Question Number 193079 by Mastermind last updated on 03/Jun/23
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R}\:\mathrm{that}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{inequalities}: \\ $$$$\left.\mathrm{a}\right)\:\mid\mathrm{4x}−\mathrm{3}\mid\leqslant\mathrm{11} \\ $$$$\left.\mathrm{b}\right)\:\mid\mathrm{x}−\mathrm{2}\mid>\mid\mathrm{x}+\mathrm{1}\mid \\ $$$$\left.\mathrm{c}\right)\:\mid\mathrm{x}\mid\:+\:\mid\mathrm{x}+\mathrm{2}\mid\:+\:\mid\mathrm{2}−\mathrm{x}\mid\leqslant\mathrm{8} \\ $$
Commented by MM42 last updated on 04/Jun/23
![a)−11≤4x−3≤11⇒−8≤4x≤14⇒−2≤x≤3.5 b)if x≤−1⇒−x+2>−x−1⇒2>−1 is always correct ⇒ x≤−1 (i) if −1≤x≤2⇒−x+2>x+1⇒x<0.5⇒−1≤x<0.5 (ii) if 2≤x⇒x−2>x+1⇒−2>1⇒ is incorrect ⇒ans=(i)∪(ii)= (−∞,0.5) c)x≤−2⇒−x−x−2+2−x≤8⇒x≥−(8/3)⇒−(8/3)≤x≤−2 (i) −2≤x≤0⇒−x+x+2+2−x≤8⇒x≥−4⇒−2≤x≤0 (ii) 0≤x≤2⇒x+x+2+2−x≤8⇒x≤4⇒0≤x≤2 (iii) 2≤x⇒x+x+2−2+x≤8⇒x≤(8/3)⇒2≤x≤(8/3) (iv) ans=(i)∪(ii)∪(iii)∪(iv)=[−(8/3),(8/3)]](https://www.tinkutara.com/question/Q193088.png)
$$\left.{a}\right)−\mathrm{11}\leqslant\mathrm{4}{x}−\mathrm{3}\leqslant\mathrm{11}\Rightarrow−\mathrm{8}\leqslant\mathrm{4}{x}\leqslant\mathrm{14}\Rightarrow−\mathrm{2}\leqslant{x}\leqslant\mathrm{3}.\mathrm{5} \\ $$$$\left.{b}\right){if}\:\:{x}\leqslant−\mathrm{1}\Rightarrow−{x}+\mathrm{2}>−{x}−\mathrm{1}\Rightarrow\mathrm{2}>−\mathrm{1}\:{is}\:{always}\:{correct}\:\Rightarrow\:{x}\leqslant−\mathrm{1}\:\left({i}\right) \\ $$$${if}\:\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{2}\Rightarrow−{x}+\mathrm{2}>{x}+\mathrm{1}\Rightarrow{x}<\mathrm{0}.\mathrm{5}\Rightarrow−\mathrm{1}\leqslant{x}<\mathrm{0}.\mathrm{5}\:\:\:\left({ii}\right) \\ $$$${if}\:\:\mathrm{2}\leqslant{x}\Rightarrow{x}−\mathrm{2}>{x}+\mathrm{1}\Rightarrow−\mathrm{2}>\mathrm{1}\Rightarrow\:{is}\:{incorrect} \\ $$$$\Rightarrow{ans}=\left({i}\right)\cup\left({ii}\right)=\:\left(−\infty,\mathrm{0}.\mathrm{5}\right) \\ $$$$\left.{c}\right){x}\leqslant−\mathrm{2}\Rightarrow−{x}−{x}−\mathrm{2}+\mathrm{2}−{x}\leqslant\mathrm{8}\Rightarrow{x}\geqslant−\frac{\mathrm{8}}{\mathrm{3}}\Rightarrow−\frac{\mathrm{8}}{\mathrm{3}}\leqslant{x}\leqslant−\mathrm{2}\:\:\left({i}\right)\: \\ $$$$−\mathrm{2}\leqslant{x}\leqslant\mathrm{0}\Rightarrow−{x}+{x}+\mathrm{2}+\mathrm{2}−{x}\leqslant\mathrm{8}\Rightarrow{x}\geqslant−\mathrm{4}\Rightarrow−\mathrm{2}\leqslant{x}\leqslant\mathrm{0}\:\:\left({ii}\right) \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\Rightarrow{x}+{x}+\mathrm{2}+\mathrm{2}−{x}\leqslant\mathrm{8}\Rightarrow{x}\leqslant\mathrm{4}\Rightarrow\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\:\:\left({iii}\right) \\ $$$$\mathrm{2}\leqslant{x}\Rightarrow{x}+{x}+\mathrm{2}−\mathrm{2}+{x}\leqslant\mathrm{8}\Rightarrow{x}\leqslant\frac{\mathrm{8}}{\mathrm{3}}\Rightarrow\mathrm{2}\leqslant{x}\leqslant\frac{\mathrm{8}}{\mathrm{3}}\:\:\left({iv}\right)\:\: \\ $$$${ans}=\left({i}\right)\cup\left({ii}\right)\cup\left({iii}\right)\cup\left({iv}\right)=\left[−\frac{\mathrm{8}}{\mathrm{3}},\frac{\mathrm{8}}{\mathrm{3}}\right] \\ $$$$ \\ $$
Commented by Subhi last updated on 03/Jun/23
$$\left({C}\right)\:\mid{a}+{b}+{c}\mid\leqslant\mid{a}\mid+\mid{b}\mid+\mid{c}\mid \\ $$$$\mid{x}\mid+\mid{x}+\mathrm{2}\mid+\mid{x}−\mathrm{2}\mid\geqslant\mid{x}+{x}+{x}+\mathrm{2}−\mathrm{2}\mid=\mid\mathrm{3}{x}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\downarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{2}−{x}\mid \\ $$$$\mid\mathrm{3}{x}\mid\leqslant\mathrm{8} \\ $$$$\frac{−\mathrm{8}}{\mathrm{3}}\leqslant{x}\leqslant\frac{\mathrm{8}}{\mathrm{3}} \\ $$