Problem:
What is the minimum value of f(x)=∣x−1∣+∣2x−1∣+∣3x−1∣+⋯+∣119x−1∣?
Answer Choices:
A. 49
B. 50
C. 51
D. 52
E. 53
Solution:
Note that
f(x)=⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧−(x−1)−(2x−1)−⋯−(119x−1), if x≤1191−(x−1)−(2x−1)−⋯−((m−1)x−1)+(mx−1)+⋯+(119x−1), if m1≤x≤m−11;2≤m≤(x−1)+(2x−1)+⋯+(119x−1), if x≥1
The graph of f(x) consists of a negatively sloped ray for x≤1191, a positively sloped ray for x≥1, and for 1191≤x≤1 a sequence of line segments whose slopes increase as x increases. The minimum value of f(x) occurs at the right\
endpoint of the rightmost interval in which the graph has a non-positive slope. The slope on the interval [m1,m−11] is
k=m∑119k−k=1∑m−1k=k=1∑119k−2k=1∑m−1k=7140−(m−1)(m)
The inequality 7140+m−m2≤0 is satisfied in the interval [−84,85] with equality at the endpoints. Therefore on the interval [851,841] the graph of f(x) has a slope of 0 and a constant value of (84)(1)+(119−84)(−1)=49.
The problems on this page are the property of the MAA's American Mathematics Competitions