1.

The solution set of the inequation `|(1)/(x)-2| lt 4`, isA. `(-oo, -1//2)`B. `(1//6, oo)`C. `(-1//2, 1//6)`D. `(-oo, -1//2) cup (1//6, oo)`

Answer» Correct Answer - D
We have, `|(1)/(x)-2| lt 4, x ne 0`
`rArr (|1-2x|)/(|x|) lt 4 rArr |2x-1| lt 4 |x|`
Now, three cases arise:
CASE I When `x ge 1//2`. In this case, we have `|2x-1| = 2x-1 " and " |x| =x`
`therefore |2x-1| lt 4|x|`
`rArr 2x-1 lt 4x rArr 2x gt -1 rArr x gt -(1)/(2)`
But, `x ge (1)/(2), " Therefore," x ge (1)/(2)` is the solution in this case.
CASE II When `0 lt x lt 1//2:` In this case, we have
`|2x-1| = -(2x-1) " and "|x|=x`
`therefore |2x-1| lt 4 |x|`
`rArr -(2x-1) lt 4x rArr 6x gt 1 rArr x gt (1)/(6)`
But, `0 lt x lt 1//2." Therefore, " x in (1//6, 1//2)`
CASE III When `x lt 0`: In this case, we have
`|x| = -x " and "|2x-1|=-(2x-1)`
`therefore |2x-1|lt 4|x|`
`rArr -(2x-1) lt -4x rArr 2x lt -1 rArr x lt -(1)/(2)`
As `x lt 0. " Therefore, " x in (-oo, -1//2)` is the solution set in this case. Hence, `x in (-oo, -1//2) cup(1//6, oo)`.


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