復(fù)習(xí)筆記Day108：武漢大學(xué)2023數(shù)學(xué)分析參考答案

2023-02-28 19:59 作者:間宮_卓司 0人讀過 | 我要投稿

這張考卷上面有不少計算題，建議大家自己算一下

一、計算題

1.求極限 $%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Clim%7D%5Cfrac%7B%5Cint_0%5E%7Bn%5Cpi%7D%7Bx%5Cleft%7C%20%5Csin%20x%20%5Cright%7C%5Cmathrm%7Bd%7Dx%7D%7D%7Bn%5Cleft(%20n%2B1%20%5Cright)%7D$ ? ?

$%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Clim%7D%5Cfrac%7B%5Cint_0%5E%7Bn%5Cpi%7D%7Bx%5Cleft%7C%20%5Csin%20x%20%5Cright%7C%5Cmathrm%7Bd%7Dx%7D%7D%7Bn%5Cleft(%20n%2B1%20%5Cright)%7D%5Cxlongequal%7B%5Ctext%7Bstolz%E5%AE%9A%E7%90%86%7D%7D%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Clim%7D%5Cfrac%7B%5Cint_%7B%5Cleft(%20n-1%20%5Cright)%20%5Cpi%7D%5E%7Bn%5Cpi%7D%7Bx%5Cleft%7C%20%5Csin%20x%20%5Cright%7C%5Cmathrm%7Bd%7Dx%7D%7D%7B2n%7D$

而

$%5Cbegin%7Baligned%7D%0A%09%5Cint_%7B%5Cleft(%20n-1%20%5Cright)%20%5Cpi%7D%5E%7Bn%5Cpi%7D%7Bx%5Cleft%7C%20%5Csin%20x%20%5Cright%7C%5Cmathrm%7Bd%7Dx%7D%26%3D%5Cint_0%5E%7B%5Cpi%7D%7B%5Cleft(%20x%2B%5Cleft(%20n-1%20%5Cright)%20%5Cpi%20%5Cright)%20%5Cleft%7C%20%5Csin%20%5Cleft(%20x%2B%5Cleft(%20n-1%20%5Cright)%20%5Cpi%20%5Cright)%20%5Cright%7C%5Cmathrm%7Bd%7Dx%7D%5C%5C%0A%09%26%3D%5Cint_0%5E%7B%5Cpi%7D%7Bx%5Csin%20x%5Cmathrm%7Bd%7Dx%7D%2B%5Cleft(%20n-1%20%5Cright)%20%5Cpi%20%5Cint_0%5E%7B%5Cpi%7D%7B%5Csin%20x%5Cmathrm%7Bd%7Dx%7D%5C%5C%0A%09%26%3D%5Cpi%20%2B2%5Cleft(%20n-1%20%5Cright)%20%5Cpi%5C%5C%0A%09%26%3D%5Cleft(%202n-1%20%5Cright)%20%5Cpi%5C%5C%0A%5Cend%7Baligned%7D$

所以

$%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Clim%7D%5Cfrac%7B%5Cint_0%5E%7Bn%5Cpi%7D%7Bx%5Cleft%7C%20%5Csin%20x%20%5Cright%7C%5Cmathrm%7Bd%7Dx%7D%7D%7Bn%5Cleft(%20n%2B1%20%5Cright)%7D%3D%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Clim%7D%5Cfrac%7B%5Cleft(%202n-1%20%5Cright)%20%5Cpi%7D%7B2n%7D%3D%5Cpi%20$

2.求極限

$%5Cunderset%7Bx%5Crightarrow%200%7D%7B%5Clim%7D%5Cfrac%7B%5Cln%20%5Cleft(%20e%5E%7B%5Ctan%20x%7D%2B%5Csqrt%5B3%5D%7B1-%5Ccos%20x%7D%20%5Cright)%20%2B%5Ctan%20x%7D%7B%5Cmathrm%7Barc%7D%5Csin%20%5Cleft(%202023%5Csqrt%5B3%5D%7B1-%5Ccos%20x%7D%20%5Cright)%7D$

對于分母，有

$%5Cmathrm%7Barc%7D%5Csin%20%5Cleft(%202023%5Csqrt%5B3%5D%7B1-%5Ccos%20x%7D%20%5Cright)%20%5Csim%202023%5Csqrt%5B3%5D%7B1-%5Ccos%20x%7D%5Cleft(%20x%5Crightarrow%200%20%5Cright)%20$

對于分子，有

$%5Cbegin%7Baligned%7D%0A%09%5Cln%20%5Cleft(%20e%5E%7B%5Ctan%20x%7D%2B%5Csqrt%5B3%5D%7B1-%5Ccos%20x%7D%20%5Cright)%20%2B%5Ctan%20x%26%3D%5Cln%20%5Cleft(%20e%5E%7B2%5Ctan%20x%7D%2Be%5E%7B%5Ctan%20x%7D%5Csqrt%5B3%5D%7B1-%5Ccos%20x%7D%20%5Cright)%5C%5C%0A%09%26%5Csim%20e%5E%7B2%5Ctan%20x%7D%2Be%5E%7B%5Ctan%20x%7D%5Csqrt%5B3%5D%7B1-%5Ccos%20x%7D-1%5C%5C%0A%5Cend%7Baligned%7D$

所以

$%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%3D%5Cunderset%7Bx%5Crightarrow%200%7D%7B%5Clim%7D%5Cfrac%7Be%5E%7B2%5Ctan%20x%7D-1%2Be%5E%7B%5Ctan%20x%7D%5Csqrt%5B3%5D%7B1-%5Ccos%20x%7D%7D%7B2023%5Csqrt%5B3%5D%7B1-%5Ccos%20x%7D%7D%3D%5Cunderset%7Bx%5Crightarrow%200%7D%7B%5Clim%7D%5Cfrac%7Be%5E%7B2%5Ctan%20x%7D-1%7D%7B2023%5Csqrt%5B3%5D%7B1-%5Ccos%20x%7D%7D%2B%5Cfrac%7B1%7D%7B2023%7D$

而

$e%5E%7B2%5Ctan%20x%7D-1%5Csim%202%5Ctan%20x%5Csim%202x$

$%5Csqrt%5B3%5D%7B1-%5Ccos%20x%7D%5Csim%20%5Cleft(%20%5Cfrac%7Bx%5E2%7D%7B2%7D%20%5Cright)%20%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D2%5E%7B-%5Cfrac%7B1%7D%7B3%7D%7Dx%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D$

所以

$%5Cunderset%7Bx%5Crightarrow%200%7D%7B%5Clim%7D%5Cfrac%7Be%5E%7B2%5Ctan%20x%7D-1%7D%7B2023%5Csqrt%5B3%5D%7B1-%5Ccos%20x%7D%7D%3D0$

3.垃圾題目，題都懶得抄上來了

二、計算題

1.求 $%5Cint_0%5E%7B%5Cpi%7D%7B%5Cfrac%7B1%7D%7B3%2B%5Csin%20%5E2x%7D%5Cmathrm%7Bd%7Dx%7D$

$%5Cbegin%7Baligned%7D%0A%09%5Cint_0%5E%7B%5Cpi%7D%7B%5Cfrac%7B1%7D%7B3%2B%5Csin%20%5E2x%7D%5Cmathrm%7Bd%7Dx%7D%26%3D%5Cint_0%5E%7B%5Cpi%7D%7B%5Cfrac%7B%5Csin%20%5E2x%2B%5Ccos%20%5E2x%7D%7B4%5Csin%20%5E2x%2B3%5Ccos%20%5E2x%7D%5Cmathrm%7Bd%7Dx%7D%5C%5C%0A%09%26%3D%5Cint_0%5E%7B%5Cpi%7D%7B%5Cfrac%7B%5Ctan%20%5E2x%2B1%7D%7B4%5Ctan%20%5E2x%2B3%7D%5Cmathrm%7Bd%7Dx%7D%5C%5C%0A%09%26%3D%5Cint_0%5E%7B%5Cpi%7D%7B%5Cfrac%7B%5Csec%20%5E2x%7D%7B4%5Ctan%20%5E2x%2B3%7D%5Cmathrm%7Bd%7Dx%7D%5C%5C%0A%09%26%3D%5Cint_0%5E%7B%5Cpi%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7D%5Ctan%20x%7D%7B4%5Ctan%20%5E2x%2B3%7D%7D%5C%5C%0A%09%26%3D%5Cint_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7D%5Ctan%20x%7D%7B4%5Ctan%20%5E2x%2B3%7D%7D%2B%5Cint_%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%5E%7B%5Cpi%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7D%5Ctan%20x%7D%7B4%5Ctan%20%5E2x%2B3%7D%7D%5C%5C%0A%09%26%3D%5Cint_0%5E%7B%2B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7Dx%7D%7B4x%5E2%2B3%7D%7D%2B%5Cint_%7B-%5Cinfty%7D%5E0%7B%5Cfrac%7B%5Cmathrm%7Bd%7Dx%7D%7B4x%5E2%2B3%7D%7D%5C%5C%0A%09%26%3D%5Cint_%7B-%5Cinfty%7D%5E%7B%2B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7Dx%7D%7B4x%5E2%2B3%7D%7D%5C%5C%0A%5Cend%7Baligned%7D$

（好像寫啰嗦了，算了懶得該了）

下面來計算最后一個積分

$%5Cint_%7B-%5Cinfty%7D%5E%7B%2B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7Dx%7D%7B4x%5E2%2B3%7D%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Cint_%7B-%5Cinfty%7D%5E%7B%2B%5Cinfty%7D%7B%5Cfrac%7B%5Cmathrm%7Bd%7D%5Cleft(%20%5Cfrac%7B2%7D%7B%5Csqrt%7B3%7D%7Dx%20%5Cright)%7D%7B%5Cleft(%20%5Cfrac%7B2%7D%7B%5Csqrt%7B3%7D%7Dx%20%5Cright)%20%5E2%2B1%7D%7D%5Ccdot%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D%5Cfrac%7B1%7D%7B2%5Csqrt%7B3%7D%7D%5Ccdot%20%5Cleft(%20%5Cfrac%7B%5Cpi%7D%7B2%7D-%5Cleft(%20-%5Cfrac%7B%5Cpi%7D%7B2%7D%20%5Cright)%20%5Cright)%20%3D%5Cfrac%7B%5Cpi%7D%7B2%5Csqrt%7B3%7D%7D$

2.已知 $f$ 連續(xù)可微，且 $f(1)%3D2%2Cf(4)%3D3$ ，求 $%5Coint_L%7B%5Cfrac%7Bf%5Cleft(%20xy%20%5Cright)%7D%7By%7D%5Cmathrm%7Bd%7Dy%7D$ ， $L$ 是 $y%3Dx%2Cy%3D4x$ , $xy%3D1%2Cxy%3D4$ 所圍區(qū)域的邊界，取逆時針方向

$%5Cint_L%7B%5Cfrac%7Bf%5Cleft(%20xy%20%5Cright)%7D%7By%7D%5Cmathrm%7Bd%7Dy%7D%5Cxlongequal%7B%5Ctext%7BGreen%E5%85%AC%E5%BC%8F%7D%7D%5Ciint_S%7B%5Cleft%7C%20%5Cbegin%7Bmatrix%7D%0A%09%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20x%7D%26%09%09%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20y%7D%5C%5C%0A%090%26%09%09%5Cfrac%7Bf%5Cleft(%20xy%20%5Cright)%7D%7By%7D%5C%5C%0A%5Cend%7Bmatrix%7D%20%5Cright%7C%5Cmathrm%7Bd%7Dx%5Cmathrm%7Bd%7Dy%7D%3D%5Ciint_S%7Bf'%5Cleft(%20xy%20%5Cright)%20%5Cmathrm%7Bd%7Dx%5Cmathrm%7Bd%7Dy%7D$

而

$%5Ciint_S%7Bf'%5Cleft(%20xy%20%5Cright)%20%5Cmathrm%7Bd%7Dx%5Cmathrm%7Bd%7Dy%7D%5Cxlongequal%5B%5Cfrac%7By%7D%7Bx%7D%3Dv%5D%7Bxy%3Du%7D%5Cint_1%5E4%7B%5Cmathrm%7Bd%7Dv%5Cint_1%5E4%7Bf'%5Cleft(%20u%20%5Cright)%20%5Cleft%7C%20%5Cfrac%7B%5Cpartial%20%5Cleft(%20x%2Cy%20%5Cright)%7D%7B%5Cpartial%20%5Cleft(%20u%2Cv%20%5Cright)%7D%20%5Cright%7C%5Cmathrm%7Bd%7Du%7D%7D$

計算可得

$%5Cleft%7C%20%5Cfrac%7B%5Cpartial%20%5Cleft(%20u%2Cv%20%5Cright)%7D%7B%5Cpartial%20%5Cleft(%20x%2Cy%20%5Cright)%7D%20%5Cright%7C%3D%5Cfrac%7B2y%7D%7Bx%7D%3D2v$

然后根據(jù)Day32的結(jié)論，有 $%5Cleft%7C%20%5Cfrac%7B%5Cpartial%20%5Cleft(%20x%2Cy%20%5Cright)%7D%7B%5Cpartial%20%5Cleft(%20u%2Cv%20%5Cright)%7D%20%5Cright%7C%3D%5Cfrac%7B1%7D%7B2v%7D$

所以

$%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%3D%5Cint_1%5E4%7B%5Cfrac%7B1%7D%7B2v%7D%5Cmathrm%7Bd%7Dv%5Cint_1%5E4%7Bf'%5Cleft(%20u%20%5Cright)%20%5Cmathrm%7Bd%7Du%7D%7D%3D%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20%5Cln%204-%5Cln%201%20%5Cright)%20%5Cright)%20%5Cleft(%20f%5Cleft(%204%20%5Cright)%20-f%5Cleft(%201%20%5Cright)%20%5Cright)%20%3D%5Cln%202$

3.求曲面積分 $%5Ciint_S%7Bz%5Cleft(%20%5Cfrac%7B%5Calpha%20x%7D%7Ba%5E2%7D%2B%5Cfrac%7B%5Cbeta%20y%7D%7Bb%5E2%7D%2B%5Cfrac%7B%5Cgamma%20z%7D%7Bc%5E2%7D%20%5Cright)%20%5Cmathrm%7Bd%7DS%7D$ ，其中 $S$ 為 $%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%2B%5Cfrac%7Bz%5E2%7D%7Bc%5E2%7D%3D1$ 的上半平面， $%5Calpha%2C%5Cbeta%2C%5Cgamma$ 為 $S$ 的外方向余弦

$%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%3D%5Ciint_S%7B%5Cfrac%7Bxz%7D%7Ba%5E2%7D%5Cmathrm%7Bd%7Dy%5Cmathrm%7Bd%7Dz%2B%5Cfrac%7Byz%7D%7Bb%5E2%7D%5Cmathrm%7Bd%7Dx%5Cmathrm%7Bd%7Dz%2B%5Cfrac%7Bz%5E2%7D%7Bc%5E2%7D%5Cmathrm%7Bd%7Dx%5Cmathrm%7Bd%7Dy%7D$

記 $S_1%3D%5C%7B(x%2Cy%2Cz)%7Cz%3D0%2C%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1%20%20%5C%7D$ ，那么

$%5Ciint_%7BS%2BS_1%7D%7B%5Cfrac%7Bxz%7D%7Ba%5E2%7D%5Cmathrm%7Bd%7Dy%5Cmathrm%7Bd%7Dz%2B%5Cfrac%7Byz%7D%7Bb%5E2%7D%5Cmathrm%7Bd%7Dx%5Cmathrm%7Bd%7Dz%2B%5Cfrac%7Bz%5E2%7D%7Bc%5E2%7D%5Cmathrm%7Bd%7Dx%5Cmathrm%7Bd%7Dy%7D%5Cxlongequal%7B%5Ctext%7BGauss%E5%85%AC%E5%BC%8F%7D%7D%5Cleft(%20%5Cfrac%7B1%7D%7Ba%5E2%7D%2B%5Cfrac%7B1%7D%7Bb%5E2%7D%2B%5Cfrac%7B2%7D%7Bc%5E2%7D%20%5Cright)%20%5Ciiint_%7B%5COmega%7D%7Bz%5Cmathrm%7Bd%7Dx%5Cmathrm%7Bd%7Dy%5Cmathrm%7Bd%7Dz%7D$

做代換 $x%3Dar%5Ccos%20%5Ctheta%20%5Csin%20%5Cvarphi%20%2Cy%3Dbr%5Csin%20%5Ctheta%20%5Csin%20%5Cvarphi%20%2Cz%3Dcr%5Ccos%20%5Cvarphi%20$ ，可得

$%5Cbegin%7Baligned%7D%0A%09%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%26%3D%5Cleft(%20%5Cfrac%7B1%7D%7Ba%5E2%7D%2B%5Cfrac%7B1%7D%7Bb%5E2%7D%2B%5Cfrac%7B2%7D%7Bc%5E2%7D%20%5Cright)%20%5Cint_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%7B%5Cmathrm%7Bd%7D%5Cvarphi%20%5Cint_0%5E%7B2%5Cpi%7D%7B%5Cmathrm%7Bd%7D%5Ctheta%20%5Cint_0%5E1%7B%5Cleft(%20abcr%5E2%5Csin%20%5Cvarphi%20%5Cright)%20cr%5Ccos%20%5Cvarphi%20%5Cmathrm%7Bd%7Dr%7D%7D%7D%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B4%7D%5Cleft(%20%5Cfrac%7B1%7D%7Ba%5E2%7D%2B%5Cfrac%7B1%7D%7Bb%5E2%7D%2B%5Cfrac%7B2%7D%7Bc%5E2%7D%20%5Cright)%20abc%5E2%5Cpi%5C%5C%0A%5Cend%7Baligned%7D$

三、解答題

1.求 $F%5Cleft(%20%5Calpha%20%5Cright)%20%3D%5Cint_0%5E%7B%2B%5Cinfty%7D%7B%5Cfrac%7B%5Cln%20%5Cleft(%201%2Bx%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20%5Cright)%7D%7Bx%5E%7B%5Calpha%7D%7D%5Cmathrm%7Bd%7Dx%7D$ 的連續(xù)區(qū)間

依比較判別法，要使這個積分收斂，首先要有

$%5Cbegin%7Bcases%7D%0A%09%5Calpha%20-%5Cfrac%7B5%7D%7B2%7D%3C1%5C%5C%0A%09%5Calpha%20%3E1%5C%5C%0A%5Cend%7Bcases%7D$

即 $1%3C%5Calpha%20%3C%5Cfrac%7B7%7D%7B2%7D$ 成立，而此時， $%5Cforall%20%5Calpha%20%5Cin%20%5Cleft%5B%20%5Calpha%20_1%2C%5Calpha%20_2%20%5Cright%5D%20%5Csubset%20%5Cleft(%201%2C%5Cfrac%7B7%7D%7B2%7D%20%5Cright)%20$ ，總成立

$F%5Cleft(%20%5Calpha%20%5Cright)%20%3C%5Cint_0%5E%7B%2B%5Cinfty%7D%7B%5Cmax%20%5Cleft%5C%7B%20%5Cfrac%7B%5Cln%20%5Cleft(%201%2Bx%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20%5Cright)%7D%7Bx%5E%7B%5Calpha%20_1%7D%7D%2C%5Cfrac%7B%5Cln%20%5Cleft(%201%2Bx%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20%5Cright)%7D%7Bx%5E%7B%5Calpha%20_2%7D%7D%20%5Cright%5C%7D%20%5Cmathrm%7Bd%7Dx%7D%5Cle%20%5Cint_0%5E%7B%2B%5Cinfty%7D%7B%5Cfrac%7B%5Cln%20%5Cleft(%201%2Bx%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20%5Cright)%7D%7Bx%5E%7B%5Calpha%20_1%7D%7D%5Cmathrm%7Bd%7Dx%7D%2B%5Cint_0%5E%7B%2B%5Cinfty%7D%7B%5Cfrac%7B%5Cln%20%5Cleft(%201%2Bx%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20%5Cright)%7D%7Bx%5E%7B%5Calpha%20_2%7D%7D%5Cmathrm%7Bd%7Dx%7D$

依魏爾斯特拉斯判別法， $F%5Cleft(%20%5Calpha%20%5Cright)%20$ 在定義域上內(nèi)閉一致收斂，依一致收斂的性質(zhì)， $F%5Cleft(%20%5Calpha%20%5Cright)%20$ 在定義域上連續(xù)

2.對點列 $z_n%3D(x_n%2Cy_n)(n%3D1%2C2%2C%5Ccdots)$ ，有

$%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Cunderline%7B%5Clim%20%7D%7D%5Cleft%5C%7C%20z_n%20%5Cright%5C%7C%20%3D%5Calpha%20%2C%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Coverline%7B%5Clim%20%7D%7D%5Cleft%5C%7C%20z_n%20%5Cright%5C%7C%20%3D%5Cbeta%20%2C%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Clim%7D%5Cleft%5C%7C%20z_%7Bn%2B1%7D-z_n%20%5Cright%5C%7C%20%3D0$

證明：對任意的 $%5Cgamma%20%5Cin%20%5Cleft(%20%5Calpha%20%2C%5Cbeta%20%5Cright)%20%2Cx%5E2%2By%5E2%3Dr%5E2$ 上至少存在 $%5C%7Bz_n%5C%7D$ 的一個聚點

記 $d_n%3D%5Cleft%5C%7C%20z_n%20%5Cright%5C%7C%20$ ，那么

$%5Cleft%7C%20d_%7Bn%2B1%7D-d_n%20%5Cright%7C%3D%5Cleft%7C%20%5Cleft%5C%7C%20z_%7Bn%2B1%7D%20%5Cright%5C%7C%20-%5Cleft%5C%7C%20z_n%20%5Cright%5C%7C%20%5Cright%7C%5Cle%20%5Cleft%5C%7C%20z_%7Bn%2B1%7D-z_n%20%5Cright%5C%7C%20%5Crightarrow%200$

依72.1，可知 $d_n$ 以任意的 $%5Cgamma%20%5Cin%20%5Cleft(%20%5Calpha%20%2C%5Cbeta%20%5Cright)%20$ 為聚點，這和要證明的結(jié)論是等價的

3.問是否存在[0,5]上的函數(shù)f(x)滿足下列條件，并說明

(1) $f(x)$ 連續(xù)可微

(2) $f(0)%3Df(5)%3D1$

(3) $%5Cleft%7C%20f'%5Cleft(%20x%20%5Cright)%20%5Cright%7C%5Cle%20%5Cfrac%7B2%7D%7B5%7D$

(4) $%5Cleft%7C%20%5Cint_0%5E5%7Bf%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%20%5Cright%7C%5Cle%20%5Cfrac%7B5%7D%7B2%7D$

先來大概判斷一下這個函數(shù)存不存在，考慮如下這種最極端的情況

這個時候面積就正好是5/2，而如果面積想要更小的話，曲線就要向下凹，這樣第三個條件就不滿足了，所以這樣的函數(shù)可能是不存在的，下面來證明這件事

當(dāng) $x%5Cin(0%2C%5Cfrac%7B5%7D%7B2%7D)%20$ 時，依泰勒展開，有

$f%5Cleft(%20x%20%5Cright)%20%3Df%5Cleft(%200%20%5Cright)%20%2B%5Cleft(%20x-%5Cfrac%7B5%7D%7B2%7D%20%5Cright)%20f'%5Cleft(%20%5Cxi%20%5Cright)%20%5Cge%20f%5Cleft(%200%20%5Cright)%20%2B%5Cfrac%7B2%7D%7B5%7D%5Cleft(%20x-%5Cfrac%7B5%7D%7B2%7D%20%5Cright)%20%3D%5Cfrac%7B2%7D%7B5%7Dx$

故 $%5Cint_0%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%7Bf%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%5Cge%20%5Cint_0%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%7B%5Cfrac%7B2%7D%7B5%7Dx%5Cmathrm%7Bd%7Dx%7D%3D%5Cfrac%7B5%7D%7B4%7D$

同理可證， $%5Cint_%7B%5Cfrac%7B5%7D%7B2%7D%7D%5E5%7Bf%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%5Cge%20%5Cfrac%7B5%7D%7B4%7D$ ，并且兩個等號同時成立當(dāng)且僅當(dāng)

$f'%5Cleft(%20x%20%5Cright)%20%3D%5Cbegin%7Bcases%7D%0A%09-%5Cfrac%7B2%7D%7B5%7D%2Cx%5Cin%20%5Cleft(%200%2C%5Cfrac%7B5%7D%7B2%7D%20%5Cright)%5C%5C%0A%09%5Cfrac%7B2%7D%7B5%7D%2Cx%5Cin%20%5Cleft(%20%5Cfrac%7B5%7D%7B2%7D%2C5%20%5Cright)%5C%5C%0A%5Cend%7Bcases%7D$

（嚴格來說是這個式子幾乎處處成立吧，但是這樣說就越來越說不清楚了）此時 $f(x)$ 不可導(dǎo)，故等號不成立，故第四個條件不成立，即這樣的函數(shù)是不存在的

4.已知 $u(x%2Cy)$ 在 $D%3Ax%5E2%2By%5E2%5Cle1$ 上連續(xù)，且在 $x%5E2%2By%5E2%3C1$ 滿足 $%5Cfrac%7B%5Cpartial%20%5E2u%7D%7B%5Cpartial%20x%5E2%7D%2B%5Cfrac%7B%5Cpartial%20%5E2u%7D%7B%5Cpartial%20y%5E2%7D%3Du$

(1)證明：存在 $x%5E2%2By%5E2%3D1$ 上有 $u(x%2Cy)%5Cge0$ ，則在 $x%5E2%2By%5E2%5Cle1$ 上也有 $u(x%2Cy)%5Cge0$

(2)證明：存在 $x%5E2%2By%5E2%3D1$ 上有 $u(x%2Cy)%3E0$ ，則在 $x%5E2%2By%5E2%5Cle1$ 上也有 $u(x%2Cy)%3E0$

(1)設(shè)在 $x%5E2%2By%5E2%3C1$ 上有極小值點 $(x%5E*%2Cy%5E*)$ ，那么依Day34的結(jié)論，在那個點，有 $%5Cfrac%7B%5Cpartial%20%5E2u%7D%7B%5Cpartial%20x%5E2%7D%2C%5Cfrac%7B%5Cpartial%20%5E2u%7D%7B%5Cpartial%20y%5E2%7D%5Cge%200$ ，故

$u%5Cleft(%20x%5E*%2Cy%5E*%20%5Cright)%20%3D%5Cleft(%20%5Cfrac%7B%5Cpartial%20%5E2u%7D%7B%5Cpartial%20x%5E2%7D%2B%5Cfrac%7B%5Cpartial%20%5E2u%7D%7B%5Cpartial%20y%5E2%7D%20%5Cright)%20%5Cleft(%20x%5E*%2Cy%5E*%20%5Cright)%20%5Cge%200$

(2)設(shè) $x%5E2%2By%5E2%3D1$ 上 $u(x%2Cy)$ 有最小值 $c%3E0$ ，記 $v%3Du-c%5Cleft(%20x%5E2%2By%5E2%20%5Cright)%20$ ，那么

$%5Cfrac%7B%5Cpartial%20%5E2v%7D%7B%5Cpartial%20x%5E2%7D%2B%5Cfrac%7B%5Cpartial%20%5E2v%7D%7B%5Cpartial%20y%5E2%7D%3D%5Cfrac%7B%5Cpartial%20%5E2u%7D%7B%5Cpartial%20x%5E2%7D%2B%5Cfrac%7B%5Cpartial%20%5E2u%7D%7B%5Cpartial%20y%5E2%7D-4c%3Du-4c%3Cv%3Du-c%5Cleft(%20x%5E2%2By%5E2%20%5Cright)%20$

那么 $v%3E%5Cfrac%7B%5Cpartial%20%5E2v%7D%7B%5Cpartial%20x%5E2%7D%2B%5Cfrac%7B%5Cpartial%20%5E2v%7D%7B%5Cpartial%20y%5E2%7D$ 且 $v%5Cge%200$ ， $x%5E2%2By%5E2%3D1$ ，和(1)一樣可以證明 $v%3E0$ ， $x%5E2%2By%5E2%3C1$ ，即 $u%3Ec%5Cleft(%20x%5E2%2By%5E2%20%5Cright)%20$ ， $x%5E2%2By%5E2%3C1$ ，故結(jié)論得證

5.設(shè) $f(x)$ 是僅有正實根的多項式，且 $%5Cfrac%7Bf'%5Cleft(%20x%20%5Cright)%7D%7Bf%5Cleft(%20x%20%5Cright)%7D%3D-%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7Bc_nx%5En%7D$

(1)證明 $c_n%3E0(n%3E0)$

(2)證明極限 $%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Clim%7D%5Cfrac%7B1%7D%7B%5Csqrt%5Bn%5D%7Bc_n%7D%7D$ 存在，且為 $f(x)$ 的根的最小值

(1)設(shè) $f%5Cleft(%20x%20%5Cright)%20%3D%5Cleft(%20x-x_1%20%5Cright)%20%5Cleft(%20x-x_2%20%5Cright)%20%5Ccdots%20%5Cleft(%20x-x_m%20%5Cright)%20$ ，那么

$%5Cfrac%7Bf'%5Cleft(%20x%20%5Cright)%7D%7Bf%5Cleft(%20x%20%5Cright)%7D%3D%5Cfrac%7B1%7D%7Bx-x_1%7D%2B%5Cfrac%7B1%7D%7Bx-x_2%7D%2B%5Ccdots%20%2B%5Cfrac%7B1%7D%7Bx-x_m%7D$

進而

$%5Cleft(%20%5Cfrac%7Bf'%5Cleft(%20x%20%5Cright)%7D%7Bf%5Cleft(%20x%20%5Cright)%7D%20%5Cright)%20%5E%7B%5Cleft(%20k%20%5Cright)%7D%3D%5Csum_%7Bn%3D1%7D%5Em%7B%5Cleft(%20-1%20%5Cright)%20%5Ekk!%5Cleft(%20x-x_n%20%5Cright)%20%5E%7B-1-k%7D%7D%3D-%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7Bk!c_%7Bn%2Bk%7Dx%5En%7D%2Ck%3D0%2C1%2C%5Ccdots%20$

令x=0可得 $%5Csum_%7Bn%3D1%7D%5Em%7B%5Cfrac%7B1%7D%7Bx_%7Bn%7D%5E%7Bk%7D%7D%7D%3Dc_k%3E0$

(2)不妨設(shè) $x_1%5Cle%20x_2%5Cle%20%5Ccdots%20%5Cle%20x_m$ ，那么

$%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Clim%7D%5Cfrac%7B1%7D%7B%5Csqrt%5Bn%5D%7Bc_n%7D%7D%3D%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Clim%7D%5Cfrac%7B1%7D%7B%5Csqrt%5Bn%5D%7B%5Csum_%7Bk%3D1%7D%5Em%7B%5Cfrac%7B1%7D%7Bx_%7Bm%7D%5E%7Bn%7D%7D%7D%7D%7D%3Dx_1$

（這個需要嚴格證明一下，但是我懶得寫了）

這套考卷的答案在公眾號考研競賽數(shù)學(xué)上面也有，我寫到一半才發(fā)現(xiàn)···大家可以結(jié)合著看

標(biāo)簽：考研數(shù)學(xué)分析

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