Let f   :   R → R be a continuous function. Then lim x → π / 4 π 4 ∫ 2…

Let f : R→R be a continuous function. Then limx→π/4π4∫2sec2xf(x)dxx2-π216 is equal to:
  1. f(2)
  2. 2f(2)
  3. 2f(2)
  4. 4f(2)

Solution

 limx→π4π4∫2sec2xfxdxx2-π216

Using L-Hospital's rule, we get

 =limx→π4π4fsec2x·2·secx·secx·tanx-02x

=π4·2(2)2·(1)·f(2)2·π4

=2f2

Asked in: JEE Main 2021 (01 Sep Shift 2)

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