Mathematics - Functions Question with Solution | TestHub

To check injectivity, assume $f(x_1)=f(x_2)$. If $x_1,x_2\ge 0$, then $\frac{x_1}{1+x_1}=\frac{x_2}{1+x_2}⟹x_1+x_1{x}_2=x_2+x_1{x}_2⟹x_1=x_2$. If $x_1,x_2<0$, then $\frac{x_1}{1-x_1}=\frac{x_2}{1-x_2}⟹x_1-x_1{x}_2=x_2-x_1{x}_2⟹x_1=x_2$. If $x_1\ge 0$ and $x_2<0$, then $f(x_1)\ge 0$ and $f(x_2)<0$, so $f(x_1)\ne f(x_2)$. Thus, $f$ is injective. To check surjectivity, we find the range. If $x\ge 0$, $f(x)=\frac{x}{1+x}=1-\frac{1}{1+x}$. As $x\in [0,\infty )$, $1+x\in [1,\infty )$, so $\frac{1}{1+x}\in (0,1]$. Thus $f(x)\in [0,1)$. If $x<0$, let $x=-t$ for $t>0$. Then $f(x)=\frac{-t}{1+t}=-\frac{t}{1+t}=-(1-\frac{1}{1+t})$. As $t\in (0,\infty )$, $1+t\in (1,\infty )$, so $\frac{1}{1+t}\in (0,1)$. Thus $f(x)\in (-1,0)$. Combining these, the range of $f(x)$ is $(-1,1)$. Since the codomain is $\mathbb{R}$ and the range is $(-1,1)\ne \mathbb{R}$, $f$ is not surjective. Therefore, $f$ is injective but not surjective. Thus, option A is correct.