InterviewSolution
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InterviewSolution
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Solve `(2)/(|x-3|)gt5,x inR.` |
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Answer» Clearly, `x-3ne0 "and therefore",xne3`. We have, `(2)/(|x-3|)gt5" "...(i)` Since `|x-3|` is positive, we may multiply both sides of (i) by `|x-3|`. This gives `2gt5|x-3|` `iff (2)/(5)gt|x-3|` `iff|x-3|gt(2)/(5)` `iff (-2)/(5)ltx-3lt(2)/(5)" " [therefore|x|lta iff-altxlta]` `iff (-2)/(5)ltx-3and x-3lt(2)/(5)` `iff (-2)/(5)+3ltx and xlt(2)/(5)+3` `iff (13)/(5)ltx and xlt(17)/(5)`. `iff (13)/(5)ltxlt(17)/(5)`. Also, as shown above, `x ne 3`. `therefore` solution set `={x inR :(13)/(5)ltxlt(17)/(5)}-{3}` `=(2.6,3.4)-{3}=(2.6,3)uu(3,3.4)`. |
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