2023新高考Ⅰ卷數(shù)學(xué)逐題解析（1）

2023-06-15 12:52 作者:CHN_ZCY 0人讀過(guò) | 我要投稿

封面：幸せ空間

作畫(huà)：カフェ

https://www.pixiv.net/artworks/105861620

一、選擇題：本大題共 8 小題，每小題 5 分，共計(jì) 40 分．每小題給出的四個(gè)選項(xiàng)中，只有一

個(gè)選項(xiàng)是正確的．請(qǐng)把正確的選項(xiàng)填涂在答題卡相應(yīng)的位置上．

已知集合 $M%20%3D%20%5Cleft%5C%7B%20-2%2C-1%2C0%2C1%2C2%20%5Cright%5C%7D$ ， $N%20%3D%20%5Cleft%5C%7B%20x%20%5Cvert%20x%5E2-x-6%20%5Cgeq%200%20%5Cright%5C%7D$ ，則 $M%20%5Ccap%20N%20%3D$

A.? $%5Cleft%5C%7B%20-2%2C-1%2C0%2C1%20%5Cright%5C%7D$

B.? $%5Cleft%5C%7B%200%2C1%2C2%20%5Cright%5C%7D$

C.? $%5Cleft%5C%7B%20-2%20%5Cright%5C%7D$

D.? $%5Cleft%5C%7B%202%20%5Cright%5C%7D$

答案? C

解析? 本題考察基本的集合交集概念，屬于簡(jiǎn)單題.

$x%5E2-x-6%20%5Cgeq%200$ ? $%5CLeftrightarrow%20$ ? $x%20%5Cleq%20-2$ ?或? $x%20%5Cgeq%203$ .

所以 $N%20%3D%20%5Cleft(%20-%5Cinfty%2C-2%20%5Cright%5D%20%5Ccup%20%5Cleft%5B%203%2C%2B%5Cinfty%20%5Cright)$ .

所以 $M%20%5Ccap%20N%20%3D%20%5Cleft%5C%7B%20-2%20%5Cright%5C%7D$ .

故選：C.
已知 $z%20%3D%20%5Cfrac%20%7B1-%5Cmathrm%7Bi%7D%7D%20%7B2%2B2%5Cmathrm%7Bi%7D%7D$ ，則 $z%20-%20%5Cbar%7Bz%7D%20%3D$

A.? $-%5Cmathrm%7Bi%7D$

B.? $%5Cmathrm%7Bi%7D$

C.? $0$

D.? $1$

答案? A

解析? 本題考察復(fù)數(shù)及共軛復(fù)數(shù)的概念，屬于簡(jiǎn)單題.

$z%20%3D%20%5Cfrac%20%7B1-%5Cmathrm%7Bi%7D%7D%20%7B2%2B2%5Cmathrm%7Bi%7D%7D%20%3D%20%5Cfrac%20%7B%5Cleft(%201-%5Cmathrm%7Bi%7D%20%5Cright)%5E2%7D%20%7B2%5Cleft(%201%2B%5Cmathrm%7Bi%7D%20%5Cright)%20%5Cleft(%201-%5Cmathrm%7Bi%7D%20%5Cright)%7D%20%3D%20%5Cfrac%20%7B-2%5Cmathrm%7Bi%7D%7D%20%7B4%7D%20%3D-%5Cfrac%20%7B%5Cmathrm%7Bi%7D%7D%20%7B2%7D$ ，

$z%20-%20%5Cbar%7Bz%7D%20%3D%20-%5Cfrac%20%7B%5Cmathrm%7Bi%7D%7D%20%7B2%7D%20%20-%5Cfrac%20%7B%5Cmathrm%7Bi%7D%7D%20%7B2%7D%20%3D%20-%5Cmathrm%7Bi%7D$ .

故選：A.
已知向量 $%5Cboldsymbol%7Ba%7D%20%3D%20%5Cleft(%201%2C1%20%5Cright)$ ， $%5Cboldsymbol%7Bb%7D%20%3D%20%5Cleft(%201%2C-1%20%5Cright)$ . 若 $%5Cleft(%20%5Cboldsymbol%7Ba%7D%20%2B%20%5Clambda%20%5Cboldsymbol%7Bb%7D%20%5Cright)%20%5Cbot%20%5Cleft(%20%5Cboldsymbol%7Ba%7D%20%2B%20%5Cmu%20%5Cboldsymbol%7Bb%7D%20%5Cright)$ ，則

A.? $%5Clambda%20%2B%20%5Cmu%20%3D%201$

B.? $%5Clambda%20%2B%20%5Cmu%20%3D%20-1$

C.? $%5Clambda%20%5Cmu%20%3D%201$

D.? $%5Clambda%20%5Cmu%20%3D%20-1$

答案? D

解析? 本題考察坐標(biāo)形式下向量的加法以及向量垂直的條件，屬于簡(jiǎn)單題.

$%5Cboldsymbol%7Ba%7D%20%2B%20%5Clambda%20%5Cboldsymbol%7Bb%7D%20%3D%20%5Cleft(%201%2B%20%5Clambda%2C%201-%20%5Clambda%20%5Cright)$ ， $%5Cboldsymbol%7Ba%7D%20%2B%20%5Cmu%20%5Cboldsymbol%7Bb%7D%20%3D%20%5Cleft(%201%2B%20%5Cmu%2C%201-%20%5Cmu%20%5Cright)$ .

由于 $%5Cleft(%20%5Cboldsymbol%7Ba%7D%20%2B%20%5Clambda%20%5Cboldsymbol%7Bb%7D%20%5Cright)%20%5Cbot%20%5Cleft(%20%5Cboldsymbol%7Ba%7D%20%2B%20%5Cmu%20%5Cboldsymbol%7Bb%7D%20%5Cright)$ ，所以

$%5Cleft(%20%5Cboldsymbol%7Ba%7D%20%2B%20%5Clambda%20%5Cboldsymbol%7Bb%7D%20%5Cright)%5Ccdot%20%5Cleft(%20%5Cboldsymbol%7Ba%7D%20%2B%20%5Cmu%20%5Cboldsymbol%7Bb%7D%20%5Cright)%3D%5Cleft(1%2B%5Clambda%5Cright)%5Cleft(1%2B%5Cmu%5Cright)%2B%5Cleft(1-%5Clambda%5Cright)%5Cleft(1-%5Cmu%5Cright)%3D0$
得 $%5Clambda%5Cmu%3D-1$ .

故選：D.
設(shè)函數(shù) $f%5Cleft(x%5Cright)%3D2%5E%7Bx%5Cleft(x-a%5Cright)%7D$ 在區(qū)間 $%5Cleft(0%2C1%5Cright)$ 單調(diào)遞減，則 $a$ 的取值范圍是

A.? $%5Cleft(-%5Cinfty%2C-2%5Cright%5D$

B.? $%5Cleft%5B-2%2C0%5Cright)$

C.? $%5Cleft(0%2C2%5Cright%5D$

D.? $%5Cleft%5B2%2C%2B%5Cinfty%5Cright)$

答案? D

解析? 本題考察基本初等函數(shù)之一的指數(shù)函數(shù)的單調(diào)性問(wèn)題，屬于簡(jiǎn)單題.

由于 $y%3D2%5Ex$ 在 $%5Cboldsymbol%7B%5Cmathrm%7BR%7D%7D$ 上單調(diào)遞增，所以 $f%5Cleft(x%5Cright)%3D2%5E%7Bx%5Cleft(x-a%5Cright)%7D$ 在區(qū)間 $%5Cleft(0%2C1%5Cright)$ 單調(diào)遞減，等價(jià)于 $y%3Dx%5Cleft(x-a%5Cright)$ 在區(qū)間 $%5Cleft(0%2C1%5Cright)$ 單調(diào)遞減.

得 $%5Cfrac%20%7Ba%7D%20%7B2%7D%20%5Cgeq%201$ ，即 $a%20%5Cgeq%202$ .

所以 $a$ 的取值范圍是 $%5Cleft%5B2%2C%2B%5Cinfty%5Cright)$ .

故選：D.
設(shè)橢圓 $C_1%3A%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%2By%5E2%3D1$ $%5Cleft(a%3E1%5Cright)$ ， $C_2%3A%5Cfrac%7Bx%5E2%7D%7B4%7D%2By%5E2%3D1$ 的離心率分別為 $e_1$ ， $e_2$ ，若 $e_2%3D%5Csqrt%7B3%7De_1$ ，則 $a%3D$

A.? $%5Cfrac%20%7B2%5Csqrt%7B3%7D%7D%20%7B3%7D$

B.? $%5Csqrt%7B2%7D$

C.? $%5Csqrt%7B3%7D$

D.? $%5Csqrt%7B6%7D$

答案? A

解析? 本題考察橢圓離心率的表達(dá)式，屬于簡(jiǎn)單題.

$e_1%3D%5Cfrac%20%7Be_2%7D%20%7B%5Csqrt%7B3%7D%7D%20%3D%5Cfrac%20%7B1%7D%20%7B2%7D$ .

由于 $a%3E1$ ，所以 $%5Cfrac%20%7B%5Csqrt%7Ba%5E2-1%7D%7D%20%7Ba%7D%20%3D%20%5Cfrac%20%7B1%7D%20%7B2%7D$ $%5Cleft(a%3E1%5Cright)$ ，得 $a%3D%5Cfrac%20%7B2%5Csqrt%7B3%7D%7D%20%7B3%7D$ .

故選：A.
過(guò) $%5Cleft(0%2C-2%5Cright)$ 與圓 $x%5E2%2By%5E2-4x-1%3D0$ 相切的兩條直線的夾角為 $%5Calpha$ ，則 $%5Csin%7B%5Calpha%7D%3D$

A.? $1$

B.? $%5Cfrac%20%7B%5Csqrt%7B15%7D%7D%20%7B4%7D$

C.? $%5Cfrac%20%7B%5Csqrt%7B10%7D%7D%20%7B4%7D$

D.? $%5Cfrac%20%7B%5Csqrt%7B6%7D%7D%20%7B4%7D$

答案? B

解析? 本題考察圓的性質(zhì)及三角函數(shù)的二倍角公式，屬于中檔題.

圓的方程即 $%7B%5Cleft(x-2%5Cright)%7D%5E2%2By%5E2%3D5$ ，圓心記為 $P%5Cleft(2%2C0%5Cright)$ ，半徑為 $%5Csqrt%7B5%7D$ .

記 $A%5Cleft(0%2C-2%5Cright)$ ，則 $%5Cvert%20PA%20%5Cvert%20%3D2%5Csqrt%7B2%7D$ .

所以 $%5Csin%7B%5Cleft(%5Cfrac%7B%5Calpha%7D%7B2%7D%5Cright)%7D%3D%5Cfrac%20%7B%5Csqrt%7B5%7D%7D%20%7B2%5Csqrt%7B2%7D%7D$ ， $%5Ccos%7B%5Cleft(%5Cfrac%7B%5Calpha%7D%7B2%7D%5Cright)%7D%3D%5Cfrac%20%7B%5Csqrt%7B3%7D%7D%20%7B2%5Csqrt%7B2%7D%7D$ .

$%5Csin%7B%5Calpha%7D%3D2%5Csin%7B%5Cleft(%5Cfrac%20%7B%5Calpha%7D%20%7B2%7D%20%5Cright)%7D%5Ccos%7B%5Cleft(%5Cfrac%20%7B%5Calpha%7D%20%7B2%7D%20%5Cright)%7D%3D%5Cfrac%20%7B%5Csqrt%7B15%7D%7D%20%7B4%7D$ .

故選：B.
記 $S_n$ 為數(shù)列 $%5Cleft%5C%7Ba_n%5Cright%5C%7D$ 的前 $n$ 項(xiàng)和，設(shè)甲： $%5Cleft%5C%7Ba_n%5Cright%5C%7D$ 為等差數(shù)列；乙： $%5Cleft%5C%7B%5Cfrac%20%7BS_n%7D%20%7Bn%7D%20%5Cright%5C%7D$ 為等差數(shù)列，則

A. 甲是乙的充分條件但不是必要條件

B. 甲是乙的必要條件但不是充分條件

C. 甲是乙的充要條件

D. 甲既不是乙的充分條件也不是乙的必要條件

答案? C

解析? 本題考察數(shù)列的性質(zhì)及命題條件，屬于中檔題.

(i) 若 $%5Cleft%5C%7Ba_n%5Cright%5C%7D$ 為等差數(shù)列，設(shè) $a_n%3Da_1%2B%5Cleft(n-1%5Cright)d$ ，則 $%5Cfrac%20%7BS_n%7D%20%7Bn%7D%20%3Da_1%2B%5Cfrac%20%7Bn-1%7D%20%7B2%7D%20d$ 為等差數(shù)列，所以甲是乙的充分條件.

(ii) 若 $%5Cleft%5C%7B%5Cfrac%20%7BS_n%7D%20%7Bn%7D%20%5Cright%5C%7D$ 為等差數(shù)列，設(shè) $%5Cfrac%20%7BS_n%7D%20%7Bn%7D%20%3Da_1%2B%5Cleft(n-1%5Cright)d$ ，則 $S_n%3Dna_1%2Bn%5Cleft(n-1%5Cright)d$ ，所以對(duì)任意 $n%20%5Cgeq%202$ 且 $n%20%5Cin%20%5Cboldsymbol%7B%5Cmathrm%7BN%5E*%7D%7D$ ，都有 $a_n%3DS_n-S_%7Bn-1%7D%3Da_1%2B2%5Cleft(n-1%5Cright)d$ ，且 $a_1%3Da_1$ 符合上式，所以 $a_n%3Da_1%2B2%5Cleft(n-1%5Cright)d$ 為等差數(shù)列，所以甲是乙的必要條件.

綜上，甲是乙的充要條件.

故選：C.