Calculus doubt........

Let a = 2007, then for the function

f(x) = a^xa^x+√a,

Prove that f(x) + f(1-x) = 1.

Using this result we have

f(1/2007) + f(2006/2007) = f(1/2007) + f(1-1/2007) =1
Similarly,
f(2/2007) + f(2005/2007) = 1

f(3/2007) + f(2004/2007) = 1

.......

f(1002/2007) + f(1005/2007) = 1

f(1003/2007) + f(1004/2007) = 1

Adding them we get the required sum as 1003. Thus, (b)

Thanxxx......sir........