JEE Main 2015MathematicsApplication of DerivativesEasyMCQ

JEE Main 2015 — Application of Derivatives Question with Solution

JEE Main 2015 (11 Apr Online)

Question

The equation of a normal to the curve, sin⁡y=xsin⁡π3+y at x=0, is:

Choose an option

Show full solutionCorrect option: D
Correct answer
D2x+3 y=0

Step-by-step explanation

sin⁡y=xsin⁡π3+y

If x=0, then y=0

So point is 0, 0

Slope of tangent =dydx=+sin⁡π3+ycos⁡y-xcos⁡π3+y

at 0, 0  dydx=sin⁡π3=32

Slope of tangent =32

Slope of Normal =-23

∴  equation of normal 2x+3y=0

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About this question

This is a previous-year question from JEE Main 2015, covering the Application of Derivatives chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.