Thank you for reporting, we will resolve it shortly
Q.
The range of the function $f \left(x\right)=\frac{x}{1+\left|x\right|}, x\,\in\,R,$ is
AIEEEAIEEE 2012Relations and Functions
Solution:
Domain:
Let $f : A \rightarrow B$ be a function, then the set $A$ is called as the domain of the function $f$.
Co-domain:
Let $f : A \rightarrow B$ be a function, then the set $B$ is called as the co-domain of the function $f$.
Range:
Let $f : A \rightarrow B$, then the range of the function $f$ consists of those elements in $B$ which have at least one pre - image in $A$. It is denoted as $f(A)$.
i.e $f(A)=\{b \in B \mid f(a)=b$ for some $a \in A\}$
Note: Range is the subset of codomain of $f$.
Calculation:
Given: $f(x)=\frac{|x|}{x}, x \neq 0$
$
\Rightarrow f(x)=\frac{|x|}{x}=\left\{\begin{array}{c}
\frac{x}{x}=1, x>0 \\
-\frac{x}{x}=-1, x<0
\end{array}\right.
$
As we know that, if $f: A \rightarrow B$, then the range of the function $f$ consists of those elements in $B$ which have at least one pre - image in $A$. It is denoted as $f(A)$.
Hence, the range of the given function is $\{1,-1\}$.