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 $$\alpha, \beta$$ and $$\gamma$$ be three positive real numbers. Let $$f(x)=\alpha x^{5}+\beta x^{3}+\gamma x, x \in \mathbf{R}$$ and $$g: \mathbf{R} \rightarrow \mathbf{R}$$ be such that $$g(f(x))=x$$ for all $$x \in \mathbf{R}$$. If $$\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}$$ be in arithmetic progression with mean zero, then the value of $$f\left(g\left(\frac{1}{\mathrm{n}} \sum\limits_{i=1}^{\mathrm{n}} f\left(\mathrm{a}_{i}\right)\right)\right)$$ is equal to :

Consider the sequence $$a_{1}, a_{2}, a_{3}, \ldots$$ such that $$a_{1}=1, a_{2}=2$$ and $$a_{n+2}=\frac{2}{a_{n+1}}+a_{n}$$ for $$\mathrm{n}=1,2,3, \ldots .$$ If $$\left(\frac{\mathrm{a}_{1}+\frac{1}{\mathrm{a}_{2}}}{\mathrm{a}_{3}}\right) \cdot\left(\frac{\mathrm{a}_{2}+\frac{1}{\mathrm{a}_{3}}}{\mathrm{a}_{4}}\right) \cdot\left(\frac{\mathrm{a}_{3}+\frac{1}{\mathrm{a}_{4}}}{\mathrm{a}_{5}}\right) \ldots\left(\frac{\mathrm{a}_{30}+\frac{1}{\mathrm{a}_{31}}}{\mathrm{a}_{32}}\right)=2^{\alpha}\left({ }^{61} \mathrm{C}_{31}\right)$$, then $$\alpha$$ is equal to :

The minimum value of the twice differentiable function $$f(x)=\int\limits_{0}^{x} \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}$$, $$x \in \mathbf{R}$$, is :