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必要条件探路的规范写法(2022新高考Ⅱ导数)

2022-10-04 23:15 作者:数学老顽童  | 我要投稿

(2022新高考Ⅱ,22)已知函数f%5Cleft(%20x%20%5Cright)%20%3Dx%5Cmathrm%7Be%7D%5E%7Bax%7D-%5Cmathrm%7Be%7D%5Ex.

(1)当a%3D1时,讨论f%5Cleft(%20x%20%5Cright)%20的单调性;

(2)当x%3E0时,f%5Cleft(%20x%20%5Cright)%20%3C-1,求a的取值范围;

(3)设n%5Cin%20%5Cmathbf%7BN%7D%5E*,证明:

%5Cfrac%7B1%7D%7B%5Csqrt%7B1%5E2%2B1%7D%7D%2B%5Cfrac%7B1%7D%7B%5Csqrt%7B2%5E2%2B2%7D%7D%2B%5Ccdots%20%2B%5Cfrac%7B1%7D%7B%5Csqrt%7Bn%5E2%2Bn%7D%7D%3E%5Cln%20%5Cleft(%20n%2B1%20%5Cright)%20

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

%5Ccolor%7Bred%7D%7Bf'%5Cleft(%20x%20%5Cright)%20%7D%3D1%5Ccdot%20%5Cmathrm%7Be%7D%5Ex%2Bx%5Ccdot%20%5Cmathrm%7Be%7D%5Ex-%5Cmathrm%7Be%7D%5Ex%3D%5Ccolor%7Bred%7D%7Bx%5Cmathrm%7Be%7D%5Ex%7D

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

x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%20-%5Cinfty%20%2C0%20%5Cright)%7D,f'%5Cleft(%20x%20%5Cright)%20%3C0,f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Csearrow%20%7D;

x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%200%2C%2B%5Cinfty%20%5Cright)%20%7D,f'%5Cleft(%20x%20%5Cright)%20%3E0,f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D.

(2)求导,得

%5Cbegin%7Baligned%7D%0A%09f'%5Cleft(%20x%20%5Cright)%20%26%3D1%5Ccdot%20%5Cmathrm%7Be%7D%5E%7Bax%7D%2Bx%5Ccdot%20%5Cmathrm%7Be%7D%5E%7Bax%7D%5Ccdot%20a-%5Cmathrm%7Be%7D%5Ex%5C%5C%0A%09%26%3D%5Cmathrm%7Be%7D%5E%7Bax%7D%5Cleft(%20ax%2B1%20%5Cright)%20-%5Cmathrm%7Be%7D%5Ex%5C%5C%0A%5Cend%7Baligned%7D

g%5Cleft(%20x%20%5Cright)%20%3D%5Cmathrm%7Be%7D%5E%7Bax%7D%5Cleft(%20ax%2B1%20%5Cright)%20-%5Cmathrm%7Be%7D%5Ex

求导,得

%5Cbegin%7Baligned%7D%0A%09g'%5Cleft(%20x%20%5Cright)%20%26%3Da%5Cmathrm%7Be%7D%5E%7Bax%7D%5Cleft(%20ax%2B1%20%5Cright)%20%2B%5Cmathrm%7Be%7D%5E%7Bax%7D%5Ccdot%20a-%5Cmathrm%7Be%7D%5Ex%5C%5C%0A%09%26%3Da%5Cmathrm%7Be%7D%5E%7Bax%7D%5Cleft(%20ax%2B2%20%5Cright)%20-%5Cmathrm%7Be%7D%5Ex%5C%5C%0A%5Cend%7Baligned%7D

g'%5Cleft(%200%20%5Cright)%20%3D2a-1.

2a-1%3E0,即a%3E%5Cfrac%7B1%7D%7B2%7D

存在正数%5Ccolor%7Bred%7D%7Bx_0%7D

使得在区间%5Ccolor%7Bred%7D%7B%5Cleft(%200%2Cx_0%20%5Cright)%20%7Dg'%5Cleft(%20x%20%5Cright)%20%3E0

从而f'%5Cleft(%20x%20%5Cright)%20%5Cnearrow%20

从而f'%5Cleft(%20x%20%5Cright)%20%3E%5Ccolor%7Bred%7D%7Bf'%5Cleft(%200%20%5Cright)%20%3D0%7D(关键一)

从而f%5Cleft(%20x%20%5Cright)%20%5Cnearrow%20

从而f%5Cleft(%20x%20%5Cright)%20%3E%5Ccolor%7Bred%7D%7Bf%5Cleft(%200%20%5Cright)%20%3D-1%7D(关键二)

不合题意.

所以%5Ccolor%7Bred%7D%7Ba%5Cleqslant%20%5Cfrac%7B1%7D%7B2%7D%7D%5Ccolor%7Bred%7D%7Bf%5Cleft(%20x%20%5Cright)%20%3C-1%7D必要条件.

下面证明a%5Cleqslant%20%5Cfrac%7B1%7D%7B2%7Df%5Cleft(%20x%20%5Cright)%20%3C-1充分条件.

a%5Cleqslant%20%5Cfrac%7B1%7D%7B2%7D时,f%5Cleft(%20x%20%5Cright)%20%5Cleqslant%20x%5Cmathrm%7Be%7D%5E%7B%5Ccolor%7Bred%7D%7B%5Cfrac%7B1%7D%7B2%7D%7Dx%7D-%5Cmathrm%7Be%7D%5Ex

故只需证x%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7B2%7Dx%7D-%5Cmathrm%7Be%7D%5Ex%3C%20-1

上式又等价于%5Ccolor%7Bred%7D%7B%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7B2%7Dx%7D-%5Cmathrm%7Be%7D%5E%7B-%5Cfrac%7B1%7D%7B2%7Dx%7D-x%3E0%7D.

h%5Cleft(%20x%20%5Cright)%20%3D%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7B2%7Dx%7D-%5Cmathrm%7Be%7D%5E%7B-%5Cfrac%7B1%7D%7B2%7Dx%7D-x,则

%5Cbegin%7Baligned%7D%0A%09h%E2%80%99%5Cleft(%20x%20%5Cright)%20%26%3D%5Cfrac%7B1%7D%7B2%7D%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7B2%7Dx%7D-%5Cleft(%20-%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5Cmathrm%7Be%7D%5E%7B-%5Cfrac%7B1%7D%7B2%7Dx%7D-1%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7B2%7Dx%7D%2B%5Cmathrm%7Be%7D%5E%7B-%5Cfrac%7B1%7D%7B2%7Dx%7D%20%5Cright)%20-1%5C%5C%0A%09%26%3E%5Cfrac%7B1%7D%7B2%7D%5Ctimes%202-1%3D0%5C%5C%0A%5Cend%7Baligned%7D

(用一下均值不等式

h%5Cleft(%20x%20%5Cright)%20%5Cnearrow%20,故h%5Cleft(%20x%20%5Cright)%20%3Eh%5Cleft(%200%20%5Cright)%20%3D0,证毕.

综上所述:%5Ccolor%7Bred%7D%7Ba%5Cin%20%5Cleft(%20-%5Cinfty%20%2C%5Cfrac%7B1%7D%7B2%7D%20%5Cright%5D%20%7D.

(3)由(2)可知:

%5Cforall%20x%5Cin%20%5Cleft(%200%2C%2B%5Cinfty%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cmathrm%7Be%7D%5E%7B%5Cfrac%7B1%7D%7B2%7Dx%7D-%5Cmathrm%7Be%7D%5E%7B-%5Cfrac%7B1%7D%7B2%7Dx%7D-x%3E0%7D

换元,令%5Cmathrm%7Be%7D%5Ex%3Dt%5Cin%20%5Cleft(%201%2C%2B%5Cinfty%20%5Cright)%20,可得

%5Ccolor%7Bred%7D%7B%5Cforall%20t%5Cin%20%5Cleft(%201%2C%2B%5Cinfty%20%5Cright)%20%2C%5Csqrt%7Bt%7D-%5Cfrac%7B1%7D%7B%5Csqrt%7Bt%7D%7D%3E%20%5Cln%20%20t%7D.

所以%5Csqrt%7B%5Ccolor%7Bred%7D%7B%5Cfrac%7Bk%2B1%7D%7Bk%7D%7D%7D-%5Cfrac%7B1%7D%7B%5Csqrt%7B%5Ccolor%7Bred%7D%7B%5Cfrac%7Bk%2B1%7D%7Bk%7D%7D%7D%7D%3E%20%5Cln%20%5Ccolor%7Bred%7D%7B%5Cfrac%7Bk%2B1%7D%7Bk%7D%7D

整理得%5Ccolor%7Bred%7D%7B%5Cfrac%7B1%7D%7B%5Csqrt%7Bk%5E2%2Bk%7D%7D%3E%20%5Cln%20%5Cfrac%7Bk%2B1%7D%7Bk%7D%7Dk%5Cin%20%5Cmathbf%7BN%7D%5E*.

从而

%5Csum_%7Bk%3D1%7D%5En%7B%5Cfrac%7B1%7D%7B%5Csqrt%7Bk%5E2%2Bk%7D%7D%7D%3E%5Csum_%7Bk%3D1%7D%5En%7B%5Cln%20%5Cfrac%7Bk%2B1%7D%7Bk%7D%7D%3D%5Cln%20%5Cleft(%20n%2B1%20%5Cright)%20

证毕.

命题背景:

%5Cbbox%5B%23def%2C5px%2Cborder%3A1px%20solid%5D%7B%0A%5Cbegin%7Baligned%7D%0A%09%5Cforall%20x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%200%2C1%20%5Cright)%7D%20%2C%5Cln%20x%26%5Ccolor%7Bred%7D%7B%3E%7D%5Csqrt%7Bx%7D-%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%7D%7D%2C%5C%5C%0A%09%5Cforall%20x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%201%2C%2B%5Cinfty%20%5Cright)%7D%20%2C%5Cln%20x%26%5Ccolor%7Bred%7D%7B%3C%7D%5Csqrt%7Bx%7D-%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%7D%7D.%5C%5C%0A%5Cend%7Baligned%7D%7D


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