If 1 < x < 2, then the value of ( x - 1 )^2 + ( 3 - x )^2 is?

Answer: Option B

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