Let a1,a2,a3,a4 be real numbers such that and a21+a22+a23+a24=1. Then the smallest possible value of the expression (a1−a2)2+(a2−a2)2+(a3−a4)2+(a4−a1)2 lies in interval .
A
(0.15,2)
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B
(−1.5,2.5)
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C
(2.5,3)
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D
(3,3.5)
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Solution
The correct option is B(−1.5,2.5) Smallest value of (a1−a2)2+(a2−a3)2+(a3−a4)2+(a4−a1)2 is 0, when a1=a2=a3=a4=12.