设a∈R,f(x)=a•2x+a-22x+1(x∈R),试确定a的值,使f(x)为奇函数.-数学

题目简介

设a∈R,f(x)=a•2x+a-22x+1(x∈R),试确定a的值,使f(x)为奇函数.-数学

题目详情

设a∈R,f(x)=
a•2x+a-2
2x+1
(x∈R)
,试确定a的值,使f(x)为奇函数.
题型:解答题难度:中档来源:不详

答案

f(x)=
a•2x+a-2
2x+1
=
a(2x+1)-2
2x+1
=a-class="stub"2
2x+1

要使函数为奇函数,则必有f(-x)=-f(x),
a-class="stub"2
2-x+1
=-a+class="stub"2
2x+1

则2a=class="stub"2
2x+1
+class="stub"2
2-x+1
=class="stub"2
2x+1
+
2•2x
1+2x
=
2(2x+1)
2x+1
=2
即a=1.
故答案为:1

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