Let A = $$\left[ {\matrix{ 2 & b & 1 \cr b & {{b^2} + 1} & b \cr 1 & b & 2 \cr } } \right]$$ where b > 0.

Then the minimum value of $${{\det \left( A \right)} \over b}$$ is -

If $$\overrightarrow \alpha $$ = $$\left( {\lambda - 2} \right)\overrightarrow a + \overrightarrow b $$  and  $$\overrightarrow \beta = \left( {4\lambda - 2} \right)\overrightarrow a + 3\overrightarrow b $$ be two given vectors $$\overrightarrow a $$ and $$\overrightarrow b $$ are non-collinear. The value of $$\lambda $$ for which vectors $$\overrightarrow \alpha $$ and $$\overrightarrow \beta $$ are collinear, is -

Let S = $$\left\{ {\left( {x,y} \right) \in {R^2}:{{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}}} \right\};r \ne \pm 1.$$ Then S represents :

The value of   $$\int\limits_{ - \pi /2}^{\pi /2} {{{dx} \over {\left[ x \right] + \left[ {\sin x} \right] + 4}}} ,$$  where [t] denotes the greatest integer less than or equal to t, is