A stationary value of $f (x) = x (ln x)^{2}$ is

2 e^{- 2}

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4 e^{- 2}

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2 e^{2}

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4 e^{2}

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Solution

The correct option is B $4 e^{- 2}$
$f (x) = x (ln x)^{2}$
$\Rightarrow f^{'} (x) = (ln x)^{2} + 2 x \frac{ln x}{x}$
$\Rightarrow f^{'} (x) = ln x (ln (x) + 2)$
For stationary points ,
$f^{'} (x) = 0$
$\Rightarrow x = 1$ or $e^{- 2}$
$f (1) = 0$
and $f (e^{- 2}) = 4 e^{- 2}$

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A stationary value of f(x)=x(lnx)2 is

The correct option is B 4e−2f(x)=x(lnx)2⇒f′(x)=(lnx)2+2xlnxx⇒f′(x)=lnx(ln(x)+2)For stationary points ,f′(x)=0 ⇒x=1 or e−2f(1)=0and f(e−2)=4e−2

A stationary value of $f (x) = x (ln x)^{2}$ is

The correct option is B $4 e^{- 2}$
$f (x) = x (ln x)^{2}$
$\Rightarrow f^{'} (x) = (ln x)^{2} + 2 x \frac{ln x}{x}$
$\Rightarrow f^{'} (x) = ln x (ln (x) + 2)$
For stationary points ,
$f^{'} (x) = 0$
$\Rightarrow x = 1$ or $e^{- 2}$
$f (1) = 0$
and $f (e^{- 2}) = 4 e^{- 2}$