Symmetric difference of two sets Let $A = \{1,2,3,4\}$ and $B = \{3,4,5,6\}$, and let $X = A\,\triangle\,B$, the symmetric difference of $A$ and $B$. Then
A. $X = \emptyset$
B. $X = \{1,2\}$
C. $X = \{3,4\}$
D. $X = \{1,2,5,6\}$
I want to confirm if the answer is D, is it correct? 
Thanks
 A: In order to avoid any confusion with edits, there are two common interpretations to what you were asking.
With $A=\{1,2,3,4\}$ and $B=\{3,4,5,6\}$
The symmetric difference of $B$ with the union of $A$ and $B$


*

*$(A\cup B)\triangle B = (\{1,2,3,4\}\cup \{3,4,5,6\})\triangle \{3,4,5,6\} = \{1,2,\color{red}{3},\color{red}{4},\color{red}{5},\color{red}{6}\}\triangle \{\color{red}{3},\color{red}{4},\color{red}{5},\color{red}{6}\} = \{1,2\}$


If this is what was intended to be asked, then no $d$ is incorrect.  The correct answer then would be $b$.
The symmetric difference of $A$ and $B$


*

*$A\triangle B = \{1,2,\color{red}{3},\color{red}{4}\}\triangle\{\color{red}{3},\color{red}{4},5,6\} = \{1,2,5,6\}$


If this is the intended question, then yes $d$ is correct.
Remember that to find the symmetric difference of two sets, you take the elements that are in exactly one of the sets.
(in general the symmetric difference of $n$ sets results in those elements that are in exactly an odd number of the sets)
A: D is indeed correct, as $A \cup B = \{1,2,3,4,5,6\}$, $A \cap B = \{3,4\}$ 
and their difference is set D. It's also the set of elements that occur exactly once among the elements of $A$ and $B$
