The function $f: \boldsymbol{R} \rightarrow \boldsymbol{R}$ 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

Given that, $f(x)=\frac{x}{\sqrt{1+x^2}}$ For injective: Let $x_1, x_2 \in \boldsymbol{R}$ such that $f\left(x_1\right)=f\left(x_2\right)$ $\Rightarrow \frac{x_1}{\sqrt{1+x_1^2}}=\frac{x_2}{\sqrt{1+x_2^2}} \Rightarrow \frac{x_1^2}{1+x_1^2}=\frac{x_2^2}{1+x_2^2}$ $\Rightarrow x_1^2+x_1^2 x_2^2=x_2^2+x_1^2 x_2^2 \Rightarrow x_1^2=x_2^2$ $\Rightarrow x_1=x_2$ So, $f(x)$ is injective. For surjective: Let $y=\frac{x}{\sqrt{1+x^2}}$ $\Rightarrow y^2\left(1+x^2\right)=x^2 \Rightarrow y^2+y^2 x^2=x^2$ $\Rightarrow x^2\left(1-y^2\right)=y^2 \Rightarrow x=\sqrt{\frac{y^2}{1-y^2}}$ $\Rightarrow \frac{y^2}{1-y^2} \geq 0$ $\therefore y \in(-1,1)$ So, $f(x)$ is not surjective.

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