Let f: R rightarrow R be the function f(x) = | vedclass

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(B) We have $f(x) = (x - a_1)(x - a_2) + (x - a_2)(x - a_3) + (x - a_3)(x - a_1)$.Expanding this,we get $f(x) = 3x^2 - 2(a_1 + a_2 + a_3)x + (a_1a_2 +

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$a_1 = a_2 = a_3$

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(B) We have $f(x) = (x - a_1)(x - a_2) + (x - a_2)(x - a_3) + (x - a_3)(x - a_1)$.Expanding this,we get $f(x) = 3x^2 - 2(a_1 + a_2 + a_3)x + (a_1a_2 + a_2a_3 + a_3a_1)$.For $f(x) \geq 0$ for all $x \in R$,the discriminant $D$ must be less than or equal to $0$.$D = [-2(a_1 + a_2 + a_3)]^2 - 4(3)(a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.$4(a_1 + a_2 + a_3)^2 - 12(a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.Dividing by $4$,we get $(a_1 + a_2 + a_3)^2 - 3(a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.$a_1^2 + a_2^2 + a_3^2 + 2(a_1a_2 + a_2a_3 + a_3a_1) - 3(a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.$a_1^2 + a_2^2 + a_3^2 - (a_1a_2 + a_2a_3 + a_3a_1) \leq 0$.Multiplying by $2$,we get $2a_1^2 + 2a_2^2 + 2a_3^2 - 2a_1a_2 - 2a_2a_3 - 2a_3a_1 \leq 0$.$(a_1 - a_2)^2 + (a_2 - a_3)^2 + (a_3 - a_1)^2 \leq 0$.Since the sum of squares is $\leq 0$,each term must be $0$.Thus,$a_1 - a_2 = 0$,$a_2 - a_3 = 0$,and $a_3 - a_1 = 0$,which implies $a_1 = a_2 = a_3$.

Let $f: R \rightarrow R$ be the function $f(x) = (x - a_1)(x - a_2) + (x - a_2)(x - a_3) + (x - a_3)(x - a_1)$ with $a_1, a_2, a_3 \in R$. Then,$f(x) \geq 0$ if and only if

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