Question

Let f :R $\to$ R be a function defined by $f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}}$. Then

• 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

Answer: (D)

We have, If x < 0 |x| = -x
$\therefore f\left(x\right) = \frac{e^{-x} - e^{-x}}{e^{x} + e^{-x}} = 0 $
$\therefore f\left(x\right)= 0 \forall x < 0 $
$\therefore f\left(x\right)$ is not one-one
Next if $ x \ge0, \left|x\right| = x$
$ \therefore f\left(x\right) = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}}$
Let $ y = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} \Rightarrow y= \frac{e^{2x}-1}{e^{2x} + 1} $
$\therefore e^{2x } = \frac{1+y}{1-y}$
For $ x \ge0 , e^{2x} \ge1$
$ \therefore \frac{1+y}{1-y} \ge1 \Rightarrow \frac{2y}{1-y} \ge0$
$ \Rightarrow y\left(y-1\right)\le0, y \ne1 \Rightarrow 0 \le y < 1 $
$\therefore $ Range of f(x) = [0 1)
$\therefore $ f(x) is not onto