MCQ
Let f : R → 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 → 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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