Let $$f\left( x \right) = \left( {1 + {b^2}} \right){x^2} + 2bx + 1$$ and let $$m(b)$$ be the minimum value of $$f(x)$$. As $$b$$ varies, the range of $$m(b)$$ is

If $${\sin ^{ - 1}}\left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 4} - ....} \right)$$ $$$ + {\cos ^{ - 1}}\left( {{x^2} - {{{x^4}} \over 2} + {{{x^6}} \over 4} - ....} \right) = {\pi \over 2}$$$
for $$0 < \left| x \right| < \sqrt 2 ,$$ then $$x$$ equals

If $$f\left( x \right) = x{e^{x\left( {1 - x} \right)}},$$ then $$f(x)$$ is

The value of $$\int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {a^x}}}dx,\,a > 0,} $$ is