Question Number 176303 by infinityaction last updated on 16/Sep/22
Answered by cortano1 last updated on 16/Sep/22
$$\:\mathrm{let}\:\begin{cases}{{a}+{b}={x}}\\{{c}+{d}={y}}\end{cases}\Rightarrow{x}+{y}=\mathrm{2}\:;\:\mathrm{y}=\mathrm{2}−\mathrm{x} \\ $$$$\:\mathrm{M}=\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{x}\left(\mathrm{2}−\mathrm{x}\right)=\mathrm{2x}−\mathrm{x}^{\mathrm{2}} \:;\:\mathrm{x}\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}\leqslant\mathrm{f}\left(\mathrm{x}\right)\leqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{0}\leqslant\:\mathrm{M}\leqslant\mathrm{1} \\ $$
Commented by cortano1 last updated on 16/Sep/22
$$\mathrm{yes}..\mathrm{sir}.\:\mathrm{thanks} \\ $$