Relations & Functions

NCERT Class 11 Maths • Exercise 2.1

1. Equality of Ordered Pairs

Q1: If $(\frac{x}{3} + 1, y – \frac{2}{3}) = (\frac{5}{3}, \frac{1}{3})$, find x and y.

Principle: Two ordered pairs are equal if and only if their corresponding first and second elements are equal.

For x:

$\frac{x}{3} + 1 = \frac{5}{3} \Rightarrow \frac{x}{3} = \frac{5}{3} – 1 \Rightarrow \frac{x}{3} = \frac{2}{3}$

$\therefore x = 2$

For y:

$y – \frac{2}{3} = \frac{1}{3} \Rightarrow y = \frac{1}{3} + \frac{2}{3} \Rightarrow y = \frac{3}{3}$

$\therefore y = 1$

Answer: $x = 2, y = 1$

2 & 3. Cartesian Products ($A \times B$)

Q2: Find number of elements in $(A \times B)$ given $n(A)=3$ and $B=\{3,4,5\}$.

Formula: $n(A \times B) = n(A) \times n(B)$

Given: $n(A) = 3$

Set $B = \{3, 4, 5\} \Rightarrow n(B) = 3$

Calculation: $n(A \times B) = 3 \times 3 = 9$

Answer: 9

Q3: If $G = \{7, 8\}$ and $H = \{5, 4, 2\}$, find $G \times H$ and $H \times G$.

$G \times H$ (First element from G, second from H):

$\{(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)\}$

$H \times G$ (First element from H, second from G):

$\{(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)\}$

4. Statement Verification

Q4: State True/False. Rewrite false statements correctly.

(i) If $P = \{m, n\}$ and $Q = \{n, m\}$, then $P \times Q = \{(m, n), (n, m)\}$

False

Reason: $n(P) \times n(Q) = 2 \times 2 = 4$ pairs. The statement lists only 2.
Correct: $P \times Q = \{(m, n), (m, m), (n, n), (n, m)\}$

(ii) If A and B are non-empty sets, then $A \times B$ is a non-empty set of ordered pairs (x, y) such that $x \in A$ and $y \in B$.

True

Definition of Cartesian Product.

(iii) If $A = \{1, 2\}, B = \{3, 4\}$, then $A \times (B \cap \phi) = \phi$.

True

$B \cap \phi = \phi$.
$A \times \phi = \phi$.

5 & 6. Operations on Sets

Q5: If $A = \{-1, 1\}$, find $A \times A \times A$.

Step 1: Find $A \times A$

$A \times A = \{(-1, -1), (-1, 1), (1, -1), (1, 1)\}$

Step 2: Find $(A \times A) \times A$

$\{ \\ (-1, -1, -1), (-1, -1, 1), \\ (-1, 1, -1), (-1, 1, 1), \\ (1, -1, -1), (1, -1, 1), \\ (1, 1, -1), (1, 1, 1) \\ \}$

Q6: If $A \times B = \{(a, x), (a, y), (b, x), (b, y)\}$, find A and B.

Set A consists of all first elements of the ordered pairs.

$A = \{a, b\}$

Set B consists of all second elements of the ordered pairs.

$B = \{x, y\}$

7. Distributive Laws

Given: $A=\{1,2\}, B=\{1,2,3,4\}, C=\{5,6\}, D=\{5,6,7,8\}$

Q7(i): Verify $A \times (B \cap C) = (A \times B) \cap (A \times C)$

LHS: $B \cap C = \phi$ (No common elements).
$A \times \phi = \phi$.

RHS:
$A \times B = \{(1,1)…(2,4)\}$
$A \times C = \{(1,5)…(2,6)\}$
Intersection contains no common pairs. Result is $\phi$.

LHS = RHS ($\phi$). Verified.

Q7(ii): Verify $A \times C$ is a subset of $B \times D$

$A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\}$

$B \times D$ contains all combinations of $\{1,2,3,4\}$ and $\{5,6,7,8\}$.

Since elements of A (1,2) are in B, and elements of C (5,6) are in D, all pairs of $A \times C$ are present in $B \times D$.

Verified.

8. Subsets of Cartesian Product

Q8: List subsets of $A \times B$ where $A=\{1,2\}, B=\{3,4\}$.

$A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$

Number of elements $n = 4$.

Number of subsets $= 2^n = 2^4 = 16$.

Subsets:
$\phi$
$\{(1,3)\}, \{(1,4)\}, \{(2,3)\}, \{(2,4)\}$
$\{(1,3), (1,4)\}, \{(1,3), (2,3)\} \dots$ (and so on)

9 & 10. Finding Sets from Partial Info

Q9: Find A and B given $n(A)=3, n(B)=2$ and pairs $(x,1), (y,2), (z,1)$.

Set A contains first components: $x, y, z$. Since $n(A)=3$ and x, y, z are distinct:

$A = \{x, y, z\}$

Set B contains second components: $1, 2$. Since $n(B)=2$:

$B = \{1, 2\}$

Q10: Find set A and remaining elements of $A \times A$.

Given $n(A \times A) = 9 \implies n(A) = 3$.

Pairs present: $(-1, 0)$ and $(0, 1)$.

Elements in A must be $-1, 0, 1$. Since $n(A)=3$:

$A = \{-1, 0, 1\}$

Remaining elements of $A \times A$:

Total pairs are $(-1,-1), (-1,0), (-1,1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1)$.

Removing given pairs $(-1, 0)$ and $(0, 1)$, the remaining are:

$\{(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)\}$