Minimum value of $f (x) = (x - 1)^{2} + (x - 2)^{2} + . . + (x - 10)^{2}$ occurs at x =

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5.5

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Solution

The correct option is C 5.5
$f^{'} (x) = 2 (\sum_{n = 1}^{10} (x - n))$
$0 = 10 x - \frac{10 \times 11}{2}$
$x = \frac{11}{2}$

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