If the function f defined on (π/6, π/3) by f(x) = √(2cos x -1) ; cot x -1, x ≠ π/4, x = π/4 is continuous, then k is equal to?

The correct option is A
1/2
∴function should be continuous at x=π/4
∴limx→π/4f(x) = f(π/4)
⇒limx→π/4√(2cos x -1) ; cot x -1 = k
⇒limx→π/4 -√(2sin x) / cosec2x = k (Using L'opital rule)
limx→π/4 √2 sin3x = k
⇒k = √2 (1/√2)3 = 1/2.