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

AP EAMCETAP EAMCET 2020

Solution:

$f(x)=\frac{x}{\sqrt{1+x^{2}}}$
$f\left(x_{1}\right)=f(x)_{2} \Rightarrow \frac{x_{1}}{\sqrt{1+x_{1}^{2}}}=\frac{x_{2}}{\sqrt{1+x_{2}^{2}}}$
$\Rightarrow x_{1}^{2}\left(1+x_{2}^{2}\right)=x_{2}^{2}\left(1+x_{1}^{2}\right)$
$\Rightarrow x_{1}^{2}=x_{2}^{2}$
$x_{1}=x_{2}$
$\therefore F$ is injective in nature
Also, Let $f(x)=y=\frac{x}{\sqrt{1+x^{2}}}$
$\Rightarrow y^{2}\left(1+x^{2}\right)=x^{2}$
$\Rightarrow \frac{y^{2}}{1-y^{2}}=x^{2}$
$\Rightarrow x=\frac{y}{\sqrt{1-y^{2}}}$
As $y^{2} \leq 1 \Rightarrow -1 \leq|y| \leq 1$
So, $f$ is non surjective.