问题
解答题
设a∈R,f(x)=
|
答案
∵f(x)=
=a•2x+a-2 2x+1
=a-a(2x+1)-2 2x+1
,2 2x+1
要使函数为奇函数,则必有f(-x)=-f(x),
即a-
=-a+2 2-x+1
,2 2x+1
则2a=
+2 2x+1
=2 2-x+1
+2 2x+1
=2•2x 1+2x
=22(2x+1) 2x+1
即a=1.
故答案为:1
设a∈R,f(x)=
|
∵f(x)=
=a•2x+a-2 2x+1
=a-a(2x+1)-2 2x+1
,2 2x+1
要使函数为奇函数,则必有f(-x)=-f(x),
即a-
=-a+2 2-x+1
,2 2x+1
则2a=
+2 2x+1
=2 2-x+1
+2 2x+1
=2•2x 1+2x
=22(2x+1) 2x+1
即a=1.
故答案为:1