The function $x (x - 1) (x - 2)$ attains its maximum value when x is

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1 + \frac{1}{\sqrt{3}}

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1 - \frac{1}{\sqrt{3}}

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1 \pm \sqrt{3}

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Solution

The correct option is C $1 - \frac{1}{\sqrt{3}}$
$f^{'} (x) = (x - 1) (x - 2) + x (x - 2) + x (x - 1)$
$= x^{2} - 3 x + 2 x^{2} - 2 x + x^{2} - x$
$0 = 3 x^{2} - 6 x + 2$
$x = \frac{6 \pm \sqrt{36 - 24}}{6}$
$x = 1 \pm \frac{1}{\sqrt{3}}$
$f (x)$ is maximum at $x = 1 - \frac{1}{\sqrt{3}}$

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