JEE Main 2023MathematicsFunctionsEasyMCQ

JEE Main 2023 — Functions Question with Solution

JEE Main 2023 (13 Apr Shift 2)

Question

The range of fx=4sin-1x2x2+1 is

Choose an option

Show full solutionCorrect option: C
Correct answer
C[0,2π)

Step-by-step explanation

Given,

fx=4sin-1x2x2+1

Now taking, x21+x2=1-11+x2<1

⇒0≤x21+x2<1

Now taking sin-1 we get,

⇒0≤sin-1x21+x2<π2

⇒0≤4sin-1x21+x2<2π

Hence, the range of 4sin-1x2x2+1 is [0,2π)

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About this question

This is a previous-year question from JEE Main 2023, covering the Functions 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.