高一数学题，求解，急！ 100
3个回答
展开全部
(1)
f(x) = log<2>[ (1-mx)/(x-1)]
f(-x) =-f(x)
log<2>[ (1+mx)/(-x-1)] = -log<2>[ (1-mx)/(x-1)]
= log<2>[ (x-1)/(1-mx)]
(1+mx)/(-x-1) = (x-1)/(1-mx)
-x^2 +1 = 1- m^2.x^2
m^2 =1
m=1(rej) or -1
ie
m=-1
(2)
m=-1
f(x) = log<2>[ (1+x)/(x-1)]
定义域
(1+x)/(x-1) >0
x>1 or x<-1
f(x) = log<2>[ (x+1)/(x-1)]
f'(x) = [1/(ln2)] [(x-1)/(x+1)] [ -2/(x-1)^2]
= -[2/(ln2)] [ 1/(x-1)(x+1)] <0 ( x>1)
max f(x) = f(2) = log<2>[ (1+2)/(2-1)] = log<2>3
min f(x) = f(3) = log<2>[ (1+3)/(3-1)] = log<2>2 =1
f(x) = log<2>[ (1-mx)/(x-1)]
f(-x) =-f(x)
log<2>[ (1+mx)/(-x-1)] = -log<2>[ (1-mx)/(x-1)]
= log<2>[ (x-1)/(1-mx)]
(1+mx)/(-x-1) = (x-1)/(1-mx)
-x^2 +1 = 1- m^2.x^2
m^2 =1
m=1(rej) or -1
ie
m=-1
(2)
m=-1
f(x) = log<2>[ (1+x)/(x-1)]
定义域
(1+x)/(x-1) >0
x>1 or x<-1
f(x) = log<2>[ (x+1)/(x-1)]
f'(x) = [1/(ln2)] [(x-1)/(x+1)] [ -2/(x-1)^2]
= -[2/(ln2)] [ 1/(x-1)(x+1)] <0 ( x>1)
max f(x) = f(2) = log<2>[ (1+2)/(2-1)] = log<2>3
min f(x) = f(3) = log<2>[ (1+3)/(3-1)] = log<2>2 =1
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