JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Application +4 | -1

The equation of a normal to the curve,

\sin \mathrm{y}=\mathrm{x} \sin \left(\frac{\pi}{3}+y\right)

\mathrm{x}=0

, is :

View answer and explanation

Correct option: B

2 x+\sqrt{3} y=0

Option B (

2 x+\sqrt{3} y=0

) is correct because it represents the equation of the normal to the curve at x=0, derived from the derivative and normal slope.

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a

y=x^3-a x

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\mathrm{C}

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

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