3 votes 3 votes Let $A$ and $B$ be sets in a finite universal set $U$. Given the following : $|A-B|,|A \oplus B|,|A|+|B|$, and $|A \cup B|$ Which of the following is in order of increasing size ? $|A-B|<|A \oplus B|<|A|+|B|<|A \cup B|$ $|A \oplus B|<|A-B|<|A \cup B|<|A|+|B|$ $|A \oplus B|<|A|+|B|<|A-B|<|A \cup B|$ $|A-B|<|A \oplus B|<|A \cup B|<|A|+|B|$ Set Theory & Algebra goclasses2025_cs_wq5 goclasses discrete-mathematics set-theory&algebra set-theory 2-marks + – GO Classes asked Apr 10 • edited Apr 10 by shadymademe GO Classes 327 views answer comment Share Follow See 1 comment See all 1 1 comment reply Deepak Poonia commented Apr 11 i edited by Deepak Poonia Apr 11 reply Follow Share Detailed Video Solution: https://youtu.be/CND7s_mwqDw Detailed Video Solution of Whole Quiz: https://www.youtube.com/watch?v=v5v-AfpwicQ&list=PLIPZ2_p3RNHige8KFyCc4uWDx2uXEq1f0&index=6 1 votes 1 votes Please log in or register to add a comment.
2 votes 2 votes Take example, A={1,2,3} and B={2,3,4} |A−B| = |{1}| = 1 $\left | A\oplus B \right |$ = |{1,4}| = 2 $\left | A\cup B \right |$ = |{1,2,3,4}| = 4. |A| + |B| = 3+3 = 6 Correct Option: D https_guru answered Apr 10 https_guru comment Share Follow See all 5 Comments See all 5 5 Comments reply Kratarth Khare commented Apr 11 reply Follow Share Lets say A = {1,2,3,4,5,6} and B = {1,2,3}. In this case - |A| + |B| = 9 |AUB| = 6 |A-B| = 3 |A sym. diff. B| = 3 We can see |A-B| = |A sym. diff. B| in this case. That means the ordering in Option D isn't always correct, right? Or Since the que. is talking about increasing order so we can neglect the equal case/sign? 0 votes 0 votes https_guru commented Apr 11 reply Follow Share The question is not asking about the "always" case. So, we have to choose the best possible option. 0 votes 0 votes Deepak Poonia commented Apr 11 reply Follow Share No option is correct, But Option $D$ is the closest option & best choice among all. Detailed Video Solution: https://youtu.be/CND7s_mwqDw For correct answer, in option D we must have $\leq$ instead of $<.$ 1 votes 1 votes Srken commented May 3 reply Follow Share in the solution video the explanation with proof was so good 0 votes 0 votes theabhi100 commented May 12 reply Follow Share nice 0 votes 0 votes Please log in or register to add a comment.
