In this chapter, we will discuss Relations and Mapping (Function) mathematical problems and their step-by-step solutions as part of our S.N. Dey Math Solution Series (Class XI). We hope this guide helps students master the core concepts for WBCHSE exams.
Very Short Answer Type Questions [Exercise: 2A]
Define and write the cartesian product of two sets $P = (a,b,c)$ and $Q = \{2,3\}$. Is $P \times Q = Q \times P$ true?
Let $P$ and $Q$ be two given sets. The cartesian product of $P$ and $Q$ is the set of all ordered pairs $(x,y)$ where $x \in P$ and $y \in Q$, denoted by $P \times Q$.
$$\text{Symbolically, } P \times Q = \{(x,y): x \in P \wedge y \in Q\}$$
Now, $P \times Q = \{(a,2),(b,2),(c,2),(a,3),(b,3),(c,3)\} \quad \dots(1)$
whereas, $Q \times P = \{(2,a),(2,b),(2,c),(3,a),(3,b),(3,c)\} \quad \dots(2)$
Hence, from $(1)$ and $(2)$, we can conclude that $P \times Q \neq Q \times P$.
If $A = \{2,3\}$, $B = \{3,4\}$, $C = \{4,6\}$, find:
(i) $(A \times B) \cup (B \times C)$
(ii) $(A \times B) \cap (B \times C)$
(i) $A \times B = \{2,3\} \times \{3,4\} = \{(2,3),(2,4),(3,3),(3,4)\}$
$B \times C = \{3,4\} \times \{4,6\} = \{(3,4),(3,6),(4,4),(4,6)\}$
Hence, $(A \times B) \cup (B \times C) = \{(2,3),(2,4),(3,3),(3,4),(3,6),(4,4),(4,6)\}$
(ii) $(A \times B) \cap (B \times C) = \{(3,4)\}$
If $A = \{1,4\}$, $B = \{4,3\}$, $C = \{3,6\}$, show that:
$$A \times (B \cup C) = (A \times B) \cup (A \times C)$$
$A \times B = \{1,4\} \times \{4,3\} = \{(1,4),(1,3),(4,4),(4,3)\}$
$A \times C = \{1,4\} \times \{3,6\} = \{(1,3),(1,6),(4,3),(4,6)\}$
Hence, $(A \times B) \cup (A \times C) = \{(1,4),(1,3),(4,4),(4,3),(1,6),(4,6)\} \quad \dots(1)$
Again, $B \cup C = \{4,3\} \cup \{3,6\} = \{3,4,6\}$
$A \times (B \cup C) = \{1,4\} \times \{3,4,6\} = \{(1,3),(1,4),(1,6),(4,3),(4,4),(4,6)\} \quad \dots(2)$
From $(1)$ and $(2)$, the result follows.
If $A = \{1,2,3\}$, $B = \{2,3,4\}$, $S = \{1,3,4\}$, $T = \{2,4,5\}$, verify that:
$$(A \times B) \cap (S \times T) = (A \cap S) \times (B \cap T)$$
$A \times B = \{(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,2),(3,3),(3,4)\}$
$S \times T = \{(1,2),(1,4),(1,5),(3,2),(3,4),(3,5),(4,2),(4,4),(4,5)\}$
$(A \times B) \cap (S \times T) = \{(1,2),(1,4),(3,2),(3,4)\} \quad \dots(1)$
Again, $A \cap S = \{1,3\}$ and $B \cap T = \{2,4\}$
$(A \cap S) \times (B \cap T) = \{1,3\} \times \{2,4\} = \{(1,2),(1,4),(3,2),(3,4)\} \quad \dots(2)$
From $(1)$ and $(2)$, the result is verified.
If $A = \{a,b\}$, $B = \{m,n\}$, $C = \{p,q\}$, show that $A \times (B \cap C) = (A \times B) \cap (A \times C)$.
$B \cap C = \{m,n\} \cap \{p,q\} = \phi$
$\therefore A \times (B \cap C) = \phi \quad \dots(1)$
Similarly, $A \times B = \{(a,m),(a,n),(b,m),(b,n)\}$
$A \times C = \{(a,p),(a,q),(b,p),(b,q)\}$
So, $(A \times B) \cap (A \times C) = \phi \quad \dots(2)$
Hence, from $(1)$ and $(2)$, the result follows.
If $A = \{0,1\}$, find (i) $A \times A$ and (ii) $A \times A \times A$.
(i) $A \times A = (0,1) \times (0,1) = \{(0,0),(0,1),(1,0),(1,1)\}$
(ii) $A \times A \times A = (0,1) \times \{(0,0),(0,1),(1,0),(1,1)\}$
$= \{(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)\}$
Short Answer Type Questions [Exercise: 2A]
If $A \times B = \{(1,2),(3,4),(5,2),(1,4),(3,2),(5,4)\}$, find $B \times A$.
$B \times A = \{(2,1),(4,3),(2,5),(4,1),(2,3),(4,5)\}$
If $P \times Q = \{(2,-1),(3,0),(2,1),(3,1),(2,0),(3,-1)\}$, find $P$ and $Q$.
$P = \{x: (x,y) \in P \times Q\} \implies P = \{2,3\}$
$Q = \{y: (x,y) \in P \times Q\} \implies Q = \{-1,0,1\}$
If $A = \{x : x \in \mathbb{N} \wedge 1 < x \leq 3\}$ and $B = \{x : x \in \mathbb{Z} \wedge -2 < x < 2\}$, find $B \times A$.
Since $A = \{x : x \in \mathbb{N} \wedge 1 < x \leq 3\} \implies A = \{2,3\}$
And $B = \{x : x \in \mathbb{Z} \wedge -2 < x < 2\} \implies B = \{-1,0,1\}$
Now, $B \times A = \{(x,y): x \in B \wedge y \in A\}$
$\implies B \times A = \{(-1,2),(-1,3),(0,2),(0,3),(1,2),(1,3)\}$
Let $A = \{x: x \in \mathbb Z \wedge -1 < x \leq 1\}$, $B = \{x: x \in \mathbb N \wedge 1 < x < 5\}$, and $C = \{x: x \text{ is an odd positive integer and } 1 < x \leq 6\}$.
Then show that: (i) $A \times (B \cap C) = (A \times B) \cap (A \times C)$ (ii) $(A \times B) \cup (A \times C) = A \times (B \cup C)$
$A = \{0,1\}$, $B = \{2,3,4\}$, $C = \{3,5\}$
(i) $B \cap C = \{3\}$
$A \times (B \cap C) = \{0,1\} \times \{3\} = \{(0,3),(1,3)\} \quad \dots(1)$
$A \times B = \{(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)\}$
$A \times C = \{(0,3),(0,5),(1,3),(1,5)\}$
$(A \times B) \cap (A \times C) = \{(0,3),(1,3)\} \quad \dots(2)$
Hence, from $(1)$ and $(2)$, $(i)$ is proved.
(ii) $(A \times B) \cup (A \times C) = \{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\} \quad \dots(3)$
$B \cup C = \{2,3,4,5\}$
$A \times (B \cup C) = \{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\} \quad \dots(4)$
Hence, from $(3)$ and $(4)$, $(ii)$ is proved.
If $A = \{1, 2, 3\}$ and $B = \{6, 7\}$, find the number of subsets of the set $A \times B$.
$A \times B = \{(1,6),(1,7),(2,6),(2,7),(3,6),(3,7)\}$
So, $n(A \times B) = 6$.
Hence, total number of subsets $= 2^6 = 64$.
The cartesian product $P \times P$ has $9$ elements; if two of its elements are $(-3,-2)$ and $(-2,-1)$, find the remaining terms of $P \times P$.
$n(P \times P) = 9 \implies n(P) = 3$.
Since $(-3,-2), (-2,-1) \in P \times P \implies \{-1,-2,-3\} \subseteq P$.
Therefore, $P = \{-1,-2,-3\}$.
Full product $P \times P = \{(-1,-1),(-1,-2),(-1,-3),(-2,-1),(-2,-2),(-2,-3),(-3,-1),(-3,-2),(-3,-3)\}$.
Remaining terms = $\{(-1,-1),(-1,-2),(-1,-3),(-2,-2),(-2,-3),(-3,-1),(-3,-3)\}$.
Let $A = \{x:x \in \mathbb N \wedge x \text{ is a prime } \in [10, 19]\}$ and $B=\{2, 3\}$; find $A \times B$.
$A = \{11,13,17,19\}$ and $B = \{2,3\}$.
$A \times B = \{(11,2),(13,2),(17,2),(19,2),(11,3),(13,3),(17,3),(19,3)\}$.
Two sets $A$ and $B$ have $4$ common elements. If $n(A) = 6$ and $n(B)=7$, find $n(A \times B)$ and $n[(A \times B) \cap (B \times A)]$.
$n(A \times B) = n(A) \cdot n(B) = 6 \times 7 = 42$.
Since $A$ and $B$ have $4$ common elements, $A \times B$ and $B \times A$ have $4 \times 4 = 16$ common elements.
Hence, $n[(A \times B) \cap (B \times A)] = 16$.
State whether the following statements are True or False:
(i) If $A = \{1, 2, 3\}, B = \{4, 5\}$, then $A \times (B \cup \phi) = \phi$.
$\rightarrow$ FALSE. ($B \cup \phi = \{4,5\}$, so $A \times (B \cup \phi) \neq \phi$)
(ii) If $X=\{a,b,c\}, Y=\{c,a,b\}$, then $X \times Y = Y \times X$.
$\rightarrow$ TRUE. ($X$ and $Y$ are identical sets)
(iii) If $A=\{3,4,5\}, B=\{1,2\}$, then $A \times (B \cap \phi) = \phi$.
$\rightarrow$ TRUE. ($B \cap \phi = \phi \implies A \times \phi = \phi$)
(iv) If $A=\{1,0,-1\}$, then $n(A \times A \times A) = \phi$.
$\rightarrow$ FALSE. ($n(A \times A \times A) = 3^3 = 27$)
For any three sets $A, B, C$, prove that: $$A \times (B-C) = (A \times B) - (A \times C)$$
Let $(x,y) \in A \times (B-C)$
$\implies x \in A \wedge y \in (B-C)$
$\implies x \in A \wedge (y \in B \wedge y \notin C)$
$\implies (x \in A \wedge y \in B) \wedge (x \in A \wedge y \notin C)$
$\implies (x,y) \in (A \times B) \wedge (x,y) \notin (A \times C)$
$\implies (x,y) \in (A \times B) - (A \times C)$
$\therefore A \times (B-C) \subseteq (A \times B) - (A \times C) \quad \dots(1)$
Conversely, let $(a,b) \in (A \times B) - (A \times C)$
$\implies (a,b) \in A \times B \wedge (a,b) \notin A \times C$
$\implies (a \in A \wedge b \in B) \wedge (a \in A \wedge b \notin C)$
$\implies a \in A \wedge (b \in B \wedge b \notin C) \implies (a,b) \in A \times (B-C)$
$\therefore (A \times B) - (A \times C) \subseteq A \times (B-C) \quad \dots(2)$
From $(1)$ and $(2)$, $A \times (B-C) = (A \times B) - (A \times C)$ (Proved).
If $n(A \times B \times C) = 60$, $n(B) = 4$, $n(C) = 3$, find the value of $n(A)$.
$n(A \times B \times C) = n(A) \cdot n(B) \cdot n(C) = 60$
$\implies n(A) \times 4 \times 3 = 60$
$\implies n(A) = \frac{60}{12} = 5$
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