Sets And Relations

Step 1: Classify the elements of S based on their remainder modulo 3.
This helps in analyzing the sum of elements in any subset. - Set of elements with remainder 0 mod 3: . Number of elements . - Set of elements with remainder 1 mod 3: . Number of elements . - Set of elements with remainder 2 mod 3: . Number of elements .
Step 2: Use a generating function (polynomial) approach.
Let be a complex cube root of unity. We know . Consider the product: The total number of subsets (including the empty set) whose sum is a multiple of 3 is given by . |
Step 3: Evaluate P(1), P( ), and P( ).
- . This is the total number of subsets.
- . We group terms based on their class mod 3.
- For , , so . The product is .
- For , , so .
The product is .
- For , , so .
The product is .
- will be the complex conjugate of , which is also 8. Or we can calculate it:
Step 4: Calculate the number of subsets with sum divisible by 3.
Let be the number of subsets (including ) whose sum is a multiple of 3.
This count includes the empty set (whose sum is 0). The number of non-empty subsets with sum divisible by 3 is .
Step 5: Calculate the final answer.
The total number of non-empty subsets of S is .
The number of subsets in T is the total number of non-empty subsets minus the number of non-empty subsets whose sum is a multiple of 3.

First, let's find the set A. If , then: So, .

If , then: which is always true. So, .

If , then: So, .

Combining these cases, we get .

Now, let's analyze set B: The summation is a geometric series: So, Since the base is 3 (greater than 1), we compare the exponents: If we let , where is an integer and , then: If , then: which is always true since .

So, .

Since is an integer, the range of will be .

✅ Comparing A and B: We have

R has "8" elements For

Step-by-Step Solution

Step 1: Understand the Set and Relation

The set has 7 elements, so . The relation is defined as . We need to find all pairs in .

Step 2: Find Elements of

Let’s compute for each :

• For , , so pair is .
• For , , so pair is .
• For , , so pair is .
• For , , so pair is .
• For , , so pair is .
• For , , so pair is .
• For , , so pair is .

So, . The number of pairs in is 7, so .

Step 3: Find (Elements to Make Reflexive)

A relation is reflexive if for every , the pair . The required pairs are .

Check which are in :

• , , , : Not in .
• , , : Already in .

We need to add 4 pairs: . So, .

Step 4: Find (Elements to Make Symmetric)

A relation is symmetric if for every , the pair . Check each pair:

• : Need . Not in . Add .
• : Need . Not in . Add .
• : Need . Not in . Add .
• : Need . Not in . Add .
• , , : Already symmetric.

We need to add 4 pairs: . So, .

Step 5: Compute

We have , , .

.

Final Answer:

Conclusion

This step-by-step solution shows how to determine the elements needed to make a relation reflexive and symmetric, a key concept in Class 12 math and JEE Main EXam Understanding relations is crucial for discrete math and CBSE exams. Practice more problems like this to master the topic!