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

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

Let $$\overrightarrow a = \overrightarrow i - \overrightarrow k ,\overrightarrow b = x\overrightarrow i + \overrightarrow j + \left( {1 - x} \right)\overrightarrow k $$ and
$$\overrightarrow c = y\overrightarrow i - x\overrightarrow j + \left( {1 + x - y} \right)\overrightarrow k .$$ Then $$\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$$ depends on

If $$\overrightarrow a \,,\,\overrightarrow b $$ and $$\overrightarrow c $$ are unit vectors, then $${\left| {\overrightarrow a - \overrightarrow b } \right|^2} + {\left| {\overrightarrow b - \overrightarrow c } \right|^2} + {\left| {\overrightarrow c - \overrightarrow a } \right|^2}$$ does NOT exceed