The value of k ( k > 0 ) , for which the function f x = e x - 1 4 sin x 2 k 2 log 1 + x 2 2 , where x ≠ 0 and f ( 0 ) = 8 , is continuous at x = 0 , is

Step-by-step solution

Here, we can know for continuity, lim x → 0 f x = 8 ⇒ lim x → 0 e x - 1 4 sin x 2 k 2 log 1 + x 2 2 = 8 Now, lim x → 0 e x - 1 4 sin x 2 k 2 log 1 + x 2 2 = lim x → 0 e x - 1 x 4 sin x 2 k 2 x 2 × log 1 + x 2 2 x 2 (dividing x 4 , to the numerator and denominator) = lim x → 0 e x - 1 x 4 sin x 2 k 2 k × x 2 k 2 × log 1 + x 2 2 x 2 2 × 2 = lim x → 0 e x - 1 x 4 sin x 2 k 2 k 2 × x 2 k 2 × log 1 + x 2 2 x 2 2 × 2 = 1 1 k 2 × 1 2 = 2 k 2