CUET Preparation Today
The slope of the normal to the curve $y=x^3-4 \sin x$ at x = 0 is: |
-4 $\frac{1}{4}$ 4 $-\frac{1}{4}$ |
$\frac{1}{4}$ |
The correct answer is Option (2) → $\frac{1}{4}$ $y=x^3-4 \sin x$ $⇒\frac{dy}{dx}=3x^2-4\cos x$ and, $\left.\frac{dy}{dx}\right|_{x=0}=3(0)^2-4\cos (0)$ $=-4$ ∴ Slope of the normal, $m=\frac{1}{4}$ $[m×\frac{dy}{dx}=-1]$ |
