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

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

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

Let $$f:\left( {0,\infty } \right) \to R$$ and $$F\left( x \right) = \int\limits_0^x {f\left( t \right)dt.} $$ If $$F\left( {{x^2}} \right) = {x^2}\left( {1 + x} \right)$$, then $$f(4)$$ equals