Question

The function $f:R\to [-\frac{1}{2},\frac{1}{2}]$ defined as $f\left(x\right)=\frac{x}{1+x^2},$ is

• A. invertible
• B. injective but not surjective
• C. surjective but not injective
• D. neither injective nor surjective

Answer 1

The correct option is C surjective but not injective

$f\left(x\right)=\frac{x}{1+x^2}$
Let $f\left(x_1\right)=f\left(x_2\right),x_1,x_2\in R$
$\Rightarrow \frac{x_1}{1+x_1^2}=\frac{x_2}{1+x_2^2}$
$\Rightarrow \frac{x_1}{1+x_1^2}-\frac{x_2}{1+x_2^2}=0$
$\Rightarrow \frac{(x_1-x_2)(1-x_1{x}_2)}{(1+x_1^2)(1+x_1^2)}=0$
$\Rightarrow x_1=x_2,x_1=\frac{1}{x_2}$
Thus, $f$ is not injective.

Let $y=\frac{x}{1+x^2}$
$\Rightarrow y{x}^2-x+y=0$
$\Rightarrow x=\frac{1\pm \sqrt{1-4y^2}}{2y}$
$\because x\in R,1-4y^2\ge 0\&y\ne 0$
$\Rightarrow y\in [-\frac{1}{2},\frac{1}{2}]-\left\{0\right\}$
Also, for $x=0,y=0$
$\therefore R_f=[-\frac{1}{2},\frac{1}{2}]=\text{Co-domain of}f$

Alternatively, put $x=\tan \theta ,\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$
Then $y=\frac{1}{2}\sin 2\theta$
$-1\le \sin 2\theta \le 1$
$\Rightarrow -\frac{1}{2}\le y\le \frac{1}{2}$
Thus, $f$ is surjective.