Question Number 16519 by tawa tawa last updated on 23/Jun/17
$$\mathrm{Solve}: \\ $$$$\mid\mathrm{2}\:−\:\mathrm{x}\mid\:−\:\mathrm{2}\:\mid\mathrm{x}\:+\:\mathrm{1}\mid\:<\:\mathrm{1} \\ $$
Answered by ajfour last updated on 23/Jun/17
Commented by ajfour last updated on 23/Jun/17
![∣2−x∣−2∣x+1∣<1 whenever y=∣2−x∣ is below y= 2∣x+1∣+1 this occurs for x<x_1 (see graph) and x>x_2 to obtain x_1 (wbich is <−1): 2−x_1 = 1−2(x_1 +1) x_1 = −3 to obtain x_2 (−1<x_2 <2) 2−x_2 = 1+2(x_2 +1) 3x_2 = −1 ⇒ x_2 = −1/3 so, for given condition to remain true x ∉ [−3, ((−1)/3)] that is x∈ (−∞, −3)∪(((−1)/3), ∞).](https://www.tinkutara.com/question/Q16527.png)
$$\mid\mathrm{2}−{x}\mid−\mathrm{2}\mid{x}+\mathrm{1}\mid<\mathrm{1} \\ $$$${whenever}\:{y}=\mid\mathrm{2}−{x}\mid\:{is}\:{below} \\ $$$$\:{y}=\:\mathrm{2}\mid{x}+\mathrm{1}\mid+\mathrm{1} \\ $$$$\:{this}\:{occurs}\:{for}\:{x}<{x}_{\mathrm{1}} \:\left({see}\:{graph}\right) \\ $$$$\:{and}\:\:{x}>{x}_{\mathrm{2}} \\ $$$${to}\:{obtain}\:{x}_{\mathrm{1}} \left({wbich}\:{is}\:<−\mathrm{1}\right): \\ $$$$\:\mathrm{2}−{x}_{\mathrm{1}} \:=\:\mathrm{1}−\mathrm{2}\left({x}_{\mathrm{1}} +\mathrm{1}\right) \\ $$$$\:{x}_{\mathrm{1}} =\:−\mathrm{3}\: \\ $$$${to}\:{obtain}\:{x}_{\mathrm{2}} \:\:\left(−\mathrm{1}<{x}_{\mathrm{2}} \:<\mathrm{2}\right) \\ $$$$\:\mathrm{2}−{x}_{\mathrm{2}} \:=\:\mathrm{1}+\mathrm{2}\left({x}_{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:\mathrm{3}{x}_{\mathrm{2}} =\:−\mathrm{1}\:\:\Rightarrow\:\:\:{x}_{\mathrm{2}} \:=\:−\mathrm{1}/\mathrm{3}\:\: \\ $$$${so},\:{for}\:{given}\:{condition}\:{to} \\ $$$${remain}\:{true} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}\:\notin\:\left[−\mathrm{3},\:\frac{−\mathrm{1}}{\mathrm{3}}\right] \\ $$$$\:\:\:{that}\:{is}\:{x}\in\:\left(−\infty,\:−\mathrm{3}\right)\cup\left(\frac{−\mathrm{1}}{\mathrm{3}},\:\infty\right). \\ $$
Commented by tawa tawa last updated on 23/Jun/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 24/Jun/17
$$\left.\mathrm{1}\right){x}\geqslant\mathrm{2}\Rightarrow−\left(\mathrm{2}−{x}\right)−\mathrm{2}\left({x}+\mathrm{1}\right)<\mathrm{1}\Rightarrow \\ $$$$−{x}−\mathrm{4}<\mathrm{1}\Rightarrow−{x}<−\mathrm{5}\Rightarrow{x}>\mathrm{5}\Rightarrow{x}\in\left[\mathrm{5},\infty\right) \\ $$$$\left.\mathrm{2}\right)\:−\mathrm{1}\leqslant{x}<\mathrm{2}\Rightarrow\mathrm{2}−{x}−\mathrm{2}\left({x}+\mathrm{1}\right)<\mathrm{1}\Rightarrow \\ $$$$−\mathrm{3}{x}<\mathrm{1}\Rightarrow{x}>−\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{x}\in\left[−\frac{\mathrm{1}}{\mathrm{3}},\mathrm{2}\right) \\ $$$$\Rightarrow{x}\in\left[−\frac{\mathrm{1}}{\mathrm{3}},\mathrm{2}\right)\cup\left[\mathrm{5},\infty\right). \\ $$
Commented by tawa tawa last updated on 24/Jun/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by ajfour last updated on 24/Jun/17
$${if}\:{x}=\mathrm{3}\:>\mathrm{2} \\ $$$$\:\mid\mathrm{2}−{x}\mid=\:\mathrm{1} \\ $$$$\:\:\mathrm{2}\mid{x}+\mathrm{1}\mid=\mathrm{8} \\ $$$$\:\:\mid\mathrm{2}−{x}\mid−\mathrm{2}\mid{x}+\mathrm{1}\mid=−\mathrm{7}\:<\mathrm{1} \\ $$$${if}\:\:{x}=−\mathrm{4} \\ $$$$\:\:\mid\mathrm{2}−{x}\mid=\:\mathrm{6} \\ $$$$\:\mathrm{2}\mid{x}+\mathrm{1}\mid\:=\:\mathrm{6} \\ $$$$\mid\mathrm{2}−{x}\mid−\mathrm{2}\mid{x}+\mathrm{1}\mid=\mathrm{0}\:<\:\mathrm{1} \\ $$$$\:{if}\:{x}\rightarrow−\infty \\ $$$$\:\mid\mathrm{2}−{x}\mid−\mathrm{2}\mid{x}+\mathrm{1}\mid=\:\mathrm{2}−{x}+\mathrm{2}\left({x}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{x}+\mathrm{4}\:\:=\:−\infty+\mathrm{4}\:<\mathrm{1}\: \\ $$
Commented by tawa tawa last updated on 24/Jun/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$