已知函数f(x)是偶函数,且f(x)在[0,+∞)上是增函数,如果x∈[1/2,1]时,不等式f(ax+1)≤f(x-2)恒成立,则实
已知函数f(x)是偶函数,且f(x)在[0,+∞)上是增函数,如果x∈[1/2,1]时,不等式f(ax+1)≤f(x-2)恒成立,则实数a的取值范围是____,答案[-2...
已知函数f(x)是偶函数,且f(x)在[0,+∞)上是增函数,如果x∈[1/2,1]时,不等式f(ax+1)≤f(x-2)恒成立,则实数a的取值范围是____,答案[-2,0]求详解,谢谢。
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2013-08-06 · 知道合伙人教育行家
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|ax+1|<=|x-2|
(ax+1)^2<=(x-2)^2
[(a+1)x-1][(a-1)x+3]<=0
a=1时,x<=1/2(舍去)
a=-1时,x<=3/2,符合条件
a>1时,-3/(a-1)<=x<=1/(a+1)<1/2(舍去)
a<-1时,1/(a+1)<=a<=-3/(a-1)
-3/(a-1)>=1
(-2-a)(a-1)>=0
-2<=a<=1
-2<=a<-1
-1<a<1时,x1=1/(a+1),x2=-3/(a-1),x解集在两根之外包括两根
1/(a+1)<=1/2且-3/(a-1)<=1/2 或 1/(a+1)>=1且-3/(a-1)>=1
(1-a)(a+1)<=0且(-5-a)(a-1)<=0(舍去) 或 -a(a+1)>=0且(-2-a)(a-1)>=0
-1<=a<=0
-1<a<=0
故-2<=a<=0
(ax+1)^2<=(x-2)^2
[(a+1)x-1][(a-1)x+3]<=0
a=1时,x<=1/2(舍去)
a=-1时,x<=3/2,符合条件
a>1时,-3/(a-1)<=x<=1/(a+1)<1/2(舍去)
a<-1时,1/(a+1)<=a<=-3/(a-1)
-3/(a-1)>=1
(-2-a)(a-1)>=0
-2<=a<=1
-2<=a<-1
-1<a<1时,x1=1/(a+1),x2=-3/(a-1),x解集在两根之外包括两根
1/(a+1)<=1/2且-3/(a-1)<=1/2 或 1/(a+1)>=1且-3/(a-1)>=1
(1-a)(a+1)<=0且(-5-a)(a-1)<=0(舍去) 或 -a(a+1)>=0且(-2-a)(a-1)>=0
-1<=a<=0
-1<a<=0
故-2<=a<=0
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