JEE Main 2021MathematicsDifferential EquationsMediumMCQ

JEE Main 2021 — Differential Equations Question with Solution

JEE Main 2021 (24 Feb Shift 2)

Question

Let f be a twice differentiable function defined on R such that f0=1, f'0=2 and f'x≠0 for all x∈R. If fxf'xf'xf''x=0, for all x∈R, then the value of f1 lies in the interval

Choose an option

Show full solutionCorrect option: D
Correct answer
D6,9

Step-by-step explanation

We have, fxf'xf'xf''x=0

⇒fxf''x-f'x2=0

⇒f''xf'x=f'xfx

On integrating both side, we get

lnf'x=lnfx+lnc

⇒f'x=cfx

⇒f'xfx=c

Again integrating, we get

ln fx=cx+k1

⇒fx=kecx

Since, f0=1=k

Therefore, f'0=c=2

Now, fx=e2x

Hence, f1=e2∈6,9

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About this question

This is a previous-year question from JEE Main 2021, covering the Differential Equations 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.