If the minimum value of $$f(x)=\frac{5 x^{2}}{2}+\frac{\alpha}{x^{5}}, x>0$$, is 14 , then the value of $$\alpha$$ is equal to :

Let $$f(x) = \left\{ {\matrix{ {{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr { - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr } } \right.$$.

Then the set of all values of b, for which f(x) has maximum value at x = 1, is :

The curve $$y(x)=a x^{3}+b x^{2}+c x+5$$ touches the $$x$$-axis at the point $$\mathrm{P}(-2,0)$$ and cuts the $$y$$-axis at the point $$Q$$, where $$y^{\prime}$$ is equal to 3 . Then the local maximum value of $$y(x)$$ is:

If xy⁴ attains maximum value at the point (x, y) on the line passing through the points (50 + $$\alpha$$, 0) and (0, 50 + $$\alpha$$), $$\alpha$$ > 0, then (x, y) also lies on the line :