JEE Main 2019 — Binomial Theorem Question with Solution
From: JEE Main 2019 (Online) 11th January Evening Slot
Question
Let (x + 10)50 + (x 10)50 = a0 + a1x + a2x2 + . . . . + a50x50, for all x R; then is equal to
Choose an option
Show full solutionCorrect option: A
Correct answer
A12.25
Step-by-step explanation
(10 + x)50 + (10 x)50
a2 = 2.50C2 1048, a0 = 2.1050
a2 = 2.50C2 1048, a0 = 2.1050
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This is a previous-year question from JEE Main 2019, covering the Binomial Theorem 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.