《2025年金版教程高中新课程创新导学案高中数学必修第一册人教版》

(2)已知集合$A = \{ x \mid x = a^{2} - 4a + 5,a \in \mathbf{R}\}$，$B = \{ y \mid y = 4b^{2} + 4b + 3,b \in \mathbf{R}\}$，则下列关系正确的是 (B)

A.$A = B$
B.$B\subsetneqq A$
C.$A \subseteq B$
D.$B \subsetneqq A$

答案：
(2)[解析]　$\because A = \{x \mid x = (a - 2)^2 + 1, a \in \mathbb{R}\} = \{x \mid x \geq 1\}$，$B = \{y \mid y = (2b + 1)^2 + 2, b \in \mathbb{R}\} = \{y \mid y \geq 2\}$，$\therefore B \subsetneq A$。
[答案]　B

(3)设集合$A = \{ x \mid x + 1 \leq 0$或$x - 4 \geq 0\}$，$B = \{ x \mid 2a \leq x \leq a + 2\}$.若$B \subseteq A$，则实数$a$的取值范围是$\{a \mid a \leq -3$，或$a \geq 2\}$.

答案：
(3)[解析]　由题意得$A = \{x \mid x \leq -1$，或$x \geq 4\}$。因为$B \subseteq A$，①当$B = \varnothing$时，满足$B \subseteq A$，则$2a ＞ a + 2 \Rightarrow a ＞ 2$；②当$B \neq \varnothing$时，则$\begin{cases} 2a \leq a + 2, \\ a + 2 \leq -1 \end{cases}$或$\begin{cases} 2a \leq a + 2, \\ 2a \geq 4, \end{cases}$即$a \leq -3$或$a = 2$。综上所述，实数$a$的取值范围为$\{a \mid a \leq -3$，或$a \geq 2\}$。
[答案]　$\{a \mid a \leq -3$，或$a \geq 2\}$

[素养训练1] (1)如果集合$P = \{ x \mid x = 2k,k \in \mathbf{N}\}$，$M = \{ x \mid x = 2^{2k + 1},k \in \mathbf{N}\}$，那么集合$P$与$M$之间的关系是 (A)

A.$M \subseteq P$
B.$P \subseteq M$
C.$P = M$
D.$P$，$M$互不包含

答案： [素养训练1]　
(1)A [由$P = \{x \mid x = 2k, k \in \mathbb{N}\}$可得集合$P$是由全体非负偶数构成的，即$P = \{0, 2, 4, 6, ·s\}$，$M = \{x \mid x = 2^{2k + 1}, k \in \mathbb{N}\} = \{x \mid x = 2 × 4^k, k \in \mathbb{N}\}$，集合$M$是由$4^k (k \in \mathbb{N})$的$2$倍构成的，即$M = \{2, 8, 32, 128, ·s\}$，$\therefore M \subseteq P$。故选A。]

(2)已知集合$A = \{ x \mid ax = 1,a \in \mathbf{R}\}$，$B = \{ x \mid x^{2} + x - 2 = 0\}$，若$A \subseteq B$，则所有$a$的取值构成的集合为 ( )

A.$\left\{ - \frac{1}{2} \right\}$
B.$\left\{ - \frac{1}{2},1 \right\}$
C.$\{ 0,1\}$
D.$\{ -\frac {1}{2},0,1\}$

答案：
(2)D [因为$B = \{x \mid x^2 + x - 2 = 0\} = \{1, -2\}$，当$a = 0$时，$A = \varnothing$，满足题意；当$a \neq 0$时，由$ax = 1$，得$x = \frac{1}{a}$，所以$\frac{1}{a} = 1$或$\frac{1}{a} = -2$，即$a = 1$或$a = -\frac{1}{2}$。故所求集合为$\left\{ -\frac{1}{2}, 0, 1 \right\}$。故选D。]

典例2 (1)设集合$M = \{ x \mid 0 \leq x ＜ 4\}$，$N = \left\{ x \mid \frac{1}{3} \leq x \leq 5 \right\}$，则$M \cap N =$ (B)

A.$\left\{ x \mid 0 ＜ x \leq \frac{1}{3} \right\}$
B.$\left\{ x \mid \frac{1}{3} \leq x ＜ 4 \right\}$
C.$\{ x \mid 4 \leq x ＜ 5\}$
D.$\{ x \mid 0 ＜ x \leq 5\}$

答案：
典例2　
(1)[解析]　集合$M$，$N$在数轴上表示如图所示，由图可得$M \cap N = \left\{ x \mid \frac{1}{3} \leq x ＜ 4 \right\}$。故选B。

[答案]　B

(2)如图，已知全集$U = \{ - 2, - 1,3,4,5\}$，集合$A = \{ - 1,3,5\}$，$B = \{ - 2,5\}$，则图中阴影部分表示的集合是 (D)


A.$\{ - 2, - 1,3,5\}$
B.$\{ - 2,5\}$
C.$\{ 5\}$
D.$\{ - 2\}$

答案：
(2)[解析]　由图可知阴影部分表示的集合是$B \cap (\complement_U A)$，因为$U = \{-2, -1, 3, 4, 5\}$，$A = \{-1, 3, 5\}$，$B = \{-2, 5\}$，所以$\complement_U A = \{-2, 4\}$，故$B \cap (\complement_U A) = \{-2\}$。故选D。
[答案]　D

(3)已知$A = \{ x \mid x^{2} - 2x - 8 = 0\}$，$B = \{ x \mid x^{2} + ax + a^{2} - 12 = 0\}$.若$B \cup A \neq A$，求实数$a$的取值范围.$\{a \mid -4 \leq a ＜ 4$，且$a \neq -2\}$

答案：
(3)[解析]　若$B \cup A = A$，则$B \subseteq A$。$\because A = \{x \mid x^2 - 2x - 8 = 0\} = \{-2, 4\}$，$\therefore$集合$B$有以下三种情况：①当$B = \varnothing$时，$\Delta = a^2 - 4(a^2 - 12) ＜ 0$，即$a^2 ＞ 16$，$\therefore a ＜ -4$或$a ＞ 4$。②当$B$是单元素集时，$\Delta = a^2 - 4(a^2 - 12) = 0$，$\therefore a = -4$或$a = 4$。若$a = -4$，则$B = \{2\}$，不符合题意；若$a = 4$，则$B = \{-2\} \subseteq A$，符合题意。③当$B = \{-2, 4\}$时，$-2$，$4$是方程$x^2 + ax + a^2 - 12 = 0$的两根，$\therefore \begin{cases} -a = -2 + 4, \\ a^2 - 12 = -2 × 4, \end{cases} \therefore a = -2$。综上所述，当$B \cup A = A$时，实数$a$的取值范围为$\{a \mid a ＜ -4$，或$a = -2$，或$a \geq 4\}$，$\therefore$当$B \cup A \neq A$时，实数$a$的取值范围为$\{a \mid -4 \leq a ＜ 4$，且$a \neq -2\}$。
[答案]　$\{a \mid -4 \leq a ＜ 4$，且$a \neq -2\}$

[素养训练2] (1)设全集$U = \{ 0,1,2,4,6,8\}$，集合$M = \{ 0,4,6\}$，$N = \{ 0,1,6\}$，则$M \cup (\complement_{U}N) =$ (A)

A.$\{ 0,2,4,6,8\}$
B.$\{ 0,1,4,6,8\}$
C.$\{ 1,2,4,6,8\}$
D.$U$

答案： [素养训练2]　
(1)A [由题意可得$\complement_U N = \{2, 4, 8\}$，则$M \cup (\complement_U N) = \{0, 2, 4, 6, 8\}$。故选A。]

(2)已知集合$A = \{ x \mid - 1 \leq x \leq 3\}$，$B = \{ x \mid x ＜ 2\}$，则$A \cap (\complement_{\mathbf{R}}B) =$ (D)

A.$\{ x \mid x ＞ 2\}$
B.$\{ x \mid x \geq 3\}$
C.$\{ x \mid - 1 ＜ x \leq 2\}$
D.$\{ x \mid 2 \leq x \leq 3\}$

答案：
(2)D [$\because B = \{x \mid x ＜ 2\}$，$\therefore \complement_R B = \{x \mid x \geq 2\}$，$\therefore A \cap (\complement_R B) = \{x \mid 2 \leq x \leq 3\}$。]