Question

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Answer 1

\[\left| \left| x \right|-1 \right|<\left| 1-x \right| \]
x belongs to R

Answer 2

The sollution of this
(a) (-1,1)
(b)(0,infinity)
(c)(-1,infinity)
(d)None of these

Answer 3

You could try plugging in the answers and see if you can't reverse-engineer the logic to fit.

Answer 4

||x|-1|<|1-x|<|1-|x||

Answer 5

Is there a algebraic method

Answer 6

3 cases :
x = 0
x < 0
x > 0

Answer 7

looks Bswan has something :) lets see that first maybe..

Answer 8

continue the 3 cases

Answer 9

Yes i saw that

Answer 10

Won't mod of x be just "x"

Answer 11

when x = 0, clearly |0-1| < |1-0| is false so \( 0 \not \in S\)

\(S = \)solution set

Answer 12

when \(x > 0, |x| = x\)
\(\implies |x-1| \lt |1-x| ~~?\)
\(\implies |1-x| \lt |1-x| ~~?\)
\(\implies FALSE\) so \(x> 0 \) also will not work

Answer 13

you wana try the \(x < 0\) case ? it may work...

Answer 14

Yes the wolfram says the same

Answer 15

i would say
(d)None of these
its ovs that
|x|>=x
|x|-1>= x-1
|x|-1>=-(1-x)
||x|-1|>=-|1-x|
mmm something confusing meh >.>

Answer 16

i will leave this to ask my professor probably

Answer 17

take examples!

like
x= 0, -1,-2,-3, 3,4,5....

Answer 18

let ganeshie continue

Answer 19

wym you will leave this ?
its a trivial problem lol, just pen down and work case3

Answer 20

when x<0

Answer 21

case3 : when \(x \lt 0, |x| = -x = a\)
\(\implies |a-1| \lt |1+a| ~~?\)
\(\implies TRUE \) so \(x \lt 0\) is the solution