We have f(x) = 2 + cosx for all values of x. f ′(x) = −sin x Statement-1: We have f ′(t) = −sint f ′(π + t) = sint Also, since f ′(t) ⋅ f ′(π + t) = − sin2 t which is negative, the equation f′(t) = 0 has at least one solution in [t, t + π]. Statement-2: Since f (x) = 2 + cos x is a periodic function with period 2π, we get f (2π + t) = f (t)