JEE Main 2019MathematicsDifferentiationMediumMCQ

JEE Main 2019 — Differentiation Question with Solution

JEE Main 2019 (12 Apr Shift 1)

Question

If ey+xy=e, the ordered pair dydx,d2ydx2 at x=0 is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B-1e,1e2

Step-by-step explanation

ey+xy=e

differentiate w. r. t. x

eydydx+xdydx+y=0

⇒dydxx+ey=-y,  dydx(0,1)=-1e

Again differentiate w. r. t. x

ey.d2ydx2+dydx.ey.dydx+x.d2ydx2+dydx+dydx=0

⇒x+eyd2ydx2+dydx2.ey+2dydx=0

Now, at 0,1

ed2ydx2+1e2e+2-1e=0

∴ d2ydx2=1e2

Hence, ordered pair dydx,d2ydx2=-1e,1e2

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About this question

This is a previous-year question from JEE Main 2019, 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.