IIT JEE Advanced 2007 Paper 2

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)