InterviewSolution
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InterviewSolution
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Solve for x: a) ||x-1|+2|le4|, b) (x-4)/(x+2) le |(x+2)/(x-1)| |
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Answer» Solution :a) `||x-1|+2|le4| rArr -4 le|x-1|+2le4` `rArr -6 le|x-1|le2` `rArr |x-1|le2 rArr -2 le-1 le2` `rArr -1 lexle3 rArr x in [-1.3]` b) Case I: Given inequation will be SATISFIED for all x such that `(x-4)/(x+2) le0 rArr x in (-2,4)-{1}`............(i) (Note: {1} is not in domain of RHS) Case 2: `(x-4)/(x+2) lt 0 rArr x in(-infty, -2) cup (4, infty)`.................(ii) Given inequation becomes `(x-2)/(x-1) ge (x-4)/(x+2)` or `(x-2)/(x-1) le (x-4)/(x+2)` On solving we geton solving we get `x in (-2,4//5) cup (1, infty)``x in (-2,0) cup (1,5//2)` taking intersection with (ii) we gettaking intersection with (ii), we get `x in (4, infty)`...........(III) `x in PHI` Hence, solution of the original inequation : `x in (-2,infty)-{1}` (taking union of (i) and (ii) |
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