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, $f^{\prime }(x)=\frac{8x}{(x^{x^2}+e^{-x^2}{)}^2}$
and $f^{\prime }(x)=0$ has real values of $x$ so, $f(x)$ is many one.
Hence, $f(x)$ is neither one-one nor onto.