f(x) = \frac{2cosx-sin2x}{(\pi-2x)^2} , g(x) = {e^{-cosx}-1}/{8x-4\pi} and h(x) = \{\table f(x) , x \pi\/2 then

Correct Answer is : h(x) is continuous at x = π/4

h(x) has removable discontinuity at x = π/2

f (\pi^{-}/2) + g (\pi^{\+}/2) = 0