(1)对R上的奇函数来说,f(0)=0,即-1+b=0,b=1.
F(x)=(-2^x+1)/(2^(x+1)+a)
又有F(-x)=- F(x)
(-2^(-x)+1)/(2^(-x+1)+a)= -(-2^x+1)/(2^(x+1)+a)……左边式子的分子分母同乘以2^x
(-1+2^x)/(2+a•2^x)= (2^x-1)/(2^(x+1)+a)
所以2+a•2^x=2^(x+1)+a
a(2^x-1)= 2^(x+1)-2,
a(2^x-1)= 2(2^x-1)
所以a=2.
(2)所以 f(x) = (-2^x+1)/[2^(x+1)+2]= (-2^x+1)/{2[2^x + 1]}
= (-2^x - 1 + 2)/{2[2^x + 1]}
= -1/2 + 1/(2^x + 1)
设 x1 < x2
则 f(x2) - f(x1) = 1/(2^x2 + 1) - 1/(2^x1 + 1)
= (2^x1 - 2^x2)/[(2^x2 + 1)(2^x1 + 1)] < 0
所以 f(x2) < f(x1)
所以 f(x)是减函数
(3)f(t²-2t)+f(2 t²-k) k-2 t²
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