JEE Main 2022MathematicsContinuity and DifferentiabilityEasyMCQ

JEE Main 2022 — Continuity and Differentiability Question with Solution

JEE Main 2022 (29 Jul Shift 2)

Question

Let the function fx=loge1+5x-loge1+αxxif x≠010if x=0 be continuous at x=0. Then α is equal to

Choose an option

Show full solutionCorrect option: D
Correct answer
D-5

Step-by-step explanation

fx=ln1+5x-ln1+αxx x≠010 x=0

For fx to be continuous

 limx→0fx=f0

limx→0ln1+5x-ln1+αxx=10

Using expansion of ln1+x=x-x22+x33- ...., we get

limx→05x-25x22+……-αx-α2x22+……x=10

i.e. 5-α=10⇒α=-5

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About this question

This is a previous-year question from JEE Main 2022, 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.