Let f(x) = $$\left\{ {\matrix{ {{{\left( {x - 1} \right)}^{{1 \over {2 - x}}}},} & {x > 1,x \ne 2} \cr {k\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {,x = 2} \cr } } \right.$$

Thevaue of k for which f s continuous at x = 2 is :

$$\mathop {\lim }\limits_{x \to 0} {{x\tan 2x - 2x\tan x} \over {{{\left( {1 - \cos 2x} \right)}^2}}}$$ equals :

Let S = {($$\lambda $$, $$\mu $$) $$ \in $$ R $$ \times $$ R : f(t) = (|$$\lambda $$| e^|t| $$-$$ $$\mu $$). sin (2|t|), t $$ \in $$ R, is a differentiable function}. Then S is a subset of :

The value of k for which the function

$$f\left( x \right) = \left\{ {\matrix{ {{{\left( {{4 \over 5}} \right)}^{{{\tan \,4x} \over {\tan \,5x}}}}\,\,,} & {0 < x < {\pi \over 2}} \cr {k + {2 \over 5}\,\,\,,} & {x = {\pi \over 2}} \cr } } \right.$$

is continuous at x = $${\pi \over 2},$$ is :