The slope of the normal to the curve $y=2 x^2+3 \sin x$ at $x=0$ is

JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

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Application +4 | -1

The slope of the normal to the curve

y=2 x^2+3 \sin x

x=0

View answer and explanation

Correct option: D

-\frac{1}{3}

Option D (

-\frac{1}{3}

) is correct because it is the negative reciprocal of the slope of the tangent to the curve at x=0, which represents the slope of the normal.

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a

y=x^3-a x

x^2 y=1-y

x y=1-y

C

y^3-3 x y+2=0

H

\mathrm{C}

x

V

C

y

y-e^{x y}+x=0

y = f\left( x \right)

x

x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)

\theta = \left( {2k + 1} \right)\pi ,\;k \in Z