JEE Main 2021MathematicsContinuity and DifferentiabilityMediumNumerical

JEE Main 2021 — Continuity and Differentiability Question with Solution

JEE Main 2021 (17 Mar Shift 1)

Question

If the function fx=cos(sinx)-cosxx4 is continuous at each point in its domain and f0=1k, then k is _________.

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Show full solutionCorrect answer: 6
Correct answer
6

Step-by-step explanation

Given limx→0cos(sinx)-cosxx4=f0

⇒limx→02sinsinx+x2sinx-sinx2x4=1k

⇒limx→02sinsinx+x2sinx-sinx2sinx+x2x-sinx2x4×sinx+x2x-sinx2=1k

⇒limx→02sinx+x2xx-sinx2x3=1k, ∵limx→0sinxx=1

⇒2limx→0sinx+x2xlimx→0x-sinx2x3=1k

Apply L'Hospital rule for an indeterminate form of 00, we get, 

⇒2×(1+1)2×12×16=1 k

⇒k=6

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About this question

This is a previous-year question from JEE Main 2021, covering the Continuity and Differentiability 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.