If the function \(f\) defined on \(\left( {\frac{\pi }{6},\frac{\pi }{3}} \right)\) by\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\sqrt 2 {\rm{cos}}x - 1}}{{{\rm{cot}}x - 1}},}&{x \ne \frac{\pi }{4}}\\ {{\rm{k}},}&{x = \frac{\pi }{4}} \end{array}} \right.\) is continuous, then k is equal to:

If the function \(f\) defined on \(\left( {\frac{\pi }{6},\frac{\pi

1.	If the function \(f\) defined on \(\left( {\frac{\pi }{6},\frac{\pi }{3}} \right)\) by\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\sqrt 2 {\rm{cos}}x - 1}}{{{\rm{cot}}x - 1}},}&{x \ne \frac{\pi }{4}}\\ {{\rm{k}},}&{x = \frac{\pi }{4}} \end{array}} \right.\) is continuous, then k is equal to:
A.	2
B.	\(\frac{1}{2}\)
C.	1
D.	\(\frac{1}{{\sqrt 2 }}\)
Answer» C. 1