HMMT 二月 2010 · CALC 赛 · 第 6 题
HMMT February 2010 — CALC Round — Problem 6
题目详情
英文原题
- [ 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
- [ 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.
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Answer: − Observe that f ( x ) can be written as ( x − ) − ( x − ) − , which has 180
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symmetry around the point ( , − ). Suppose the graph of f cuts the line y = c + x into two segments
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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
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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
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P . For that line, we can find that c = y − x = − − = − .
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