Question

The function $f: R \rightarrow R$ defined by $f(x)=\frac{x}{\sqrt{1+x^{2}}}$ is

• A. surjective but not injective
• B. bijective
• C. injective but not surjective
• D. neither injective nor surjective

Answer 1

Answer: (C)

We have,
$f(x)=\frac{x}{\sqrt{1+x^{2}}}$
Injective
Let $x_{1}, x_{2} \in R$ such that
$\Rightarrow \frac{x_{1}}{\sqrt{1+x_{1}^{2}}}=\frac{x_{2}}{\sqrt{1+x_{2}^{2}}}$
$\Rightarrow \frac{x_{1}^{2}}{1+x_{1}^{2}}=\frac{x_{2}^{2}}{1+x_{2}^{2}} $
$\Rightarrow x_{1}^{2}=x_{2}^{2} $
$\Rightarrow x_{1}=x_{2}$
$\therefore f(x)$ is injective
Surjective
Let $ y=\frac{x}{\sqrt{1+x^{2}}} $
$\Rightarrow y^{2}\left(1+x^{2}\right)=x^{2} $
$\Rightarrow y^{2}+y^{2} x^{2}=x^{2}$
$\Rightarrow x^{2}\left(1-y^{2}\right)=y^{2} $
$\Rightarrow x=\sqrt{\frac{y^{2}}{1-y^{2}}} $
$\Rightarrow \frac{y^{2}}{1-y^{2}} \geq 0$

$\therefore y \in(-\infty,-1) \cup[0,1)$
$\therefore $ Range of $f(x)=(-\infty,-1) \cup[0,1)$
So, $f(x)$ is not surjective.