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Question: Answered & Verified by Expert
The range of fx=4sin-1x2x2+1 is
MathematicsFunctionsJEE MainJEE Main 2023 (13 Apr Shift 2)
Options:
  • A [0,2π]
  • B [0,π]
  • C [0,2π)
  • D [0,π)
Solution:
2703 Upvotes Verified Answer
The correct answer is: [0,2π)

Given,

fx=4sin-1x2x2+1

Now taking, x21+x2=1-11+x2<1

0x21+x2<1

Now taking sin-1 we get,

0sin-1x21+x2<π2

04sin-1x21+x2<2π

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

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