Question

If $y=ta{n}^{-1}\left(\frac{log\left(e/x^2\right)}{log\left(e{x}^2\right)}\right)$$+ta{n}^{-1}\left(\frac{3+2logx}{1-6logx}\right)$, then $\frac{d^{2}y}{d{x}^2}$ is equal to

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

Answer: (C)

$\because y=ta{n}^{-1}\left(\frac{log\left(\frac{e}{x^2}\right)}{log\left(e{x}^2\right)}\right)+ta{n}^{-1}\left(\frac{3+2logx}{1-6logx}\right)$
$=ta{n}^{-1}\left(\frac{1-log{x}^2}{1+log{x}^2}\right)+ta{n}^{-1}\left(\frac{3+2logx}{1-6logx}\right)$
$=ta{n}^{-1}\left(1\right)-ta{n}^{-1}\left(2logx\right)+ta{n}^{-1}\left(3\right)+ta{n}^{-1}\left(2logx\right)$
$\therefore y=ta{n}^{-1}\left(1\right)+ta{n}^{-1}\left(3\right)$
$\Rightarrow \frac{dy}{dx}=0$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=0$