Let f(x) = (x-p)^2 + (x-q)^2 + (x-r)^2. Then f(x) has a minimum at x = , where is equal to

JEE Advanced Mathematics question

Let . Then has a minimum at , where is equal to

Answer: (A)

. $f'(x) = 6x - 2(p+q+r) = 0 ⇒ x =

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Question ID VK-109404