已知命题P函数y=log
a(1-2x)在定义域上单调递增;命题Q不等式(a-2)x
2+2(a-2)x-4<0对任意实数x恒成立若P∨Q是真命题,求实数a的取值范围.
考点分析:
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计算:
(1)
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102651667824262/SYS201311031026516678242016_ST/0.png)
;
(2)已知3
x=4
y=36,求
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102651667824262/SYS201311031026516678242016_ST/1.png)
的值.
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函数y=log
a(x+3)-1(a>0,a≠1)的图象恒过定点A,若点A在直线mx+ny+1=0上,其中mn>0,则
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102651667824262/SYS201311031026516678242015_ST/0.png)
+
![manfen5.com 满分网](http://img.manfen5.com/res/GZSX/web/STSource/20131103102651667824262/SYS201311031026516678242015_ST/1.png)
的最小值为
.
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函数y=-x
2-4mx+1在[2,+∞)上是减函数,则m的取值范围是
.
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