Let f x = 1 − cos x 2 π − x 2 ⋅ tan 2 x ln 1 + 4 π 2 − 4 π x + x 2 : x ≠ 2 π                               λ : x = 2 π is continuous at x = 2 π , then the value of λ is equal to

Step-by-step solution

For f x to be continuous at x = 2 π , lim x → 2 π ⁡ f x = f 2 π Now, lim x → 2 π ⁡ 1 - cos ⁡ x tan 2 ⁡ x 2 π - x 2 ln ⁡ 1 + 2 π - x 2 = λ Putting x = 2 π + t , we get, lim t → 0 1 − cos t t 2 ⋅ tan 2 t ln 1 + t 2 = λ lim t → 0 1 2 ⋅ tan 2 t t 2 ⋅ t 2 ln 1 + t 2 = λ So, 1 2 × 1 × 1 = λ ⇒ λ = 1 2