Question

The function $f:R\to R$ 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)

$f(x)=\frac{x}{\sqrt{1+x^2}}$
$f\left(x_1\right)=f(x{)}_2\Rightarrow \frac{x_1}{\sqrt{1+x_1^2}}=\frac{x_2}{\sqrt{1+x_2^2}}$
$\Rightarrow x_1^2\left(1+x_2^2\right)=x_2^2\left(1+x_1^2\right)$
$\Rightarrow x_1^2=x_2^2$
$x_1=x_2$
$\therefore F$ 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 $y^2\le 1\Rightarrow -1\le |y|\le 1$
So, $f$ is non surjective.