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)

We have,
$f(x)=\frac{x}{\sqrt{1+x^2}}$
Injective
Let $x_1,x_2\in R$ such that
$\Rightarrow \frac{x_1}{\sqrt{1+x_1^2}}=\frac{x_2}{\sqrt{1+x_2^2}}$
$\Rightarrow \frac{x_1^2}{1+x_1^2}=\frac{x_2^2}{1+x_2^2}$
$\Rightarrow x_1^2=x_2^2$
$\Rightarrow x_1=x_2$
$\therefore f(x)$ is injective
Surjective
Let $y=\frac{x}{\sqrt{1+x^2}}$
$\Rightarrow y^2\left(1+x^2\right)=x^2$
$\Rightarrow y^2+y^2{x}^2=x^2$
$\Rightarrow x^2\left(1-y^2\right)=y^2$
$\Rightarrow x=\sqrt{\frac{y^2}{1-y^2}}$
$\Rightarrow \frac{y^2}{1-y^2}\ge 0$

$\therefore y\in (-\infty ,-1)\cup [0,1)$
$\therefore$ Range of $f(x)=(-\infty ,-1)\cup [0,1)$
So, $f(x)$ is not surjective.