Let `F(x)=int_x^[x^2+pi/6](2cos^2t)dt` for all `x in R` and `f:[0,1/2] -> [0,oo)` be a continuous function.For `a in [0,1/2]`, 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

Explanation

`F(x)=int_(x)^(x^(2)+(pi)/6)2cos^(2)tdt`
`:.F'(x)=2(cos(x^(2)+(pi)/6))^(2)2x-2cos^(2)x`
According to the question
`:.F'(a)+2int_(0)^(a)f(x)dx`
`implies2(cos(a^(2)+(pi)/a))^(2)2a-2cos^(2)a+2=int_(0)^(a)f(x)dx`
Defferentiating w.r.t `a` we get
`4cos^(2)(a^(2)+(pi)/6)+4axx2(a^(2)(pi)/6)(-sin(a^(2)+(pi)/6))`
`xx2a+4cos a sin a=f(a)`
`:. f(0)=4((sqrt(3))/2)^(2)=3`