JEE Main 2020 — Differentiation Question with Solution
From: JEE Main 2020 (Online) 7th January Morning Slot
Question
Let xk + yk = ak, (a, k > 0 ) and , then k is:
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
xk + yk = ak
kxk - 1 + kyk - 1 = 0
= 0 ...(1)
Given ...(2)
Comparing (1) and (2), we get
k - 1 =
k =
kxk - 1 + kyk - 1 = 0
= 0 ...(1)
Given ...(2)
Comparing (1) and (2), we get
k - 1 =
k =
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This is a previous-year question from JEE Main 2020, 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.