The greatest value of $f (x) = x - 2 l n x$ in [1, e] is attained at x =

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\sqrt{e}

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\frac{e}{2}

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Solution

The correct option is A 1
$f (x) = x - 2 l n x$
$f^{'} (x) = 1 - \frac{2}{x}$
For maxima or minima,
$f^{'} (x) = 0$
$\Rightarrow x = 2$
$f (2) = 2 - 2 l n 2$
$f (1) = 1$ and $f (e) = e - 2$
Hence, the greatest value is attained at $x = 1$

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The greatest value of f(x)=x−2lnx in [1, e] is attained at x =

The correct option is A 1f(x)=x−2lnx f′(x)=1−2xFor maxima or minima,f′(x)=0⇒x=2f(2)=2−2ln2f(1)=1 and f(e)=e−2Hence, the greatest value is attained at x=1

The greatest value of $f (x) = x - 2 l n x$ in [1, e] is attained at x =

The correct option is A 1
$f (x) = x - 2 l n x$
$f^{'} (x) = 1 - \frac{2}{x}$
For maxima or minima,
$f^{'} (x) = 0$
$\Rightarrow x = 2$
$f (2) = 2 - 2 l n 2$
$f (1) = 1$ and $f (e) = e - 2$
Hence, the greatest value is attained at $x = 1$