JEE Main 2019MathematicsContinuity and DifferentiabilityEasyMCQ

JEE Main 2019 — Continuity and Differentiability Question with Solution

JEE Main 2019 (09 Apr Shift 1)

Question

If the function f defined on π6,π3 by fx=2cosx-1cotx-1, x≠π4k,             x=π4 is continuous, then k is equal to

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Show full solutionCorrect option: A
Correct answer
A12

Step-by-step explanation

∵ fx  is continuous at x=π4

∴ fπ4=k=limx→π42cosx-1cotx-1→00 form 

Applying L' Hospital Rule

k=limx→π42-sinx-cosec2x

⇒k=2sinπ4cosec2π4=2×1222=12.

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About this question

This is a previous-year question from JEE Main 2019, 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.