Question

已知a＞1,命題p:a(x-2)+1＞0,命題q:(x-1)²＞a(x-2)+1＞0.若命題p、q同時成立,求x的取值范圍.

Answer 1

解:依題意,有                  

①若1＜a＜2時,則有

而a-＞0,即a＞2-,

∴x＞2或2-＜x＜a.

故此時x的取值范圍為(2-,a)∪(2,+∞).                                       

②若a=2時,則x＞且x≠2,

此時x的取值范圍為(,2)∪(2,+∞).                                           

③若a＞2時,則有x＞a或2-＜x＜2.

此時x的取值范圍為(2-,2)∪(a,+∞).                                         

綜上所述,當(dāng)1＜a＜2時,x的取值范圍為(2-,a)∪(2,+∞);

當(dāng)a=2時,x的取值范圍為(,2)∪(2,+∞);

當(dāng)a＞2時,x的取值范圍為(2-,2)∪(a,?+∞).