Question Number 213467 by efronzo1 last updated on 06/Nov/24
$$\:\:\:\underbrace{\boldsymbol{{B}}} \\ $$
Commented by mr W last updated on 06/Nov/24
��
Commented by mr W last updated on 06/Nov/24
$${question}\:{is}\:{strange}. \\ $$$${if}\:{X}\geqslant\mathrm{1},\:{then}\:{X}\geqslant\mathrm{2},\:{X}\geqslant\mathrm{3},\:{etc}.\:{are}\:{all} \\ $$$${true}.\:{that}\:{means}\:{for}\:{X}\geqslant{a},\:{there}\:{is} \\ $$$${no}\:{maximum}\:{value}\:{of}\:{a}. \\ $$
Commented by A5T last updated on 06/Nov/24
$${Say}\:{X}=\mid{x}−\mathrm{1}\mid+\mid{x}−\mathrm{2}\mid+\mid{x}−\mathrm{3}\mid+…+\mid{x}−\mathrm{2022}\mid\geqslant{a}^{\mathrm{2}} \\ $$$${must}\:{be}\:{true}\:{for}\:{all}\:{x}. \\ $$$${I}\:{showed}\:{that}\:{X}\geqslant\mathrm{1011}^{\mathrm{2}} \:{and}\:{equality}\:{holds}\:{when} \\ $$$${x}=\frac{\mathrm{2023}}{\mathrm{2}}. \\ $$$${So},\:{the}\:{maximum}\:{a}\:{for}\:{which}\:{X}\geqslant{a}^{\mathrm{2}} \:{will}\:{be}\:{true} \\ $$$${for}\:{all}\:{x}\:{is}\:\mathrm{1011}.\:{So},\:{if}\:{a}>\mathrm{1011};\:{then}\:{X}\geqslant{a}^{\mathrm{2}} \:{will} \\ $$$${no}\:{longer}\:{be}\:{true}\:{for}\:{x}=\frac{\mathrm{2023}}{\mathrm{2}}\:{for}\:{instance}. \\ $$
Answered by A5T last updated on 06/Nov/24
$$\mid{x}−\mathrm{1}\mid+\mid\mathrm{2022}−{x}\mid+\mid{x}−\mathrm{2}\mid+\mid\mathrm{2021}−{x}\mid+…+ \\ $$$$\mid{x}−\mathrm{1011}\mid+\mid\mathrm{1012}−{x}\mid\geqslant\mathrm{2021}+\mathrm{2019}+\mathrm{2017}+…+\mathrm{1} \\ $$$$=\mathrm{1011}^{\mathrm{2}} \Rightarrow{Maximum}\:{value}\:{of}\:{a}=\mathrm{1011} \\ $$