$Iff:R\to (0,1]definedbyf\left(x\right)=\frac{1}{x^2+1},thenfis$

$\begin{array}{l}f\left(x\right)=\frac{1}{x^2+1}andf:R\to (0,1]\\ Letf\left(x_1\right)=f\left(x_2\right)\forall x_1,x_2\in R\\ \Rightarrow \frac{1}{x_1^2+1}=\frac{1}{x_2^2+1}\\ \Rightarrow x_1^2+1=x_2^2+1\\ \Rightarrow x_1^2=x_2^2\\ \Rightarrow x_1=\pm x_2\\ \therefore fisnotone-one\\ Let\frac{1}{x^2+1}=y\\ \Rightarrow x^2+1=\frac{1}{y}\\ \Rightarrow x^2=\frac{1}{y}-1\\ \Rightarrow x=\sqrt{\frac{1}{y}-1}\end{array}$

$\begin{array}{l}isdefinedif\frac{1}{y}-1\ge 0\\ \Rightarrow \frac{1}{y}\ge 1\\ \Rightarrow \frac{1}{y}\ge 1\\ \Rightarrow y\le 1andy>0\\ \therefore Range=(0,1]\\ \therefore Range=Codomain\\ \therefore fisonto\end{array}$