Practicing Success
If y = loge\((\frac{2x}{1-x})\), then \(\frac{d^2y}{dx^2}\) at \(x = \frac{1}{2}\) is |
\(\frac{1}{2}\) \(\frac{1}{4}\) 0 \(\frac{3}{5}\) |
0 |
$t=log_e(\frac{2x}{1-x})$ ⇒ So, $\frac{dy}{dx}=\frac{\frac{1}{2x}}{1-x}×\frac{d}{dx}(\frac{2x}{1-x})$ $\Rightarrow \frac{dy}{dx}=\frac{1-x}{2x} \frac{(1-x)2+2x}{(1-x)^2}=\frac{2}{2x(1-x)}=\frac{1}{1-x}+\frac{1}{x}$ $\frac{d^2y}{dx^2}=\frac{1}{(1-x)^2}-\frac{1}{x^2}$ at $x=\frac{1}{2}$ $\frac{d^2y}{dx^2}=\frac{1}{(1-\frac{1}{2})^2}-\frac{1}{(\frac{1}{2})^2}⇒\frac{1}{(\frac{2-1}{2})^2}-\frac{1}{(\frac{1}{4})}$ $\frac{1}{(\frac{1}{2})^2}-\frac{1}{\frac{1}{4}}=\frac{1}{(\frac{1}{4})}-\frac{1}{(\frac{1}{4})}⇒4-4=0$ Correct option is 3rd. |