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Question-176303




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

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