If f : R → π 3 , π defined by f x = c o s - 1 λ - x 2 x 2 + 3 is a surjective function, then λ is equal to

Step-by-step solution

Range of f ( x ) ∈ π 3 , π π 3 ≤ c o s - 1 λ - x 2 x 2 + 3 < π ⇒ 1 2 ≥ λ - x 2 x 2 + 3 > - 1 ⇒ 1 2 ≥ λ + 3 - x 2 + 3 x 2 + 3 > - 1 1 2 ≥ λ + 3 x 2 + 3 - 1 > - 1 ⇒ 3 2 ≥ λ + 3 x 2 + 3 > 0        … … . i Now, 3 ≤ x 2 + 3 < ∞ ⇒ 1 3 ≥ 1 x 2 + 3 > 0 … … . i i On comparing i and i i , we get, λ + 3 = 9 2 &#86