公切线问题的通解通法（2022全国甲（文）导数）

2022-09-22 21:10 作者:数学老顽童 0人读过 | 我要投稿

（2022全国甲文，20）已知函数 $f%5Cleft(%20x%20%5Cright)%20%3Dx%5E3-x$ ， $g%5Cleft(%20x%20%5Cright)%20%3Dx%5E2%2Ba$ ，曲线 $y%3Df%5Cleft(%20x%20%5Cright)%20$ 在点 $%5Cleft(%20x_1%2Cf%5Cleft(%20x_1%20%5Cright)%20%5Cright)%20$ 处的切线也是曲线 $y%3Dg%5Cleft(%20x%20%5Cright)%20$ 的切线.

（1）若 $x_1%3D-1$ ，求 $a$ ；

（2）求 $a$ 的取值范围.

解：（1） $f%5Cleft(%20-1%20%5Cright)%20%3D%5Cleft(%20-1%20%5Cright)%20%5E3-%5Cleft(%20-1%20%5Cright)%20%3D0$ ，

所以切点坐标为 $%5Ccolor%7Bred%7D%7B%5Cleft(%20-1%2C0%20%5Cright)%20%7D$ ，

$f'%5Cleft(%20x%20%5Cright)%20%3D3x%5E2-1$ ，

故该点处的切线斜率为

$f'%5Cleft(%20-1%20%5Cright)%20%3D3%5Ctimes%20%5Cleft(%20-1%20%5Cright)%20%5E2-1%3D%5Ccolor%7Bred%7D%7B2%7D$ ，

故切线方程为 $y-0%3D2%5Cleft(%20x%2B1%20%5Cright)%20$ ，

整理得 $%5Ccolor%7Bred%7D%7By%3D2x%2B2%7D$ .

联立 $y%3D2x%2B2$ 与 $y%3Dx%5E2%2Ba$ ，消去 $y$ ，得

$2x%2B2%3Dx%5E2%2Ba$ ,

整理得 $x%5E2-2x%2Ba-2%3D0$ ，

依题意可知 $%5CDelta%3D%5Cleft(-2%5Cright)%5E2-4%5Cleft(a-2%5Cright)%3D0$ ,

解得 $%5Ccolor%7Bred%7D%7Ba%3D3%7D$ .

（2）函数 $f%5Cleft(%20x%20%5Cright)$ 在 $x%3Dx_1$ 处的切线斜率为

$f'%5Cleft(%20x_1%20%5Cright)%20%3D3x_%7B1%7D%5E%7B2%7D-1$ ，

故该处的切线方程为

$y-%5Cleft(%20x_%7B1%7D%5E%7B3%7D-x_1%20%5Cright)%20%3D%5Cleft(%203x_%7B1%7D%5E%7B2%7D-1%20%5Cright)%20%5Cleft(%20x-x_1%20%5Cright)%20$ ，

整理得 $y%3D%5Ccolor%7Bred%7D%7B%5Cleft(%203x_%7B1%7D%5E%7B2%7D-1%20%5Cright)%7D%20x%5Ccolor%7Bred%7D%7B-2x_%7B1%7D%5E%7B3%7D%7D$ …… $%5Cotimes%20$

设曲线 $y%3Dg%5Cleft(%20x%20%5Cright)%20$ 上的切点为 $%5Cleft(%20x_2%2Cx_%7B2%7D%5E%7B2%7D%2Ba%20%5Cright)%20$ ，

因 $g'%5Cleft(%20x%20%5Cright)%20%3D2x$ ，

故 $x%3Dx_2$ 处的切线斜率为 $g'%5Cleft(%20x_2%20%5Cright)%20%3D2x_2$ ，

所以此处的切线方程为

$y-%5Cleft(%20x_%7B2%7D%5E%7B2%7D%2Ba%20%5Cright)%20%3D2x_2%5Cleft(%20x-x_2%20%5Cright)%20$ ，

整理得 $y%3D%5Ccolor%7Bred%7D%7B2x_2%7D%5Ccdot%20x%2B%5Ccolor%7Bred%7D%7Ba-x_%7B2%7D%5E%7B2%7D%7D$ …… $%5Coplus%20$

因为 $%5Cotimes%20$ 与 $%5Coplus%20$ 为同一条直线，所以

$%5Cbegin%7Bcases%7D%093x_%7B1%7D%5E%7B2%7D-1%3D2x_2%2C%5C%5C%09-2x_%7B1%7D%5E%7B3%7D%3Da-x_%7B2%7D%5E%7B2%7D%2C%5C%5C%5Cend%7Bcases%7D$ 所以

$%5Cbegin%7Baligned%7D%0A%09%5Ccolor%7Bred%7D%7Ba%7D%26%3D-2x_%7B1%7D%5E%7B3%7D%2Bx_%7B2%7D%5E%7B2%7D%5C%5C%0A%09%26%3D-2x_%7B1%7D%5E%7B3%7D%2B%5Cleft(%20%5Cfrac%7B3x_%7B1%7D%5E%7B2%7D-1%7D%7B2%7D%20%5Cright)%20%5E2%5C%5C%0A%09%26%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B9%7D%7B4%7Dx_%7B1%7D%5E%7B4%7D-2x_%7B1%7D%5E%7B3%7D-%5Cfrac%7B3%7D%7B2%7Dx_%7B1%7D%5E%7B2%7D%2B%5Cfrac%7B1%7D%7B4%7D%7D%5C%5C%0A%5Cend%7Baligned%7D$

令 $h%5Cleft(%20x%20%5Cright)%20%3D%5Cfrac%7B9%7D%7B4%7Dx%5E4-2x%5E3-%5Cfrac%7B3%7D%7B2%7Dx%5E2%2B%5Cfrac%7B1%7D%7B4%7D$ ，则

$%5Cbegin%7Baligned%7D%0A%09h'%5Cleft(%20x%20%5Cright)%20%26%3D9x%5E3-6x%5E2-3x%5C%5C%0A%09%26%3D3x%5Cleft(%203x%2B1%20%5Cright)%20%5Cleft(%20x-1%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$

令 $h'%5Cleft(%20x%20%5Cright)%20%3D0$ ，

解得 $%5Ccolor%7Bred%7D%7Bx%3D-%5Cfrac%7B1%7D%7B3%7D%7D$ 、 $%5Ccolor%7Bred%7D%7Bx%3D0%7D$ 、 $%5Ccolor%7Bred%7D%7Bx%3D1%7D$ ，

故 $h%5Cleft(%20x%20%5Cright)$ 的单调区间如下表所示：

$%5Cbegin%7Barray%7D%7Bc%7Cc%7Cc%7Cc%7Cc%7D%0A%09x%5Cin%26%09%09%5Cleft(%20-%5Cinfty%20%2C-%5Cdfrac%7B1%7D%7B3%7D%20%5Cright)%26%09%09%5Cleft(%20-%5Cdfrac%7B1%7D%7B3%7D%2C0%20%5Cright)%26%09%09%5Cleft(%200%2C1%20%5Cright)%26%09%09%5Cleft(%201%2C%2B%5Cinfty%20%5Cright)%5C%5C%0A%09%5Chline%0A%09h'%5Cleft(%20x%20%5Cright)%26%09%09-%26%09%09%2B%26%09%09-%26%09%09%2B%5C%5C%0A%09%5Chline%0A%09h%5Cleft(%20x%20%5Cright)%26%09%09%5Csearrow%26%09%09%5Cnearrow%26%09%09%5Csearrow%26%09%09%5Cnearrow%5C%5C%0A%5Cend%7Barray%7D$

故 $h%5Cleft(%20x%20%5Cright)%20$ 在 $%5Ccolor%7Bred%7D%7Bx%3D-%5Cfrac%7B1%7D%7B3%7D%7D$ 和 $%5Ccolor%7Bred%7D%7Bx%3D1%7D$ 处取得极小值，

$%5Ccolor%7Bred%7D%7Bh%5Cleft(%20-%5Cfrac%7B1%7D%7B3%7D%20%5Cright)%7D%20%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B5%7D%7B27%7D%7D$ ， $%5Ccolor%7Bred%7D%7Bh%5Cleft(%201%20%5Cright)%7D%20%3D%5Ccolor%7Bred%7D%7B-1%7D%3C%5Cfrac%7B5%7D%7B27%7D$ ,