设f(x)=log1/2((1-ax)/(x-1))为奇函数,a为常数,(1)求a的值(2)证明：函数f(x)在区间(1,+无穷大)内单调递增(3)若对于区间[3,4]上的每一个x的值,不等式f(x)>0.5∧2+m恒成立,求实数m的取值范

(1)f(x)为奇函数,那么f(-x)=-f(x)f(-x)+f(x)=0所以log(1/2)[(1+ax)/(-x-1)]+log(1/2)[(1-ax)/(x-1)]=0即log(1/2)[(1+ax)/(-x-1)*(1-ax)/(x-1)]=0所以(1+ax)/(-x-1)*(1-ax)/(x-1)=11-a^2x^2=1-x^2(a^2-1)x^2=0因为x是...f(x)=log(1/2)[(1+x)/(x-1)] 任取1x1-x2∴1+x2-x1+x1x2>1+x1-x2+x1x2>0∴(1+x2-x1+x1x2)/(1+x1-x2-x1x2)>1因为底数为1/2∴log(1/2)[(1+x2-x1+x1x2)/(1+x1-x2-x1x2)]<0即f(x1)-f(x2)<0∴f(x1)0.5∧2+m恒成立，只需f(x)min>(1/2)^2+m即可因为f(x)是[3,5]上的增函数∴f(x)min=f(3)=log(1/2)(2)=-1∴-1>0.5∧2+m题目有问题0.5^2+m是什么东西若是log(1/2)(2+m)的话那么log(1/2)(2)>log(1/2)(2+m)得到m<0且m>-2-2