Class 11 RD Sharma Solutions - Chapter 1 Sets - Exercise 1.6 | Set 2

Exercise 1.6 | Set 2 in RD Sharma's Class 11 mathematics textbook continues the exploration of Mathematical Induction within the broader context of Set Theory. This set builds upon the foundational concepts introduced in Set 1, offering students a more advanced set of problems to tackle.

The exercises in this set are designed to challenge students' understanding and application of Mathematical Induction, pushing them to apply the technique to increasingly complex mathematical statements. By working through these problems, students will further develop their logical reasoning skills, enhance their ability to construct rigorous proofs and gain a deeper appreciation for the power and versatility of Mathematical Induction.

Class 11 RD Sharma Solutions - Chapter 1 Sets - Exercise 1.6

Question 8. Find sets A, B and C, such that A ∩ B and B ∩ C and A ∩ C are non-empty sets, and A ∩ B ∩ C = ϕ

Solution:

Let us consider the sets, 

A = {5, 6, 10}

B = {6, 8, 9}

C = {9, 10, 11}

Now, we have, 

A ∩ B = 6 ≠ ϕ

B ∩ C = 9 ≠ ϕ

A ∩ C = 10 ≠ ϕ

And, A ∩ B ∩ C = ϕ

Now, we have A ∩ B and B ∩ C and A ∩ C as non-empty sets, but A ∩ B ∩ C is empty set. 

Question 9. For any two sets A and B, prove that A ∩ B = ϕ => A ⊂ B'

Solution:

Let, a ∈ A => a ∉ B

Thus, 

A ∩ B  = ϕ

=> a ∈ B'

Thus, a ∈ A and a ∈ B' => A ⊂ B'

Question 10. Prove the following:

(i) A - B and A ∩ B are disjoint sets

(ii) B - A and A ∩ B are disjoint sets

(iii) A - B and B - A are disjoint sets

Solution:

(i) A - B and A ∩ B 

Let a  ∈ A - B => a ∈ A and a ∉ B => a ∉ A ∩ B

Therefore, A - B and A ∩ B are disjoint sets.

(ii) Let a ∈ B - A => a ∈ B and a ∉ A => a ∉ A ∩ B

Hence, B - A and A ∩ B are disjoint sets.

(iii) A - B and B - A,

A - B = x, x : x ∈ A and x ∉ B

A - B and B - A are disjoint sets. 

Question 11. Using properties of sets, show that for any two sets A and B,

(A ∪ B) ∩ (A ∩ B') = A

Solution: 

We have, 

LHS = A ∪ B ∩ A ∩ B'

Solving this, we get, 

= A ∪ B ∩ A ∪ A ∪ B ∩ B'

= A ∪ A ∪ B ∩ B'

Since, B ∩ B' = ∅

= A ∪ A ∩ B'

= A 

Therefore, LHS = RHS.

Question 12.

(i)Show that for any two sets A and B,

A' U B = U => A ⊂ B 

(ii) Show that for any two sets A and B,

B' ⊂ A' = U => A ⊂ B 

Solution:

(i) Let a ∈ A

= a ∈ U

= a ∈ A' ∪ B, because, U = A' ∪ B

= a ∈ B, because a ∉ A'

Hence, A ⊂ B

(ii) Let a ∈ A

= a ∉ A'

= a ∉ B', because, B' is a subset of A'

= a ∈ B

Hence, A ⊂ B

Question 13. Is it true that for any set A and B, P(A) ∪ P(B) = P(A ∪ B)? Justify your answer.

Solution: 

Result is False.

Proof:

Let X ∈ P(A) ∪ P(B)

= X ∈ P(A) or X ∈ P(B) 

= X ⊂ A or X ⊂ B

= X ⊂ A U B

= X ∈ P(A ∩ B)

Thus, P(A) ∪ P(B) ⊂ P(A ∪ B)

Also, Let us assume, 

X ∈ P(A ∪ B). But, X ∉ P(A) or X ∉ P(B) 

For instance, we have X = 1, 2, 3, 4 and A = 2, 5 and B = 1, 3, 4.

So, X ∉ P(A) ∪ P(B) 

Therefore, P(A ∪ B) doesn't necessarily have to be a subset of P(A) ∪ P(B).

Question 14. 

(i) Show that For any sets A and B,

A = (A ∩ B) ∩ (A - B)

(ii) Show that For any sets A and B,

A ∪ (B - A) = A ∪ B

Solution:

(i) We have,

RHS = (A ∩ B) ∪ (A - B)

= (A ∩ B) ∪ (A ∩ B)'

= (A ∩ B) ∪ (A ∩ A) ∩ (B ∪ B)'

= A ∩ (A ∪ B)' ∩ (B ∪ B)'

= A ∩ (A ∪ B)' ∩ U

= A ∩ (A ∪ B)'

= A

Therefore, RHS = LHS

(ii) We have,\bigcup\limits_{r=1}^{20} X_{r} = S

LHS = A ∪ (B - A)

= A ∪ (B ∩ A)'

= (A ∪ B) ∩ (A ∪ A)'

= (A ∪ B) ∩ U

LHS = A ∪ B = RHS

Question 15. Each set X, contains 5 elements and each set Y, contains 2 elements and  \bigcup\limits_{r=1}^{20} X_{r} = S = \bigcup\limits_{r=1}^{n} Y_{r}  each element of S belongs to exactly 10 of the Xr's and to exactly 4 of Yr's, then find the value of n.

Solution:

We have, Each set X contains 5 elements, and \bigcup\limits_{r=1}^{20} X_{r} = S

Therefore, n(S) = 20 x 5 = 100

But, we know, that each of the element of S belong to exactly 10 of the X_r's.

Therefore, n(S) = 100/10 = 10        -(1)

Also, Y contains 2 elements and

\bigcup\limits_{r=1}^{n} Y_{r} = S

Therefore, n(S) = n x 2 = 2n 

Each of the element of S belong to exactly 4 of the Yr's.

n(S) = 2n/4 = n/2         -(2)

From equation (1) and (2)

10 = n/2

n = 20.

Summary

Exercise 1.6 | Set 2 in RD Sharma's Class 11 mathematics textbook offers an advanced exploration of Mathematical Induction, building upon the foundational concepts established in Set 1. This set presents students with a diverse array of challenging problems that require sophisticated applications of inductive reasoning. By engaging with these exercises, students refine their skills in constructing complex mathematical proofs, deepen their understanding of algebraic and numeric patterns, and develop a more nuanced appreciation for the power of Mathematical Induction. The problems in this set cover a wide range of mathematical concepts, including series summations, divisibility properties, inequalities, and exponential relationships, providing a comprehensive training ground for students to hone their analytical and problem-solving abilities. Mastery of these advanced induction problems not only strengthens students' mathematical foundation but also prepares them for tackling more complex theoretical concepts and real-world applications in their future academic and professional endeavors.