Functions Ques 17
- 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
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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.
BETA
