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Question: Answered & Verified by Expert
Let f : RR be a continuous function. Then limxπ/4π42sec2xf(x)dxx2-π216 is equal to:
MathematicsDefinite IntegrationJEE MainJEE Main 2021 (01 Sep Shift 2)
Options:
  • A f(2)
  • B 2f(2)
  • C 2f(2)
  • D 4f(2)
Solution:
1914 Upvotes Verified Answer
The correct answer is: 2f(2)

 limxπ4π42sec2xfxdxx2-π216

Using L-Hospital's rule, we get

 =limxπ4π4fsec2x·2·secx·secx·tanx-02x

=π4·2(2)2·(1)·f(2)2·π4

=2f2

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