Let $$R_{1}$$ and $$R_{2}$$ be two relations defined on $$\mathbb{R}$$ by

$$a \,R_{1} \,b \Leftrightarrow a b \geq 0$$ and $$a \,R_{2} \,b \Leftrightarrow a \geq b$$

Then,

Let $$f, g: \mathbb{N}-\{1\} \rightarrow \mathbb{N}$$ be functions defined by $$f(a)=\alpha$$, where $$\alpha$$ is the maximum of the powers of those primes $$p$$ such that $$p^{\alpha}$$ divides $$a$$, and $$g(a)=a+1$$, for all $$a \in \mathbb{N}-\{1\}$$. Then, the function $$f+g$$ is

Let the minimum value $$v_{0}$$ of $$v=|z|^{2}+|z-3|^{2}+|z-6 i|^{2}, z \in \mathbb{C}$$ is attained at $${ }{z}=z_{0}$$. Then $$\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2}$$ is equal to :

Let $$A=\left(\begin{array}{cc}1 & 2 \\ -2 & -5\end{array}\right)$$. Let $$\alpha, \beta \in \mathbb{R}$$ be such that $$\alpha A^{2}+\beta A=2 I$$. Then $$\alpha+\beta$$ is equal to