若函数f(x)(x∈R)为奇函数,且存在反函数f-1(x)(与f(x)不同),F(x)=2f(x)?2f?1(x)2f(x)+2f?1(

2025-06-26 05:04:56
推荐回答(1个)
回答1:

∵f(x)(x∈R)为奇函数∴f(-x)=-f(x)
∴f-1(-x)=-f-1(x)
F(?x)=

2f(?x)?2f?1(?x)
2f(?x)+2f?1(?x)

=
2?f(x)?2?f?1(x)
2?f(x)+2?f?1(x)

=
1
2f(x)
?
2f?1(x)
1
2f(x)
+
1
2f?1(x)

=?
2f(x)?2f?1(x)
2f(x)+2f?1(x)
=-F(x)
∴F(x)是奇函数.
故选A.