f（x）=1-2的x次方/2的x次方+1为奇函数,求对于任意的t属于R,不等式f(t^2-2t)+f(2t^2-k)

因为定义域为R,根据奇函数的定义有两种算法,1.f（0）=0 2.f（x）+f（-x）=0 算出a的值.
证明函数的单调性要用定义就是任取x1,x2属于R且x1

y = 2的X次方/(2的X次方+1)
定义域x∈R
y = 2的X次方/(2的X次方+1) = 1/{1+1/(2的X次方)}
0＜2的X次方＜+∞
+∞＞1/(2的X次方)＞0
1＜1+1/(2的X次方)＜+∞
0＜1/{1+1/(2的X次方)}＜1
值域（0,1）

1、y=2sinx,x属于闭区间-π/6,π/6
先确定函数的值域,这也是所求反函数的定义域.
y = 2sin(-π/6) = -1
y = 2sin(π/6) = 1
y=2sinx
sinx = y/2
x = arcsin(y/2)
反函数为
y = arcsin(x/2) ,x 属于 闭区间 [-1,1]
2、y=1+ln(x+2)
y - 1 =ln(x+2)
x + 2 = e ^(y-1)
x = e ^(y-1) -2
所求的反函数为
y = e ^(x-1) -2
3、y =2^x / 2^(x+1)
此题括号在应打在哪?不清楚,请把题目明确一下.

(1)
f(x) = [1-2^x]/ [2^(x+1) + a]
f(1) = -1/(4+a)
f(-1) = (1/2)/(1+a) = 1/2(1+a)
f(1) = -f(-1)
1/(4+a) = 1/2(1+a)
2(1+a) = 4+a
a = 2 #

f(x) = [1-2^x]/ [2^(x+1) + 2]
= [1-2^x]/ 2[1+ 2^x]
f(t^2-6t-1)+f(2t^2-k) < 0
put t=0
f(-1) + f(-k) < 0
(1/2)/ 2(3/2) + (1-2^-k)/ 2(1+2^-k) 4
2^(-k+1) > 2^2
-k+1 > 2
k < -1

记t=2^x>0
则y=t/(t+1/t)=1/(1+1/t^2),由此得0