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

• A

  surjective but not injective
• B

  bijective
• Correct answer
  C

  injective but not surjective
• D

  neither injective nor surjective

Solution

$f(x)=\frac{x}{\sqrt{1+x^2}}$ Injective Let $x_1, x_2 \in R$ such that $\begin{array}{ll} & f\left(x_1\right)=f\left(x_2\right) \\ \Rightarrow & \frac{x_1}{\sqrt{1+x_1^2}}=\frac{x_2}{\sqrt{1+x_2^2}} \\ \Rightarrow & \frac{x_1^2}{1+x_1^2}=\frac{x_2^2}{1+x_2^2} \\ \Rightarrow & x_1^2+x_1^2 x_2^2=x_2^2+x_1^2 x_2^2 \\ \Rightarrow & \quad x_1^2=x_2^2 \\ \Rightarrow & \quad x_1=x_2 \\ \therefore & \text { Surjective Let } y=\frac{x}{\sqrt{1+x^2}}\end{array}$ $\begin{array}{ll}\Rightarrow & y^2\left(1+x^2\right)=x^2 \Rightarrow y^2+y^2 x^2=x^2 \\ \Rightarrow & x^2\left(1-y^2\right)=y^2 \Rightarrow x=\sqrt{\frac{y^2}{1-y^2}} \\ \Rightarrow & \frac{y^2}{1-y^2} \geq 0\end{array}$
$\therefore y \in(-\infty,-1) \cup[0,1)$ $\therefore$ Range of $f(x)=(-\infty,-1) \cup[0,1)$ So, $f(x)$ is not surjective.

Hence, the correct answer is option: 'C'.