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趣味數學問題懸賞6 初三、高一組 參考答案

2023-03-24 13:16 作者:数学只因  | 我要投稿

不知道啥是趣味數學問題懸賞看這裏

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第一题解答:

1)分情况讨论

    况一:x%5Cin%20%5B%5Cspace%200%2C%5Cspace2-%5Csqrt%7B3%7D%5Cspace%5D

         %5Cbegin%7Baligned%7D%0Af(x)%26%3D1-%7C1-2x%7C%5C%5C%0A%26%3D1-(1-2x)%5C%5C%0A%26%3D2x%5C%5C%0A%26%5Cleq2(2-%5Csqrt%7B3%7D)%5C%5C%0A%26%3D4-2%5Csqrt%7B3%7D%5C%5C%0A%26%3D(1-%5Csqrt%7B3%7D)%5E2%5C%5C%0A%26%3D(2-%5Csqrt%7B3%7D-1)%5E2%5C%5C%0A%26%5Cleq(x-1)%5E2%5C%5C%0A%26%3Dg(x)%0A%5Cend%7Baligned%7D

     况二:x%5Cin%20(%5Cspace2-%5Csqrt%7B3%7D%2C%20%5Cspace%200.5%5Cspace%5D

     %5Cbegin%7Baligned%7D%0Af(x)%26%3D1-%7C1-2x%7C%5C%5C%0A%26%3D1-(1-2x)%5C%5C%0A%26%3D2x%5C%5C%0A%26%5Cgeq2(2-%5Csqrt%7B3%7D)%5C%5C%0A%26%3D4-2%5Csqrt%7B3%7D%5C%5C%0A%26%3D(1-%5Csqrt%7B3%7D)%5E2%5C%5C%0A%26%3D(2-%5Csqrt%7B3%7D-1)%5E2%5C%5C%0A%26%5Cgeq(x-1)%5E2%5C%5C%0A%26%3Dg(x)%0A%5Cend%7Baligned%7D

     况三:x%5Cin%20(%5Cspace%200.5%2C%20%5Cspace%201%5Cspace%5D

     %5Cbegin%7Baligned%7D%0Af(x)%26%3D1-%7C1-2x%7C%5C%5C%0A%26%3D1-(2x-1)%5C%5C%0A%26%3D2-2x%5C%5C%0A%26%5Cgeq1%2Bx%5E2-2x%5C%5C%0A%26%3D(x-1)%5E2%5C%5C%0A%26%3Dg(x)%0A%0A%5Cend%7Baligned%7D

     综上可得 F(x) 表达式:

     %5Cbegin%7Bequation%7D%0AF(x)%3D%20%5Cleft%5C%7B%20%0A%5Cbegin%7Baligned%7D%20%0A2x%26%20%26%20x%20%5Cin%5B%5Cspace0%2C%5Cspace2-%5Csqrt%7B3%7D%5Cspace%20%5D%5C%5C%0A(x-1)%5E2%26%26x%20%5Cin(%5Cspace2-%5Csqrt%7B3%7D%2C%5Cspace1%5Cspace%20%5D%5C%5C%0A%5Cend%7Baligned%7D%0A%5Cright.%0A%0A%5Cend%7Bequation%7D

2)分情况讨论

    况一:x%5Cin%20%5B%5Cspace%200%2C%5Cspace2-%5Csqrt%7B3%7D%5Cspace%5D

    注意到此时 F(x) 严格单调递增

    %5Ctherefore 此时 F(x) 最大值为 F(2-%5Csqrt%7B3%7D)%3D4-2%5Csqrt%7B3%7D  

    况二:x%5Cin%20(%5Cspace%202-%5Csqrt%7B3%7D%2C%20%5Cspace%201%5Cspace%5D

    %5Cbegin%7Baligned%7D%0AF(x)%26%3D(x-1)%5E2%5C%5C%0A%26%3C(2-%5Csqrt%7B3%7D-1)%5E2%5C%5C%0A%26%3D4-2%5Csqrt%7B3%7D%0A%5Cend%7Baligned%7D

    %5Ctherefore  F(x) 最大值为 4-2%5Csqrt%7B3%7D  

3) 分情况讨论

    况一:x%5Cin%20%5B%5Cspace%200%2C%5Cspace2-%5Csqrt%7B3%7D%5Cspace%5D 

    %5Cbegin%7Baligned%7D%0AF(x)%26%3D%5Cfrac%7B1%7D%7B3%7D%5C%5C%0A2x%26%3D%5Cfrac%7B1%7D%7B3%7D%5C%5C%0A%5Cend%7Baligned%7D

    %5Ctherefore%20x%3D%5Cfrac%7B1%7D%7B6%7D

    况二:x%5Cin%20(%5Cspace%202-%5Csqrt%7B3%7D%2C%20%5Cspace%201%5Cspace%5D

    %5Cbegin%7Baligned%7D%0AF(x)%26%3D%5Cfrac%7B1%7D%7B3%7D%5C%5C%0A(x-1)%5E2%26%3D%5Cfrac%7B1%7D%7B3%7D%5C%5C%0A%5Cend%7Baligned%7D

     %5Ctherefore%20x%3D%5Cfrac%7B3-%5Csqrt%7B3%7D%7D%7B3%7D%20%5Cspace%20or%20%5Cspace%20%20x%3D%5Cfrac%7B3%2B%5Csqrt%7B3%7D%7D%7B3%7D

    验算后得 %20x%3D%5Cfrac%7B3-%5Csqrt%7B3%7D%7D%7B3%7D%20%5Cspace%20or%20%5Cspace%20x%20%3D%20%5Cfrac%7B1%7D%7B6%7D

第二题解答:

    注意到一颗骰子最多能表示 7 个不同数位

    可得两颗骰子可表示不超过 7%5Ccdot7%5Ccdot2%3D98

    但由于共有 99 个不同得分数要表示,所以不能只用两颗骰子表示所有分数

    同理可得一颗骰子也不能表示所有分数

    考虑三个骰子的情况,将骰子命名为 a , b , c ,并且它们的点数以以下方式安排

    a : 0,1,2,3,4,5

    b : 0,1,2,6,7,8

    c : 3,4,5,6,7,8

    由于 6 可用来表示 9 ,易证明可以在任两骰子中择一来表示 0 - 9 中的其中一个整数

    因此,易证明只需三颗骰子便可表示所有分数

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