If the function ƒ defined on , $$\left( {{\pi \over 6},{\pi \over 3}} \right)$$ by $$$f(x) = \left\{ {\matrix{ {{{\sqrt 2 {\mathop{\rm cosx}\nolimits} - 1} \over {\cot x - 1}},} & {x \ne {\pi \over 4}} \cr {k,} & {x = {\pi \over 4}} \cr } } \right.$$$ is continuous, then k is equal to

Let ƒ : R $$ \to $$ R be a differentiable function satisfying ƒ'(3) + ƒ'(2) = 0.
Then $$\mathop {\lim }\limits_{x \to 0} {\left( {{{1 + f(3 + x) - f(3)} \over {1 + f(2 - x) - f(2)}}} \right)^{{1 \over x}}}$$ is equal to

Let ƒ : [–1,3] $$ \to $$ R be defined as

$$f(x) = \left\{ {\matrix{ {\left| x \right| + \left[ x \right]} & , & { - 1 \le x < 1} \cr {x + \left| x \right|} & , & {1 \le x < 2} \cr {x + \left[ x \right]} & , & {2 \le x \le 3} \cr } } \right.$$

where [t] denotes the greatest integer less than or equal to t. Then, ƒ is discontinuous at:

$$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}$$ equals: