Question

If the function $f:R\to A\mathrm{given}\mathrm{by}f\left(x\right)=\frac{x^2}{x^2+1}$ is a surjection, then A =
(a) R
(b) [0, 1]
(c) [0, 1)
(d) [0, 1)

Answer 1

$$\begin{gathered} \mathrm{As}f\mathrm{is}\mathrm{surjective},\text{r}\mathrm{ange}\mathrm{of}f=\mathrm{co}-\mathrm{domain}\mathrm{of}f \\ \Rightarrow A=\mathrm{range}\mathrm{of}f \\ \because f\left(x\right)=\frac{x^2}{x^2+1}, \\ y=\frac{x^2}{x^2+1} \\ \Rightarrow y\left(x^2+1\right)=x^2 \\ \Rightarrow \left(y-1\right)x^2+y=0 \\ \Rightarrow x^2=\frac{-y}{\left(y-1\right)} \\ \Rightarrow x=\sqrt{\frac{y}{\left(1-y\right)}} \\ \Rightarrow \frac{y}{\left(1-y\right)}\ge 0 \\ \Rightarrow y\in [0,1) \\ \Rightarrow \mathrm{Range}\mathrm{of}f=[0,1) \\ \Rightarrow A=[0,1) \end{gathered}$$
So, the answer is (d).