设对所有实数x,不等式x2log24(a+1)a+2xlog22aa+1+log2(a+1)24a2>0恒成立,则a的取值范围为______.-数学

题目简介

设对所有实数x,不等式x2log24(a+1)a+2xlog22aa+1+log2(a+1)24a2>0恒成立,则a的取值范围为______.-数学

题目详情

设对所有实数x,不等式x2log2
4(a+1)
a
+2xlog2
2a
a+1
+log2
(a+1)2
4a2
>0恒成立,则a的取值范围为______.
题型:填空题难度:中档来源:闸北区二模

答案

∵不等式x2log2
4(a+1)
a
+2xlog2class="stub"2a
a+1
+log2
(a+1)2
4a2
>0恒成立
由二次不等式的性质可得,log2
4(a+1)
a
>0
△=4(log2class="stub"2a
a+1
)2-log2
4(a+1)
a
•log2
(a+1)2
4a2
×4<0
令t=log2class="stub"a+1
a

(1+log2class="stub"a
a+1
)2-(2+log2class="stub"a+1
a
)
(2log2class="stub"a+1
a
-2)<0

整理可得,(log2class="stub"a+1
a
+5)(log2class="stub"a+1
a
-1)>0

log2
4(a+1)
a
>0

log2class="stub"a+1
a
>1

解可得,0<a<1
故答案为:0<a<1

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