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2023新高考Ⅰ卷數(shù)學逐題解析(5)

2023-06-15 19:36 作者:CHN_ZCY  | 我要投稿

封面:由崎司(《總之就是非??蓯邸罚?/p>


18. 如圖,在正四棱柱ABCD-A_1B_1C_1D_1中,AB%3D2,AA_1%3D4,點A_2B_2,C_2,D_2分別在棱AA_1,BB_1CC_1,DD_1上,AA_2%3D1,BB_2%3DDD_2%3D2,CC_2%3D3.

(1)證明:B_2C_2%2F%2FA_2D_2

(2)點PBB_1上,當二面角P-A_2C_2-D_2150%5E%7B%5Ccirc%7D時,求B_2P.

第18題圖

答案? (1)見解析;

(2)1

解析? 本題考察空間中線與線的位置關(guān)系和二面角,屬于中檔題.

(1)以C為原點,CDx軸正方向,CBy軸正方向,CC_1z軸正方向,建立空間直角坐標系.
B_2%20%5Cleft(%200%2C2%2C2%20%5Cright),C_2%20%5Cleft(%200%2C0%2C3%20%5Cright)A_2%20%5Cleft(%202%2C2%2C1%20%5Cright),D_2%20%5Cleft(%202%2C0%2C2%20%5Cright).

%5Coverrightarrow%7BB_2%20C_2%7D%3D%5Coverrightarrow%7BA_2%20D_2%7D%3D%5Cleft(%200%2C-2%2C1%20%5Cright).

所以%5Coverrightarrow%7BB_2%20C_2%7D%2F%2F%5Coverrightarrow%7BA_2%20D_2%7D.

由于B_2%20C_2A_2%20D_2不重合,所以B_2%20C_2%20%2F%2F%20A_2%20D_2.

(2)設(shè)P%20%5Cleft(%200%2C2%2Ct%20%5Cright)%5Cleft(%200%20%5Cleq%20t%20%5Cleq%204%20%5Cright).

%5Coverrightarrow%7BA_2%20C_2%7D%20%3D%20%5Cleft(-2%2C-2%2C2%20%5Cright),%5Coverrightarrow%7BA_2%20D_2%7D%20%3D%20%5Cleft(0%2C-2%2C1%20%5Cright)%5Coverrightarrow%7BA_2%20P%7D%20%3D%20%5Cleft(-2%2C0%2Ct-1%20%5Cright).

設(shè)平面A_2C_2D_2的法向量為%5Cboldsymbol%7Bn_1%7D%3D%5Cleft(x_1%2Cy_1%2Cz_1%5Cright).

%5Cleft%5C%7B%5Cbegin%7Baligned%7D%0A-2x_1-2y_1%2B2z_1%3D0%5C%5C%0A-2y_1%2Bz_1%3D0%0A%5Cend%7Baligned%7D%5Cright.,可取%5Cboldsymbol%7Bn_1%7D%3D%5Cleft(1%2C1%2C2%5Cright).

設(shè)平面A_2C_2P的法向量為%5Cboldsymbol%7Bn_2%7D%3D%5Cleft(x_2%2Cy_2%2Cz_2%5Cright).

%5Cleft%5C%7B%5Cbegin%7Baligned%7D%0A-2x_2-2y_2%2B2z_2%3D0%5C%5C%0A-2x_2%2B%5Cleft(t-1%5Cright)z_2%3D0%0A%5Cend%7Baligned%7D%5Cright.,可取%5Cboldsymbol%7Bn_2%7D%3D%5Cleft(-t%2B1%2Ct-3%2C-2%5Cright).

-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D%5Ccos150%5E%7B%5Ccirc%7D%3D%5Cfrac%7B%5Cboldsymbol%7Bn_1%7D%5Ccdot%5Cboldsymbol%7Bn_2%7D%7D%7B%5Cvert%5Cboldsymbol%7Bn_1%7D%5Cvert%5Ccdot%5Cvert%5Cboldsymbol%7Bn_2%7D%5Cvert%7D%3D-%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B%5Csqrt%7Bt%5E2-4t%2B7%7D%7D%5Cleft(%200%20%5Cleq%20t%20%5Cleq%204%20%5Cright)

即?t%5E2-4t%2B3%3D0%5Cleft(%200%20%5Cleq%20t%20%5Cleq%204%20%5Cright).

即?t%3D1%E6%88%963.

所以B_2P%3D%5Cvert%20t-2%20%5Cvert%20%3D%201.

19. 已知函數(shù)f%5Cleft(x%5Cright)%3Da%5Cleft(%7B%5Cmathrm%7Be%7D%7D%5Ex%2Ba%5Cright)-x.

(1)討論f%5Cleft(x%5Cright)的單調(diào)性;

(2)證明:當a%3E0時,f%5Cleft(x%5Cright)%3E2%5Cln%20a%20%2B%20%5Cfrac%7B3%7D%7B2%7D.

答案? (1)見解析;

(2)見解析.

解析? 本題考察函數(shù)的單調(diào)性、零點,導數(shù)的運算及其應用,屬于中檔題.

(1)f'%5Cleft(x%5Cright)%3Da%5Cmathrm%7Be%7D%5Ex-1.

a%5Cleq0,f%5Cleft(x%5Cright)%5Cboldsymbol%7B%5Cmathrm%7BR%7D%7D上單調(diào)遞減;

a%3E0,f%5Cleft(x%5Cright)%5Cleft(-%5Cinfty%2C-%5Cln%20a%20%5Cright%5D上單調(diào)遞減,在%5Cleft%5B-%5Cln%20a%2C%2B%5Cinfty%5Cright)上單調(diào)遞增.

(2)當a%3E0時,

f%5Cleft(x%5Cright)%5Cgeq%20f%5Cleft(-%5Cln%20a%5Cright)%3Da%5E2%2B1%2B%5Cln%20a

%5Cleft(a%5E2%2B1%2B%5Cln%20a%5Cright)-%5Cleft(2%5Cln%20a%2B%5Cfrac%7B3%7D%7B2%7D%5Cright)%3Da%5E2-%5Cfrac%7B1%7D%7B2%7D-%5Cln%20a.

設(shè)g%5Cleft(a%5Cright)%3Da%5E2-%5Cfrac%7B1%7D%7B2%7D-%5Cln%20a?%5Cleft(a%3E0%5Cright).

g'%5Cleft(a%5Cright)%3D2a-%5Cfrac%7B1%7D%7Ba%7D%3D%5Cfrac%7B2a%5E2-1%7D%7Ba%7D?%5Cleft(a%3E0%5Cright).

g%5Cleft(a%5Cright)%5Cgeq%20g%5Cleft(%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Cright)%3D%5Cln%5Csqrt%7B2%7D%3E0?%5Cleft(a%3E0%5Cright).

所以a%5E2%2B1%2B%5Cln%20a%3E2%5Cln%20a%2B%5Cfrac%7B3%7D%7B2%7D?%5Cleft(a%3E0%5Cright).

所以當a%3E0時,f%5Cleft(x%5Cright)%5Cgeq%20a%5E2%2B1%2B%5Cln%20a%3E2%5Cln%20a%2B%5Cfrac%7B3%7D%7B2%7D.







2023新高考Ⅰ卷數(shù)學逐題解析(5)的評論 (共 條)

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