Question #228620

, The solutions of the inequality IIxI-1I <=1/2 are ?



Expert's answer

∣∣x∣−1∣≤12\lvert \lvert x \rvert - 1\rvert \leq \cfrac {1}{2}


There are 3 cases:


1) ∣x∣−1=0\lvert x \rvert - 1 = 0

2) ∣x∣−1>0\lvert x \rvert - 1 > 0

3) ∣x∣−1<0\lvert x \rvert - 1 < 0


1) ∣x∣−1=0\lvert x \rvert - 1 = 0

∣x∣=1x∈{−1;1}\lvert x \rvert = 1 \\ x \in \lbrace -1; 1\rbrace - satisfies the inequality.


2) ∣x∣−1>0:\lvert x \rvert - 1 > 0 :


0<∣x∣−1≤121<∣x∣≤320 < \lvert x \rvert - 1 \leq \cfrac{1}{2} \\ 1 < \lvert x \rvert \leq \cfrac{3}{2}


Answer: x∈[−32;−1}⋃{1;32]x \in \Big [ - \cfrac {3}{2} ; -1 \Big \} \bigcup \Big \{ 1; \cfrac{3}{2} \Big ]


3) ∣x∣−1<0:\lvert x \rvert - 1 < 0 :


−12≤∣x∣−1<012≤∣x∣<1-\cfrac{1}{2} \leq \lvert x \rvert - 1 < 0 \\ \cfrac{1}{2} \leq \lvert x \rvert < 1


Answer: x∈{−1;−12]⋃[12;1}x \in \Big \{ - 1 ; - \cfrac{1}{2} \Big ] \bigcup \Big [ \cfrac{1}{2} ; 1 \Big \}



Final answer: x∈[−32;−12]⋃[12;32]x \in \Big [ - \cfrac{3}{2}; - \cfrac{1}{2} \Big ] \bigcup \Big [ \cfrac{1}{2} ; \cfrac {3}{2} \Big ]

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