∫(-π->π) [cosnx]^2 dx
= (1/2)∫(-π->π) [1+cos2nx] dx
= (1/2)[x+ (1/(2n))sin2nx] |(-π->π)
=π
F(a)= ∫(-π->π) [f(x) -acosnx]^2 dx
F(t)= ∫(-π->π) [f(x) -tcosnx]^2 dx
F'(t) = -2∫(-π->π) f(x)cosnx dx + 2t∫(-π->π) [cosnx]^2 dx
F'(a) =0
-2∫(-π->π) f(x).cosnx dx + 2a∫(-π->π) [cosnx]^2 dx =0
a=∫(-π->π) f(x).cosnx dx/∫(-π->π) [cosnx]^2 dx
=(1/π)∫(-π->π) f(x).cosnx dx
Ans:B
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