Q67 / 155 · WBJEE 2019 · Mathematics
Let $f: R \rightarrow R$ be defined by $f(x)=x^{2}-\frac{x^{2}}{1+x^{2}}$ 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
- Correct answerD$f$ is neither one-one nor onto
Solution
We have,
$$
\begin{aligned}
f(x) &=x^{2}-\frac{x^{2}}{1+x^{2}} \\
f(-x) &=(-x)^{2}-\frac{(-x)^{2}}{1+(-x)^{2}} \\
&=x^{2}-\frac{x^{2}}{1+x^{2}}=f(x) \\
\therefore f(-x)=f(x) &
\end{aligned}
$$
So, $f(x)$ is many one Again, $f(x)=x^{2}-\frac{x^{2}}{1+x^{2}}$
$$
=\frac{x^{2}+x^{4}-x^{2}}{1+x^{2}}=\frac{x^{4}}{1+x^{2}}
$$
$\therefore$ Range of $f(x)$ will be $[0, \infty)$ But, co-domain of $f(x)=R$
$\therefore$ Range $\neq$ co-domain. So, $f(x)$ is onto.
Hence, the correct answer is option: 'D'.