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1/(|x|-3) le1/2...

`1/(|x|-3) le1/2`

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Given, `1/(|x|-3) le1/2`
`rArr |x|-3ge2 [:. 1/alt1/b rArr agtb]`
`rArr |x|ge5` [adding 3 to both sides]
`rArr xle-5 or x ge5 [:. |x|ge a rArr |x|le-a rArr |x|gea]`

`rArr x in (-infty, -5]cup[5,infty)`....(i)
But ` |x|-3 ne 0`
Either ` |x|-3lt0 or |x|-3 gt0`
`rArr |x| lt3 or |x|gt3`
`rArr -3lt x lt 3 or x lt-3 or x gt3`....(ii)
`[:' |x|lta rArr -a lt x lt and |x| gt a rArr x lt - a xgta]`
On combining results of Eqs. (i) and (ii), we get
`x in (-infty, - 5] cup (-3, 3) cup[5, infty)`
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