Functions Ques 17

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

(2017 Main)

(a) invertible

(b) injective but not surjective

(c) surjective but not injective

(d) neither injective nor surjective

Show Answer

Answer:

Correct Answer: 17.(c)

Solution: (c) We have, $f(x)=\frac{x}{1+x^2}$

$\therefore \quad f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}}{1+\frac{1}{x^2}}=\frac{x}{1+x^2}=f(x)$

$\therefore \quad f\left(\frac{1}{2}\right)=f(2)$ or $f\left(\frac{1}{3}\right)=f(3)$ and so on.

So, $f(x)$ is many-one function.

Again, let $\quad y=f(x) \Rightarrow y=\frac{x}{1+x^2}$

$\Rightarrow \quad y+x^2 y=x \Rightarrow y x^2-x+y=0$

As, $\quad x \in R$

$\therefore \quad(-1)^2-4(y)(y) \geq 0$

$\Rightarrow \quad 1-4 y^2 \geq 0$

$\Rightarrow \quad y \in\left[\frac{-1}{2}, \frac{1}{2}\right]$

$\therefore$ Range $=$ Codomain $=\left[\frac{-1}{2}, \frac{1}{2}\right]$

So, $f(x)$ is surjective.

Hence, $f(x)$ is surjective but not injective.



Table of Contents