Question:
The slope of the normal to the curve $y=2 x^{2}+3 \sin x$ at $x=0$ is
(A) 3
(B) $\frac{1}{3}$
(C) $-3$
(D) $-\frac{1}{3}$
Solution:
The equation of the given curve is $y=2 x^{2}+3 \sin x$.
Slope of the tangent to the given curve at x = 0 is given by,
$\left.\left.\frac{d y}{d x}\right]_{x-0}=4 x+3 \cos x\right]_{x=0}=0+3 \cos 0=3$
Hence, the slope of the normal to the given curve at x = 0 is
$\frac{-1}{\text { Slope of the tangent at } x=0}=\frac{-1}{3}$
The correct answer is D.
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