JEE Main 2023MathematicsDifferentiationMediumMCQ

JEE Main 2023 — Differentiation Question with Solution

JEE Main 2023 (31 Jan Shift 1)

Question

Let y=fx=sin3π3cosπ32-4x3+5x2+132. Then, at x=1,

Choose an option

Show full solutionCorrect option: B
Correct answer
B2y'+3π2y=0

Step-by-step explanation

The given equation can be written as

y=sin3π3cosgx 

where,

gx=π32-4x3+5x2+13/2

⇒g'x=π22-4x3+5x2+11/2-12x2+10x

⇒g'1=π222-2=-π

And,

g1=2π3=π-π3

Now,

y'=3sin2π3cosgx×cosπ3cosgx×π3-singxg'x

y'1=3sin2-π6·cosπ6·π3-sin2π3g'1

y'1=34·32·π3-32-π=3π216

y1=sin3π3cos2π3=-18

Therefore,

2y'1+3π2y1=0

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About this question

This is a previous-year question from JEE Main 2023, covering the Differentiation 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.