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,求實(shí)數(shù)a的取值范圍.

(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,求實(shí)數(shù)a的取值范圍.

Answer 1

(理)解：(1)因?yàn)閍=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)當(dāng)P≠Q(mào)時(shí),由PQ,

得

解得0≤a≤2;                                                             

當(dāng)P=,即2a+1＜a+1時(shí),得a＜0,此時(shí)有P=Q,

所以a＜0為所求.

綜上,實(shí)數(shù)a的取值范圍是(-∞,2］.                                           

(文)解:(1)因?yàn)閍=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,即實(shí)數(shù)a的取值范圍是［0,2］.