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2023浙江大學(xué)強(qiáng)基數(shù)學(xué)逐題解析（1）

2023-06-18 20:42 作者:CHN_ZCY 0人讀過 | 我要投稿

封面：Date wtih 二乃

作畫：耕太

https://www.pixiv.net/artworks/75459895

已知 $%5Calpha$ ， $%5Cbeta$ $%5Cin$ $%5Cleft(0%2C%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright)$ ，則

$%5Cfrac%7B%5Cleft(1-%5Csqrt%7B%5Ctan%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ctan%7B%5Cfrac%7B%5Cbeta%7D%7B2%7D%7D%7D%5Cright)%5E2%7D%7B%5Ccot%5Calpha%2B%5Ccot%5Cbeta%7D$
的最大值為___________.

答案?? $3-2%5Csqrt%7B2%7D$

解析??

設(shè) $m%3D%5Csqrt%7B%5Ctan%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%7D%5Cin%5Cleft(0%2C1%5Cright)$ ， $n%3D%5Csqrt%7B%5Ctan%5Cfrac%7B%5Cbeta%7D%7B2%7D%7D%5Cin%5Cleft(0%2C1%5Cright)$ .

則

$%5Cbegin%7Baligned%7D%0A%26%5Cfrac%7B%5Cleft(1-%5Csqrt%7B%5Ctan%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ctan%7B%5Cfrac%7B%5Cbeta%7D%7B2%7D%7D%7D%5Cright)%5E2%7D%7B%5Ccot%5Calpha%2B%5Ccot%5Cbeta%7D%5C%5C%26%3D%5Cfrac%7B2m%5E2n%5E2%5Cleft(1-mn%5Cright)%7D%7B%5Cleft(m%5E2%2Bn%5E2%5Cright)%5Cleft(1%2Bmn%5Cright)%7D%5C%5C%26%5Cleq%5Cfrac%7Bmn%5Cleft(1-mn%5Cright)%7D%7B1%2Bmn%7D%5C%5C%26%3D3-%5Cleft(1%2Bmn%5Cright)-%5Cfrac%7B2%7D%7B1%2Bmn%7D%5C%5C%26%5Cleq3-2%5Csqrt%7B2%7D%0A%5Cend%7Baligned%7D$
當(dāng)且僅當(dāng) $%5Ctan%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%3D%5Ctan%7B%5Cfrac%7B%5Cbeta%7D%7B2%7D%7D%3D%5Csqrt%7B2%7D-1$ 時取等.

所以 $%5Cfrac%7B%5Cleft(1-%5Csqrt%7B%5Ctan%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ctan%7B%5Cfrac%7B%5Cbeta%7D%7B2%7D%7D%7D%5Cright)%5E2%7D%7B%5Ccot%5Calpha%2B%5Ccot%5Cbeta%7D$ 的最大值是 $3-2%5Csqrt%7B2%7D$ .
$%5Cleft%7C%202%5Ex3%5Ey-2%5Eu5%5Ev%20%5Cright%7C%20%3D2$ 的正整數(shù)解 $%5Cleft(x%2Cy%2Cu%2Cv%5Cright)$ 個數(shù)為

A. 4

B. 6

C. 7

D. 以上三項(xiàng)均不正確

答案? A

解析??

(1) 若 $x%3D1$ ，方程化為

$3%5Ey-2%5E%7Bu-1%7D5%5Ev%3D1%E6%88%96-1$
兩邊對5取模，得

$3%5Ey%5Cequiv%201%E6%88%964%5Cpmod%7B5%7D$
(i) 若 $3%5Ey%5Cequiv%201%5Cpmod%7B5%7D$ ，則 $y%3D4k%5Cleft(k%20%5Cin%20%5Cmathbb%7BN%7D%5E*%5Cright)$ ， $3%5E%7B4k%7D-2%5E%7Bu-1%7D5%5Ev%3D1$ .

于是

$2%5E%7Bu-1%7D5%5Ev%3D3%5E%7B4k%7D-1%3D80%5Cleft(1%2B81%2B81%5E2%2B%5Ccdots%2B81%5E%7Bk-1%7D%5Cright)$
即

$1%2B81%2B81%5E2%2B%5Ccdots%2B81%5E%7Bk-1%7D%3D2%5E%7Bu-5%7D5%5E%7Bv-1%7D$
由于 $%5Cforall%20n%5Cin%20%5Cmathbb%7BN%7D%2C81%5En%20%5Cequiv%201%20%5Cpmod5$ ，

$2%5E%7Bu-5%7D5%5E%7Bv-1%7D%3D1%2B81%2B81%5E2%2B%5Ccdots%2B81%5E%7Bk-1%7D%5Cequiv%20k%20%5Cpmod5$
若 $v%5Cgeq2$ ，則 $5%20%5Cmid%20k$ ，得 $3%5E5-1%5Cmid3%5E%7B4k%7D-1$ .

而 $3%5E5-1%3D242%3D2%5Ccdot11%5E2$ ， $2%5E%7Bu-1%7D5%5Ev%3D3%5E%7B4k%7D-1$ ，

從而 $11%5Cmid2%5E%7Bu-1%7D5%5Ev$ ，這是不可能的，所以 $v%5Cgeq2$ 不符合題意.

若 $v%3D1$ ，

$2%5E%7Bu-1%7D%5Ccdot5%3D3%5E%7B4k%7D-1%3D%5Cfrac%7B9%5Ek-1%7D%7B2%7D%5Ccdot%5Cfrac%7B9%5Ek%2B1%7D%7B2%7D$
由于 $9%5Ek%2B1%5Cequiv2%5Cpmod4$ ，所以 $%5Cfrac%7B9%5Ek%2B1%7D%7B2%7D$ 是奇數(shù)，又因?yàn)槠洳恍∮?，所以

$%5Cfrac%7B9%5Ek%2B1%7D%7B2%7D%3D5$ ，即 $k%3D1$ .

從而 $%5Cleft(x%2Cy%2Cu%2Cv%5Cright)%3D%5Cleft(1%2C4%2C5%2C1%5Cright)$ .

(ii) 若 $3%5Ey%5Cequiv%204%5Cpmod%7B5%7D$ ，則 $y%3D4k-2%5Cleft(k%5Cin%5Cmathbb%7BN%7D%5E*%5Cright)$ ， $3%5E%7B4k-2%7D-2%5E%7Bu-1%7D5%5Ev%3D-1$ .

兩邊對4取模，得

$1-2%5E%7Bu-1%7D%20%5Cequiv%203%20%5Cpmod%204$
即 $2%5E%7Bu-1%7D%20%5Cequiv%202%20%5Cpmod%204$ ，即 $u%3D2$ .
所以 $3%5E%7B4k-2%7D-2%5Ccdot5%5Ev%3D-1$ ，即
$2%5Ccdot5%5Ev%3D3%5E%7B4k-2%7D%2B1%3D10%5Cleft%5B1%2B%5Cleft(-9%5Cright)%2B%5Cleft(-9%5Cright)%5E2%2B%5Ccdots%2B%5Cleft(-9%5Cright)%5E%7B2k-2%7D%5Cright%5D$
即
$1%2B%5Cleft(-9%5Cright)%2B%5Cleft(-9%5Cright)%5E2%2B%5Ccdots%2B%5Cleft(-9%5Cright)%5E%7B2k-2%7D%3D5%5E%7Bv-1%7D$
由于 $%5Cforall%20n%5Cin%20%5Cmathbb%7BN%7D%2C%5Cleft(-9%5Cright)%5En%20%5Cequiv%201%20%5Cpmod5$ ，
$5%5E%7Bv-1%7D%3D1%2B%5Cleft(-9%5Cright)%2B%5Cleft(-9%5Cright)%5E2%2B%5Ccdots%2B%5Cleft(-9%5Cright)%5E%7B2k-2%7D%5Cequiv2k-1%5Cpmod%205$
若 $v%5Cgeq2$ ，則 $5%20%5Cmid%202k-1$ ，得 $3%5E5-1%5Cmid3%5E%7B4k-2%7D-1$ .
而 $3%5E5-1%3D242%3D2%5Ccdot11%5E2$ ， $2%5Ccdot5%5E%7Bv%7D%3D3%5E%7B4k-2%7D-1$ ，
從而 $11%5Cmid%202%5Ccdot5%5Ev$ ，這是不可能的.?所以 $v%5Cgeq2$ 不符合題意.
若 $v%3D1$ ，得 $%5Cleft(x%2Cy%2Cu%2Cv%5Cright)%3D%5Cleft(1%2C2%2C2%2C1%5Cright)$ .
(2) 若 $x%5Cgeq2$ ，則 $%5Cleft%7C%202%5E%7Bx-1%7D3%5Ey-2%5E%7Bu-1%7D5%5Ev%5Cright%7C%3D1$ ，所以 $u%3D1$ .
方程化為
$2%5E%7Bx-1%7D3%5Ey-5%5Ev%3D1%E6%88%96-1$
(i) 若 $x%3D2$ ，方程化為
$2%5Ccdot3%5Ey-5%5Ev%3D1%E6%88%96-1$
若 $2%5Ccdot3%5Ey-5%5Ev%3D-1$ ，即 $2%5Ccdot3%5Ey%3D5%5Ev-1$ .
由于 $4%5Cmid5%5Ev-1$ ，所以 $4%5Cmid2%5Ccdot3%5Ey$ ，這是不可能的，所以該情況不符合題意.
若 $2%5Ccdot3%5Ey-5%5Ev%3D1$ ，兩邊對3取模得 $-5%5Ev%5Cequiv1%5Cpmod3$ ，即 $5%5Ev%5Cequiv2%5Cpmod3$ .
所以 $v$ 為奇數(shù).
于是
$2%5Ccdot3%5Ey%3D5%5Ev%2B1%3D6%5Cleft%5B1%2B%5Cleft(-5%5Cright)%2B%5Cleft(-5%5Cright)%5E2%2B%5Ccdots%2B%5Cleft(-5%5Cright)%5E%7Bv-1%7D%5Cright%5D$
即
$1%2B%5Cleft(-5%5Cright)%2B%5Cleft(-5%5Cright)%5E2%2B%5Ccdots%2B%5Cleft(-5%5Cright)%5E%7Bv-1%7D%3D3%5E%7By-1%7D$
由于 $%5Cforall%20n%5Cin%20%5Cmathbb%7BN%7D%2C%5Cleft(-5%5Cright)%5En%20%5Cequiv%201%20%5Cpmod3$ ，
$3%5E%7By-1%7D%3D1%2B%5Cleft(-5%5Cright)%2B%5Cleft(-5%5Cright)%5E2%2B%5Ccdots%2B%5Cleft(-5%5Cright)%5E%7Bv-1%7D%5Cequiv%20v%5Cpmod%203$
若 $y%5Cgeq2$ ，則 $3%5Cmid%20v$ ，得 $5%5E3%2B1%20%5Cmid%205%5Ev%2B1$ .
而 $5%5E3%2B1%3D126%3D2%5Ccdot3%5E2%5Ccdot7$ ， $2%5Ccdot3%5Ey%3D5%5Ev%2B1$ ，
從而 $7%5Cmid2%5Ccdot3%5Ey$ ，這是不可能的，所以 $y%5Cgeq2$ 不符合題意.
若 $y%3D1$ ，則 $%5Cleft(x%2Cy%2Cu%2Cv%5Cright)%3D%5Cleft(2%2C1%2C1%2C1%5Cright)$ .
(ii) 若 $x%5Cgeq3$ ，則
$2%5E%7Bx-1%7D3%5Ey-5%5Ev%5Cequiv3%5Cpmod4$
所以 $2%5E%7Bx-1%7D3%5Ey-5%5Ev%3D-1$ .
兩邊對3取模得 $-5%5Ev%5Cequiv2%5Cpmod%203$ ，即 $5%5Ev%5Cequiv1%5Cpmod%203$
所以 $v$ 為偶數(shù). 則
$2%5E%7Bx-1%7D3%5Ey%3D5%5Ev-1%3D%5Cfrac%7B5%5E%7B%5Cfrac%7Bv%7D%7B2%7D%7D-1%7D%7B2%7D%5Ccdot%5Cfrac%7B5%5E%7B%5Cfrac%7Bv%7D%7B2%7D%7D%2B1%7D%7B2%7D$
由于 $%5Cleft(%5Cfrac%7B5%5E%7B%5Cfrac%7Bv%7D%7B2%7D%7D-1%7D%7B2%7D%2C%5Cfrac%7B5%5E%7B%5Cfrac%7Bv%7D%7B2%7D%7D%2B1%7D%7B2%7D%5Cright)%3D1$ ，且 $%5Cfrac%7B5%5E%7B%5Cfrac%7Bv%7D%7B2%7D%7D%2B1%7D%7B2%7D$ 為奇數(shù)，所以
$%5Cfrac%7B5%5E%7B%5Cfrac%7Bv%7D%7B2%7D%7D%2B1%7D%7B2%7D%3D3%5Ey$
即
$2%5Ccdot3%5Ey-5%5E%7B%5Cfrac%7Bv%7D%7B2%7D%7D%3D1$
由(i)知 $%5Cbegin%7Bcases%7D%0Ay%3D1%5C%5C%0Av%3D2%0A%5Cend%7Bcases%7D$ ，得 $%5Cleft(x%2Cy%2Cu%2Cv%5Cright)%3D%5Cleft(4%2C1%2C1%2C2%5Cright)$ .
綜上，
$%5Cleft(x%2Cy%2Cu%2Cv%5Cright)%3D%5Cleft(1%2C4%2C5%2C1%5Cright)%E6%88%96%5Cleft(1%2C2%2C2%2C1%5Cright)%E6%88%96%5Cleft(2%2C1%2C1%2C1%5Cright)%E6%88%96%5Cleft(4%2C1%2C1%2C2%5Cright)$
所以該方程的正整數(shù)解的個數(shù)為4.
故選：A.

標(biāo)簽：

2023浙江大學(xué)強(qiáng)基數(shù)學(xué)逐題解析（1）的評論 (共條)

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