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Question

Solve 1|x|21, x±2
  1. [1,2) 
  2. (2,1][1,2)  
  3. [1,1] 
  4. (2,1] 


Solution

The correct option is B (2,1][1,2)  
1|x|211+1|x|20|x|1|x|20
|x|10 and |x|2>0
|x|1 and |x|>2 Not possible

Or, |x|10 and |x|2<0
|x|1 and |x|<2
x(,1][1,) and x(2,2)

x(2,1][1,2)

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