If the minimum and the maximum values of the function f : π 4 , π 2 → R , defined by f θ = - sin 2 θ - 1 - sin 2 θ 1 - cos 2 θ - 1 - cos 2 θ 1 12 10 - 2 are m and M respectively, then the ordered pair ( m , M ) is equal to :

Step-by-step solution

C 2 → C 2 - C 1 f θ = - sin 2 θ - 1 1 - cos 2 θ - 1 1 12 - 2 - 2 = 4 cos 2 θ − sin 2 θ = 4 cos 2 θ Since cos 2 θ = cos 2 θ - sin 2 θ Again, π 4 ≤ θ ≤ π 2 ⇒ 2 π 4 ≤ 2 θ ≤ 2 π 2 ⇒ π 2 ≤ θ ≤ π ∴   θ ∈ π 4 , π 2 ⇒ 2 θ ∈ π 2 , π i.e., Second Quadrant. ⇒ - 1 ≤ cos 2 θ ≤ 0 f θ max = M = 0