设f(x)=log1
设f(x)=log1/2(1-ax/x-1)为奇函数,a为常数,求a值
f(x)=log[(1-ax)/(x-1)] =log[(x-1)/(1-ax)] 则,f(-x)=log[(-x-1)/(1+ax)] 已知f(x)为奇函数,所以:f(-x)=-f(x) 即:log[(-x-1)/(1+ax)]=-log[(x-1)/(1-ax)] =log[(1-ax)/(x-1)] ===> (-x-1)/(1+ax)=(1-ax)/(x-1) ===> (1+ax)*(1-ax)=(-x-1)*(x-1) ===> 1-a^2*x^2=1-x^2 ===> (1-a^2)*x^2=0 上式对于定义域内的x均成立 所以,1-a^2=0 所以,a=±1 而,当a=1时,f(x)=log[(1-x)/(x-1)]无意义,舍去 所以,a=-1.
介绍一种简捷方法 f(x)=log[(1-ax)/(x-1)]为奇函数, f(x)+f(-x)=0 设u(x)=(1-ax)/(x-1), u(x)*u(-x)=1 [(1-ax)/(x-1)][(1+ax)(-x-1)]=1 1-(ax)^2=1-x^2, (ax)^2=x^2 a^1=1, a=±1 a=1时,u(x)=-1, f(x)无意义,舍去; a=-1时,u(x)=(1+x)/(x-1)>0, x0 所以a=-1
答:已知函数f(x)=log1/2|sinx|判断奇偶性和周期性 f(x)=log<1/2>|sinx| 则,f(-x)=log<1/2>|sin(-x)|=log...详情>>
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