MCQ
Let $f : R \rightarrow R$ be a function defined by $\text{f(x)}=\frac{\text{e}^{|\text{x}|}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}.$ Then,
  • A
    $f$ is a bijection.
  • B
    $f$ is an injection only.
  • C
    $f$ is surjection on only.
  • $f$ is neither an injection nor a surjection.

Answer

Correct option: D.
$f$ is neither an injection nor a surjection.
$f : R \rightarrow R$
$\text{f(x)}=\frac{\text{e}^{|\text{x}|}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
For $x = -2$ and $-3 \in\text{R}$
$\text{f(-2)}=\frac{\text{e}^{|-2|}-\text{e}^2}{\text{e}^{-2}+\text{e}^2}$
$=\frac{\text{e}^2-\text{e}^2}{\text{e}^{-2}+\text{e}^2}$
$=0$
Hence, for different values of $x$ we are getting same values of $f(x)$
That means, the given function is many one.
Therefore, this function is not injective.
For $x < 0$
$f(x) = 0$
For $x > 0$
$\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
$=\frac{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{}e^{-\text{x}}}-\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
$=1-\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
The value of $\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$ is always positive.
Therefore, the value of $f(x)$ is always less than $1.$
Numbers more than $1$ are not included in the range but they are included in co$-$domain.
As the codomain is $R.$
$\therefore\ \text{Co-domain}\neq\text{Range}$
Hence, the given function is not onto.
Therefore, this function is not surjective.

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