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Question: Answered & Verified by Expert
Let $f: R \rightarrow R$ be the function $f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right)+\left(x-a_{2}\right)\left(x-a_{3}\right)+\left(x-a_{3}\right)\left(x-a_{1}\right)$ with $a_{1}, a_{2}, a_{3} \in R .$ Then $f(x) \geq 0$ if and only if -
MathematicsFunctionsKVPYKVPY 2012 (SB/SX)
Options:
  • A At least two of $a_{i}, a_{7}, a_{7}$ are equal
  • B $a_{i}=a_{i}=a_{7}$
  • C $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}$ are all distinct
  • D $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}$, are all positive and distinct
Solution:
1736 Upvotes Verified Answer
The correct answer is: $a_{i}=a_{i}=a_{7}$
Only when $\mathrm{a}_{1}=\mathrm{a}_{2}=\mathrm{a}_{3}$
In other cases $\mathrm{f}(\mathrm{x})$ will take both positive and negative values

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