Question Number 10973 by ridwan balatif last updated on 05/Mar/17
$$\mid\mid{x}\mid+\mathrm{2x}\mid\leqslant\mathrm{3},\:\mathrm{interval}\:\mathrm{x}=...? \\ $$$$\mathrm{A}.−\mathrm{3}\leqslant{x}\leqslant\mathrm{3} \\ $$$$\mathrm{B}.\:{x}\geqslant\mathrm{0} \\ $$$$\mathrm{C}.\:{x}\leqslant\mathrm{0} \\ $$$$\mathrm{D}.\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$$\mathrm{E}.\:{x}\leqslant\mathrm{2} \\ $$
Commented by ajfour last updated on 05/Mar/17
$$ \\ $$$$ \\ $$$${D}.\: \\ $$$${When}\:{x}\geqslant\mathrm{0}\:,\:\mid\mid{x}\mid+\mathrm{2}{x}\mid=\mid\mathrm{3}{x}\mid\:=\mathrm{3}{x} \\ $$$${while}\:{when}\:{x}\leqslant\mathrm{0},\:{the}\:{same}\:{is}\:\mid−{x}+\mathrm{2}{x}\mid \\ $$$$=\:\mid{x}\mid\:=−{x}\:. \\ $$$${so}\:{the}\:{entire}\:{range}\:{is}\:\:−\mathrm{3}\leqslant\:{x}\leqslant\:\mathrm{1}\:. \\ $$$$..{as}\:{can}\:{be}\:{viewed}\:{in}\:{the}\:{graph}\:{below}. \\ $$$$ \\ $$$$ \\ $$
Answered by ajfour last updated on 05/Mar/17
Commented by ridwan balatif last updated on 05/Mar/17
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by sandy_suhendra last updated on 05/Mar/17
$$\mathrm{if}\:\mathrm{x}\geqslant\mathrm{0}\:\Rightarrow\:\mid\mathrm{x}+\mathrm{2x}\mid\leqslant\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{3x}\mid\leqslant\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{x}\mid\leqslant\mathrm{1}\:\Rightarrow\:−\:\mathrm{1}\leqslant\mathrm{x}\leqslant\mathrm{1} \\ $$$$\mathrm{so}\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{1}\:...\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{x}<\mathrm{0}\:\Rightarrow\:\mid\mathrm{x}−\mathrm{2x}\mid\leqslant\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid−\mathrm{x}\mid\leqslant\mathrm{3}\:\Rightarrow\:−\mathrm{3}\leqslant\mathrm{x}\leqslant\mathrm{3} \\ $$$$\mathrm{so}\:−\mathrm{3}\leqslant\mathrm{x}<\mathrm{0}\:...\:\left(\mathrm{2}\right) \\ $$$$\mathrm{from}\:\left(\mathrm{1}\right)\cup\left(\mathrm{2}\right)=\left\{−\mathrm{3}\leqslant\mathrm{x}\leqslant\mathrm{1}\right\} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{choose}\:\mathrm{D} \\ $$
Commented by ridwan balatif last updated on 05/Mar/17
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by Mechas88 last updated on 06/Mar/17
$$ \\ $$$$ \\ $$$$−\mathrm{3}\leqslant\mid{x}\mid+\mathrm{2}{x}\leqslant\mathrm{3} \\ $$$$−\mathrm{3}\leqslant\mathrm{3}{x}\leqslant\mathrm{3} \\ $$$$−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$$ \\ $$$${x}\leqslant\mathrm{1} \\ $$$${x}\geqslant−\mathrm{1} \\ $$$$ \\ $$$${Rta}\:\:{D} \\ $$