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  • Let <img alt="\mathrm{A=\left \{ x \in \mathbb{R} : \left | x+1 \right |< 2 \right \}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BA%3D%5Cleft%20%5C%7B%20x%20%5Cin%20%5Cmat

Let \mathrm{A=\left \{ x \in \mathbb{R} : \left | x+1 \right |< 2 \right \}} and \mathrm{ B =\left \{ x \in \mathbb{R} : \left | x-1 \right |\geq 2 \right \}}. Then which one of the following statements is NOT true ?

Option: 1

\mathrm{A-B=\left ( -1,1 \right )}


Option: 2

\mathrm{B-A=\mathbb{R}-\left ( -3,1 \right )}


Option: 3

\mathrm{A\cap B=(-3,-1 ] }


Option: 4

\mathrm{A\cup B=\mathbb{R}-{[1,3)}


Answers (1)

best_answer

\mathrm{|x+1|<2 \Rightarrow-2<x+1<2 \Rightarrow x \in(-3,1)=A}\\

\mathrm{|x-1| \geq 2 \Rightarrow x-1 \leq-2 \vee x-1 \geq 2}\\

\mathrm{ \Rightarrow x \leq-1 \cup x \geq 3 \Rightarrow x \in(-\infty-1] \cup [3, \infty)}

\mathrm{A=(-3,1) \quad B=(-\infty,-1] \cup[3, \infty)}\\

\mathrm{A^{\prime}=(-\infty,-3] \cup[1, \infty) \quad B^{\prime}=(-1,3)}\\

\mathrm{A-B=A \cap B^{\prime}=(-1,1)}

\mathrm{B-A=A^{\prime} \cap B=(-\infty,-3] \cup[3, \infty)}\\

\mathrm{A \cup B=(-\infty, 1] \cup[3, \infty)}\\

\mathrm{A \cap B=(-3,-1]

 

Posted by

Suraj Bhandari

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