Let f (x) = x² -bx + c, b is a odd positive integer, f (x) = 0 have two prime numbers as
roots and b+c = 35. Then the global minimum value of f (x) is

$$\begin{gathered} \text{ Given, function is }f\left(x\right)=x^2-bx+c\text{. } \\ \text{ Consider }\alpha ,\beta \text{ be the roots of the function }\mathrm{f}\left(\mathrm{x}\right)\text{. } \\ \Rightarrow \alpha +\beta =\mathrm{b} \\ \text{ Here, }\mathrm{f}\left(\mathrm{x}\right)=0\text{ has to prime numbers as roots and }\mathrm{b}\text{ is an odd positive integer. } \\ \Rightarrow \text{ One root of the function }\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-\mathrm{bx}+\mathrm{c}\text{ as }2 \\ \begin{array}{l}\Rightarrow \mathrm{f}\left(2\right)=0\\ \Rightarrow 2^2-\mathrm{b}\left(2\right)+\mathrm{c}=0\\ \Rightarrow 2\mathrm{b}-\mathrm{c}=4\dots \left(1\right) \\ \text{ and }b+c=35\dots \text{ (2) } \\ \Rightarrow b=13\text{ and }c=22\end{array} \\ \text{ The global minimum value of }f\left(x\right)\text{ is attained }x=\frac{13}{2} \\ \begin{array}{l}it\text{'}s value=f\left(\frac{13}{2}\right)={\left(\frac{13}{2}\right)}^2-13\cdot \left(\frac{13}{2}\right)+22\\ =-\frac{81}{4}\end{array} \end{gathered}$$