The outcome of each of 30 items was observed; 10 items gave an outcome $${1 \over 2}$$ – d each, 10 items gave outcome $${1 \over 2}$$ each and the remaining 10 items gave outcome $${1 \over 2}$$+ d each. If the variance of this outcome data is $${4 \over 3}$$ then |d| equals :

The maximum value of the function f(x) = 3x³ – 18x² + 27x – 40 on the set S = $$\left\{ {x\, \in R:{x^2} + 30 \le 11x} \right\}$$ is :

Let f : R $$ \to $$ R be defined by f(x) = $${x \over {1 + {x^2}}},x \in R$$.   Then the range of f is :

Let  $$\overrightarrow a = \widehat i + 2\widehat j + 4\widehat k,$$ $$\overrightarrow b = \widehat i + \lambda \widehat j + 4\widehat k$$ and $$\overrightarrow c = 2\widehat i + 4\widehat j + \left( {{\lambda ^2} - 1} \right)\widehat k$$ be coplanar vectors. Then the non-zero vector $$\overrightarrow a \times \overrightarrow c $$ is :