If the function f x = log e 1 - x + x 2 + log e 1 + x + x 2 s e c x - cos x , x ∈ - π 2 , π 2 - 0 k , x = 0 is continuous at x = 0 , then k is equal to:

Step-by-step solution

Given, f x = log e 1 - x + x 2 + log e 1 + x + x 2 s e c   x - cos   x , x ∈ - π 2 , π 2 - 0 k ,   x = 0 ⇒ lim x → 0 ln 1 + x 2 + x 4 cos x 1 - cos 2 x = k Taking L.H.S = lim x → 0 ln 1 + x 2 + x 4 cos x 1 - cos 2 x = lim x → 0 ln 1 + x 2 + x 4 x 2 + x 4 x 2 1 + x 2 cos x sin 2 x x 2 x 2 = lim x → 0 ln 1 + x 2 + x 4 x 2 + x 4 1 + x 2 cos x sin 2 x x 2 = 1 1 + 0 1 1 = 1 Now equating with R.H.S we get, k = 1