Question

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

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

The correct option is C $0$

$y=ta{n}^{-1}\left(\frac{1-2Logx}{1+2Logx}\right)+ta{n}^{-1}\left(\frac{3+2Logx}{1-6Logx}\right)$

$y=ta{n}^{-1}\frac{\frac{1-2Logx}{1+2Logx}+\frac{3+2Logx}{1-6Logx}}{1-\left(\frac{1-2Logx}{1+2Logx}\right)\left(\frac{3+2Logx}{1-6Logx}\right)}$

$=ta{n}^{-1}\left(\frac{(1-6Logx)(1-2Logx)+(3+2Logx)(1+2Logx)}{(1+2Logx)(1-6Logx)-(1-2Logx)(3+2Logx)}\right)$

$=ta{n}^{-1}\frac{(1-8Logx+12Lo{g}^{2}x+3+8Logx+4Lo{g}^{2}x)}{1-4Logx-12Lo{g}^{2}x-(3-4Logx-4Lo{g}^{2}x)}$

$=ta{n}^{-1}\left(\frac{4+16Lo{g}^{2}x}{-2-8Lo{g}^{2}x}\right)$

$y=ta{n}^{-1}(-\frac{1}{2})$

$y=c$
$\frac{dy}{dx}=0$
$\frac{d^{2}y}{d{x}^2}=0$