x-2x-3-interval-x-A-3-x-3-B-x-0-C-x-0-D-1-x-1-E-x-2-

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
$$\\$$$$\\$$$$−\mathrm{3}\leqslant\mid{x}\mid+\mathrm{2}{x}\leqslant\mathrm{3} \\$$$$−\mathrm{3}\leqslant\mathrm{3}{x}\leqslant\mathrm{3} \\$$$$−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\$$$$\\$$$${x}\leqslant\mathrm{1} \\$$$${x}\geqslant−\mathrm{1} \\$$$$\\$$$${Rta}\:\:{D} \\$$