①若p真q假,则{■("-" 2

　　②若p假q真,则{■(a≤"-" 2"或" a≥2"," @a<2"," )┤解得a≤-2.

　　综上所述,所求实数a的取值范围为(-∞,-2].

★11.已知p:|1"-" (x"-" 1)/3|≤2,q:x2-2x+1-m2≤0(m>0),若􀱑p是􀱑q的充分不必要条件,求实数m的取值范围.

解由|1"-" (x"-" 1)/3|≤2,得-2≤x≤10,

　　∴􀱑p:A={x|x>10或x<-2}.

　　由x2-2x+1-m2≤0,得1-m≤x≤1+m(m>0),

　　∴􀱑q:B={x|x>1+m或x<1-m}(m>0).

　　∵􀱑p是􀱑q的充分不必要条件,

　　∴A⫋B.

　　∴{■(m>0"," @1+m<10"," @1"-" m≥"-" 2)┤或{■(m>0"," @1+m≤10"," @1"-" m>"-" 2"," )┤

　　解得0

　　故实数m的取值范围是(0,3].

★12.已知a>0,且a≠1,设p:函数y=loga(x+1)在(0,+∞)内是减少的;q:曲线y=x2+(2a-3)x+1与x轴交于不同的两点.如果p和q有且只有一个为真命题,求实数a的取值范围.

分析利用对数函数与二次函数的单调性及简易逻辑,结合分类讨论的数学思想解答该题.

解当01时,y=loga(x+1)在(0,+∞)内是增加的.

　　曲线y=x2+(2a-3)x+1与x轴交于不同的两点等价于(2a-3)2-4>0,即a<1/2 或a>5/2.

　　(1)若p为真命题,q为假命题,则函数y=loga(x+1)在(0,+∞)内是减少的,曲线y=x2+(2a-3)x+1与x轴不交于两点,

　　所以a∈(0,1)∩([1/2 "," 1)⋃(1"," 5/2]),

　　即a∈[1/2 "," 1).

　　(2)若p为假命题,q为真命题,则函数y=loga(x+1)在(0,+∞)内不是减少的,曲线y=x2+(2a-3)x+1与x轴交于不同的两点,

　　所以a∈(1,+∞)∩((0"," 1/2)⋃(5/2 "," +"∞" )),即a∈(5/2 "," +"∞" ).

综上所述,实数a的取值范围是[1/2 "," 1)∪(5/2 "," +"∞" ).