Question

$f\left(x\right)=\frac{1}{x+|x-1|},g\left(x\right)=\frac{1}{x+|x+1|}$

• A. Domain of $f$ is $R$ but domain of $g$ is not $R$
• B. Domain of $g$ is $R$ but domain of $f$ is not $R$
• C. Domain of both $f$ and $g$ are $R$.
• D. Domain of neither $f$ nor $g$ is $R$

Answer 1

The correct option is A

Domain of $f$ is $R$ but domain of $g$ is not $R$

$f\left(x\right)=\frac{1}{x+|x-1|}x+|x-1|=\{\begin{array}{cc}2x-1,& x\ge 1\\ 1,& x<1\end{array}$
$Forx\ge 1,x+|x-1|=0\Rightarrow 2x-1=0\Rightarrow x=\frac{1}{2}$
Since $\frac{1}{2}<1$, this is not possible.
$Forx<1,x+|x-1|=0\Rightarrow 1=0$ - not possible
$\Rightarrow x+|x-1|\ne 0$
Hence, Domain of $f$ is $R$.
$g\left(x\right)=\frac{1}{x+|x+1|}x+|x+1|=\{\begin{array}{cc}2x+1,& x\ge -1\\ -1,& x<1\end{array}Forx\ge -1,x+|x+1|=2x+1=0\Rightarrow x=\frac{-1}{2}\frac{-1}{2}>-1-possible$
g(x) is not defined at $x=\frac{-1}{2}$
Hence, domain of $g$ is not $R$.