Let F(x)=∫x2+π6x2cos2tdt for all xϵR and f:[0,12]→[0,∞) be a continuous function. For αϵ[0,12], if F′(a)+2 is the area of the region bounded by x=0,y=0,y=f(x) and x=a, then f(0) is
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F(x)=∫x2+π6x2cos2tdt Differentiating both side w.r.t x F′(x)=2cos2(x2+π6)⋅2x−2cos2x ∴F′(a)=4acos2(a2+π6)−2cos2a Now given, F′(a)+2=∫a0f(x)dx 4acos2(a2+π6)−2cos2a+2=∫a0f(x)dx Differentiating both side w.r.t a 4cos2(a2+π6)−8acos(a2+π6)sin(a2+π6)+4cosasina=f(a) ∴f(0)=4cos2(π6)=4×34=3
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