較為常規(guī)的必要條件探路（2023全國(guó)乙，21）

2023-06-09 22:32 作者:數(shù)學(xué)老頑童 0人讀過(guò) | 我要投稿

（2023全國(guó)乙，21）已知函數(shù) $f%5Cleft(%20x%20%5Cright)%20%3D%5Cleft(%20%5Cfrac%7B1%7D%7Bx%7D%2Ba%20%5Cright)%20%5Cln%20%5Cleft(%201%2Bx%20%5Cright)%20$ .

（1）當(dāng) $a%3D-1$ 時(shí)，求曲線 $y%3Df%5Cleft(%20x%20%5Cright)%20$ 在點(diǎn) $%5Cleft(%201%2Cf%5Cleft(%201%20%5Cright)%20%5Cright)%20$ 處的切線方程；

（2）是否存在 $a$ 、 $b$ ，使得曲線 $y%3Df%5Cleft(%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright)%20$ 關(guān)于直線 $x%3Db$ 對(duì)稱，若存在，求 $a$ 、 $b$ 的值，若不存在，說(shuō)明理由；

（3）若 $f%5Cleft(%20x%20%5Cright)%20$ 存在極值，求 $a$ 的取值范圍.

解：（1）當(dāng) $a%3D-1$ 時(shí)，

$f%5Cleft(%20x%20%5Cright)%20%3D%5Cleft(%20%5Cfrac%7B1%7D%7Bx%7D-1%20%5Cright)%20%5Cln%20%5Cleft(%201%2Bx%20%5Cright)%20$ ，

$f%5Cleft(%201%20%5Cright)%20%3D0$ ，故切點(diǎn)為 $%5Ccolor%7Bred%7D%7B%5Cleft(%201%2C0%20%5Cright)%20%7D$ ，

$%5Cbegin%7Baligned%7D%0A%09f'%5Cleft(%20x%20%5Cright)%20%26%3D-%5Cfrac%7B1%7D%7Bx%5E2%7D%5Ccdot%20%5Cln%20%5Cleft(%201%2Bx%20%5Cright)%20%2B%5Cleft(%20%5Cfrac%7B1%7D%7Bx%7D-1%20%5Cright)%20%5Ccdot%20%5Cfrac%7B1%7D%7B1%2Bx%7D%5Ccdot%201%5C%5C%0A%09%26%3D%5Cfrac%7B1-x%7D%7Bx%5Cleft(%201%2Bx%20%5Cright)%7D-%5Cfrac%7B%5Cln%20%5Cleft(%201%2Bx%20%5Cright)%7D%7Bx%5E2%7D%5C%5C%0A%5Cend%7Baligned%7D$

$f'%5Cleft(%201%20%5Cright)%20%3D%5Ccolor%7Bred%7D%7B-%5Cln%20%202%7D$ ，

故所求切線為 $y-0%3D-%5Cln%20%202%5Ccdot%20%5Cleft(%20x-1%20%5Cright)%20$ ，

即 $%5Ccolor%7Bred%7D%7Bx%5Cln%20%202%2By-%5Cln%20%202%3D0%7D$ .

（2）令

$g%5Cleft(%20x%20%5Cright)%20%3Df%5Cleft(%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright)%20%3D%5Cleft(%20x%2Ba%20%5Cright)%20%5Cln%20%5Cleft(%201%2B%5Cfrac%7B1%7D%7Bx%7D%20%5Cright)%20$

由于 $1%2B%5Cfrac%7B1%7D%7Bx%7D%3E0$ ，故 $g%5Cleft(%20x%20%5Cright)%20$ 的定義域為

$x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%20-%5Cinfty%20%2C-1%20%5Cright)%20%5Ccup%20%5Cleft(%200%2C%2B%5Cinfty%20%5Cright)%20%7D$ ，

若存在 $b$ ，使得曲線 $y%3Dg%5Cleft(%20x%20%5Cright)%20$ 關(guān)于直線 $x%3Db$ 對(duì)稱，則 $%5Ccolor%7Bred%7D%7Bb%3D-%5Cfrac%7B1%7D%7B2%7D%7D$ ；

即 $%5Cforall%20x%5Cin%20%5Cleft(%20-%5Cinfty%20%2C-1%20%5Cright)%20%5Ccup%20%5Cleft(%200%2C%2B%5Cinfty%20%5Cright)%20$

皆有 $%5Ccolor%7Bred%7D%7Bg%5Cleft(%20-1-x%20%5Cright)%20%3Dg%5Cleft(%20x%20%5Cright)%20%7D$ ，即

$%5Cleft(%20-1-x%2Ba%20%5Cright)%20%5Cln%20%5Cleft(%201%2B%5Cfrac%7B1%7D%7B-1-x%7D%20%5Cright)%20%3D%5Cleft(%20x%2Ba%20%5Cright)%20%5Cln%20%5Cleft(%201%2B%5Cfrac%7B1%7D%7Bx%7D%20%5Cright)%20$

整理得 $%5Ccolor%7Bred%7D%7B%5Cleft(%202a-1%20%5Cright)%20%7D%5Cln%20%5Cleft(%201%2B%5Cfrac%7B1%7D%7Bx%7D%20%5Cright)%20%3D0$ ，

故 $2a-1%3D0$ ，解得 $%5Ccolor%7Bred%7D%7Ba%3D%5Cfrac%7B1%7D%7B2%7D%7D$ .

綜上所述： $a%3D%5Cfrac%7B1%7D%7B2%7D$ ， $b%3D-%5Cfrac%7B1%7D%7B2%7D$ .

（3）求導(dǎo)，得

$%5Cbegin%7Baligned%7D%0A%09f'%5Cleft(%20x%20%5Cright)%20%26%3D-%5Cfrac%7B1%7D%7Bx%5E2%7D%5Ccdot%20%5Cln%20%5Cleft(%201%2Bx%20%5Cright)%20%2B%5Cleft(%20%5Cfrac%7B1%7D%7Bx%7D%2Ba%20%5Cright)%20%5Ccdot%20%5Cfrac%7B1%7D%7B1%2Bx%7D%5Ccdot%201%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7Bx%5E2%7D%5Cleft%5B%5Ccolor%7Bred%7D%7B%20%5Cfrac%7Bax%5E2%2Bx%7D%7B1%2Bx%7D-%5Cln%20%5Cleft(%201%2Bx%20%5Cright)%20%7D%5Cright%5D%5C%5C%0A%5Cend%7Baligned%7D$

令 $h%5Cleft(%20x%20%5Cright)%20%3D%5Cfrac%7Bax%5E2%2Bx%7D%7B1%2Bx%7D-%5Cln%20%5Cleft(%201%2Bx%20%5Cright)%20$ ，

注意到 $%5Ccolor%7Bred%7D%7Bh%5Cleft(%200%20%5Cright)%20%3D0%7D$ ，求導(dǎo)，得

$%5Cbegin%7Baligned%7D%0A%09h'%5Cleft(%20x%20%5Cright)%20%26%3D%5Cfrac%7B%5Cleft(%202ax%2B1%20%5Cright)%20%5Cleft(%201%2Bx%20%5Cright)%20-%5Cleft(%20ax%5E2%2Bx%20%5Cright)%20%5Ccdot%201%7D%7B%5Cleft(%201%2Bx%20%5Cright)%20%5E2%7D-%5Cfrac%7B1%7D%7B1%2Bx%7D%5Ccdot%201%5C%5C%0A%09%26%3D%5Cfrac%7Bx%5Cleft(%20%5Ccolor%7Bred%7D%7Bax%2B2a-1%7D%20%5Cright)%7D%7B%5Cleft(%201%2Bx%20%5Cright)%20%5E2%7D%5C%5C%0A%5Cend%7Baligned%7D$

1.若 $%5Ccolor%7Bred%7D%7Ba%5Cleqslant%200%7D$ ，

則 $h'%5Cleft(%20x%20%5Cright)%20%3C0$ ，則 $h%5Cleft(%20x%20%5Cright)%20%5Csearrow%20$ ，

則 $h%5Cleft(%20x%20%5Cright)%20%3Ch%5Cleft(%200%20%5Cright)%20%3D0$ ，則 $f%5Cleft(%20x%20%5Cright)%20%5Csearrow%20$ ，

無(wú)極值，不合題意；

2.若 $%5Cbegin%7Bcases%7D%09a%3E0%2C%5C%5C%09a%5Ccdot%200%2B2a-1%5Cgeqslant%200%2C%5C%5C%5Cend%7Bcases%7D$ 即 $%5Ccolor%7Bred%7D%7Ba%5Cgeqslant%20%5Cfrac%7B1%7D%7B2%7D%7D$ ，

則 $h'%5Cleft(%20x%20%5Cright)%20%3E0$ ，則 $h%5Cleft(%20x%20%5Cright)%20%5Cnearrow%20$ ，

則 $h%5Cleft(%20x%20%5Cright)%20%3Eh%5Cleft(%200%20%5Cright)%20%3D0$ ，則 $f%5Cleft(%20x%20%5Cright)%20%5Cnearrow%20$ ，

無(wú)極值，不合題意；

3.若 $%5Cbegin%7Bcases%7D%09a%3E0%2C%5C%5C%09a%5Ccdot%200%2B2a-1%3C0%2C%5C%5C%5Cend%7Bcases%7D$ 即 $%5Ccolor%7Bred%7D%7B0%3Ca%3C%5Cfrac%7B1%7D%7B2%7D%7D$ ，

令 $h'%5Cleft(%20x%20%5Cright)%20%3D0$ ，得 $x%3D%5Cfrac%7B1%7D%7Ba%7D-2$ ，

當(dāng) $x%5Cin%20%5Cleft(%200%2C%5Cfrac%7B1%7D%7Ba%7D-2%20%5Cright)%20$ ，

$h'%5Cleft(%20x%20%5Cright)%20%3C0$ ， $h%5Cleft(%20x%20%5Cright)%20%5Csearrow%20$ ；

當(dāng) $x%5Cin%20%5Cleft(%20%5Cfrac%7B1%7D%7Ba%7D-2%2C%2B%5Cinfty%20%5Cright)%20$ ，

$h'%5Cleft(%20x%20%5Cright)%20%3E0$ ， $h%5Cleft(%20x%20%5Cright)%20%5Cnearrow%20$ ，

故 $h%5Cleft(%20x%20%5Cright)%20_%7B%5Cmin%7D%3D%5Ccolor%7Bred%7D%7Bh%5Cleft(%20%5Cfrac%7B1%7D%7Ba%7D-2%20%5Cright)%20%3C%7Dh%5Cleft(%200%20%5Cright)%20%3D%5Ccolor%7Bred%7D%7B0%7D$

又因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=%5Ccolor%7Bred%7D%7Bh%5Cleft(%20%5Cfrac%7B4%7D%7Ba%5E2%7D%20%5Cright)%20%3E0%7D" alt="%5Ccolor%7Bred%7D%7Bh%5Cleft(%20%5Cfrac%7B4%7D%7Ba%5E2%7D%20%5Cright)%20%3E0%7D">，

故 $%5Cexists%20x_0%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%20%5Cfrac%7B1%7D%7Ba%7D-2%2C%5Cfrac%7B4%7D%7Ba%5E2%7D%20%5Cright)%20%7D$ ，使得 $h%5Cleft(%20x_0%20%5Cright)%20%3D0$