f : R → R is defined by f(x)= e^ x^2 -e^ -x^2 e^ x^2 +e^ -x^2 is: - Vidyadip

MCQ

f : R → R is defined by $\text{f(x)}=\frac{\text{e}^{\text{x}^2}-\text{e}^{-\text{x}^2}}{\text{e}^{\text{x}^2}+\text{e}^{-\text{x}^2}}$ is:

A
One-one but not onto.
B
Many-one but onto.
C
One-one and onto.
✓
Neither one-one nor onto.

✓

Answer

Correct option: D.

Neither one-one nor onto.

We have,
$\text{f(x)}=\frac{\text{e}^{\text{x}^2}-\text{e}^{-\text{x}^2}}{\text{e}^{\text{x}^2}+\text{e}^{-\text{x}^2}}$

Here, $-2,2\in\text{R}$

Now, $2\neq-2$

But, f(2) = f(-2)

Therefore, function is not one-one.

And,

The minimum value of the function is 0 and maximum value is 1.

That is range of the function is [0, 1] but the co-domain of the function is given R.

Therefore, function is not onto.

$\therefore$ function is neither one-one nor onto.

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