设f(x)=log1/2(1-ax/x-1)为奇函数 .
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设f(x)=log1/2(1-ax/x-1)为奇函数 .
(1)f(x)=log(1/2)[(1-ax)/(x-1)]是奇函数,
1/2是底数,后面方括号内是真数
f(-x)=-f(x),所以f(x)+f(-x)=0,
log(1/2) [(1-ax)/(x-1)]+ log(1/2) [(1+ax)/(-x-1)]
=log(1/2) [(1-ax)/(x-1)]*
[(1+ax)/(-x-1)]=0,
(1-ax)/(x-1)]*
[(1+ax)/(-x-1)=1
(1-a^2x^2)/(1-x^2)=1,
1-a^2x^2=1-x^2,
a^2=1,a=-1,a=1(不合题意,舍)
(2)f(x)=log(1/2)[(x+1)/(x-1)],令(x+1)/(x-1)>0 x1,为定义域
设t=(x+1)/(x-1)=[(x-1)+2]/(x-1)=1 +2/(x-1),作出这个图象,
可知,当x>1时,t=1 +2/(x-1),是递减的
而y=log(1/2)t,是递减的,
所以函数f(x)=log(1/2)[(x+1)/(x-1)],在(1,+∞)是单调递增的
(3) log(1/2)[(x+1)/(x-1)]>(1/2)^x+m,x∈[3,4]
左边 f(x)=log(1/2)[(x+1)/(x-1)],的最小值=f(3)=log(1/2)2=-1
设右边=g(x)=(1/2)^x+m,的最大值=g(3)=(1/2)^3 +m=1/8 +m
要满足题意,必须 f(3)>g(3)
所以-1>1/8 +m 得 m
1/2是底数,后面方括号内是真数
f(-x)=-f(x),所以f(x)+f(-x)=0,
log(1/2) [(1-ax)/(x-1)]+ log(1/2) [(1+ax)/(-x-1)]
=log(1/2) [(1-ax)/(x-1)]*
[(1+ax)/(-x-1)]=0,
(1-ax)/(x-1)]*
[(1+ax)/(-x-1)=1
(1-a^2x^2)/(1-x^2)=1,
1-a^2x^2=1-x^2,
a^2=1,a=-1,a=1(不合题意,舍)
(2)f(x)=log(1/2)[(x+1)/(x-1)],令(x+1)/(x-1)>0 x1,为定义域
设t=(x+1)/(x-1)=[(x-1)+2]/(x-1)=1 +2/(x-1),作出这个图象,
可知,当x>1时,t=1 +2/(x-1),是递减的
而y=log(1/2)t,是递减的,
所以函数f(x)=log(1/2)[(x+1)/(x-1)],在(1,+∞)是单调递增的
(3) log(1/2)[(x+1)/(x-1)]>(1/2)^x+m,x∈[3,4]
左边 f(x)=log(1/2)[(x+1)/(x-1)],的最小值=f(3)=log(1/2)2=-1
设右边=g(x)=(1/2)^x+m,的最大值=g(3)=(1/2)^3 +m=1/8 +m
要满足题意,必须 f(3)>g(3)
所以-1>1/8 +m 得 m
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