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2023浙江大学强基数学逐题解析(4)

2023-06-20 21:44 作者:CHN_ZCY  | 我要投稿

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9. 已知椭圆%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1%5Cleft(a%3Eb%3E0%5Cright),过右焦点F作相互垂直的弦ACBD,已知四边形ABCD的面积的取值范围为%5Cleft%5B8%2C%5Cfrac%7B25%7D%7B2%7D%5Cright%5D,则%5Cfrac%7Ba%7D%7Bb%7D%3D___________.

答案  2

解析  设FAx轴正方向的夹角(逆时针为正)为%5Ctheta%20%5Cin%20%5Cleft%5B0%2C2%5Cpi%5Cright),记c%3D%5Csqrt%7Ba%5E2-b%5E2%7D

%5Cleft%7CFA%5Cright%7C%3D%5Cfrac%7Bb%5E2%7D%7Ba%2Bc%5Ccos%5Ctheta%7D

%5Cleft%7CFC%5Cright%7C%3D%5Cfrac%7Bb%5E2%7D%7Ba-c%5Ccos%5Ctheta%7D

所以

%5Cleft%7CAC%5Cright%7C%3D%5Cfrac%7B2ab%5E2%7D%7Ba%5E2-c%5E2%5Ccos%5E2%5Ctheta%7D

同理

%5Cleft%7CBD%5Cright%7C%3D%5Cfrac%7B2ab%5E2%7D%7Ba%5E2-c%5E2%5Csin%5E2%5Ctheta%7D

%5Cfrac%7B1%7D%7B%5Cleft%7CAC%5Cright%7C%7D%2B%5Cfrac%7B1%7D%7B%5Cleft%7CBD%5Cright%7C%7D%3D%5Cfrac%7B2a%5E2-c%5E2%7D%7B2ab%5E2%7D%3D%5Cfrac%7Ba%5E2%2Bb%5E2%7D%7B2ab%5E2%7D

于是

%5Cfrac%7B1%7D%7B%5Cleft%7CAC%5Cright%7C%5Ccdot%5Cleft%7CBD%5Cright%7C%7D%3D%5Cfrac%7B1%7D%7B%5Cleft%7CAC%5Cright%7C%7D%5Cleft(%5Cfrac%7Ba%5E2%2Bb%5E2%7D%7B2ab%5E2%7D-%5Cfrac%7B1%7D%7B%5Cleft%7CAC%5Cright%7C%7D%5Cright)

其中

%5Cfrac%7B1%7D%7B%5Cleft%7CAC%5Cright%7C%7D%5Cin%5Cleft%5B%5Cfrac%7B1%7D%7B2a%7D%2C%5Cfrac%7Ba%7D%7B2b%5E2%7D%5Cright%5D

所以

%5Cfrac%7B1%7D%7B%5Cleft%7CAC%5Cright%7C%5Ccdot%5Cleft%7CBD%5Cright%7C%7D%5Cin%5Cleft%5B%5Cfrac%7B1%7D%7B4b%5E2%7D%2C%5Cfrac%7B%5Cleft(a%5E2%2Bb%5E2%5Cright)%5E2%7D%7B16a%5E2b%5E4%7D%5Cright%5D

因此

S_%7B%E5%9B%9B%E8%BE%B9%E5%BD%A2ABCD%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft%7CAC%5Cright%7C%5Ccdot%5Cleft%7CBD%5Cright%7C%5Cin%5Cleft%5B%5Cfrac%7B8a%5E2b%5E4%7D%7B%5Cleft(a%5E2%2Bb%5E2%5Cright)%5E2%7D%2C2b%5E2%5Cright%5D

%5Cbegin%7Bcases%7D%0Aa%3Eb%3E0%5C%5C%0A%5Cfrac%7B8a%5E2b%5E4%7D%7B%5Cleft(a%5E2%2Bb%5E2%5Cright)%5E2%7D%3D8%5C%5C%0A2b%5E2%3D%5Cfrac%7B25%7D%7B2%7D%0A%5Cend%7Bcases%7D

%5Cbegin%7Bcases%7D%0Aa%3D5%5C%5C%0Ab%3D%5Cfrac%7B5%7D%7B2%7D%0A%5Cend%7Bcases%7D,所以%5Cfrac%7Ba%7D%7Bb%7D%3D2.

10.

 f%5Cleft(x%5Cright)%3Dx%5E2-2x-14%5Csqrt%7Bx-1%7D%2Bx%5Csqrt%7Bx%5E2-4x-28%5Csqrt%7Bx-1%7D%2B61%7D%5Cleft(x%5Cgeq1%5Cright)

的最小值为___________.

答案  -4

解析  

%5Cbegin%7Baligned%7D%0Af%5Cleft(x%5Cright)%26%3Dx%5E2-2x-14%5Csqrt%7Bx-1%7D%2Bx%5Csqrt%7Bx%5E2-4x-28%5Csqrt%7Bx-1%7D%2B61%7D%5C%5C%26%0A%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft(x%2B%5Csqrt%7Bx%5E2-4x-28%5Csqrt%7Bx-1%7D%2B61%7D%5Cright)%5E2-%5Cfrac%7B61%7D%7B2%7D%0A%5Cend%7Baligned%7D

根据几何意义

%5Cbegin%7Baligned%7D%0Ax%2B%5Csqrt%7Bx%5E2-4x-28%5Csqrt%7Bx-1%7D%2B61%7D%26%3D%5Csqrt%7B%5Cleft(x-2%5Cright)%5E2%2B%5Cleft(2%5Csqrt%7Bx-1%7D-0%5Cright)%5E2%7D%2B%5Csqrt%7B%5Cleft(x-4%5Cright)%5E2%2B%5Cleft(2%5Csqrt%7Bx-1%7D-7%5Cright)%5E2%7D%5C%5C%26%5Cgeq%5Csqrt%7B%5Cleft(4-2%5Cright)%5E2%2B%5Cleft(7-0%5Cright)%5E2%7D%5C%5C%26%3D%5Csqrt%7B53%7D%0A%5Cend%7Baligned%7D

当且仅当x%3D%5Cfrac%7B4%5Csqrt%7B53%7D%2B106%7D%7B49%7D时取等.

所以

f%5Cleft(x%5Cright)%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft(x%2B%5Csqrt%7Bx%5E2-4x-28%5Csqrt%7Bx-1%7D%2B61%7D%5Cright)%5E2-%5Cfrac%7B61%7D%7B2%7D%5Cgeq-4

当且仅当x%3D%5Cfrac%7B4%5Csqrt%7B53%7D%2B106%7D%7B49%7D时取等.

因此f%5Cleft(x%5Cright)的最小值为-4.

11. 设复数abc%5Cleft%7Ca%5Cright%7C%5E2%2B%5Cleft%7Cb%5Cright%7C%5E2%2B%5Cleft%7Cc%5Cright%7C%5E2%3D1,则

%5Cleft%7Cab%5Cleft(a%5E2-b%5E2%5Cright)%2Bbc%5Cleft(b%5E2-c%5E2%5Cright)%2Bca%5Cleft(c%5E2-a%5E2%5Cright)%5Cright%7C

的最大值为___________.

答案  %5Cfrac%7B9%7D%7B16%7D

解析  

a%3Dx_1%2By_1%5Cmathrm%7Bi%7D

b%3Dx_2%2By_2%5Cmathrm%7Bi%7D

c%3Dx_3%2By_3%5Cmathrm%7Bi%7D

%5Cbegin%7Baligned%7D%0A%26%5Cleft%7Cab%5Cleft(a%5E2-b%5E2%5Cright)%2Bbc%5Cleft(b%5E2-c%5E2%5Cright)%2Bca%5Cleft(c%5E2-a%5E2%5Cright)%5Cright%7C%5C%5C%26%3D%5Cleft%7C%5Cleft(a-b%5Cright)%5Cleft(b-c%5Cright)%5Cleft(c-a%5Cright)%5Cleft(a%2Bb%2Bc%5Cright)%5Cright%7C%5C%5C%26%0A%3D%5Cleft%7Ca-b%5Cright%7C%5Cleft%7Cb-c%5Cright%7C%5Cleft%7Cc-a%5Cright%7C%5Cleft%7Ca%2Bb%2Bc%5Cright%7C%5C%5C%26%0A%3D%5Csqrt%7B%5Cleft(x_1-x_2%5Cright)%5E2%2B%5Cleft(y_1-y_2%5Cright)%5E2%7D%5Ccdot%5Csqrt%7B%5Cleft(x_2-x_3%5Cright)%5E2%2B%5Cleft(y_2-y_3%5Cright)%5E2%7D%5Ccdot%5Csqrt%7B%5Cleft(x_3-x_1%5Cright)%5E2%2B%5Cleft(y_3-y_1%5Cright)%5E2%7D%20%5Ccdot%20%5Csqrt%7B%5Cleft(x_1%2Bx_2%2Bx_3%5Cright)%5E2%2B%5Cleft(y_1%2By_2%2By_3%5Cright)%5E2%7D%5C%5C%26%0A%5Cleq%5Cleft%5B%5Cfrac%7B%5Cleft(x_1-x_2%5Cright)%5E2%2B%5Cleft(y_1-y_2%5Cright)%5E2%2B%5Cleft(x_2-x_3%5Cright)%5E2%2B%5Cleft(y_2-y_3%5Cright)%5E2%2B%5Cleft(x_3-x_1%5Cright)%5E2%2B%5Cleft(y_3-y_1%5Cright)%5E2%2B%5Cleft(x_1%2Bx_2%2Bx_3%5Cright)%5E2%2B%5Cleft(y_1%2By_2%2By_3%5Cright)%5E2%7D%7B4%7D%5Cright%5D%5E2%5C%5C%26%0A%3D%5Cleft%5B%5Cfrac%7B3%5Cleft(x_1%5E2%2By_1%5E2%2Bx_2%5E2%2By_2%5E2%2Bx_3%5E2%2By_3%5E2%5Cright)%7D%7B4%7D%5Cright%5D%5E2%5C%5C%26%0A%3D%5Cleft%5B%5Cfrac%7B3%5Cleft(%5Cleft%7Ca%5Cright%7C%5E2%2B%5Cleft%7Cb%5Cright%7C%5E2%2B%5Cleft%7Cc%5Cright%7C%5E2%5Cright)%7D%7B4%7D%5Cright%5D%5E2%5C%5C%26%0A%3D%5Cfrac%7B9%7D%7B16%7D%0A%5Cend%7Baligned%7D

当且仅当abc对应的以原点为起点的向量的终点构成边长为%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D,重心到原点距离为%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B6%7D的等边三角形时取等.

所以

%5Cleft%7Cab%5Cleft(a%5E2-b%5E2%5Cright)%2Bbc%5Cleft(b%5E2-c%5E2%5Cright)%2Bca%5Cleft(c%5E2-a%5E2%5Cright)%5Cright%7C

的最大值为%5Cfrac%7B9%7D%7B16%7D.

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