Let F x =   ∫ x x 2 + π 6 2 cos 2 ⁡ t   d t for all x   ∈ R   a n…

Let Fx= xx2+π62cos2t dt for all x R and f : 0,120,  be a continuous function. For a0,12, if Fa+2 is the area of the region bounded by x=0, y=0, y=fx and x=a, then f(0) is

Solution

Fa+2=0afxdx
Fa=fa ...(i)
Fx=xx2+π62cos2t dt
Fx=2cos2x2+π6.2x-2cos2x
Fx=4cos2x2+π6-8xcosx2+π6.sinx2+π62x+4cosxsinx
f0=F0=4×34=3

Asked in: JEE Advanced 2015 (Paper 1)

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