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摘除“定根”,及時降冪(2020天津,9)

2022-09-10 10:50 作者:數(shù)學(xué)老頑童  | 我要投稿

(2020天津,9)已知函數(shù)f%5Cleft(%20x%20%5Cright)%20%3D%5Cbegin%7Bcases%7D%09x%5E3%2Cx%5Cgeq%20%200%2C%5C%5C%09-x%2Cx%3C0%2C%5C%5C%5Cend%7Bcases%7D若函數(shù)g%5Cleft(%20x%20%5Cright)%20%3Df%5Cleft(%20x%20%5Cright)%20-%5Cleft%7C%20kx%5E2-2x%20%5Cright%7Ck%5Cin%20%5Cmathbf%7BR%7D)恰有4個零點,則k的取值范圍是(? ? )

A.%5Cleft(%20-%5Cinfty%20%2C-%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5Ccup%20%5Cleft(%202%5Csqrt%7B2%7D%2C%2B%5Cinfty%20%5Cright)%20

B.%5Cleft(%20-%5Cinfty%20%2C-%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5Ccup%20%5Cleft(%200%2C2%5Csqrt%7B2%7D%20%5Cright)%20

C.%5Cleft(%20-%5Cinfty%20%2C0%20%5Cright)%20%5Ccup%20%5Cleft(%200%2C2%5Csqrt%7B2%7D%20%5Cright)%20

D.%5Cleft(%20-%5Cinfty%20%2C0%20%5Cright)%20%5Ccup%20%5Cleft(%202%5Csqrt%7B2%7D%2C%2B%5Cinfty%20%5Cright)%20


解:注意到無論%5Ccolor%7Bred%7D%7Bk%7D取何值,%5Ccolor%7Bred%7D%7Bx%3D0%7D恒為%5Ccolor%7Bred%7D%7Bg%5Cleft(%20x%20%5Cright)%7D的一個零點.故問題本質(zhì)為

若方程%5Ccolor%7Bred%7D%7B%5Cfrac%7B%20f%5Cleft(%20x%20%5Cright)%20%7D%7B%5Cleft%7Cx%20%5Cright%7C%7D%3D%5Cleft%7C%20kx-2%20%5Cright%7C%7D恰有%5Ccolor%7Bred%7D%7B3%7D個實根,求%5Ccolor%7Bred%7D%7Bk%7D的取值范圍.

m%5Cleft(%20x%20%5Cright)%20%3D%5Cfrac%7Bf%5Cleft(%20x%20%5Cright)%7D%7B%5Cleft%7C%20x%20%5Cright%7C%7D%3D%5Cbegin%7Bcases%7D%09x%5E2%2Cx%3E0%2C%5C%5C%091%2Cx%3C0%2C%5C%5C%5Cend%7Bcases%7D

n%5Cleft(%20x%20%5Cright)%3D%5Cleft%7C%20kx-2%20%5Cright%7C,

數(shù)形結(jié)合:

1.當(dāng)k%3C0時,

顯然符合題意.

2.當(dāng)k%3E0時,

y%3D-%5Cleft(%20kx-2%20%5Cright)%20x%5E2恒有%5Ccolor%7Bred%7D%7B1%7D交點,

故只需使y%3Dkx-2x%5E2%5Ccolor%7Bred%7D%7B2%7D個交點即可,

x%5E2%3Dkx-2,

x%5E2-kx%2B2%3D0

%5CvarDelta%20%3D%5Cleft(%20-k%20%5Cright)%20%5E2-4%5Ctimes%202%3Dk%5E2-8%3E0,

解得k%3C-2%5Csqrt%7B2%7D(舍)或k%3E2%5Csqrt%7B2%7D.

綜上所述:k%5Cin%20%5Cleft(%20-%5Cinfty%20%2C0%20%5Cright)%20%5Ccup%20%5Cleft(%202%5Csqrt%7B2%7D%2C%2B%5Cinfty%20%5Cright)%20,

D.

摘除“定根”,及時降冪(2020天津,9)的評論 (共 條)

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