Let f x = 1 - cos x 2 π - x 2 sin 2 x log 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

Question

Let f x = 1 - cos x 2 π - x 2 sin 2 x log 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,

MARKS App · Accepted Answer

For the function f x to be continuous at x = 2 π l i m x → 2 π f x = f 2 π Now, l i m x → 2 π 1 - cos ⁡ x 2 π - x 2 s i n 2 x l o g 1 + 2 π - x 2 = λ Putting x = 2 π + t , we get, l i m t → 0 1 - c o s t t 2 ⋅ s i n 2 t l o g 1 + t 2 = λ l i m t → 0 1 2 ⋅ s i n 2 t t 2 ⋅ t 2 l o g 1 + t 2 = λ 1 2 × 1 × 1 = λ ⇒ λ = 1 2

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