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必要條件探路(2020新高考Ⅰ&Ⅱ?qū)?shù))

2022-10-17 10:56 作者:數(shù)學老頑童  | 我要投稿

(2020新高考Ⅰ,21;新高考Ⅱ,22)已知函數(shù)f%5Cleft(%20x%20%5Cright)%20%3Da%5Cmathrm%7Be%7D%5E%7Bx-1%7D-%5Cln%20%20x%2B%5Cln%20%20a.

(1)當a%3D%5Cmathrm%7Be%7D時,求曲線y%3Df%5Cleft(%20x%20%5Cright)%20在點%5Cleft(%201%2Cf%5Cleft(%201%20%5Cright)%20%5Cright)%20處的切線與坐標軸圍成的三角形的面積;

(2)若f%5Cleft(%20x%20%5Cright)%20%5Cgeqslant%201,求a的取值范圍.

解:(1)當a%3D%5Cmathrm%7Be%7D時,

f%5Cleft(%20x%20%5Cright)%20%3D%5Cmathrm%7Be%7D%5Ex-%5Cln%20%20x%2B1,

f%5Cleft(%201%20%5Cright)%20%3D%5Cmathrm%7Be%7D%2B1,故切點坐標為%5Ccolor%7Bred%7D%7B%5Cleft(%201%2C%5Cmathrm%7Be%7D%2B1%20%5Cright)%20%7D,

f'%5Cleft(%20x%20%5Cright)%20%3D%5Cmathrm%7Be%7D%5Ex-%5Cfrac%7B1%7D%7Bx%7D,

故切線斜率為f'%5Cleft(%201%20%5Cright)%20%3D%5Ccolor%7Bred%7D%7B%5Cmathrm%7Be%7D-1%7D,

故切線方程為

y-%5Cleft(%20%5Cmathrm%7Be%7D%2B1%20%5Cright)%20%3D%5Cleft(%20%5Cmathrm%7Be%7D-1%20%5Cright)%20%5Cleft(%20x-1%20%5Cright)%20

整理得%5Ccolor%7Bred%7D%7By%3D%5Cleft(%20%5Cmathrm%7Be%7D-1%20%5Cright)%20x%2B2%7D.

x%3D0,得y%3D2

y%3D0,得x%3D%5Cfrac%7B2%7D%7B1-%5Cmathrm%7Be%7D%7D

故所求三角形得面積

%5Ccolor%7Bred%7D%7BS_%7B%5Cbigtriangleup%7D%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cleft%7C%202%20%5Cright%7C%5Ctimes%20%5Cleft%7C%20%5Cfrac%7B2%7D%7B1-%5Cmathrm%7Be%7D%7D%20%5Cright%7C%3D%5Ccolor%7Bred%7D%7B%5Cfrac%7B2%7D%7B%5Cmathrm%7Be%7D-1%7D%7D.

(2)%5Ccolor%7Bred%7D%7Bf%5Cleft(%201%20%5Cright)%7D%20%3Da%2B%5Cln%20%20a%5Cgeqslant1,

g%5Cleft(%20a%20%5Cright)%20%3Da%2B%5Cln%20%20a,

易知g%5Cleft(%20a%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D,且注意到%5Ccolor%7Bred%7D%7Bg%5Cleft(%201%20%5Cright)%20%3D1%7D,所以

a%5Cgeqslant1,g%5Cleft(%20a%20%5Cright)%20%5Cgeqslant1,

所以a%5Cgeqslant%201f%5Cleft(%20x%20%5Cright)%20%5Cgeqslant%201必要條件.

下面證明:a%5Cgeqslant%201f%5Cleft(%20x%20%5Cright)%20%5Cgeqslant%201充分條件.

a%5Cgeqslant%201時,f%5Cleft(%20x%20%5Cright)%20%5Cgeqslant%20%5Cmathrm%7Be%7D%5E%7Bx-1%7D-%5Cln%20%20x

故只需證%5Ccolor%7Bred%7D%7B%5Cmathrm%7Be%7D%5E%7Bx-1%7D-%5Cln%20%20x%5Cgeqslant%201%7D.

h%5Cleft(%20x%20%5Cright)%20%3D%5Cmathrm%7Be%7D%5E%7Bx-1%7D-%5Cln%20%20x,

h'%5Cleft(%20x%20%5Cright)%20%3D%5Cmathrm%7Be%7D%5E%7Bx-1%7D-%5Cfrac%7B1%7D%7Bx%7D,

易知h'%5Cleft(%20x%20%5Cright)%20%5Cnearrow%20,且注意到h'%5Cleft(%201%20%5Cright)%20%3D0,所以

x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%200%2C1%20%5Cright)%20%7D,h'%5Cleft(%20x%20%5Cright)%20%3C0,h%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Csearrow%20%7D;

x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%201%2C%2B%5Cinfty%20%5Cright)%7D%20,h'%5Cleft(%20x%20%5Cright)%20%3E0,h%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D

h%5Cleft(%20x%20%5Cright)%20%5Cgeqslant%20h%5Cleft(%20x%20%5Cright)%20_%7B%5Cmin%7D%3Dh%5Cleft(%201%20%5Cright)%20%3D1,證畢.

綜上所述:%5Ccolor%7Bred%7D%7Ba%5Cin%20%5Cleft%5B%201%2C%2B%5Cinfty%20%5Cright)%20%7D.

命題背景:%5Ccolor%7Bred%7D%7B%5Cmathrm%7Be%7D%5E%7Bx-1%7D%5Cgeqslant%20%5Cln%20%20x%2B1%7D.


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