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Question: Answered & Verified by Expert
limxπ482-cosx+sinx72-2sin2x is equal to
MathematicsLimitsJEE MainJEE Main 2022 (25 Jul Shift 2)
Options:
  • A 14
  • B 7
  • C 142
  • D 72
Solution:
2554 Upvotes Verified Answer
The correct answer is: 14

limxπ482-cosx+sinx72-2sin2x   00form

=limxπ4-7cosx+sinx6-sinx+cosx-22cos2x  (using L'Hospital Rule)

=limxπ456cosx-sinx22cos2x 00 form

=limxπ4-56sinx+cosx-42sin2x  (using L'Hospital Rule)

=56242=14

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