Question

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

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

Answer: (C)

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