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MIE 2016/Day 1/Problem 8

Revision as of 21:45, 7 January 2018 by Anishanne (talk | contribs) (Created page with "===Problem 8=== Let <math>f(x)=\sqrt{|x-1|+|x-2|+|x-3|+...+|x-2017|}</math>. The minimum value of <math>f(x)</math> is in the interval: (a) <math>(-\infty,1008]</math> (b) <...")
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Problem 8

Let $f(x)=\sqrt{|x-1|+|x-2|+|x-3|+...+|x-2017|}$. The minimum value of $f(x)$ is in the interval:

(a) $(-\infty,1008]$

(b) $(1008,1009]$

(c) $(1009,1010]$

(d) $(1010,1011]$

(e) $(1011,+\infty)$

Solution

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