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

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 :

Let f(x) be a polynomial of degree $$4$$ having extreme values at $$x = 1$$ and $$x = 2.$$

If   $$\mathop {lim}\limits_{x \to 0} \left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right) = 3$$   then f($$-$$1) is equal to :

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 :