Given that, $f(x)=2x−5$, and $g(x)=7−2x$.
Now, $|f(x)+g(x)|=|f(x)|+|g(x)| $
$\textbf{Case 1:}\; f(x)\geq0$ and $g(x)\geq 0$
$\Rightarrow |2x−5+7−2x|=|2x−5|+|7−2x|$
$\Rightarrow 2=2x−5+7−2x$
$\Rightarrow \boxed{2 = 2\;\text{(True)}}$
So, $2x−5\geq0$ and $7−2x\geq0$
$\Rightarrow x\geq\frac{5}{2}$ and $x\leq\frac{7}{2}$
$\Rightarrow \boxed{\frac{5}{2}\leq x\leq \frac{7}{2}}$
$\textbf{Case 2:}\;f(x)\geq0$ and $g(x)\leq0$
$\Rightarrow |2x−5+7−2x| = |2x−5|+|7−2x|$
$\Rightarrow 2=2x−5−(7−2x)$
$\Rightarrow 2=2x−5−7+2x$
$\Rightarrow \boxed{2=4x−12\;\text{(False)}}$
This is an invalid case.
$\textbf{Case 3:}\;f(x)\leq0$ and $g(x)\geq0$
$\Rightarrow |2x−5+7−2x|=|2x−5|+|7−2x|$
$\Rightarrow 2=−(2x−5)+7−2x$
$\Rightarrow 2=−2x+5+7−2x$
$\Rightarrow \boxed{2=−4x+12\;\text{(False)}}$
This is an invalid case.
$\textbf{Case 4:}\;f(x)\leq0$ and $g(x)\leq0$
$\Rightarrow |2x−5+7−2x|=|2x−5|+|7−2x|$
$\Rightarrow 2=−(2x−5)−(7−2x)$
$\Rightarrow 2=−2x+5−7+2x$
$\Rightarrow \boxed{2=−2\;\text{(False)}}$
$\therefore$ $\boxed{\frac{5}{2}\leq x\leq \frac{7}{2}} \quad [\because \text{From case 1}]$
Correct Answer $:\text{D}$