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Question

The solution set of the inequality ||x|1<1x,2xϵR is equal to

A
(0,)
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B
(1,)
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C
(1,1)
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D
(,0)
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Solution

The correct option is D (,0)
(,0)
Here because of |x| which is equalto either x or -x we shall consider the cases x>0 or x<0
Then we will have |x - 1| = |x + 1| when x < 0
Case 1: x > 0, then |x-1|<(1-x)
or |1-x| < (1-x) i.e. |t|<t
Above is not true for any value of t.
Case 2: x<0,|x1|<1x
or |x+1|<(1x)(1)
If x+10i.ex1 and x<0
i.e. 1x<0 i.e. xϵ[1,0) then (1) reduces to
x+1=1x or 2x<0 or x<0
If x+1<0 i.e. x<1 then (1) reduce to
(x+1)<(1x) or 2<0 is true for all x<1(2)
Hence from (1) and (2) xϵ(,0)

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