NCERT Solutions Class 11 Maths Chapter 1 Miscellaneous Exercise

1. Decide which sets are subsets of one another.

Step 1: Decode the Sets

Set A: $\{x \in R : x^2 – 8x + 12 = 0\}$
$(x-6)(x-2) = 0 \Rightarrow x = 2, 6$
$\therefore A = \{2, 6\}$

Set B: $\{2, 4, 6\}$

Set C: $\{2, 4, 6, 8, \dots\}$ (Even natural numbers)

Set D: $\{6\}$

Step 2: Compare

$D \subset A$ (Since $\{6\} \subset \{2, 6\}$)
$D \subset B$ (Since $\{6\} \subset \{2, 4, 6\}$)
$D \subset C$ (Since $\{6\} \subset \{2, 4, 6, \dots\}$)
$A \subset B$ (Since $\{2, 6\} \subset \{2, 4, 6\}$)
$A \subset C$ (Since $\{2, 6\} \subset \{2, 4, 6, \dots\}$)
$B \subset C$ (Since $\{2, 4, 6\} \subset \{2, 4, 6, \dots\}$)

2. Determine True/False. Prove or give counter-example.

(i) If $x \in A$ and $A \in B$, then $x \in B$

False

Counter-Example:
Let $A = \{1\}$ and $B = \{ \{1\}, 2 \}$.
Here, $1 \in A$ and $A \in B$.
But $1 \notin B$ (The element of B is $\{1\}$, not $1$).

(ii) If $A \subset B$ and $B \in C$, then $A \in C$

False

Counter-Example:
Let $A = \{1\}$, $B = \{1, 2\}$, $C = \{ \{1, 2\}, 3 \}$.
Here $A \subset B$ and $B \in C$.
But $A \notin C$ (Only $\{1, 2\}$ is in C, not $\{1\}$).

(iii) If $A \subset B$ and $B \subset C$, then $A \subset C$

True

Proof:
Let $x \in A$.
Since $A \subset B$, $x \in B$.
Since $B \subset C$, $x \in C$.
Therefore, $A \subset C$.

(iv) If $A \not\subset B$ and $B \not\subset C$, then $A \not\subset C$

False

Counter-Example:
Let $A = \{1, 2\}$, $B = \{0, 6, 8\}$, $C = \{0, 1, 2, 6, 9\}$.
$A \not\subset B$ and $B \not\subset C$.
But $A \subset C$.

(v) If $x \in A$ and $A \not\subset B$, then $x \in B$

False

Counter-Example:
Let $A = \{1, 2, 3\}$ and $B = \{1, 2\}$.
$3 \in A$ and $A \not\subset B$.
But $3 \notin B$.

(vi) If $A \subset B$ and $x \notin B$, then $x \notin A$

True

Proof:
Suppose $x \in A$.
Since $A \subset B$, then $x \in B$.
This contradicts the given $x \notin B$.
Therefore, assumption is wrong. $x \notin A$.

3. Show that $B = C$ given set conditions.

Given: $A \cup B = A \cup C$ and $A \cap B = A \cap C$.

Proof:
Let $x \in B$.
$\implies x \in A \cup B$
$\implies x \in A \cup C$ (Since $A \cup B = A \cup C$)
This means $x \in A$ or $x \in C$.

Case 1: If $x \in C$, we are done.
Case 2: If $x \in A$, then since $x \in B$, we have $x \in A \cap B$.
Since $A \cap B = A \cap C$, then $x \in A \cap C$.
$\implies x \in C$.

In both cases, $x \in C$. Thus, $B \subset C$.
Similarly, we can prove $C \subset B$.
$\therefore B = C$.

4. Show conditions are equivalent.

(i) $A \subset B$ (ii) $A – B = \phi$ (iii) $A \cup B = B$ (iv) $A \cap B = A$

Logic: All these statements fundamentally describe the same relationship where “Set A is entirely contained within Set B”.

1. If A is inside B ($A \subset B$), then removing B removes all of A ($A-B = \phi$).
2. If A is inside B, combining them gives nothing new ($A \cup B = B$).
3. If A is inside B, their common part is just A ($A \cap B = A$).

5 & 6. Proofs regarding Subsets and Differences.

Q5: Show if $A \subset B$, then $C – B \subset C – A$.

Let $x \in (C – B) \implies x \in C \text{ and } x \notin B$.
Since $A \subset B$, if $x$ is not in B, it cannot be in A ($x \notin A$).
So, $x \in C \text{ and } x \notin A \implies x \in (C – A)$.
$\therefore (C – B) \subset (C – A)$.

Q6: Show $A = (A \cap B) \cup (A – B)$.

RHS $= (A \cap B) \cup (A \cap B’)$ (Definition of difference)
$= A \cap (B \cup B’)$ (Distributive Law)
$= A \cap U$ (Union with complement is Universal)
$= A$ = LHS.

7. Properties of Sets (Absorption Laws).

(i) $A \cup (A \cap B) = A$
Since $(A \cap B) \subset A$, their union is just A.
(ii) $A \cap (A \cup B) = A$
Since $A \subset (A \cup B)$, their intersection is just A.

8. Show that $A \cap B = A \cap C$ need not imply $B = C$.

Counter-Example:
Let $A = \{1, 2\}$, $B = \{2, 3\}$, $C = \{2, 4\}$.
$A \cap B = \{2\}$
$A \cap C = \{2\}$
But $B \neq C$ ($\{2, 3\} \neq \{2, 4\}$).

9. Show $A = B$ using set X properties.

Given: $A \cap X = B \cap X = \phi$ and $A \cup X = B \cup X$.

$A = A \cap (A \cup X)$ (Absorption Law)
$= A \cap (B \cup X)$ (Given $A \cup X = B \cup X$)
$= (A \cap B) \cup (A \cap X)$ (Distributive Law)
$= (A \cap B) \cup \phi$ (Given $A \cap X = \phi$)
$= A \cap B$

Similarly, $B = B \cap (B \cup X)$
$= B \cap (A \cup X)$
$= (B \cap A) \cup (B \cap X)$
$= (B \cap A) \cup \phi$
$= B \cap A$

$\therefore A = A \cap B = B$. So, $A = B$.

10. Find sets A, B, C with specific intersections.

Condition: Pairs ($A \cap B$, $B \cap C$, $A \cap C$) are non-empty, but triple intersection ($A \cap B \cap C$) is empty.

Let $A = \{1, 2\}$
Let $B = \{2, 3\}$
Let $C = \{1, 3\}$

Check:
$A \cap B = \{2\}$ (Non-empty)
$B \cap C = \{3\}$ (Non-empty)
$A \cap C = \{1\}$ (Non-empty)
$A \cap B \cap C = \phi$ (No element is common to all three).