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【一眼瞬秒】2023年新高考数学19题

2023-06-07 21:07 作者:_谨诺  | 我要投稿

19. 已知函数f%5Cleft(%20x%20%5Cright)%20%3Da%5Cleft(%20e%5Ex%2Ba%20%5Cright)%20-x%0A

(1) 讨论f(x)的单调性;

(2) 证明:当a%3E0时,f%5Cleft(%20x%20%5Cright)%20%3E2%5Cln%20a%2B%5Cfrac%7B3%7D%7B2%7D%0A

解:

(1) f%5Cleft(%20x%20%5Cright)%20%3Dae%5Ex-x%2Ba%5E2%5CRightarrow%20f'%5Cleft(%20x%20%5Cright)%20%3Dae%5Ex-1%0A%0A

a%5Cleq0时,f'(x)%3C0,即f(x)%5Cleft(%20-%5Cinfty%20%2C%2B%5Cinfty%20%5Cright)%20%0A上单调递减;

a%3E0时,f'%5Cleft(%20x%20%5Cright)%20%3D0%5CRightarrow%20x%3D-%5Cln%20a%0A

f(x)%5Cleft(%20-%5Cinfty%20%2C-%5Cln%20a%20%5Cright)%20%0A上单调递减,在%5Cleft%5B%20-%5Cln%20a%2C%2B%5Cinfty%20%5Cright)%20%0A上单调递增

(2) 由(1)知,当a%3E0时,f(x)%5Cleft(%20-%5Cinfty%20%2C-%5Cln%20a%20%5Cright)%20%0A上单调递减,在%5Cleft%5B%20-%5Cln%20a%2C%2B%5Cinfty%20%5Cright)%20%0A上单调递增

那么f_%7B%5Cmin%7D%5Cleft(%20x%20%5Cright)%20%3Df%5Cleft(%20-%5Cln%20a%20%5Cright)%20%3Da%5E2%2B%5Cln%20a%2B1%0A

只需要证明a%5E2%2B%5Cln%20a%2B1%3E2%5Cln%20a%2B%5Cfrac%7B3%7D%7B2%7D%5CRightarrow%20a%5E2-%5Cln%20a-%5Cfrac%7B1%7D%7B2%7D%3E0%2C%20%5Cforall%20a%3E0%0A

g%5Cleft(%20a%20%5Cright)%20%3Da%5E2-%5Cln%20a-%5Cfrac%7B1%7D%7B2%7D%2C%20a%3E0%0A

那么g'%5Cleft(%20a%20%5Cright)%20%3D2a-%5Cfrac%7B1%7D%7Ba%7D%0A,由g'%5Cleft(%20a%20%5Cright)%20%3D0%0Aa%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%0A

因此g(a)%5Cleft(%200%2C%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%5Cright)%20%0A上单调递减,在%5Cleft%5B%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2C%2B%5Cinfty%20%5Cright)%20%0A上单调递增

g%5Cleft(%20a%20%5Cright)%20%5Cge%20g_%7B%5Cmin%7D%5Cleft(%20a%20%5Cright)%20%3Dg%5Cleft(%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%5Cright)%20%3D%5Cfrac%7B1%7D%7B2%7D%5Cln%202%3E0%0A

原问题得证。

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