已知函数f(x)=log21+x1-x.(1)判断函数f(x)的奇偶性;(2)求证f(x1)+f(x2)=f(x1+x21+x1x2)(3)若f(a+b1+ab)=1,f(-b)=12,求f(a)的值

题目简介

已知函数f(x)=log21+x1-x.(1)判断函数f(x)的奇偶性;(2)求证f(x1)+f(x2)=f(x1+x21+x1x2)(3)若f(a+b1+ab)=1,f(-b)=12,求f(a)的值

题目详情

已知函数f(x)=log2
1+x
1-x

(1)判断函数f(x)的奇偶性;
(2)求证f(x1)+f(x2)=f(
x1+x2
1+x1x2
)

(3)若f(
a+b
1+ab
)=1
f(-b)=
1
2
,求f(a)的值.
题型:解答题难度:中档来源:不详

答案

(1)由class="stub"1+x
1-x
>0
得函数f(x)的定义域为{x|-1<x<1},
f(x)+f(-x)=log2class="stub"1+x
1-x
+log2class="stub"1-x
1+x
=0

所以函数f(x)为奇函数
(2)证明:f(x1)+f(x2)=log2
1+x1
1-x1
+log2
1+x2
1-x2
=log2(
1+x1
1-x1
1+x2
1-x2
)
=log2
1+x1+x2+x1x2
1-x1-x2+x1x2
f(
x1+x2
1+x1x2
)=log2
1+
x1+x2
1+x1x2
1-
x1+x2
x1x2
=log2
1+x1+x2+x1x2
1-x1-x2+x1x2

f(x1)+f(x2)=f(
x1+x2
1+x1x2
)

(3)由(2)的结论知f(a)+f(b)=f(class="stub"a+b
1+ab
)=1

又由(1)知f(b)=-f(-b)=-class="stub"1
2

f(a)=1-f(b)=1+class="stub"1
2
=class="stub"3
2

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