《2025年全效核心素养测评高中数学必修第一册人教版》

13. 已知集合$A = \{x|-1\lt x\lt 1\}$，$B = \{x|x\lt a\}$.
(1)若$A\cap B\neq \varnothing$，求$a$的取值范围；
(2)若$A\cup B = \{x|x\lt 1\}$，求$a$的取值范围.

答案： 13.解：
(1)$A = \{x\mid -1 ＜ x ＜ 1\}$,$B = \{x\mid x ＜ a\}$,且$A\cap B\neq\varnothing$,
$\therefore a ＞ -1$,即$a$的取值范围是$\{a\mid a ＞ -1\}$.
(2)$A = \{x\mid -1 ＜ x ＜ 1\}$,$B = \{x\mid x ＜ a\}$,且$A\cup B = \{x\mid x ＜ 1\}$,$\therefore a$的取值范围是$\{a\mid -1 ＜ a\leqslant 1\}$.

14. 已知集合$A = \{x|x^{2}+ax + b = 0\}$，$B = \{x|x^{2}+cx + 6 = 0\}$，$A\cup B = \{2,3\}$，$A\cap B = \{3\}$，求$a$，$b$，$c$的值.

答案： 14.解：$\because A\cap B = \{3\}$,$\therefore 3$是方程$x^{2}+cx + 6 = 0$的根，则$3^{2}+3c + 6 = 0$,解得$c = -5$,$\therefore B = \{x\mid x^{2}-5x + 6 = 0\} = \{3,2\}$.
$\because A\cup B = \{2,3\}$,$A\cap B = \{3\}$,$\therefore A\neq B$,则$A = \{3\}$,$\therefore$方程$x^{2}+ax + b = 0$有两个相等的实数根$3$,$\therefore\begin{cases}\Delta = a^{2}-4b = 0\\9 + 3a + b = 0\end{cases}$,解得$\begin{cases}a = -6\\b = 9\end{cases}$.综上,$a = -6$,$b = 9$,$c = -5$.

15. (2024·集美中学高一检测)设集合$A$，$B$，$C$均为非空集合，下列命题中，为真命题的是(D)

A.若$A\cap B = B\cap C$，则$A = C$
B.若$A\cup B = B\cup C$，则$A = C$
C.若$A\cup B = B\cap C$，则$C\subseteq B$
D.若$A\cap B = B\cup C$，则$C\subseteq B$

答案： 15.D

16. 已知集合$A = \{x|x^{2}+4x - 5 = 0\}$，$B = \{x|x^{2}+2(a + 2)x + a^{2}+2a - 2 = 0\}$.
(1)若$A\cap B = \{1\}$，求实数$a$的值；
(2)若$A\cup B = A$，求实数$a$的取值范围.

答案： 16.
(1)解：由题可知，$A = \{-5,1\}$,$\because A\cap B = \{1\}$,$\therefore 1\in B$,即$x = 1$是方程$x^{2}+2(a + 2)x + a^{2}+2a - 2 = 0$的根，则有$1 + 2(a + 2)+a^{2}+2a - 2 = 0$,即$a^{2}+4a + 3 = 0$,解得$a = -1$,或$a = -3$.当$a = -1$时，$B = \{x\mid x^{2}+2x - 3 = 0\} = \{1,-3\}$,满足$A\cap B = \{1\}$;当$a = -3$时，$B = \{x\mid x^{2}-2x + 1 = 0\} = \{1\}$,满足$A\cap B = \{1\}$,均符合题意.综上,实数$a$的值为$-1$或$-3$.
(2)$\because A\cup B = A$,$\therefore B\subseteq A$.当$B = \varnothing$时，$\Delta = [2(a + 2)]^{2}-4(a^{2}+2a - 2)=8a + 24 ＜ 0$,解得$a ＜ -3$;
若$B\neq\varnothing$,当$B = \{1\}$时，由
(1)知，$a = -3$,符合题意;
当$B = \{-5\}$时，可得$\begin{cases}(-5)^{2}+2(a + 2)×(-5)+a^{2}+2a - 2 = 0\\\Delta = [2(a + 2)]^{2}-4(a^{2}+2a - 2)=8a + 24 = 0\end{cases}$,无解;
当$B = \{1,-5\}$时，可得$\begin{cases}(-5)^{2}+2(a + 2)×(-5)+a^{2}+2a - 2 = 0\\1^{2}+2(a + 2)+a^{2}+2a - 2 = 0\end{cases}$,无解.
综上,实数$a$的取值范围是$a\leqslant -3$.