+2\log x}\right)+{\tan }^{-1}\left(\frac{3+2\log x}{1-6\log x}\right)$
$={\tan }^{-1}1-{\tan }^{-1}\left(2\log x\right)+{\tan }^{-1}3+{\tan }^{-1}\left(2\log x\right)$
$={\tan }^{-1}3-{\tan }^{-1}1=k$ (constant)
$\Rightarrow \frac{dy}{dx}=0\therefore \frac{d^{2}y}{d{x}^2}=0$