For any two sets A and B, \[A=B\] is equivalent to
(A) \[A-B=B-A\]
(B) \[A\cup B=A\cap B\]
(C) \[A\cup C=B\cup C\] and \[A\cap C=B\cap C\] for any set C
(D) \[A\cap B=\phi \]
Answer
616.2k+ views
Hint: It is given that \[A=B\] . First of all, check option (A). The LHS is equal to \[\left( A-B \right)\] . Now, replace A by B and B by A in the equation \[\left( A-B \right)\] and check whether the LHS is equal to RHS or not. Now, check option (B). The LHS and RHS is equal to \[\left( A\cup B \right)\] and \[\left( A\cap B \right)\] respectively. Now, replace A by B in the LHS and RHS. We know the property that the intersection and union of a set with the same set is the set itself. Now, use this property and check whether the LHS and RHS hold the same result or not. Then, check option (C). It has two parts. The first part is \[A\cup C=B\cup C\] and the second part is \[A\cap C=B\cap C\] . Replace A by B in the LHS of the first part and check whether it is equal to RHS or not. Similarly, replace A by B in the LHS of the second part and check whether it is equal to RHS or not. Now, check option (D). In LHS, we have \[A\cap B\] and in RHS we have, \[\phi \] . Replace A by B in the LHS. We know the property that the intersection of a set with the same set is the set itself. Now, check whether the LHS and RHS hold the same result or not.
Complete step-by-step answer:
According to the question, we have two sets A and B,
\[A=B\] ……………………..(1)
We have four options A, B, C, and D,
In option (A), we have \[\left( A-B \right)\] as LHS and \[\left( B-A \right)\] as RHS.
LHS= \[\left( A-B \right)\] ………………..(2)
RHS = \[\left( B-A \right)\] ……………………(3)
From equation (1), we have \[A=B\] .
Now, replacing A by B and B by A in equation (2), we get
LHS = \[A-B=B-A\] ……………………..(4)
From equation (3), we have RHS equal to \[\left( B-A \right)\] .
So, LHS=RHS.
Hence option (A) is the correct one.
In option (B), we have \[\left( A\cup B \right)\] as LHS and \[\left( A\cap B \right)\] as RHS.
LHS= \[\left( A\cup B \right)\] ………………..(5)
RHS = \[\left( A\cap B \right)\] ……………………(6)
From equation (1), we have \[A=B\] .
Now, replacing A by B in equation (5) and equation (6), we get
LHS = \[\left( A\cup B \right)=\left( B\cup B \right)\] ……………………..(7)
RHS = \[\left( A\cap B \right)=\left( B\cap B \right)\] …………………………..(8)
We know the property that the intersection and union of a set with the same set is the set itself.
Using this property in equation (7) and equation (8), we get
LHS = \[\left( B\cup B \right)=B\] ………………………..(9)
RHS = \[\left( B\cap B \right)=B\] ……………………………(10)
So, LHS=RHS.
Hence option (B) is the correct one.
In option (C), we have two parts. One is \[A\cup C=B\cup C\] and the other part is \[A\cap C=B\cap C\] .
In the first part, we have \[A\cup C=B\cup C\] .
LHS = \[\left( A\cup C \right)\] …………………………..(11)
RHS = \[\left( B\cup C \right)\] ……………………………(12)
From equation (1), we have \[A=B\] .
Now, replacing A by B in equation (11), we get
LHS = \[\left( A\cup C \right)=\left( B\cup C \right)\] ……………………(13)
From equation (12), we have RHS equal to \[\left( B\cup C \right)\] .
So, LHS = RHS.
Therefore, the first part is correct.
In the second part, we have \[A\cap C=B\cap C\] .
LHS = \[\left( A\cap C \right)\] …………………………..(14)
RHS = \[\left( B\cap C \right)\] ……………………………(15)
From equation (1), we have \[A=B\] .
Now, replacing A by B in equation (14), we get
LHS = \[\left( A\cap C \right)=\left( B\cap C \right)\] ……………………(16)
From equation (15), we have RHS equal to \[\left( B\cap C \right)\] .
So, LHS = RHS.
Therefore, the second part is also correct.
Hence, the option is also correct.
In option (D), we have \[A\cap B\] as LHS and \[\phi \] as RHS.
LHS = \[\left( A\cap B \right)\] ………………………………..(17)
RHS = \[\phi \] …………………………..(18)
From equation (1), we have \[A=B\] .
Now, replacing A by B in equation (17), we get
LHS = \[\left( A\cap B \right)=\left( A\cap A \right)\] ……………………………(19)
We know the property that the intersection of a set with the same set is the set itself.
Using this property in equation (19), we get
LHS = \[\left( A\cap A \right)=A\] ………………………………(20)
From equation (18), we have the RHS equal to \[\phi \] .
So, \[LHS\ne RHS\] .
Therefore, option (D) is not correct.
Hence, the correct options are (A), (B), and (C).
Note: The best way to solve this type of question is to check each and every option given. If the LHS and RHS of an option holds the same result then that option is correct. Here, one might make a mistake and take the union and intersection of the same set as \[\phi \] . This is wrong because the union and intersection of the same set is the set itself.
Complete step-by-step answer:
According to the question, we have two sets A and B,
\[A=B\] ……………………..(1)
We have four options A, B, C, and D,
In option (A), we have \[\left( A-B \right)\] as LHS and \[\left( B-A \right)\] as RHS.
LHS= \[\left( A-B \right)\] ………………..(2)
RHS = \[\left( B-A \right)\] ……………………(3)
From equation (1), we have \[A=B\] .
Now, replacing A by B and B by A in equation (2), we get
LHS = \[A-B=B-A\] ……………………..(4)
From equation (3), we have RHS equal to \[\left( B-A \right)\] .
So, LHS=RHS.
Hence option (A) is the correct one.
In option (B), we have \[\left( A\cup B \right)\] as LHS and \[\left( A\cap B \right)\] as RHS.
LHS= \[\left( A\cup B \right)\] ………………..(5)
RHS = \[\left( A\cap B \right)\] ……………………(6)
From equation (1), we have \[A=B\] .
Now, replacing A by B in equation (5) and equation (6), we get
LHS = \[\left( A\cup B \right)=\left( B\cup B \right)\] ……………………..(7)
RHS = \[\left( A\cap B \right)=\left( B\cap B \right)\] …………………………..(8)
We know the property that the intersection and union of a set with the same set is the set itself.
Using this property in equation (7) and equation (8), we get
LHS = \[\left( B\cup B \right)=B\] ………………………..(9)
RHS = \[\left( B\cap B \right)=B\] ……………………………(10)
So, LHS=RHS.
Hence option (B) is the correct one.
In option (C), we have two parts. One is \[A\cup C=B\cup C\] and the other part is \[A\cap C=B\cap C\] .
In the first part, we have \[A\cup C=B\cup C\] .
LHS = \[\left( A\cup C \right)\] …………………………..(11)
RHS = \[\left( B\cup C \right)\] ……………………………(12)
From equation (1), we have \[A=B\] .
Now, replacing A by B in equation (11), we get
LHS = \[\left( A\cup C \right)=\left( B\cup C \right)\] ……………………(13)
From equation (12), we have RHS equal to \[\left( B\cup C \right)\] .
So, LHS = RHS.
Therefore, the first part is correct.
In the second part, we have \[A\cap C=B\cap C\] .
LHS = \[\left( A\cap C \right)\] …………………………..(14)
RHS = \[\left( B\cap C \right)\] ……………………………(15)
From equation (1), we have \[A=B\] .
Now, replacing A by B in equation (14), we get
LHS = \[\left( A\cap C \right)=\left( B\cap C \right)\] ……………………(16)
From equation (15), we have RHS equal to \[\left( B\cap C \right)\] .
So, LHS = RHS.
Therefore, the second part is also correct.
Hence, the option is also correct.
In option (D), we have \[A\cap B\] as LHS and \[\phi \] as RHS.
LHS = \[\left( A\cap B \right)\] ………………………………..(17)
RHS = \[\phi \] …………………………..(18)
From equation (1), we have \[A=B\] .
Now, replacing A by B in equation (17), we get
LHS = \[\left( A\cap B \right)=\left( A\cap A \right)\] ……………………………(19)
We know the property that the intersection of a set with the same set is the set itself.
Using this property in equation (19), we get
LHS = \[\left( A\cap A \right)=A\] ………………………………(20)
From equation (18), we have the RHS equal to \[\phi \] .
So, \[LHS\ne RHS\] .
Therefore, option (D) is not correct.
Hence, the correct options are (A), (B), and (C).
Note: The best way to solve this type of question is to check each and every option given. If the LHS and RHS of an option holds the same result then that option is correct. Here, one might make a mistake and take the union and intersection of the same set as \[\phi \] . This is wrong because the union and intersection of the same set is the set itself.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Which among the following are examples of coming together class 11 social science CBSE

