The value of the integral $$\int\limits_0^1 {\sqrt {{{1 - x} \over {1 + x}}} dx} $$ is

If $$\omega $$ $$\left( { \ne 1} \right)$$ be a cube root of unity and $${\left( {1 + {\omega ^2}} \right)^n} = {\left( {1 + {\omega ^4}} \right)^n},$$ then the least positive value of n is

If $$f\left( x \right) = {x^a}\log x$$ and $$f\left( 0 \right) = 0,$$ then the value of $$\alpha $$ for which Rolle's theorem can be applied in $$\left[ {0,1} \right]$$ is

If $$f\left( x \right) = {x^3} + b{x^2} + cx + d$$ and $$0 < {b^2} < c,$$ then in $$\left( { - \infty ,\infty } \right)$$