f : R→R is defined by f(x)
Question:

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

(a) one-one but not onto
(b) many-one but onto
(c) one-one and onto
(d) neither one-one nor onto

Solution:

(d) neither one-one nor onto

We have,

$f(x)=\frac{e^{x^{2}}-e^{-x^{2}}}{e^{x^{2}+e^{-x^{2}}}}$

Here, $-2,2 \in R$

Now, $2 \neq-2$

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

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.

Administrator

Leave a comment

Please enter comment.
Please enter your name.