对于函数f(x)=(x-1)/(x+1),设f1(x)=f(x),f2(x)=f[f1(x)],f3(x)=f[f2(x)],…fn+1(x)=f[fn(x)],n∈N*且n ≥2领集合M={x/f2008(x)=x^2,x∈R}则集合M为A空集B实数集C单元素集D二元素集
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![对于函数f(x)=(x-1)/(x+1),设f1(x)=f(x),f2(x)=f[f1(x)],f3(x)=f[f2(x)],…fn+1(x)=f[fn(x)],n∈N*且n ≥2领集合M={x/f2008(x)=x^2,x∈R}则集合M为A空集B实数集C单元素集D二元素集](/uploads/image/z/1843100-44-0.jpg?t=%E5%AF%B9%E4%BA%8E%E5%87%BD%E6%95%B0f%28x%29%3D%28x-1%29%2F%28x%2B1%29%2C%E8%AE%BEf1%28x%29%3Df%28x%29%2Cf2%28x%29%3Df%5Bf1%28x%29%5D%2Cf3%28x%29%3Df%5Bf2%28x%29%5D%2C%E2%80%A6fn%2B1%28x%29%3Df%5Bfn%28x%29%5D%2Cn%E2%88%88N%2A%E4%B8%94n+%E2%89%A52%E9%A2%86%E9%9B%86%E5%90%88M%3D%7Bx%2Ff2008%28x%29%3Dx%5E2%2Cx%E2%88%88R%7D%E5%88%99%E9%9B%86%E5%90%88M%E4%B8%BAA%E7%A9%BA%E9%9B%86B%E5%AE%9E%E6%95%B0%E9%9B%86C%E5%8D%95%E5%85%83%E7%B4%A0%E9%9B%86D%E4%BA%8C%E5%85%83%E7%B4%A0%E9%9B%86)
对于函数f(x)=(x-1)/(x+1),设f1(x)=f(x),f2(x)=f[f1(x)],f3(x)=f[f2(x)],…fn+1(x)=f[fn(x)],n∈N*且n ≥2领集合M={x/f2008(x)=x^2,x∈R}则集合M为A空集B实数集C单元素集D二元素集
对于函数f(x)=(x-1)/(x+1),设f1(x)=f(x),f2(x)=f[f1(x)],f3(x)=f[f2(x)],…fn+1(x)=f[fn(x)],n∈N*且n ≥2
领集合M={x/f2008(x)=x^2,x∈R}则集合M为A空集B实数集C单元素集D二元素集
对于函数f(x)=(x-1)/(x+1),设f1(x)=f(x),f2(x)=f[f1(x)],f3(x)=f[f2(x)],…fn+1(x)=f[fn(x)],n∈N*且n ≥2领集合M={x/f2008(x)=x^2,x∈R}则集合M为A空集B实数集C单元素集D二元素集
应该有规律:
由题意先推规律
f1(x)=f(x)=1-2/(x+1),
f2(x)=f[f1(x)]=-1/x,
f3(x)=f[f2(x)]=-1-2/(x-1),
f4(x)=f[f3(x)]=x,
∴f5(x)=f[f4(x)]=f1(x)=f(x)=1-2/(x+1),
.
四个一循环
故f2008(x)=f4(x)=x=x^2
∴x=0或1
但经观察,fn(x)系列定义域不能有:-1,0,1,而f2008(x)正是由前面推导而来,
要有意义,需全舍去
∴集合M为∅,选A