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MIE 2016/Day 1/Problem 8

Revision as of 12:57, 9 May 2024 by Zhenghua (talk | contribs) (→‎Solution)

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

The minimum value of this function is when $x$ is in the middle between 1 and 2017. $\frac{2017+1}{2}=1009$

See Also

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