JEE Main 2019 — Application Of Derivatives Question with Solution
From: JEE Main 2019 (Online) 10th January Evening Slot
Question
The tangent to the curve, y = xex2 passing through the point (1, e) also passes through the point
Choose an option
Show full solutionCorrect option: A
Correct answer
A
Step-by-step explanation
y = xex2
T : y e = 3e (x 1)
y = 3ex 3e + e
y =
lies on it
T : y e = 3e (x 1)
y = 3ex 3e + e
y =
lies on it
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This is a previous-year question from JEE Main 2019, 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.