Question

If $|2x-3|+|x-1|=|x-2|$, then $x\in$

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

Answer 1

The correct option is B $[1,\frac{3}{2}]$

$|2x-3|+|x-1|=|(2x-3)-(x-1)|$
This is of form $|a|+|b|=|a-b|$
$\Rightarrow ab\le 0\left(a\text{and}b\text{should have opposite sign}\right)$
$\Rightarrow (2x-3)(x-1)\le 0$
$\Rightarrow 1\le x\le \frac{3}{2}$