返回题库

HMMT 二月 2010 · CALC 赛 · 第 6 题

HMMT February 2010 — CALC Round — Problem 6

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 5 ] Let f ( x ) = x − x . For a given value of c , the graph of f ( x ), together with the graph of the linec + x , split the plane up into regions. Suppose that c is such that exactly two of these regions havefinite area. Find the value of c that minimizes the sum of the areas of these two regions.
解析

英文解析

  1. [ 5 ] Let f ( x ) = x − x . For a given value of c , the graph of f ( x ), together with the graph of the linec + x , split the plane up into regions. Suppose that c is such that exactly two of these regions havefinite 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 alongthe 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 ofthe 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