If f : R → π 3 , π defined by f x = c o s - 1 λ - x 2 x 2 + 2 is a surjective function, then the value of 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 + 2 < π 1 2 ≥ λ - x 2 x 2 + 2 > - 1 ⇒ 1 2 ≥ λ + 2 - x 2 + 2 x 2 + 2 > - 1 1 2 ≥ λ + 2 x 2 + 2 - 1     >   - 1 ⇒ 3 2 ≥ λ + 2 x 2 + 2 > 0 … 1 Now, 2 ≤ x 2 + 2 < ∞ 1 2 ≥ 1 x 2 + 2 > 0 … 2 On comparing 1 and 2 , we get, λ + 2 = 3 ⇒ λ = 1