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Q.

The point on the curve $y = x^{3}$ the tangents at which are inclined at an angle of $\frac{π}{3}$ to x-axis are

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a

(1, 1), (–1, –1)

b

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

c

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

d

(1, 1), (2, 2)

answer is B.

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Detailed Solution

$\begin{array}{rcl} \frac{dy}{dx} & = & \tan \frac{π}{3} = \sqrt{3} \\ y & = & x^{3} \Rightarrow \frac{dy}{dx} = 3 x^{2} = \sqrt{3} \\ x^{2} & = & \frac{1}{\sqrt{3}} \Rightarrow x = {(\frac{1}{3})}^{\frac{1}{4}} = \pm 3^{\frac{- 1}{4}} \\ x & = & 3^{\frac{- 1}{4}} \Rightarrow y = (3^{\frac{- 1}{4}}) = 3^{\frac{- 3}{4}} \\ x & = & - 3^{\frac{- 1}{4}} \Rightarrow y = {(- 3^{\frac{- 1}{4}})}^{3} = - 3^{\frac{- 3}{4}} \end{array}$

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