A maximum point of $3 x^{4} - 2 x^{3} - 6 x^{2} + 6 x + 1$ in $[0, 2]$ is-

x = 0

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x = 1

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x = \frac{1}{2}

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does not exist

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Solution

The correct option is C $x = \frac{1}{2}$
$f^{'} (x)$
$= 12 x^{3} - 6 x^{2} - 12 x + 6$
$= 6 (2 x^{3} - x^{2} - 2 x + 1)$
$= 0$ ... to find the critical point.
Hence
$2 x^{3} - x^{2} - 2 x + 1 = 0$
$2 x^{3} - 2 x - (x^{2} - 1) = 0$
$2 x (x^{2} - 1) - (x^{2} - 1) = 0$
$(2 x - 1) (x^{2} - 1) = 0$
Or
$x = \frac{1}{2}$ and $x = \pm 1$
Now
$x ϵ [0, 2]$ so only two values are applicable $x = 1, \frac{1}{2}$ .
Now
$f " (x)$
$= 36 x^{2} - 12 x - 12$
$= 12 (3 x^{2} - x - 1)$
Now
$f " (x)_{x = 1} > 0$ and
$f " (x)_{x = \frac{1}{2}} < 0$
Hence it attains a maximum at $x = \frac{1}{2}$ .

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