Let a > b > 0 and I(n) = a^1/n $$-$$ b^1/n, J(n) = (a $$-$$ b)^1/n for all n $$ \ge $$ 2, then

The domain of definition of $$f(x) = \sqrt {{{1 - |x|} \over {2 - |x|}}} $$ is

If f : R $$ \to $$ R be defined by f (x) = e^x and g : R $$ \to $$ R be defined by g(x) = x². The mapping gof : R $$ \to $$ R be defined by (gof) (x) = g[f(x)] $$\forall $$x$$ \in $$R. Then,

For 0 $$ \le $$ p $$ \le $$ 1 and for any positive a, b; let I(p) = (a + b)^p, J(p) = a^p + b^p, then