Question

Let $f:R\to R$ be a function defined by $f\left(x\right)=\frac{\left[e^x-e^{-x}\right]}{\left[e^x+e^{-x}\right]}$ then, $\left(1\right)$ is both one-one and onto $\left(2\right)f$ is one-one but onto

• A. $f$ is both one-one and onto
• B. $f$ is one-one but not onto
• C. $f$ is onto but not one-one
• D. $f$ is neither one-one nor onto

Answer 1

The correct option is D

$f$ is neither one-one nor onto

Step 1: Given information

A function is injective (one-one), when

$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow x=y \end{gathered}$$

A function is subjective(onto) if the range of $f$ equals the codomain of $f$.

Step 2: Checking if the function is injective:

$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow \frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^y-e^{-y}}{e^y+e^{-y}} \\ \Rightarrow 2e^{2x}=2e^{2y} \\ \Rightarrow e^{2(x-y)}=0 \\ \therefore x\ne y \end{gathered}$$

Therefore, $f$ is not injective.

Step 3: Checking if the function is subjective:

$$\begin{gathered} y=\frac{\left[e^x-e^{-x}\right]}{\left[e^x+e^{-x}\right]} \\ \text{Applying componendo and dividendo:} \\ \frac{y+1}{y-1}=\frac{2e^x}{-2e^{-x}} \\ \Rightarrow \frac{y+1}{1-y}=\frac{2e^x}{2e^{-x}} \\ \Rightarrow \frac{y+1}{1-y}=e^{2x} \\ \Rightarrow 2x=\log \left(\frac{y+1}{1-y}\right) \\ \Rightarrow x=\frac{1}{2}\log \left(\frac{y+1}{1-y}\right) \end{gathered}$$

Here, $\frac{1+y}{1-y}$ must be greater than $1$

Therefore, for positive values $y\in \left[-1,1\right]$

$f$ is not surjective.

Hence, the correct answer is Option (D).