Question

The function $f:[0,\infty )\to [0,\infty )$ defined by $f\left(x\right)=\frac{2x}{1+2x}$ is

• A. One-One and Onto
• B. One-One but not Onto
• C. not One-One but Onto
• D. neither One-One not Onto

Answer 1

The correct option is B

One-One but not Onto

Explanation for the correct option

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

Differentiate the given function with respect to $x$.

$$\begin{gathered} \frac{\mathrm{d}}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left(\frac{2x}{1+2x}\right) \\ f\text{'}\left(x\right)=\frac{\left(1+2x\right){\displaystyle \frac{d}{dx}}\left(2x\right)-\left(2x\right){\displaystyle \frac{d}{dx}}\left(1+2x\right)}{{\left(1+2x\right)}^2}\left[\because \frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{\left(}\frac{\mathbf{u}}{\mathbf{v}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{v}{\displaystyle \frac{du}{dx}}\mathbf{-}\mathbf{u}{\displaystyle \frac{dv}{dx}}}{{\mathbf{v}}^{\mathbf{2}}}\right] \\ f\text{'}\left(x\right)=\frac{\left[\left(1+2x\right)\times 2\right]-\left[\left(2x\right)\times 2\right]}{{\left(1+2x\right)}^2} \\ f\text{'}\left(x\right)=\frac{2+4x-4x}{{\left(1+2x\right)}^2} \\ f\text{'}\left(x\right)=\frac{2}{{\left(1+2x\right)}^2} \end{gathered}$$

As ${\left(1+2x\right)}^2\ge 0$, thus $\frac{2}{{\left(1+2x\right)}^2}>0$.

So, $f\text{'}\left(x\right)>0$ and the given function is always increasing.

Hence, the given function is a One-One function.

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

Replace the values of $x$ and $y$ with each other.

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

As the denominator, $\left(2-2x\right)\ne 0$.

Thus, $x\ne 1$.

Hence, $1$ is not a pre-image of the given function, whereas the given co-domain is $[0,\infty )$.

So, the given function is an Into function.

Therefore, the function $f\left(x\right)=\frac{2x}{1+2x}$ is One-One but not Onto.

Hence, the given function is One-One but not Onto, and therefore correct option is (B).