The points on the curve $y=x^3$, the tangents at which are inclined at an angle of 60° to x-axis are:
Answer & explanation
Correct answer: option 1
$\frac{dy}{dx}=3x^2⇒60°=3x^2⇒\sqrt{3}=3x^2⇒x^2=\frac{1}{\sqrt{3}}$
$⇒x=±3^{-1/4},y=±3^{-3/4}$
⇒ Point = $(3^{-1/4},3^{-3/4}),(-3^{-1/4},-3^{-3/4})$