Question

If $|4x-5|+|6x-12|=|2x-7|$, then $x$ belongs to

• A. $[\frac{5}{4},2]$
• B. $(-\infty ,2]$
• C. $(2,\infty )$
• D. $(-1,3]$

Answer 1

The correct option is A $[\frac{5}{4},2]$

$|4x-5|+|6x-12|=|2x-7|$
We know that $|a-b|=|a|+|b|$, then $ab\le 0$
Here, $2x-7=(6x-12)-(4x-5)$
So,
$(6x-12)(4x-5)\le 0\Rightarrow (x-2)(x-\frac{5}{4})\le 0$
$\therefore x\in [\frac{5}{4},2]$