Question
The solution set of the inequality ||x|−1<1−x,2∀xϵR is equal to
Answer: Option D
:
D
(−∞,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,|−x−1|<1−x
or |x+1|<(1−x)……(1)
If x+1≥0i.ex≥−1 and x<0
i.e. −1≤x<0 i.e. xϵ[−1,0) then (1) reduces to
x+1=1−x or 2x<0 or x<0
If x+1<0 i.e. x<−1 then (1) reduce to
−(x+1)<(1−x) or −2<0 is true for all x<−1……(2)
Hence from (1) and (2) xϵ(−∞,0)
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:
D
(−∞,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,|−x−1|<1−x
or |x+1|<(1−x)……(1)
If x+1≥0i.ex≥−1 and x<0
i.e. −1≤x<0 i.e. xϵ[−1,0) then (1) reduces to
x+1=1−x or 2x<0 or x<0
If x+1<0 i.e. x<−1 then (1) reduce to
−(x+1)<(1−x) or −2<0 is true for all x<−1……(2)
Hence from (1) and (2) xϵ(−∞,0)
Was this answer helpful ?
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