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2023新高考Ⅰ卷數(shù)學(xué)逐題解析(4)

2023-06-15 14:55 作者:CHN_ZCY  | 我要投稿

封面:秋晴れ

作畫(huà):jimmy

https://www.pixiv.net/artworks/93503344


14. 在正四棱臺(tái)ABCD-A_1B_1C_1D_1中,AB%3D2,A_1B_1%3D1AA_1%3D%5Csqrt%7B2%7D,則該棱臺(tái)的體積為_(kāi)__________.

答案??%5Cfrac%20%7B7%5Csqrt%7B6%7D%7D%20%7B6%7D

解析??本題考察棱臺(tái)體積的計(jì)算,屬于簡(jiǎn)單題.

延長(zhǎng)AA_1BB_1,CC_1,DD_1交于點(diǎn)P.

過(guò)點(diǎn)PPH%20%5Cbot%20%E5%B9%B3%E9%9D%A2ABCD于點(diǎn)H.

PA%3D%5Cfrac%20%7BAB%7D%20%7BAB-A_1B_1%7D%20AA_1%3D2%5Csqrt%7B2%7D,AH%3D%5Cfrac%20%7B%5Csqrt%7B2%7D%7D%20%7B2%7DAB%3D%5Csqrt%7B2%7D.

PH%3D%5Csqrt%7BPA%5E2-PH%5E2%7D%3D%5Csqrt%7B6%7D.

V_%7B%E6%AD%A3%E5%9B%9B%E6%A3%B1%E5%8F%B0ABCD-A_1B_1C_1D_1%7D%3D%5Cleft%5B1-%5Cleft(%5Cfrac%20%7BA_1B_1%7D%20%7BAB%7D%20%5Cright)%5E3%20%5Cright%5DV_%7B%E5%9B%9B%E6%A3%B1%E9%94%A5P-ABCD%7D%3D%5Cfrac%20%7B7%7D%20%7B8%7D%20V_%7B%E5%9B%9B%E6%A3%B1%E9%94%A5P-ABCD%7D%3D%5Cfrac%20%7B7%7D%20%7B8%7D%20%5Ccdot%20%5Cfrac%20%7B1%7D%20%7B3%7D%20%5Ccdot%202%5E2%20%5Ccdot%20%5Csqrt%7B6%7D%20%3D%20%5Cfrac%20%7B7%5Csqrt%7B6%7D%7D%20%7B6%7D

Remark.?本題也可利用棱臺(tái)體積公式V%3D%5Cfrac%7B1%7D%7B3%7D%5Cleft(S_1%2B%5Csqrt%7BS_1S_2%7D%2BS_2%5Cright)h求解.

15. 已知函數(shù)f%5Cleft(x%5Cright)%3D%5Ccos%7B%5Comega%20x%7D-1%5Cleft(%5Comega%20%3E0%20%5Cright)在區(qū)間%5Cleft%5B0%2C2%5Cpi%5Cright%5D有且僅有3個(gè)零點(diǎn),則%5Comega的取值范圍是___________.

答案??%5Cleft%5B2%2C3%5Cright)

解析??本題考察三角函數(shù)的零點(diǎn)與周期問(wèn)題,屬于簡(jiǎn)單題.

f%5Cleft(x%5Cright)在區(qū)間%5Cleft%5B0%2C2%5Cpi%5Cright%5D有且僅有3個(gè)零點(diǎn),得%5Cfrac%7B4%5Cpi%7D%7B%5Comega%7D%5Cleq2%5Cpi%3C%5Cfrac%7B6%5Cpi%7D%7B%5Comega%7D,即2%5Cleq%5Comega%3C3.

所以%5Comega的取值范圍是%5Cleft%5B2%2C3%5Cright).

16. 已知雙曲線(xiàn)C%3A%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2B%5Cfrac%7By%5E2%7D%7Bb%5E2%7D%3D1%5Cleft(a%3E0%2Cb%3E0%5Cright)的左、右焦點(diǎn)分別為F_1,F_2. 點(diǎn)AC上,點(diǎn)By軸上,%5Coverrightarrow%7BF_1A%7D%5Cbot%5Coverrightarrow%7BF_1B%7D,%5Coverrightarrow%7BF_2A%7D%3D-%5Cfrac%7B2%7D%7B3%7D%5Coverrightarrow%7BF_2B%7D,則C的離心率為?___________.

答案??%5Cfrac%7B3%5Csqrt%7B5%7D%7D%7B5%7D

解析??本題考察雙曲線(xiàn)的定義和性質(zhì),屬于中檔題.

設(shè)A%5Cleft(x_0%2Cy_0%5Cright)B%5Cleft(0%2Cy_1%5Cright).

%5Cleft%5C%7B%5Cbegin%7Baligned%7D%0Ac%5Cleft(x_0%2Bc%5Cright)%2By_0y_1%3D0%5C%5C%0Ax_0-c%3D-%5Cfrac%7B2%7D%7B3%7D%5Cleft(-c%5Cright)%5C%5C%0Ay_0%3D-%5Cfrac%7B2%7D%7B3%7Dy_1%0A%5Cend%7Baligned%7D%0A%5Cright.,解得%5Cleft%5C%7B%5Cbegin%7Baligned%7D%0Ax_0%3D%5Cfrac%7B5%7D%7B3%7Dc%5C%5C%0A%5Cvert%20y_0%20%5Cvert%3D%5Cfrac%7B4%7D%7B3%7Dc%0A%5Cend%7Baligned%7D%0A%5Cright..

所以%5Cvert%20F_1A%20%5Cvert%3D%5Cfrac%7B4%5Csqrt%7B5%7D%7D%7B3%7Dc,%5Cvert%20F_2A%20%5Cvert%20%3D%20%5Cfrac%7B2%5Csqrt%7B5%7D%7D%7B3%7Dc

2a%3D%5Cvert%20%5Cvert%20F_1A%20%5Cvert%20-%20%5Cvert%20F_2B%20%5Cvert%20%5Cvert%20%3D%20%5Cfrac%7B2%5Csqrt%7B5%7D%7D%7B3%7Dc,得e%3D%5Cfrac%7Bc%7D%7Ba%7D%3D%5Cfrac%7B3%5Csqrt%7B5%7D%7D%7B5%7D.

四、解答題:本題共 6 小題,共 70 分。解答應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟。

17. 已知在%5Ctriangle%20ABC中,A%2BB%3D3C,2%5Csin%20%5Cleft(%20A-C%20%5Cright)%20%3D%20%5Csin%20B.

(1)求%5Csin%20A;

(2)設(shè)AB%3D5,求AB邊上的高.

答案? (1)%5Cfrac%7B3%5Csqrt%7B10%7D%7D%7B10%7D;

(2)6.

解析??考察三角函數(shù)與解三角形,屬于中檔題.

(1)由3C%3DA%2BB%3D%5Cpi-CC%3D%5Cfrac%7B%5Cpi%7D%7B4%7D,于是

2%5Csin%20%5Cleft(%20A-C%20%5Cright)%3D2%5Csin%20%5Cleft(%20A-%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Cright)%3D%5Csin%20B%3D%5Csin%20%5Cleft(%20A%2B%5Cfrac%7B%5Cpi%7D%7B4%7D%20%5Cright)

%5Csqrt%7B2%7D%20%5Csin%20A%20-%20%5Csqrt%7B2%7D%20%5Ccos%20A%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%5Csin%20A%20%2B%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%5Ccos%20A

所以%5Ccos%20A%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Csin%20A.

于是

1%3D%5Csin%5E2A%2B%5Ccos%5E2A%3D%5Cfrac%7B10%7D%7B9%7D%5Csin%5E2A

解得%5Csin%20A%3D%5Cfrac%7B3%5Csqrt%7B10%7D%7D%7B10%7D%20%E6%88%96%20-%5Cfrac%7B3%5Csqrt%7B10%7D%7D%7B10%7D.

由于0%3CA%3C%5Cpi,所以%5Csin%20A%3E0,所以%5Csin%20A%3D%5Cfrac%7B3%5Csqrt%7B10%7D%7D%7B10%7D.

(2)%5Ctan%20A%3D3,所以%5Ctan%20B%3D%20-%20%5Ctan%20%5Cleft(%20A%20%2B%20C%20%5Cright)%20%3D%20-%20%5Cfrac%20%7B%5Ctan%20A%20%2B%20%5Ctan%20C%7D%20%7B1-%5Ctan%20A%20%5Ctan%20C%7D%3D2.

所以%5Csin%20B%20%3D%20%5Cfrac%7B2%5Csqrt%7B5%7D%7D%7B5%7D.

由正弦定理,%5Cfrac%7Bc%7D%7B%5Csin%20C%7D%3D%5Cfrac%7Bb%7D%7B%5Csin%20B%7D,得b%3D2%5Csqrt%7B10%7D.

所以AB邊上的高為b%5Csin%20A%3D6.


2023新高考Ⅰ卷數(shù)學(xué)逐題解析(4)的評(píng)論 (共 條)

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