Illustration: 1. Decide, among the following sets, which sets are subsets of one another:
$A = \{x : x \in \text{ R } \text{ and } x \text{ satisfy } x^2 - 8x + 12 = 0\}$
$B = \{2, 4, 6\}, \quad C = \{2, 4, 6, 8, \dots\}, \quad D = \{6\}$
Solution: $x^2 - 8x + 12 = 0$
$\implies x^2 - 6x - 2x + 12 = 0$
$\implies x(x - 6) - 2(x - 6) = 0$
$\implies (x - 2)(x - 6) = 0$
$\implies x = 2, 6$
$A = \{2, 6\}, \quad B = \{2, 4, 6\}, \quad C = \{2, 4, 6, 8, \dots\}, \quad D = \{6\}$
अतः $A \subset B, \quad A \subset C, \quad B \subset C, \quad D \subset A, \quad D \subset B, \quad D \subset C$
Illustration: 2. In each of the following, determine whether the statement is true or false. If it is true, prove it, if it is false, give an example.
Illustration: 2 (i) If $x \in A$ and $A \in B$ , then $x \in B$
Solution: False
Because a set cannot be a member of another set in this manner; a set can be a subset, superset, or equivalent set. For example:
$C = \{1, 2, 3\}$ and $D = \{1, 2, 3, 4\}$ , then $C \subset D$ .
Illustration: 2 (ii) If $A \subset B$ and $B \in C$ , then $A \subset C$ .
Solution: False
B cannot be an element of another set, i.e., $B \notin C$ .
Illustration: 2 (iii) If $A \subset B$ and $B \subset C$ , then $A \subset C$ .
Solution: True
$x \in A \implies x \in B \quad [\because A \subset B]$
and $x \in B \implies x \in C \quad [\because B \subset C]$
$\therefore A \subset C$
Illustration: 2 (iv). If $A \not\subset B$ and $B \not\subset C$ , then $A \not\subset C$ .
Solution: False
Example: $A = \{2, 4, 6\}, \quad B = \{a, b, c\}, \quad C = \{4, 6\}$
$\therefore A \not\subset B$ and $B \not\subset C$ , but $A \not\subset C$ (A can be or not subset of C).
Illustration: 2 (v). If $x \in A$ and $A \not\subset B$ , then $x \in B$ .
Solution: False
For example: $A = \{8, 9, 10\}, \quad B = \{x, y, z\}$
$A \not\subset B$ . $\therefore$ Any element of A cannot be an element of B.
Illustration: 2 (vi) If $A \subset B$ and $x \notin B$ , then $x \notin A$ .
Solution: True
$x \notin B \implies x \notin A \quad [\because A \subset B]$
Illustration: 3. Let A, B and C be the sets such that $A \cup B = A \cup C$ and $A \cap B = A \cap C$ . Show that B = C.
Solution: $A \cup B = A \cup C$
$\implies (A \cup B) \cap C = (A \cup C) \cap C$
$\implies (A \cap C) \cup (B \cap C) = C \quad [\because (A \cup C) \cap C = C]$
$\implies (A \cap B) \cup (B \cap C) = C \quad \dots \text{(1)}$
$[\because A \cap C = A \cap B]$
Again $A \cup B = A \cup C$
$(A \cup B) \cap B = (A \cup C) \cap B$
$\implies B = (A \cap B) \cup (C \cap B)$
$[\because (A \cup B) \cap B = B]$
$\implies B = (A \cap B) \cup (B \cap C) \quad \dots \text{(2)}$
From (1) and (2), we get
$B = C$
Illustration: 4. Show that the following four conditions are equivalent:
(i) $A \subset B$
(ii) $A - B = \phi$
(iii) $A \cup B = B$
(iv) $A \cap B = A$
Solution: Let $A \subset B$ and $x \in A \implies x \in B$
$\implies x \in A \land x \in B \quad [\because A \subset B]$
$\implies x \in A \cap B$
$\therefore A \subset A \cap B \quad \dots \text{(1)}$
Let $x \in A \cap B$
Then $x \in A \cap B \implies x \in A \land x \in B$
$\implies x \in A$
$\therefore A \cap B \subset A \quad \dots \text{(2)}$
From (1) and (2), we get:
$A \cap B = A$
(ii) Let $A \cap B = A$
$x \in A \implies x \in A \cap B \quad [\because A = A \cap B]$
$\implies x \in A \land x \in B$
$\implies x \in B$
$A \subset B$
$A \subset B \implies A \cap B = A$
Again $x \in A - B \implies x \in A \land x \notin B$
$\implies x \in B \land x \notin B \quad [\because x \in A \implies x \in B \text{ because } A \subset B]$
$\implies x \in \phi$
$\therefore A - B = \phi \quad \dots \text{(3)}$
$\phi \subset A - B \quad [\because \phi \text{ is subset of every set}] \quad \dots \text{(4)}$
From (3) and (4), we get:
$A - B = \phi$
Similarly $A \subset B \implies A - B = \phi \quad \dots \text{(5)}$
$A - B = \phi \implies A - B \text{ has not any element}$
$\therefore$ A has not any element which is not in B
$\implies A \text{ has each element which is in } B$
$\implies A \subset B$
$\therefore A - B = \phi \implies A - B \dots \text{(6)}$
From (5) and (6), we get:
$A \subset B \iff A - B = \phi$
(iii) Let $A \subset B$ then
$x \in A \cup B \implies x \in A \lor x \in B$
$\implies x \in B \quad [\because A \subset B]$
$\therefore A \cup B \subset B \quad \dots \text{(1)}$
Let $x \in B$ then
$x \in B \implies x \in A \cup B \quad [\text{By def.}]$
$\therefore B \subset A \cup B \quad \dots \text{(2)}$
From (1) and (2), we get:
$A \cup B = B$
Again $A \cup B = B$ then
$x \in A \implies x \in A \cup B \quad [\because A \subset A \cup B]$
$\implies x \in B \quad [\because A \cup B = B]$
$\therefore A \subset B$
Similarly
$A \subset B \iff A \cup B = B$
Illustration: 5. Show that for any sets A and B, if $A \subset B$ , then $C - B \subset C - A$ .
Solution: If $A \subset B$
$C - B = \{x : x \in C \land x \notin B\}$
$= \{x : x \in C \land x \notin A\} \quad [\because A \subset B]$
$= \{x : x \in C - A\}$
$\implies C - B \subset C - A$
$\implies A \subset B \implies C - B \subset C - A$

Sets Miscellaneous Exercise Class 11

Illustration: 6. Show that for any sets A and B:
$A = (A \cap B) \cup (A - B)$ and $A \cup (B - A) = A \cup B$
Solution: Let $x \in (A \cap B) \cup (A - B)$
$\implies x \in (A \cap B) \lor x \in (A - B)$
$\implies (x \in A \land x \in B) \lor (x \in A \land x \notin B)$
$\implies x \in A$
$\implies (A \cap B) \cup (A - B) \subset A \quad \dots \text{(1)}$
Again let $x \in A$
$x \in (x \in A \land x \in B) \lor (x \in A \land x \notin B)$
$\implies x \in (A \cap B) \lor x \in (A - B)$
$\implies x \in (A \cap B) \cup (A - B)$
$\therefore A \subset (A \cap B) \cup (A - B) \quad \dots \text{(2)}$
From (1) and (2), we get:
$A = (A \cap B) \cup (A - B)$
Let $x \in A \cup (B - A)$
$\implies x \in A \lor x \in (B - A)$
$\implies x \in A \lor (x \in B \land x \notin A)$
$\implies x \in A \cup B \quad [\text{ By def. }]$
$A \cup (B - A) \subset A \cup B \quad \dots \text{(3)}$
$x \in A \cup B \implies x \in A \lor x \in B$
$\implies x \in A \lor (x \in B \land x \notin A)$
$\implies x \in A \lor (x \in B - A)$
$\implies x \in A \cup (B - A)$
$\therefore A \cup B \subset A \cup (B - A) \quad \dots \text{(4)}$
From (3) and (4), we get:
$A \cup (B - A) = (A \cup B)$
Illustration: 7. Using properties of sets, show that:
Illustration: 7 (i). $A \cup (A \cap B) = A$
Solution: $A \cup (A \cap B) = A$
$= (A \cup A) \cap (A \cup B) \quad [\text{By distributive law }]$
$\implies A \cap (A \cup B)$
$\implies A$
$\therefore A \cup (A \cap B) \subset A \quad \dots \text{(1)}$
$A \implies A \cap (A \cup B)$
$\implies (A \cup A) \cap (A \cup B)$
$\implies A \cup (A \cap B) \quad [\text{By distributive law}]$
$\implies A \cup (A \cap B)$
$\therefore A \subset A \cup (A \cap B) \quad \dots \text{(2)}$
From (1) and (2), we get:
$A = A \cup (A \cap B)$
Illustration: 7 (ii). $A \cap (A \cup B) = A$
Solution: $A \cap (A \cup B) = A$
$A \cap (A \cup B)$
$= (A \cap A) \cup (A \cap B)$
$\implies A \cup (A \cap B)$
$\implies A$
$\therefore A \cap (A \cup B) \subset A \quad \dots \text{(3)}$
Let A
$\implies A \cup (A \cap B)$
$\implies (A \cap A) \cup (A \cap B)$
$\implies A \cap (A \cup B)$
$A \subset A \cap (A \cup B) \quad \dots \text\{(4)\}$
From (3) and (4), we get:
$A \cap (A \cup B) = A$
Illustration: 8. Show that $A \cap B = A \cap C$ need not imply B = C.
Solution: Let $A = \{x, y\}, B = \{x, z\}, C = \{x, a\}$
$A \cap B = \{x, y\} \cap \{x, z\}$
$= \{x\}$
and $A \cap C = \{x, y\} \cap \{x, a\}$
$= \{x\}$
$\implies A \cap B = A \cap C$ , but $B \neq C$
Illustration: 9. Let A and B be sets. If $A \cap X = B \cap X = \phi$ and $A \cup X = B \cup X$ for some set X, show that A = B.
Solution: $A \cup X = B \cup X$
$(A \cup X) \cap A = (B \cup X) \cap A$
$\implies A = (B \cup X) \cap A \quad [\because (A \cup X) \cap A = A]$
$A = (B \cap A) \cup (A \cap X) \quad [\text{By distributive law}]$
$= (A \cap B) \cup \phi \quad [\because A \cap X = \phi]$
$A = A \cap B \quad \dots \text\{(1)\}$
$A \cup X = B \cup X$
$(A \cup X) \cap B = (B \cup X) \cap B$
$\implies (A \cup X) \cap B = B \quad [\because (B \cup X) \cap B = B]$
$(A \cap B) \cup (B \cap X) = B \quad [\text{By distributive law }]$
$\implies (A \cap B) \cup \phi = B \quad [\because B \cap X = \phi]$
$\implies A \cap B = B \quad \dots \text\{(2)\}$
From (1) and (2), we get:
$A = B$
Illustration: 10. Find sets A, B and C such that $A \cap B, B \cap C$ and $A \cap C$ are non-empty sets and $A \cap B \cap C = \phi$ .
Solution: $A = \{1, 2\}, \quad B = \{1, 3\}, \quad C = \{2, 3\}$
$A \cap B = \{1, 2\} \cap \{1, 3\} = \{1\}$
$B \cap C = \{1, 3\} \cap \{2, 3\} = \{3\}$
$A \cap C = \{1, 2\} \cap \{2, 3\} = \{2\}$
And $A \cap B \cap C = \{1, 2\} \cap \{1, 3\} \cap \{2, 3\} = \phi$
$\implies B \cap C = \{3\}$
$A \cap C = \{1, 2\} \cap \{2, 3\}$
$\implies A \cap C = \{2\}$
$\implies \text{But } A \cap B \cap C = \phi$
By solving the above illustrations, one can understand Sets Miscellaneous Exercise Class 11.

Also Read This Article:- Properties of Sets Class 11 with Proof

3.Practice Questions of Sets Miscellaneous Exercise Class 11 for Students:

Sets Miscellaneous Exercise Class 11

(1.)Let P(A)=P(B),Prove that A=B
(2.)For any set A and B is it true $P(A) \cup P(B) = P(A \cup B)$ ? Clarify your answer.
By solving the above questions,you can understand the Sets Miscellaneous Exercise Class 11 well because the concept is well understood when you solve it practically.
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Also Read This Article:- Basic Operation On Sets: Examples and History of Sets

4.Frequently Asked Questions Related to Sets Miscellaneous Exercise Class 11:

Q:1.What is a power set?

Ans:The collection of all subsets of any set A is called the power set of A.The power set of A is denoted by P(A).

Q:2.Define the union of sets

Ans:The union of two sets A and B is the set which consists of all these elements which are either in A or in B (including those which are in both).

Q:3.Define the intersection of sets.

Ans:The intersection of two sets A and B is the set of all those elements which belong to both A and B.
By answering the above questions,you can know about the primary terms of Sets Miscellaneous Exercise Class 11.

5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

Sets Miscellaneous Exercise Class 11

🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
यदि $x * y = \sqrt{\frac{(x + y)(y^2 - 12x)}{(x - 2)(y - 2)}}$
तो 5 * 9 का मान होगा:
(If $x * y = \sqrt{\frac{(x + y)(y^2 - 12x)}{(x - 2)(y - 2)}}$ , so the value of 5 * 9 would be)
*पिछली प्रश्नोत्तरी का हल*
$\frac{?}{32} = \frac{72}{?}$
$\therefore (?)^2 = 32 \times 72$
$\therefore ? = \sqrt{32 \times 72}$
$= \sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3}$
$= 48$
*Previous Quiz Solution*
$\frac{?}{32} = \frac{72}{?}$
$\therefore (?)^2 = 32 \times 72$
$\therefore ? = \sqrt{32 \times 72}$
$= \sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3}$
$? = 48$
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*

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