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The points on the curve $y = x^{3}$ , the tangents at which are inclined at an angle of $60^{0}$ to x- axis are.

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$(3^{.1 / 2}, 3^{- 2 / 5}) (- 3^{- 1 / 3}, - 3^{2 / 3})$

$(3^{- 1 / 2}, 3^{- 2 / 5}) (- 3^{- 1 / 3}, - 3^{- 2 / 3})$

$(3^{- 1 / 4}, 3^{- 3 / 4}) (- 3^{- 1 / 4}, - 3^{- 3 / 4})$

$(2^{1 / 4}, 2^{- 2 / 5}) (- 3^{1 / 2}, - 3^{- 1 / 2})$

answer is A.

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Detailed Solution

$\frac{d y}{d x} = 3 x^{2}, \tan 60^{o} = 3 x^{2} = 3 x^{2} \Rightarrow \sqrt{3} = 3 x^{2} \Rightarrow x^{2} = \frac{1}{\sqrt{3}} \Rightarrow x = \pm 3^{1 / 4} y = \pm 3^{- 3 / 4}$

$∴ p o int = (3^{- 1 / 4}, 3^{- 3 / 4}) (- 3^{- 1 / 4}, - 3^{3 / 4})$

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