A) 0
B) \[\frac{\sqrt{3}}{2}\]
C) 1
D) 2
Correct Answer: C
Solution :
Let \[y=|a+b\omega +c{{\omega }^{2}}|\] for y to be minimum \[{{y}^{2}}\] must be minimum. \[{{y}^{2}}=|a+b\omega +c{{\omega }^{2}}{{|}^{2}}\] \[{{y}^{2}}=(a+b\omega +c{{\omega }^{2}})\,(a+b\omega +c{{\omega }^{2}})\] = \[\frac{1}{2}[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}]\] Since a, b and c are not equal at a time so minimum value of \[{{y}^{2}}\] occurs when any two are same and third is differ by 1. Þ Minimum of \[y=1\] (as \[a,b,c\] are integers)
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