STD 12 - 6. Application of derivatives — JEE STD 11 Science — Question

$\Rightarrow \log y=-2 x^{2} \log x$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=-4 x \log x-2 x$

$\Rightarrow \frac{d y}{d x}=-2 x(2 \log x+1) y$

$\Rightarrow \frac{d y}{d x}=-2 x\left(\log x^{2}+1\right)\left(\frac{1}{x}\right)^{2 x^{2}}$

Now, the value of $y$ is maximum or minimum when, $\frac{d y}{d x}=0$ $-2 x\left(\log x^{2}+1\right)\left(\frac{1}{x}\right)^{2 x^{2}}=0$

$\Rightarrow \frac{4 x \log x+2 x}{x^{2 x^{2}}}=0$

$\Rightarrow 2 x=-4 x \log x$

$\Rightarrow \log x=\frac{-1}{2}$

$\Rightarrow x=\frac{1}{\sqrt{e}}$

Its double derivative will be negative there hence value will be maximum

Thus, the maximum value of $y$ is,

$\Rightarrow\left(\frac{1}{\frac{1}{\sqrt{e}}}\right)^{2\left(\frac{1}{\sqrt{e}}\right)^{2}}$

$\Rightarrow(\sqrt{e})^{\frac{2}{e}}$

$\Rightarrow e^{\frac{1}{2} \times \frac{2}{e}}$

$\Rightarrow e^{\frac{1}{e}}$