If ∫sin⁡x1 t2f(t)dt=1−sin⁡x,∀x∈0,π2, then f13 is

If sinx1t2f(t)dt=1sinx,x0,π2, then f13 is

  1. A
  2. B
  3. C
  4. D

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    Solution:

    ddxsinx1t2f(t)=ddx(1sinx)12f(1)(0)sin2xf(sinx)cosx=cosx

        f(sinx)=1sin2x    f13=(3)2=3  put sinx=13

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