JEE Main 2014MathematicsLimitsEasyMCQ

JEE Main 2014 — Limits Question with Solution

JEE Main 2014 (06 Apr)

Question

limx→0sinπcos2xx2 is equal to 

Choose an option

Show full solutionCorrect option: B
Correct answer
B π

Step-by-step explanation

We have  limx→0sinπcos2xx2

=limx→0sinπ-πcos2xx2                  ∵   sin π - θ = sin  θ

=limx→0sinπ sin2xπ sin2x×π sin2xx2

=limx→0πsinπ sin2xπ sin2x   sin  x x 2      

=π11     As, limx→0sin tt=1

= π .

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About this question

This is a previous-year question from JEE Main 2014, covering the Limits 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.