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

The correct option is B f(x) is neither one-one nor onto

$\text{solution:}f\left(x\right)=\frac{e^{x^2}-e^{-x^2}}{e^{x^2}+e^{-x^2}}=\frac{{\left(e^{x^2}\right)}^2-1}{{\left(e^{x^2}\right)}^2+1}$

$f^{\prime }\left(x\right)=\frac{\{{\left(e^{x^2}\right)}^2+1\}\left\{4x\left(e^{x^2}\right)\right\}-\{{\left(e^{x^2}\right)}^2-1\}\left\{4x\left(e^{x^2}\right)\right\}}{{({\left(e^{x^2}\right)}^2+1)}^2}$

$\text{Since}{f}^{\prime }\left(x\right)\text{is nither positive, nor negative}$

$\text{for all}x\text{so}f\left(x\right)\text{is not one - one function}$

$f\left(x\right)=\frac{{\left(e^{x^2}\right)}^2-1}{{\left(e^{x^2}\right)}^2+1}$

Range of $f\left(x\right)$ is $[0,1)$

since Range of $f\left(x\right)$ is not equal to co-domain

of $f\left(x\right)$ so $f\left(x\right)$ is not onto function.

Answer: option: (B)