Points at which normal to the curve y=x^3-3 x is parallel to y-axis a…

The correct answer is Option (2) → (1, -2) and (-1, 2)

Slope of the tangent line,

$\frac{dy}{dx}=\frac{d}{dx}(x^3-3x)=3x^2-3$

$⇒3x^2-3=0$

$⇒x=±1$

$y=(1)^3-3(1)=1-3=-2$

$y=(-1)^3-3(-1)=2$

$⇒(1, -2)$ and $(-1, 2)$