Question

Let $f:R\to R$ be defined by $f\left(x\right)=x^2-\frac{x^2}{1+x^2}$ for all $x\in R$. Then

• A. $f$ is one-one but not onto mapping
• B. $f$ is onto but not one-one mapping
• C. $f$ is both one-one and onto
• D. $f$ is neither one-one nor onto

Answer 1

The correct option is D $f$ is neither one-one nor onto

$f\left(x\right)=\frac{x^4}{1+x^2}$
$f(-x)=\frac{(-x{)}^4}{1+(-x{)}^2}=f\left(x\right)$
$\Rightarrow f\left(x\right)$ is an even function, hence it is many one.

Since, $f\left(x\right)\ge 0\forall x\in R$, so $f\left(x\right)$ does not take any negative value.
So, range of $f\left(x\right)$ is not $R$.
Hence, it is not onto.
$\Rightarrow f\left(x\right)$ is neither one-one nor onto.