HMMT 二月 2010 · CALC 赛 · 第 6 题

HMMT February 2010 — CALC Round — Problem 6

[ 5 ] Let f ( x ) = x − x . For a given value of c , the graph of f ( x ), together with the graph of the line c + x , split the plane up into regions. Suppose that c is such that exactly two of these regions have finite area. Find the value of c that minimizes the sum of the areas of these two regions.

解析

[ 5 ] Let f ( x ) = x − x . For a given value of c , the graph of f ( x ), together with the graph of the line c + x , split the plane up into regions. Suppose that c is such that exactly two of these regions have finite area. Find the value of c that minimizes the sum of the areas of these two regions. 11 1 1 1 2 3 ◦ Answer: − Observe that f ( x ) can be written as ( x − ) − ( x − ) − , which has 180 27 3 3 3 27 1 2 symmetry around the point ( , − ). Suppose the graph of f cuts the line y = c + x into two segments 3 27 of lengths a and b . When we move the line toward point P with a small distance ∆ x (measured along the line perpendicular to y = x + c ), the sum of the enclosed areas will increase by | a − b | (∆ x ). As long ∗ as the line x + c does not passes through P , we can find a new line x + c that increases the sum of the enclosed areas. Therefore, the sum of the areas reaches its maximum when the line passes through 2 1 11 P . For that line, we can find that c = y − x = − − = − . 27 3 27