# Aptitude Question ID : 93352

If $1 < x < 2$, then the value of $\sqrt{\left ( x-1 \right )^{2}}+\sqrt{\left ( x-3 \right )^{2}}$ is:
[A]2
[B]2x – 4
[C]1
[D]3

2
Since 1 < x < 2, we have x – 1 > 0 and x – 3 > 0 or 3 – x > 0
$\therefore \sqrt{\left ( x - 1 \right )^{2}} + \sqrt{\left ( x - 3 \right )^{2}}$
$= \sqrt{\left ( x - 1 \right )^{2}} + \sqrt{\left ( 3 - x \right )^{2}}$
$\left [ \because \left ( x-3 \right )^{2} = \left ( 3-x \right )^{2} \right ]$
$= x -1 +3-x = 2$
Hence option [A] is the right answer.