Mathematics - Functions Question with Solution | TestHub

Given that, $f(x)=\frac{x}{\sqrt{1+x^2}}$ For injective: Let for any x 1 , x2 belongd to 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}\ge 0$$\therefore y\in (-1,1)$ So, $f(x)$ is not surjective.