Correct Answer - B
We have `||x-1|-5| lt 2x -5 `
`therefore 2x- 5 gt 0 `
` rArr x gt 5/2`
Now from (1) ,
`-2 x +5 lt |x-1|-5 lt2x -5 `
`rArr -2x+ 10 lt |x-1| lt 2x`
Case I: `x ge 1`
`therefore -2x10 lt |x-1| -5 lt 2x -5 `
`rArr x gt 11//3`
Case II `x lt 1`
`therefore -2x +10 lt 1 -x lt 2x `
`rArr x gt 9 and x gt 1//3`
`rArr x gt 9 " not possible as" x lt 1`
Therefore `x in (11//3 , oo)`
