ΛΥΣΗ

α) Ισχύουν:

$$f({\displaystyle \frac{1}{2}})={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{{\displaystyle \frac{1}{2}}}}$$ $$={\displaystyle \frac{1}{2}}+2={\displaystyle \frac{1}{2}}+{\displaystyle \frac{4}{2}}={\displaystyle \frac{5}{2}}$$

$$f(1)=1+{\displaystyle \frac{1}{1}}$$ $$=1+1=2$$

$$f(2)=2+{\displaystyle \frac{1}{2}}$$ $$={\displaystyle \frac{4}{2}}+{\displaystyle \frac{1}{2}}={\displaystyle \frac{5}{2}}$$

Οπότε:

$$A=f({\displaystyle \frac{1}{2}})+f(1)-f(2)$$ $$={\displaystyle \frac{5}{2}}+2-{\displaystyle \frac{5}{2}}=2$$

β) Ισχύει ότι:

$$f(x)={\displaystyle \frac{5}{2}}$$ $$\iff x+{\displaystyle \frac{1}{x}}={\displaystyle \frac{5}{2}}$$ $$\iff {\displaystyle \frac{x^2+1}{x}}={\displaystyle \frac{5}{2}}$$ $$\iff 2(x^2+1)=5x$$ $$\iff 2x^2+2=5x$$ $$\iff 2x^2-5x+2=0$$

Το τριώνυμο $2x^2-5x+2$ έχει $\alpha =2$, $\beta =-5$, $\gamma =2$ και διακρίνουσα:

$$\Delta ={\beta }^2-4\alpha \gamma$$ $$=(-5{)}^2-4\cdot 2\cdot 2$$ $$=25-16=9>0$$

Οι ρίζες του τριωνύμου είναι οι:

$$x_{\text{1,2}}={\displaystyle \frac{-\beta \pm \sqrt{\Delta }}{2a}}$$ $$={\displaystyle \frac{-(-5)\pm \sqrt{9}}{2\cdot 2}}$$ $$={\displaystyle \frac{5\pm 3}{4}}$$ $$=\{\begin{array}{l}{\displaystyle \frac{5+3}{4}}=2\\ {\displaystyle \frac{5-3}{4}}={\displaystyle \frac{1}{2}}\end{array}$$