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扔掉一项,豁然开朗(2021新高考Ⅰ导数)

2022-10-13 22:05 作者:数学老顽童  | 我要投稿

(2021新高考Ⅰ,22)已知函数f%5Cleft(%20x%20%5Cright)%20%3Dx%5Cleft(%201-%5Cln%20%20x%20%5Cright)%20.

(1)讨论f%5Cleft(%20x%20%5Cright)%20的单调性;

(2)设ab为两个互不相等的正数,且b%5Cln%20%20a-a%5Cln%20%20b%3Da-b,证明:2%3C%5Cfrac%7B1%7D%7Ba%7D%2B%5Cfrac%7B1%7D%7Bb%7D%3C%5Cmathrm%7Be%7D.

解:(1)求导,得

%5Ccolor%7Bred%7D%7Bf%E2%80%99%5Cleft(%20x%20%5Cright)%7D%20%3D1%5Ccdot%20%5Cleft(%201-%5Cln%20%20x%20%5Cright)%20%2Bx%5Ccdot%20%5Cleft(%20-%5Cfrac%7B1%7D%7Bx%7D%20%5Cright)%20%3D%5Ccolor%7Bred%7D%7B-%5Cln%20%20x%7D

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

x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%200%2C1%20%5Cright)%20%7Df'%5Cleft(%20x%20%5Cright)%20%3E0f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D

x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%201%2C%2B%5Cinfty%20%5Cright)%20%7Df'%5Cleft(%20x%20%5Cright)%20%3C0f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Csearrow%20%7D.

(2)b%5Cln%20%20a-a%5Cln%20%20b%3Da-b等价于

%5Ccolor%7Bred%7D%7B%5Cfrac%7B1%7D%7Ba%7D%5Cleft(%201-%5Cln%20%5Cfrac%7B1%7D%7Ba%7D%20%5Cright)%20%3D%5Cfrac%7B1%7D%7Bb%7D%5Cleft(%201-%5Cln%20%5Cfrac%7B1%7D%7Bb%7D%20%5Cright)%20%7D

f%5Cleft(%20%5Cfrac%7B1%7D%7Ba%7D%20%5Cright)%20%3Df%5Cleft(%20%5Cfrac%7B1%7D%7Bb%7D%20%5Cright)%20

%5Cfrac%7B1%7D%7Ba%7D%3Dx_1%5Cfrac%7B1%7D%7Bb%7D%3Dx_2,则待证命题即为

2%3Cx_1%2Bx_2%3C%5Cmathrm%7Be%7D.

  • 先证2%3Cx_1%2Bx_2

其等价于%5Ccolor%7Bred%7D%7B2-x_1%3Cx_2%7D,(第一次转化)

又因0%3Cx_1%3C1,故2-x_1%3E1

2-x_1x_2同属%5Cleft(%201%2C%2B%5Cinfty%20%5Cright)%20

由前文已知f%5Cleft(%20x%20%5Cright)%20在此区间%5Csearrow%20,故

2-x_1%3Cx_2%5CLeftrightarrow%20%20%5Ccolor%7Bred%7D%7Bf%5Cleft(%202-x_1%20%5Cright)%20%3Ef%5Cleft(%20x_2%20%5Cright)%20%7D

(第二次转化)

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

f%5Cleft(%202-x_1%20%5Cright)%20%3Ef%5Cleft(%20x_2%20%5Cright)%20又等价于

%5Ccolor%7Bred%7D%7Bf%5Cleft(%202-x_1%20%5Cright)%20%3Ef%5Cleft(%20x_1%20%5Cright)%20%7D(第三次转化),

f%5Cleft(%202-x_1%20%5Cright)%20-f%5Cleft(%20x_1%20%5Cright)%20%3E0.

g%5Cleft(%20x%20%5Cright)%20%3Df%5Cleft(%202-x%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20

其中x%5Cin%20%5Cleft(%200%2C1%20%5Cright)%20,则

%5Cbegin%7Baligned%7D%0A%09%5Ccolor%7Bred%7D%7Bg'%5Cleft(%20x%20%5Cright)%7D%20%26%3Df'%5Cleft(%202-x%20%5Cright)%20%5Cleft(%202-x%20%5Cright)%20'-f'%5Cleft(%20x%20%5Cright)%5C%5C%0A%09%26%3D-%5Cln%20%5Cleft(%202-x%20%5Cright)%20%5Ccdot%20%5Cleft(%20-1%20%5Cright)%20-%5Cleft(%20-%5Cln%20x%20%5Cright)%5C%5C%0A%09%26%3D%5Cln%20%5Cleft%5B%20x%5Cleft(%202-x%20%5Cright)%20%5Cright%5D%20%5Ccolor%7Bred%7D%7B%3C0%7D%5C%5C%0A%5Cend%7Baligned%7D

g%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Csearrow%7D%20

g%5Cleft(%20x%20%5Cright)%20%3Eg%5Cleft(%201%20%5Cright)%20%3D0

2%3Cx_1%2Bx_2得证.

  • 再证x_1%2Bx_2%3C%5Cmathrm%7Be%7D

h%5Cleft(%20x%20%5Cright)%20%3D%5Cbegin%7Bcases%7D%09x%2Cx%5Cin%20%5Cleft(%200%2C1%20%5Cright)%20%2C%5C%5C%09%5Cmathrm%7Be%7D-x%2Cx%5Cin%20%5Cleft(%201%2C%5Cmathrm%7Be%7D%20%5Cright)%20%2C%5C%5C%5Cend%7Bcases%7D

x_3%5Cin%20%5Cleft(%200%2C1%20%5Cright)%20x_4%5Cin%20%5Cleft(%201%2C%5Cmathrm%7Be%7D%20%5Cright)%20,使得:

f%5Cleft(%20x_1%20%5Cright)%20%3Dh%5Cleft(%20x_3%20%5Cright)%20%3Df%5Cleft(%20x_2%20%5Cright)%20%3Dh%5Cleft(%20x_4%20%5Cright)%20%3Dt

易知:%5Ccolor%7Bred%7D%7Bx_3%3Dt%7D%5Ccolor%7Bred%7D%7Bx_4%3D%5Cmathrm%7Be%7D-t%7D.

构造函数:

H%5Cleft(%20x%20%5Cright)%20%3Dh%5Cleft(%20x%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%3D%5Cbegin%7Bcases%7D%09x%5Cln%20%20x%2Cx%5Cin%20%5Cleft(%200%2C1%20%5Cright)%20%2C%5C%5C%09x%5Cln%20%20x-2x%2B%5Cmathrm%7Be%7D%2Cx%5Cin%20%5Cleft(%201%2C%5Cmathrm%7Be%7D%20%5Cright)%20.%5C%5C%5Cend%7Bcases%7D

x%5Cin%20%5Cleft(%200%2C1%20%5Cright)%20,易知H%5Cleft(%20x%20%5Cright)%20%3C0

h%5Cleft(%20x%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%3C0

h%5Cleft(%20x%20%5Cright)%20%3Cf%5Cleft(%20x%20%5Cright)%20

h%5Cleft(%20x_1%20%5Cright)%20%3Cf%5Cleft(%20x_1%20%5Cright)%20%3Dh%5Cleft(%20x_3%20%5Cright)%20

又因为h%5Cleft(%20x%20%5Cright)%20%5Cnearrow%20

%5Ccolor%7Bred%7D%7Bx_1%3Cx_3%7D

x%5Cin%20%5Cleft(%201%2C%5Cmathrm%7Be%7D%20%5Cright)%20H'%5Cleft(%20x%20%5Cright)%20%3D%5Cln%20%20x-1%3C0

H%5Cleft(%20x%20%5Cright)%20%5Csearrow%20

H%5Cleft(%20x%20%5Cright)%20%3EH%5Cleft(%20%5Cmathrm%7Be%7D%20%5Cright)%20%3D0

h%5Cleft(%20x%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%3E0

h%5Cleft(%20x%20%5Cright)%20%3Ef%5Cleft(%20x%20%5Cright)%20

h%5Cleft(%20x_2%20%5Cright)%20%3Ef%5Cleft(%20x_2%20%5Cright)%20%3Dh%5Cleft(%20x_4%20%5Cright)%20

又因为h%5Cleft(%20x%20%5Cright)%20%5Csearrow%20

%5Ccolor%7Bred%7D%7Bx_2%3Cx_4%7D

所以x_1%2Bx_2%3Cx_3%2Bx_4%3Dt%2B%5Cmathrm%7Be%7D-t%3D%5Cmathrm%7Be%7D.

综上所述:2%3C%5Cfrac%7B1%7D%7Ba%7D%2B%5Cfrac%7B1%7D%7Bb%7D%3C%5Cmathrm%7Be%7D,证毕.

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