【一眼瞬秒】2023年新高考數(shù)學(xué)19題

2023-06-07 21:07 作者:_謹諾 0人讀過 | 我要投稿

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

(1) 討論 $f(x)$ 的單調(diào)性；

(2) 證明：當 $a%3E0$ 時， $f%5Cleft(%20x%20%5Cright)%20%3E2%5Cln%20a%2B%5Cfrac%7B3%7D%7B2%7D%0A$

解：

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

當 $a%5Cleq0$ 時， $f'(x)%3C0$ ，即 $f(x)$ 在 $%5Cleft(%20-%5Cinfty%20%2C%2B%5Cinfty%20%5Cright)%20%0A$ 上單調(diào)遞減；

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

即 $f(x)$ 在 $%5Cleft(%20-%5Cinfty%20%2C-%5Cln%20a%20%5Cright)%20%0A$ 上單調(diào)遞減，在 $%5Cleft%5B%20-%5Cln%20a%2C%2B%5Cinfty%20%5Cright)%20%0A$ 上單調(diào)遞增

(2) 由(1)知，當 $a%3E0$ 時， $f(x)$ 在 $%5Cleft(%20-%5Cinfty%20%2C-%5Cln%20a%20%5Cright)%20%0A$ 上單調(diào)遞減，在 $%5Cleft%5B%20-%5Cln%20a%2C%2B%5Cinfty%20%5Cright)%20%0A$ 上單調(diào)遞增

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

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

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

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

因此 $g(a)$ 在 $%5Cleft(%200%2C%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%5Cright)%20%0A$ 上單調(diào)遞減，在 $%5Cleft%5B%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2C%2B%5Cinfty%20%5Cright)%20%0A$ 上單調(diào)遞增

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