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The slope of the normal to the curve $y=x^3+3 x-2$ at its point of intersection with y-axis is: |
3 -3 $\frac{1}{3}$ $-\frac{1}{3}$ |
$-\frac{1}{3}$ |
The correct answer is Option (4) → $-\frac{1}{3}$ $\frac{dy}{dx}=3x^2+3$ when it intersects y axis ⇒ x = 0 ⇒ tangent = $\frac{dy}{dx}=3$ so normal $=-\frac{dx}{dy}=-\frac{1}{3}$ |
