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AP EAPCET 24-Aug-2021 Shift 1 Paper

Section: Mathematics

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Question : 70 of 160

Marks: +1, -0

The maximum value of

y=x(\log x)^{2}

is ______

$e^{-2}$
$2e^{-2}$
$3e^{-2}$
$4e^{-2}$

Solution:

y=x(\log x)^{2}

For maximum value

\frac{dy}{dx}=x\left(2\log x \cdot \frac{1}{x}\right)+(\log x)^{2}

=2\log x+(\log x)^{2}

Now, put

\frac{dy}{dx}=0

2\log x+(\log x)^{2}=0

(\log x)[\log x+2]=0

\log x=0 | \log x+2=0

x=1

|

\log x=-2

&

x=e^{-2}

Using second order derivative lest. Again differentiating

y

w.I.t.

x

\frac{d^{2}y}{dx^{2}}=2x+2\log x \cdot \frac{1}{x}

\frac{d^{2}y}{dx^{2}}\big|_{x=1}=2>0

∴

x=1

is point of minima

\frac{d^{2}y}{dx^{2}}\big|_{x=e^{-2}} =\frac{2}{e^{-2}}+2\cdot \log e^{-2} \cdot \frac{1}{e^{-2}}

=\frac{1}{e^{-2}}(2-4)=-\frac{2}{e^{-2}}<0

∴

x=e^{-2}

is point of maxima.

∴

y_{\max}=e^{-2}(\log e^{-2})^{2}

=e^{-2}(-2)^{2}=4e^{-2}

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