Question

(理)已知集合P={x|a+1≤x≤2a+1},Q={x|x²-3x≤10}.

(1)若a=3,求(P)∩Q;

(2)若PQ,求实数a的取值范围.

(文)已知非空集合P={x|a+1≤x≤2a+1},Q={x|x²-3x≤10}.

(1)若a=3,求(P)∩Q;

(2)若PQ,求实数a的取值范围.

Answer 1

(理)解：(1)因为a=3,所以P={x|4≤x≤7},P={x|x＜4或x＞7}.                  

又Q={x|x²-3x-10≤0}={x|-2≤x≤5},                                          

所以(P)∩Q={x|x＜4或x＞7}∩{x|-2≤x≤5}={x|-2≤x＜4}.                        

(2)当P≠Q时,由PQ,

得

解得0≤a≤2;                                                             

当P=,即2a+1＜a+1时,得a＜0,此时有P=Q,

所以a＜0为所求.

综上,实数a的取值范围是(-∞,2］.                                           

(文)解:(1)因为a=3,所以P={x|4≤x≤7},P={x|x＜4或x＞7}.                      

又Q={x|x²-3x-10≤0}={x|-2≤x≤5},                                          

所以(P)∩Q={x|x＜4或x＞7}∩{x|-2≤x≤5}={x|-2≤x≤4}.                       

(2)由PQ,得

解得0≤a≤2,即实数a的取值范围是［0,2］.