Let a1,a2,a3,a4 be real numbers such that a1+a2+a3+a4=0 and a21+a22+a23+a24=1. Then the smallest possible value of the expression (a1–a2)2+(a2–a3)2+(a3–a4)2+(a4–a1)2 lies in the interval
A
(0,1.5)
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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) Given : a1+a2+a3+a4=0 and a21+a22+a23+a24=1 Minimizing the given equation (a1–a2)2+(a2–a3)2+(a3–a4)2+(a4–a1)2=2[a21+a22+a23+a24]−2[a1a2+a2a3+a3a4+a4a1]=2−2[a1a2+a3(a2+a4)+a4a1]=2−2[a1(a2+a4)+a3(a2+a4)]=2−2[(a2+a4)(a1+a3)]=2+2[(a1+a3)2] For minimum value of the given expression, (a1+a3)=0 ∴ the minimum value of the expression =2