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Question Number 63162 by Rio Michael last updated on 29/Jun/19

$$Findthesetofvaluesofxwhichsatisfytheinequalities$$ $$\frac{2}{x-1}⩽\frac{1}{x}and{x}^2-\mid 3x\mid +2<0$$

Commented byPrithwish sen last updated on 30/Jun/19

$$\frac{2}{\mathrm{x}-1}⩽\frac{1}{\mathrm{x}}\Rightarrow \mathrm{x}⩽-1$$ $${\mathrm{x}}^2-\mid 3\mathrm{x}\mid +2<0\Rightarrow {\mathrm{x}}^2+3\mathrm{x}+2<0(\because \mathrm{x}⩽-1)$$ $$\mathrm{i}.\mathrm{e}(\mathrm{x}+1)(\mathrm{x}+2)<0$$ $$\mathrm{either}\mathrm{or}$$ $$(\mathrm{x}+1)<0(\mathrm{x}+1)>0$$ $$\mathrm{and}(\mathrm{x}+2)>0\mathrm{and}(\mathrm{x}+2)<0$$ $$\Rightarrow -2<\mathrm{x}<-1\Rightarrow -1<\mathrm{x}<-2$$ $$\because \mathrm{x}⩽-1\therefore \mathrm{the}\mathrm{only}\mathrm{possible}\mathrm{solution}\mathrm{is}$$ $$-2<\mathrm{x}<-1$$ $$\mathrm{please}\mathrm{check}\mathrm{the}\mathrm{answer}.$$

Commented byRio Michael last updated on 02/Jul/19

$$ithinkitmakessense..butexplicitforOlevelstudentsuknow..$$ $$Alevelstudentscanunderstand$$

Commented bymathmax by abdo last updated on 30/Jun/19

$$\left(e_1\right)forx\ne 0andx\ne 1\frac{2}{x-1}⩽\frac{1}{x}\iff \frac{2}{x-1}-\frac{1}{x}⩽0\Rightarrow \frac{2x-x+1}{x(x-1)}⩽0\Rightarrow$$ $$\frac{x+1}{x(x-1)}⩽0$$ $$x-\mathrm{\infty }-101+\mathrm{\infty }$$ $$x+1-0+++$$ $$x(x-1)++0-0+$$ $$\frac{x+1}{x(x-1)}-+-+$$ $$\Rightarrow D_1=]-\mathrm{\infty },-1]\cup ]0,1[$$ $$\left(e_2\right)x^2-3\mid x\mid +2<0\Rightarrow \mid x\mid -3\mid x\mid +2<0$$ $$\mathrm{\Delta }=9-8=1\Rightarrow \mid x{\mid }_1=\frac{3+1}{2}=2\Rightarrow x=\stackrel{-}{+}2$$ $$\mid x{\mid }_2=\frac{3-1}{2}=1\Rightarrow x=\stackrel{-}{+}1\Rightarrow x^2-3\mid x\mid +2=(\mid x\mid -1)(\mid x\mid -2)$$ $$\left(e_2\right)\Rightarrow (\mid x\mid -1)(\mid x\mid -2)<0\Rightarrow \mid x\mid \in ]1,2[\Rightarrow 1<\mid x\mid <2$$ $$1<\mid x\mid \Rightarrow x>1orx<-1\Rightarrow x\in ]-\mathrm{\infty },-1[\cup ]1,+\mathrm{\infty }[$$ $$\mid x\mid <2\Rightarrow -2<x<2\Rightarrow \cap =]-2,-1[\cup ]1,2[=D_2\Rightarrow$$ $$D=D_1\cap D_2=....$$

Answered by ajfour last updated on 30/Jun/19

$$1<\mid x\mid <2$$ $$\{\begin{array}{l}1<x<2\\ x⩽-1\end{array}or\{\begin{array}{l}-2<x<-1\\ x⩽-1\end{array}$$ $$\Rightarrow x\in \varphi or\Rightarrow -2<x<-1$$ $$hencex\in (-1,-2).$$

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