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丢开同构,通法解题(2022全国甲导数)

2022-10-03 19:54 作者:数学老顽童  | 我要投稿

(2022全国甲,21)已知函数f%5Cleft(%20x%20%5Cright)%20%3D%5Cfrac%7B%5Cmathrm%7Be%7D%5Ex%7D%7Bx%7D-%5Cln%20%20x%2Bx-a.

(1)若f%5Cleft(%20x%20%5Cright)%20%5Cgeqslant%200,求a的取值范围;

(2)证明:若f%5Cleft(%20x%20%5Cright)%20有两个零点,x_1x_2,则x_1x_2%3C1.

解:(1)求导,得

%5Cbegin%7Baligned%7D%0A%09f'%5Cleft(%20x%20%5Cright)%20%26%3D%5Cfrac%7B%5Cmathrm%7Be%7D%5Ex%5Ccdot%20x-%5Cmathrm%7Be%7D%5Ex%5Ccdot%201%7D%7Bx%5E2%7D-%5Cfrac%7B1%7D%7Bx%7D%2B1%5C%5C%0A%09%26%3D%5Cfrac%7B%5Cleft(%20x-1%20%5Cright)%20%5Cleft(%20%5Cmathrm%7Be%7D%5Ex%2Bx%20%5Cright)%7D%7Bx%5E2%7D%5C%5C%0A%5Cend%7Baligned%7D

f'%5Cleft(%20x%20%5Cright)%20%3D0,得x%3D1

x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%200%2C1%20%5Cright)%20%7Df'%5Cleft(%20x%20%5Cright)%20%3C0f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Csearrow%20%7D

x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%201%2C%2B%5Cinfty%20%5Cright)%20%7Df'%5Cleft(%20x%20%5Cright)%20%3E0f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D

f%5Cleft(%20x%20%5Cright)%20_%7B%5Cmin%7D%3Df%5Cleft(%201%20%5Cright)%20%3D%5Cmathrm%7Be%7D%2B1-a%5Cgeqslant%200

解得%5Ccolor%7Bred%7D%7Ba%5Cleqslant%20%5Cmathrm%7Be%7D%2B1%7D.

(2)x_1x_2%3C1等价于%5Ccolor%7Bred%7D%7Bx_2%3C%5Cfrac%7B1%7D%7Bx_1%7D%7D

第一次转化

易知0%3Cx_1%3C1,故%5Cfrac%7B1%7D%7Bx_1%7D%3E1

所以x_2%5Cfrac%7B1%7D%7Bx_1%7D%5Cin%20%5Cleft(%201%2C%2B%5Cinfty%20%5Cright)%20

由(1)知f%5Cleft(%20x%20%5Cright)%20%5Cleft(%201%2C%2B%5Cinfty%20%5Cright)%20%5Cnearrow%20

x_2%3C%5Cfrac%7B1%7D%7Bx_1%7D等价于%5Ccolor%7Bred%7D%7Bf%5Cleft(%20x_2%20%5Cright)%20%3Cf%5Cleft(%20%5Cfrac%7B1%7D%7Bx_1%7D%20%5Cright)%20%7D.

第二次转化

又因为f%5Cleft(%20x_1%20%5Cright)%20%3D0%3Df%5Cleft(%20x_2%20%5Cright)%20

所以f%5Cleft(%20x_2%20%5Cright)%20%3Cf%5Cleft(%20%5Cfrac%7B1%7D%7Bx_1%7D%20%5Cright)%20等价于

%5Ccolor%7Bred%7D%7Bf%5Cleft(%20x_1%20%5Cright)%20%3Cf%5Cleft(%20%5Cfrac%7B1%7D%7Bx_1%7D%20%5Cright)%20%7D

f%5Cleft(%20x_1%20%5Cright)%20-f%5Cleft(%20%5Cfrac%7B1%7D%7Bx_1%7D%20%5Cright)%20%3C0.

第三次转化

g%5Cleft(%20x%20%5Cright)%20%3Df%5Cleft(%20x%20%5Cright)%20-f%5Cleft(%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright)%20x%5Cin%20%5Cleft(%200%2C1%20%5Cright)%20

%5Cbegin%7Baligned%7D%0A%09g'%5Cleft(%20x%20%5Cright)%20%26%3Df'%5Cleft(%20x%20%5Cright)%20-f'%5Cleft(%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright)%20%5Ccdot%20%5Cleft(%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright)%20'%5C%5C%0A%09%26%3D%5Cfrac%7B%5Cleft(%20x-1%20%5Cright)%20%5Cleft(%20%5Cmathrm%7Be%7D%5Ex%2Bx%20%5Cright)%7D%7Bx%5E2%7D-%5Cfrac%7B%5Cleft(%20%5Cfrac%7B1%7D%7Bx%7D-1%20%5Cright)%20%5Cleft(%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7Bx%7D%7D%2B%5Cfrac%7B1%7D%7Bx%7D%20%5Cright)%7D%7B%5Cleft(%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright)%20%5E2%7D%5Ccdot%20%5Cleft(%20-%5Cfrac%7B1%7D%7Bx%5E2%7D%20%5Cright)%5C%5C%0A%09%26%3D%5Cfrac%7B%5Cleft(%20x-1%20%5Cright)%20%5Cleft(%20x-1%2B%5Ccolor%7Bred%7D%7B%5Cmathrm%7Be%7D%5Ex-x%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7Bx%7D%7D%7D%20%5Cright)%7D%7Bx%5E2%7D%5C%5C%0A%5Cend%7Baligned%7D

h%5Cleft(%20x%20%5Cright)%20%3D%5Cmathrm%7Be%7D%5Ex-x%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7Bx%7D%7Dx%5Cin%20%5Cleft(%200%2C1%20%5Cright)%20,则

%5Cbegin%7Baligned%7D%0A%09%5Ccolor%7Bred%7D%7Bh'%5Cleft(%20x%20%5Cright)%7D%20%26%3D%5Cmathrm%7Be%7D%5Ex-%5Cleft%5B%201%5Ccdot%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7Bx%7D%7D%2Bx%5Ccdot%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7Bx%7D%7D%5Ccdot%20%5Cleft(%20-%5Cfrac%7B1%7D%7Bx%5E2%7D%20%5Cright)%20%5Cright%5D%5C%5C%0A%09%26%3D%5Cmathrm%7Be%7D%5Ex%2B%5Cleft(%20%5Cfrac%7B1%7D%7Bx%7D-1%20%5Cright)%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7Bx%7D%7D%5Ccolor%7Bred%7D%7B%3E0%7D%5C%5C%0A%5Cend%7Baligned%7D

h%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D

%5Ccolor%7Bred%7D%7Bh%5Cleft(%20x%20%5Cright)%20%3C%7Dh%5Cleft(%201%20%5Cright)%20%3D%5Ccolor%7Bred%7D%7B0%7D.

%5Ccolor%7Bred%7D%7Bg'%5Cleft(%20x%20%5Cright)%20%3E0%7Dg%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D

%5Ccolor%7Bred%7D%7Bg%5Cleft(%20x%20%5Cright)%20%3C%7Dg%5Cleft(%201%20%5Cright)%20%3D%5Ccolor%7Bred%7D%7B0%7D,证毕.

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