Let a_1, a_2, a_3, a_4 be real numbers such t | vedclass

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(B) Let $S = (a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$. Expanding the squares,we get $S = 2(a_1^2+a_2^2+a_3^2+a_4^2) - 2(a_1a_2+a_2a_3+a_3a_4+a

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$(1.5, 2.5)$

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(B) Let $S = (a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$. Expanding the squares,we get $S = 2(a_1^2+a_2^2+a_3^2+a_4^2) - 2(a_1a_2+a_2a_3+a_3a_4+a_4a_1)$. Given $a_1^2+a_2^2+a_3^2+a_4^2=1$,so $S = 2 - 2(a_1+a_3)(a_2+a_4)$. Since $a_1+a_2+a_3+a_4=0$,we have $a_2+a_4 = -(a_1+a_3)$. Let $x = a_1+a_3$,then $a_2+a_4 = -x$. Thus,$S = 2 - 2(x)(-x) = 2 + 2x^2$. To minimize $S$,we need to minimize $x^2$. Since $a_1^2+a_2^2+a_3^2+a_4^2=1$,by Cauchy-Schwarz inequality,$(a_1+a_3)^2 \leq 2(a_1^2+a_3^2)$ and $(a_2+a_4)^2 \leq 2(a_2^2+a_4^2)$. Summing these,$x^2 + (-x)^2 \leq 2(a_1^2+a_2^2+a_3^2+a_4^2) = 2$,so $2x^2 \leq 2$,which means $x^2 \leq 1$. The minimum value of $x^2$ is $0$ (when $a_1+a_3=0$ and $a_2+a_4=0$). If $x=0$,$S = 2 + 2(0) = 2$. Since $2$ lies in the interval $(1.5, 2.5)$,the correct option is $B$.

Let $a_1, a_2, a_3, a_4$ be real numbers such that $a_1+a_2+a_3+a_4=0$ and $a_1^2+a_2^2+a_3^2+a_4^2=1$. Then,the smallest possible value of the expression $(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$ lies in the interval

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