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莫被因式分解卡脖子(2021全国乙(文)导数)

2022-09-26 14:01 作者:数学老顽童  | 我要投稿

(2021全国乙文,21)已知函数f%5Cleft(%20x%20%5Cright)%20%3Dx%5E3-x%5E2%2Bax%2B1.

(1)讨论f%5Cleft(%20x%20%5Cright)%20的单调性;

(2)求曲线y%3Df%5Cleft(%20x%20%5Cright)%20过坐标原点的切线与曲线y%3Df%5Cleft(%20x%20%5Cright)%20的公共点的坐标.

解:(1)f'%5Cleft(%20x%20%5Cright)%20%3D3x%5E2-2x%2Ba

%5CvarDelta%20%3D%5Cleft(%20-2%20%5Cright)%20%5E2-4%5Ctimes%203%5Ccdot%20a%3D4%5Cleft(%201-3a%20%5Cright)%20

1.1-3a%5Cleqslant%200,即%5Ccolor%7Bred%7D%7Ba%5Cgeqslant%20%5Cfrac%7B1%7D%7B3%7D%7D

f'%5Cleft(%20x%20%5Cright)%20%5Cgeqslant%200f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D

2.1-3a%3E0,即%20%5Ccolor%7Bred%7D%7Ba%3C%5Cfrac%7B1%7D%7B3%7D%7D

f'%5Cleft(%20x%20%5Cright)%20%3D0,解得

x_1%3D%5Cfrac%7B1-%5Csqrt%7B1-3a%7D%7D%7B3%7D

x_2%3D%5Cfrac%7B1%2B%5Csqrt%7B1-3a%7D%7D%7B3%7D

x%5Cin%20%20%5Ccolor%7Bred%7D%7B%5Cleft(%20-%5Cinfty%20%2C%5Cfrac%7B1-%5Csqrt%7B1-3a%7D%7D%7B3%7D%20%5Cright)%20%7D

f'%5Cleft(%20x%20%5Cright)%20%3E0f%5Cleft(%20x%20%5Cright)%20%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D

x%5Cin%20%20%5Ccolor%7Bred%7D%7B%5Cleft(%20%5Cfrac%7B1-%5Csqrt%7B1-3a%7D%7D%7B3%7D%2C%5Cfrac%7B1%2B%5Csqrt%7B1-3a%7D%7D%7B3%7D%20%5Cright)%20%7D,

f'%5Cleft(%20x%20%5Cright)%20%3C0f%5Cleft(%20x%20%5Cright)%20%20%5Ccolor%7Bred%7D%7B%5Csearrow%20%7D

x%5Cin%20%20%5Ccolor%7Bred%7D%7B%5Cleft(%20%5Cfrac%7B1%2B%5Csqrt%7B1-3a%7D%7D%7B3%7D%2C%2B%5Cinfty%20%5Cright)%20%7D

f'%5Cleft(%20x%20%5Cright)%20%3E0f%5Cleft(%20x%20%5Cright)%20%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D.

(2)设切点坐标为

%5Cleft(%20x_0%2Cx_%7B0%7D%5E%7B3%7D-x_%7B0%7D%5E%7B2%7D%2Bax_0%2B1%20%5Cright)%20

则切线斜率为f'%5Cleft(%20x_0%20%5Cright)%20%3D3x_%7B0%7D%5E%7B2%7D-2x_0%2Ba

切线方程为

y-%5Cleft(%20x_%7B0%7D%5E%7B3%7D-x_%7B0%7D%5E%7B2%7D%2Bax_0%2B1%20%5Cright)%20%3D%5Cleft(%203x_%7B0%7D%5E%7B2%7D-2x_0%2Ba%20%5Cright)%20%5Cleft(%20x-x_0%20%5Cright)%20

……%5Coplus%20

因该切线过坐标原点,故

%20%5Ccolor%7Bred%7D%7B0%7D-%5Cleft(%20x_%7B0%7D%5E%7B3%7D-x_%7B0%7D%5E%7B2%7D%2Bax_0%2B1%20%5Cright)%20%3D%5Cleft(%203x_%7B0%7D%5E%7B2%7D-2x_0%2Ba%20%5Cright)%20%5Cleft(%20%20%5Ccolor%7Bred%7D%7B0%7D-x_0%20%5Cright)%20

整理得2x_%7B0%7D%5E%7B3%7D-x_%7B0%7D%5E%7B2%7D-1%3D0.

2x_%7B0%7D%5E%7B3%7D%20%5Ccolor%7Bred%7D%7B-2x_%7B0%7D%5E%7B2%7D%2Bx_%7B0%7D%5E%7B2%7D%7D-1%3D0,即

2x_%7B0%7D%5E%7B2%7D%5Cleft(%20x_0-1%20%5Cright)%20%2B%5Cleft(%20x_0%2B1%20%5Cright)%20%5Cleft(%20x_0-1%20%5Cright)%20%3D0

%5Cleft(%20x_0-1%20%5Cright)%20%5Cleft(%202x_%7B0%7D%5E%7B2%7D%2Bx_0%2B1%20%5Cright)%20%3D0

%20%5Ccolor%7Bred%7D%7B2x_%7B0%7D%5E%7B2%7D%2Bx_0%2B1%3E0%7D

x_0-1%3D0

%20%5Ccolor%7Bred%7D%7Bx_0%3D1%7D.

x_0%3D1代入%5Coplus%20

%20%5Ccolor%7Bred%7D%7By%3D%5Cleft(%20a%2B1%20%5Cright)%20x%7D……%5Cotimes%20(此即切线方程)

y%3Dx%5E3-x%5E2%2Bax%2B1联立,消去y,得

%20%5Ccolor%7Bred%7D%7Bx%5E3-x%5E2-x%2B1%3D0%7D

x%5E2%5Cleft(%20x-1%20%5Cright)%20-%5Cleft(%20x-1%20%5Cright)%20%3D0

%5Cleft(%20x-1%20%5Cright)%20%5Cleft(%20x%5E2-1%20%5Cright)%20%3D0

%5Cleft(%20x-1%20%5Cright)%20%5Cleft(%20x-1%20%5Cright)%20%5Cleft(%20x%2B1%20%5Cright)%20%3D0

%5Cleft(%20x-1%20%5Cright)%20%5E2%5Cleft(%20x%2B1%20%5Cright)%20%3D0

解得%20%5Ccolor%7Bred%7D%7Bx%3D-1%7D,或%20%5Ccolor%7Bred%7D%7Bx%3D1%7D

分别代入%5Cotimes%20,可得

y%3D-a-1,或y%3Da%2B1

故公共点坐标为

%20%5Ccolor%7Bred%7D%7B%5Cleft(%20-1%2C-a-1%20%5Cright)%20%7D%20%5Ccolor%7Bred%7D%7B%5Cleft(%201%2Ca%2B1%20%5Cright)%20%7D.


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