259 views

Let $f\left ( x \right )=2x-5$ and $g\left ( x \right )=7-2x.$ Then $\mid f\left ( x \right ) + g\left ( x \right ) \mid=\mid f(x) \mid + \mid g\left ( x \right ) \mid$ if and only if

1. $5/2<x<7/2$
2. $x\leq 5/2$ or $x\geq 7/2$
3. $x< 5/2$ or $x\geq 7/2$
4. $5/2\leq x\leq 7/2$

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 $:\text{D}$
10.3k points

1 vote
1
201 views
1 vote