Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x $$ \in $$ R. If f(x) attains maximum value at $$\alpha $$ and g(x) attains minimum value at $$\beta $$, then $$\mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}$$ is equal to :

The derivative of $${\tan ^{ - 1}}\left( {{{\sin x - \cos x} \over {\sin x + \cos x}}} \right)$$, with respect to $${x \over 2}$$ , where $$\left( {x \in \left( {0,{\pi \over 2}} \right)} \right)$$ is :

If the area (in sq. units) bounded by the parabola y² = 4$$\lambda $$x and the line y = $$\lambda $$x, $$\lambda $$ > 0, is $${1 \over 9}$$ , then $$\lambda $$ is equal to :

A value of $$\alpha $$ such that
$$\int\limits_\alpha ^{\alpha + 1} {{{dx} \over {\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}}} = {\log _e}\left( {{9 \over 8}} \right)$$ is :