Question

The function $f:\mathrm{ℝ}\to \left[-\frac{1}{2},\frac{1}{2}\right]$ is defined as $f\left(x\right)=\frac{x}{1+x^2}$ is

• A. Injective but not Surjective.
• B. Surjective but not Injective.
• C. Neither Injective nor Surjective.
• D. Invertible.

Answer 1

The correct option is B

Surjective but not Injective.

The explanation for the correct option

Given function, $f\left(x\right)=\frac{x}{1+x^2}$.

Differentiate the given function with respect to $x$.

$$\begin{gathered} \frac{\mathrm{d}}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left(\frac{x}{1+x^2}\right) \\ \Rightarrow f\text{'}\left(x\right)=\frac{\left(1+x^2\right){\displaystyle \frac{d}{dx}}\left(x\right)-\left(x\right){\displaystyle \frac{d}{dx}}\left(1+x^2\right)}{{\left(1+x^2\right)}^2}\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\frac{\mathbf{u}}{\mathbf{v}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{v}{\displaystyle \frac{du}{dx}}\mathbf{-}\mathbf{u}{\displaystyle \frac{dv}{dx}}}{{\mathbf{v}}^{\mathbf{2}}}\right] \\ \Rightarrow f\text{'}\left(x\right)=\frac{\left[\left(1+x^2\right)\times 1\right]-\left[\left(x\right)\times 2x\right]}{{\left(1+x^2\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{1+x^2-2x^2}{{\left(1+x^2\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{1-x^2}{{\left(1+x^2\right)}^2} \end{gathered}$$

Now, for increasing intervals $\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{>}\mathbf{0}$.

$$\begin{gathered} \Rightarrow \frac{1-x^2}{{\left(1+x^2\right)}^2}>0 \\ \Rightarrow 1-x^2>0\times {\left(1+x^2\right)}^2\left[\because {\left(1+x^2\right)}^2\ge 0\right] \\ \Rightarrow \left(1-x\right)\left(1+x\right)>0 \\ \Rightarrow \left(x-1\right)\left(x+1\right)<0 \\ \Rightarrow x\in \left(-1,1\right) \end{gathered}$$

Thus, the given function is increasing in nature in $\left(-1,1\right)$ and decreasing in nature in $\mathrm{ℝ}-\left(-1,1\right)$.

Hence, the given function is not an injective function.

Now, let $y=\frac{x}{1+x^2}$.

$$\begin{gathered} \Rightarrow y\left(1+x^2\right)=x \\ \Rightarrow y{x}^2-x+y=0 \end{gathered}$$

Discriminant of the given quadratic equation, $D={\left(-1\right)}^2-4\left(y\right)\left(y\right)=1-4y^2$.

For real solutions, $\mathit{D}\mathbf{\ge }\mathbf{0}$

$$\begin{gathered} \Rightarrow 1-4y^2\ge 0 \\ \Rightarrow 4y^2-1\le 0 \\ \Rightarrow 4\left(y^2-\frac{1}{4}\right)\le 0 \\ \Rightarrow {\left(y\right)}^2-{\left(\frac{1}{2}\right)}^2\le 0 \\ \Rightarrow \left(y-\frac{1}{2}\right)\left(y+\frac{1}{2}\right)\le 0 \\ \Rightarrow y\in \left[-\frac{1}{2},\frac{1}{2}\right] \end{gathered}$$

Thus the range of the given function is $\left[-\frac{1}{2},\frac{1}{2}\right]$, which is same as the co-domain.

So, the given function is a subjective function.

Hence, the correct option is $\mathrm{Option}\left(\mathrm{B}\right)$