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Manipal Entrance Test 2018 Math Paper

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Question : 55 of 70

Marks: +1, -0

The minimum value of

x \log x

is equal to

$e$
$\frac{1}{e}$
$-\frac{1}{e}$
$\frac{2}{e}$

Solution:

Let

y=x \log _{e} x

On differentiating w.r.t.

x

, we get

\frac{dy}{dx}=x \cdot \frac{1}{x}+\log x=(1+\log x)

Again, differentiating, we get

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

Put,

\frac{dy}{dx}=0

for maxima or minima.

\Rightarrow 1+\log x=0

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

\therefore \left(\frac{d^{2}y}{dx^{2}}\right)_{\left(x=\frac{1}{c}\right)}=e

\therefore y

is minimum at

x = \frac{1}{e}

\therefore y_{\min} = \frac{1}{e} \log_e \left( \frac{1}{e} \right) = -\frac{1}{e}

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