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扔掉一項,豁然開朗(2021新高考Ⅰ導(dǎo)數(shù))

2022-10-13 22:05 作者:數(shù)學(xué)老頑童  | 我要投稿

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

(1)討論f%5Cleft(%20x%20%5Cright)%20的單調(diào)性;

(2)設(shè)ab為兩個互不相等的正數(shù),且b%5Cln%20%20a-a%5Cln%20%20b%3Da-b,證明:2%3C%5Cfrac%7B1%7D%7Ba%7D%2B%5Cfrac%7B1%7D%7Bb%7D%3C%5Cmathrm%7Be%7D.

解:(1)求導(dǎo),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%7D,f'%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%3C0,f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Csearrow%20%7D.

(2)b%5Cln%20%20a-a%5Cln%20%20b%3Da-b等價于

%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,則待證命題即為

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

  • 先證2%3Cx_1%2Bx_2,

其等價于%5Ccolor%7Bred%7D%7B2-x_1%3Cx_2%7D,(第一次轉(zhuǎn)化)

又因0%3Cx_1%3C1,故2-x_1%3E1

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

由前文已知f%5Cleft(%20x%20%5Cright)%20在此區(qū)間%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

(第二次轉(zhuǎn)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又等價于

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

%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得證.

  • 再證x_1%2Bx_2%3C%5Cmathrm%7Be%7D

設(shè)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)%20,x_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.

構(gòu)造函數(shù):

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

又因為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)%20,H'%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,

又因為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.

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

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