JEE Main 2019MathematicsLimitsMediumMCQ

JEE Main 2019 — Limits Question with Solution

JEE Main 2019 (12 Jan Shift 2)

Question

limx→1-π-2sin-1x1-x is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B2π

Step-by-step explanation

Given limit can be written as

L=limx→1-π-2sin-1x1-x×π+2sin-1xπ+2sin-1x

⇒L=limx→1-π-2sin-1x1-x π+2sin-1x=limx→1-2cos-1x1-x π+2sin-1x

Let K=limx→1-cos-1x1-x and put x=cosθ, we get

K=limθ→0θ2.22.sinθ2=limθ→02sinθ2θ2=2 ∵limx→0sinxx=1

∴ L=222 π=2π.

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About this question

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