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A point on the curve $y = x^{3} - 3 x + 5$ at which the tangent line is parallel to $y = - 2 x$ is

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(1, 3)

(0, 5)

$(\frac{1}{\sqrt{3}}, 5 - \frac{8 \sqrt{3}}{9})$

$(\frac{1}{\sqrt{2}}, 0)$

answer is C.

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Detailed Solution

$Since y^{'} (x) = 3 x^{2} - 3$ so by the condition of parallelism

$3 x^{2} - 3 = - 2$

$x = \pm \frac{1}{\sqrt{3}}$

If $x = \frac{1}{\sqrt{3}}, then y = \frac{1}{3 \sqrt{3}} - \frac{3}{\sqrt{3}} + 5$

$= 5 - \frac{8}{3 \sqrt{3}} = 5 - \frac{8 \sqrt{3}}{9}$

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