Question

The range of the function $f\left(x\right)=\frac{x}{1+|x|},x\in R,$ is

• A. $R$
• B. $(1,-1)$
• C. $R-\left\{0\right\}$
• D. $[-1,1]$

Answer 1

Answer: (B)

Domain: Let $f:A\to B$ be a function, then the set $A$ is called as the domain of the function $f$.
Co-domain:
Let $f:A\to B$ be a function, then the set $B$ is called as the co-domain of the function $f$.
Range: Let $f:A\to B$, 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: $f(x)=\frac{|x|}{x},x\ne 0$ $\Rightarrow f(x)=\frac{|x|}{x}=\{\begin{array}{c}\frac{x}{x}=1,x>0\\ -\frac{x}{x}=-1,x<0\end{array}$
As we know that, if $f:A\to B$, 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 $\{1,-1\}$.