Question

求答案及详细解析

Answer 1

解:由(x-1)f(x+1/x-1)+f(x)=x(1)令y=x+1/x-1,于是x=y+1/y-1,带入到(1)式得(2/(y-1))*f(y)+f(y+1/y-1)=y+1/y-1(2)将变量y换成x得(2/(x-1))*f(x)+f(x+1/x-1)=x+1/x-1(3)等式两边乘以x-1,得2f(x)+(x-1)*f(x+1/x-1)=x+1(4)联立(1)，(4)解得f(x)=1(x不能等于1)3.f(x)=根号下(x-1)因为x^2-x+1>0恒成立，所以x^2>x-1|x|>根号下(x-1)或根号下(x-1)<|x|取k=1,即得|f(x)|<|x|4.f(x)=x/(x^2+x+1)|f(x)|=|x/(x^2+x+1)|≤k|x|等价于:|1/(x^2+x+1)|≤k等价于:x^2+x+1≥1/kx^2+x+1=x^2+x+1/4+3/4=(x+1/2)^2+3/4≥3/4(=1/k)取k=4/3则f(x)≤4/3|x|