If $y=x \log \left(\frac{x}{2-3 x}\right)$ for $0 < x < \frac{2}{3}$, then $\frac{d^2 y}{d x^2}$ at…
If $y=x \log \left(\frac{x}{2-3 x}\right)$ for $0 < x < \frac{2}{3}$, then $\frac{d^2 y}{d x^2}$ at $x=\frac{1}{2}$ is
- 4
- 16
- 32
- 2
Solution
Given, $y=x \log \left(\frac{x}{2-3 x}\right)$, for $0 < x < \frac{2}{3}$
On differentiating w.r.t. to ' $x$ ', we are getting
$
\begin{aligned}
\frac{d y}{d x} & =x\left(\frac{1}{x}-\frac{-3}{2-3 x}\right)+\log \left(\frac{x}{2-3 x}\right) \\
\Rightarrow \quad \frac{d y}{d x} & =\frac{2}{2-3 x}+\log \left(\frac{x}{2-3 x}\right)
\end{aligned}
$
Again differentiating w.r.t. to ' $\chi$ ', we are getting
$
\begin{aligned}
\frac{d^2 y}{d x^2} & =\frac{-2}{(2-3 x)^2}(-3)+\left(\frac{1}{x}-\frac{-3}{2-3 x}\right) \\
& =\frac{6}{(2-3 x)^2}+\frac{2}{x(2-3 x)}
\end{aligned}
$
So,
$
\begin{aligned}
& \frac{d^2 y}{d x^2}\left(\text { at } x=\frac{1}{2}\right)=\frac{6}{(1 / 2)^2}+\frac{2}{1 / 2 \times \frac{1}{2}} \\
& =24+8=32 \\
&
\end{aligned}
$
Asked in: AP EAMCET 2018 (23 Apr Shift 2)
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