Question

Let $f : R \to R$ defined by $f\left(x\right) = \frac{e^{x^2} - e^{-x^2}}{e^{x^2}+ e^{-x^2}} $ then

• A. $f(x)$ is one-one but not onto
• B. $f(x)$ is neither one-one nor onto
• C. $f(x)$ is many one but onto
• D. $f(x)$ is one-one and onto

Answer 1

Answer: (B)

$f\left(x\right) = \frac{e^{x^2} - e^{-x^2}}{e^{x^2}+ e^{-x^2}} = \frac{p(x)}{q(x)}$ (say)
Now $q(x) > 0$ and $p(x) = e^{x^2} - e^{-x^2}$
$= 2 \{ x^2 + \frac{x^2}{3!} + ... \} > 0$
[Using defintion of exponential function i.e.
$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + ....]$
Hence, $f (x) > 0 \,\forall \,x \in R $
$\therefore f(x)$ is into function.
$(\because y $ can not have pre-images for $-ve$ values)
Again, $f'(x) = \frac {8x}{(x^{x^2} + e^{-x^2})^2}$
and $f'(x) = 0$ has real values of $x$ so, $f(x)$ is many one.
Hence, $f(x)$ is neither one-one nor onto.