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分參即可(2018課標(biāo)Ⅱ?qū)?shù))

2022-11-06 20:57 作者:數(shù)學(xué)老頑童  | 我要投稿

(2018課標(biāo)Ⅱ,21)已知函數(shù)f%5Cleft(%20x%20%5Cright)%20%3D%5Cmathrm%7Be%7D%5Ex-ax%5E2.

(1)若a%3D1,證明:當(dāng)x%5Cgeqslant%200時(shí),f%5Cleft(%20x%20%5Cright)%20%5Cgeqslant%201;

(2)若f%5Cleft(%20x%20%5Cright)%20%5Cleft(%200%2C%2B%5Cinfty%20%5Cright)%20只有一個(gè)零點(diǎn),求a.

解:(1)若a%3D1,f%5Cleft(%20x%20%5Cright)%20%3D%5Cmathrm%7Be%7D%5Ex-x%5E2,

f%5Cleft(%20x%20%5Cright)%20%5Cgeqslant%201即為%5Cmathrm%7Be%7D%5Ex-x%5E2%5Cgeqslant%201

其等價(jià)于%5Ccolor%7Bred%7D%7B%5Cfrac%7Bx%5E2%2B1%7D%7B%5Cmathrm%7Be%7D%5Ex%7D%5Cleqslant%201%7D.

g%5Cleft(%20x%20%5Cright)%20%3D%5Cfrac%7Bx%5E2%2B1%7D%7B%5Cmathrm%7Be%7D%5Ex%7D,則

%5Cbegin%7Baligned%7D%0A%09%5Ccolor%7Bred%7D%7Bg'%5Cleft(%20x%20%5Cright)%20%7D%26%3D%5Cfrac%7B2x%5Cmathrm%7Be%7D%5Ex-%5Cleft(%20x%5E2%2B1%20%5Cright)%20%5Cmathrm%7Be%7D%5Ex%7D%7B%5Cleft(%20%5Cmathrm%7Be%7D%5Ex%20%5Cright)%20%5E2%7D%5C%5C%0A%09%26%3D%5Cfrac%7B-%5Cleft(%20x-1%20%5Cright)%20%5E2%7D%7B%5Cmathrm%7Be%7D%5Ex%7D%5Ccolor%7Bred%7D%7B%5Cleqslant%200%7D%5C%5C%0A%5Cend%7Baligned%7D

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

所以g%5Cleft(%20x%20%5Cright)%20%5Cleqslant%20g%5Cleft(%200%20%5Cright)%20%3D1,證畢.

(2)f%5Cleft(%20x%20%5Cright)%20的零點(diǎn)即為方程a%3D%5Cfrac%7B%5Cmathrm%7Be%7D%5Ex%7D%7Bx%5E2%7D的解.

h%5Cleft(%20x%20%5Cright)%20%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B%5Cmathrm%7Be%7D%5Ex%7D%7Bx%5E2%7D%7D,則

%5Cbegin%7Baligned%7D%0A%09h'%5Cleft(%20x%20%5Cright)%20%26%3D%5Cfrac%7B%5Cmathrm%7Be%7D%5Exx%5E2-%5Cmathrm%7Be%7D%5Ex%5Ccdot%202x%7D%7Bx%5E4%7D%5C%5C%0A%09%26%3D%5Cfrac%7B%5Cmathrm%7Be%7D%5Ex%5Cleft(%20x-2%20%5Cright)%7D%7Bx%5E3%7D%5C%5C%0A%5Cend%7Baligned%7D

h'%5Cleft(%20x%20%5Cright)%20%3D0,得x%3D2,

當(dāng)x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%200%2C2%20%5Cright)%20%7D,h'%5Cleft(%20x%20%5Cright)%20%3C0h%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Csearrow%20%7D;

當(dāng)x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%202%2C%2B%5Cinfty%20%5Cright)%20%7D,h'%5Cleft(%20x%20%5Cright)%20%3E0h%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D.

所以%5Ccolor%7Bred%7D%7Bh%5Cleft(%20x%20%5Cright)%20_%7B%5Cmin%7D%7D%3Dh%5Cleft(%202%20%5Cright)%20%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B%5Cmathrm%7Be%7D%5E2%7D%7B4%7D%7D.

a%3C%5Cfrac%7B%5Cmathrm%7Be%7D%5E2%7D%7B4%7D,則a%3Dh%5Cleft(%20x%20%5Cright)%20無(wú)解,即f%5Cleft(%20x%20%5Cright)%20無(wú)零點(diǎn),不合題意;

a%3D%5Cfrac%7B%5Cmathrm%7Be%7D%5E2%7D%7B4%7D,則a%3Dh%5Cleft(%20x%20%5Cright)%20僅有唯一解2,即f%5Cleft(%20x%20%5Cright)%20僅有唯一零點(diǎn)2,符合題意;

a%3E%5Cfrac%7B%5Cmathrm%7Be%7D%5E2%7D%7B4%7D,則由于

h%5Cleft(%20%5Ccolor%7Bred%7D%7B%5Cfrac%7B1%7D%7Ba%7D%7D%20%5Cright)%20%3Ea,h%5Cleft(%20%5Ccolor%7Bred%7D%7B2%7D%20%5Cright)%20%3Cah%5Cleft(%20%5Ccolor%7Bred%7D%7B2a%7D%20%5Cright)%20%3Ea,

%5Cexists%20x_1%5Cin%20%5Cleft(%20%5Cfrac%7B1%7D%7Ba%7D%2C2%20%5Cright)%20、%5Cexists%20x_2%5Cin%20%5Cleft(%202%2C2a%20%5Cright)%20,使得h%5Cleft(%20x_1%20%5Cright)%20%3Dh%5Cleft(%20x_2%20%5Cright)%20%3Da,不合題意;

綜上所述:%5Ccolor%7Bred%7D%7Ba%3D%5Cfrac%7B%5Cmathrm%7Be%7D%5E2%7D%7B4%7D%7D.


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