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

Options

A. 1
B. 2
C. 3
D. 6

Answer

c

Explanation

It is given that F'(a) +2 is the area of the region bounded by x=0, y=0 , y=f(x) and x=a . Therefore,
`F'(a)+2=underset(0)overset(a)int f(x) dx`
Differentiating both sides with respect to a, we obtain
`F''(a)=f(a) implies f(0)=F''(0)`
Now, `F(x)=underset(0)overset(x^(2)+(pi)/(6))int 2cos^(2)t dt`
`implies F'(x) = underset(x)overset(x^(2)+(pi)/(6))int 0 dt + 2x{2cos^(2) (x^(2)+(pi)/(6))}-1(2cos^(2)x)`
`implies F''(x)=4xcos^(2)(x^(2)+(pi)/(6))-2cos^(2)x`
`implies F''(x)=4cos^(2)(x^(2)+(pi)/(6))-16x^(2) cos (x^(2)+(pi)/(6))sin (x^(2)+(pi)/(6))`
`implies F''(0)=4cos^(2)(pi)/(6)-0+4xx0=3`
`implies f(0)=3`