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If |a|=3 and -1 le k le 2, then |k a|lies in the interval |
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Answer» Solution :The smalest value of |k | will exist at NUMERICALLY SMALLEST value of k, i.e, at k =0, which gives `|ka|=|k||a|=0xx3=0` The numerically greaterst value of k is 2 at which `|k a|=6` |
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