If 1 < x < 2, then the value of $$\sqrt {{{\left( {x - 1} \right)}^2}} {\text{ + }}\sqrt {{{\left( {3 - x} \right)}^2}} {\text{ is?}}$$

A. 1

B. 2

C. 3

D. 2x - 4

Answer: Option B

Solution(By Examveda Team)

$$\eqalign{ & 1 < x < 2 \cr & \sqrt {{{\left( {x - 1} \right)}^2}} {\text{ + }}\sqrt {{{\left( {3 - x} \right)}^2}} \cr & \left( {{\text{Square root cancel with square}}} \right) \cr & \therefore x - 1 + 3 - x \cr & = 2 \cr} $$