1.

If |a|=3 and -1 le k le 2, then |k a|lies in the interval

Answer»

[0, 6]
[-3, 6]
[3, 6]
[1, 2]

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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