Lecture 06: Ordered Pair, Cartesian Product, Relation, Domain and Range
Theory – 1: Ordered Pair & Cartesian Product
An Ordered Pair consists of two elements written in a fixed order within parentheses, like $(a, b)$. The order is strictly maintained, meaning $(a, b) \neq (b, a)$ unless $a = b$.
ક્રમયુક્ત જોડ (Ordered Pair) માં બે ઘટકોને કૌંસમાં ચોક્કસ ક્રમમાં દર્શાવવામાં આવે છે, જેમ કે $(a, b)$. અહીં ક્રમનું ખૂબ મહત્ત્વ છે, એટલે કે $(a, b) \neq (b, a)$ (સિવાય કે $a = b$).
Equality of Ordered Pairs: Two ordered pairs $(a, b)$ and $(x, y)$ are equal if and only if their corresponding elements are equal. That is, $a = x$ and $b = y$.
ક્રમયુક્ત જોડની સમાનતા: જો બે ક્રમયુક્ત જોડો સમાન હોય $(a, b) = (x, y)$, તો તેમના અનુરૂપ ઘટકો પણ સમાન થાય છે. એટલે કે, $a = x$ અને $b = y$.
For two non-empty sets $A$ and $B$, the Cartesian product is defined as $A \times B = \{(x, y) : x \in A \text{ and } y \in B\}$.
બે અરિક્ત ગણો $A$ અને $B$ માટે, કાર્તેઝીય ગુણાકાર $A \times B = \{(x, y) : x \in A \text{ અને } y \in B\}$.
If $n(A) = p$ and $n(B) = q$, then $n(A \times B) = p \times q$.
જો $n(A) = p$ અને $n(B) = q$ હોય, તો $n(A \times B) = p \times q$.
Extra Question (Easy)
If $\left(\frac{x}{3} + 1, y – \frac{2}{3}\right) = \left(\frac{5}{3}, \frac{1}{3}\right)$, find the values of $x$ and $y$.
જો $\left(\frac{x}{3} + 1, y – \frac{2}{3}\right) = \left(\frac{5}{3}, \frac{1}{3}\right)$ હોય, તો $x$ અને $y$ ની કિંમત શોધો.
Solution: Since the ordered pairs are equal, we compare their corresponding elements:
$$ \frac{x}{3} + 1 = \frac{5}{3} \quad \text{and} \quad y – \frac{2}{3} = \frac{1}{3} $$
Solving for $x$:
$$ \frac{x}{3} = \frac{5}{3} – 1 \Rightarrow \frac{x}{3} = \frac{2}{3} \Rightarrow x = 2 $$
Solving for $y$:
$$ y = \frac{1}{3} + \frac{2}{3} \Rightarrow y = \frac{3}{3} \Rightarrow y = 1 $$
Answer: $x = 2, y = 1$
Classwork – 1 (Moderate)
If A & B are two sets such that $n(A \times B) = 60$ & $n(A) = 12$ also $n(A \cap B) = K$ then sum of maximum & minimum possible value of K is
(A) 17 (B) 12 (C) 5 (D) 7
જો A & B બે ગણ માટે $n(A \times B) = 60$ & $n(A) = 12$ & $n(A \cap B) = K$, તો K ની મહત્તમ અને ન્યુનતમ કિંમતોનો સરવાળો શોધો.
(A) 17 (B) 12 (C) 5 (D) 7
Solution: Given: $n(A \times B) = 60$ and $n(A) = 12$.
Using the formula $n(A \times B) = n(A) \times n(B)$:
$$ 60 = 12 \times n(B) \Rightarrow n(B) = 5 $$
Now, it is given that $n(A \cap B) = K$.
The minimum possible value of $K$ is 0 (if the sets are entirely disjoint).
The maximum possible value of $K$ is 5 (since set B only has 5 elements, at most 5 elements can be common with set A).
Sum of maximum and minimum values = $0 + 5 = 5$.
Correct Answer: (C) 5
Theory – 2: Subsets of Cartesian Product
If $n(A \times B) = m$, then the total number of subsets of $A \times B$ is $2^m$.
જો $n(A \times B) = m$ હોય, તો $A \times B$ ના કુલ ઉપગણોની સંખ્યા $2^m$ થાય છે.
Often, sets are defined by exponential equations or inequalities. Solve them to find integer elements before finding subsets.
ઘણીવાર ગણ A અને B ના ઘટકો ઘાતાંકીય સમીકરણ કે અસમતા સ્વરૂપે હોય છે. તેમાં x ની કિંમતો મેળવીને પહેલા ગણના સભ્યો નક્કી કરવા પડે છે.
Classwork – 2 (Moderate)
If the difference between the number of subsets of two sets A and B is 120, then $n(A \times B)$ is equal to
(A) 21 (B) 25 (C) 18 (D) 24
A અને B ના ભિન્ન અલગ ગણ ના ઉપગણો ની સંખ્યા 120 છે તો $n(A \times B) = $
(A) 21 (B) 25 (C) 18 (D) 24
Solution: Let $n(A) = m$ and $n(B) = n$.
The number of subsets for these sets will be $2^m$ and $2^n$ respectively.
According to the given condition:
$$ 2^m – 2^n = 120 $$
We know that $2^7 = 128$ and $2^3 = 8$.
$$ 128 – 8 = 120 \Rightarrow 2^7 – 2^3 = 120 $$
Therefore, $m = 7$ and $n = 3$.
So, $n(A \times B) = m \times n = 7 \times 3 = 21$.
Correct Answer: (A) 21
Classwork – 4 (Hard / JEE Main)
Let Z be the set of integers. If $A = \{x \in Z : 2^{(x+2)(x^2-5x+6)} = 1\}$ and $B = \{x \in Z : -3 < 2x - 1 < 9\}$ then the number of subsets of the set $A \times B$, is :
(A) $2^{18}$ (B) $2^{10}$ (C) $2^{15}$ (D) $2^{12}$
જો Z એ પૂર્ણાંક સંખ્યા ગણ છે. જો $A = \{x \in Z : 2^{(x+2)(x^2-5x+6)} = 1\}$ અને $B = \{x \in Z : -3 < 2x - 1 < 9\}$ તો $A \times B$ ના ઉપગણોની સંખ્યા મેળવો.
(A) $2^{18}$ (B) $2^{10}$ (C) $2^{15}$ (D) $2^{12}$
Solution: For Set A:
$2^{(x+2)(x^2-5x+6)} = 2^0$
Comparing the exponents: $(x+2)(x^2-5x+6) = 0 \Rightarrow (x+2)(x-2)(x-3) = 0$
Thus, $x = -2, 2, 3$. This gives $A = \{-2, 2, 3\}$ and $n(A) = 3$.
For Set B:
$-3 < 2x - 1 < 9$
$\Rightarrow -2 < 2x < 10$
$\Rightarrow -1 < x < 5$
Since $x \in Z$ (integers), $B = \{0, 1, 2, 3, 4\}$, which means $n(B) = 5$.
Now, $n(A \times B) = n(A) \times n(B) = 3 \times 5 = 15$.
The number of subsets = $2^{n(A \times B)} = 2^{15}$.
Correct Answer: (C) $2^{15}$
Linked Homework: Q17 (Same Logic / Repeat)
Theory – 3: Relation, Inverse Relation, Domain and Range
A relation $R$ from set A to set B is a subset of $A \times B$. Total number of relations = $2^{n(A) \times n(B)}$.
ગણ A થી ગણ B પરનો સંબંધ $R$ એ $A \times B$ નો ઉપગણ છે. કુલ સંબંધોની સંખ્યા = $2^{n(A) \times n(B)}$.
The set of all first elements of ordered pairs is Domain. The set of all second elements is Range.
સંબંધ $R$ ની ક્રમયુક્ત જોડોના પ્રથમ ઘટકોના ગણને પ્રદેશ (Domain) કહે છે. બીજા ઘટકોના ગણને વિસ્તાર (Range) કહે છે.
Inverse Relation ($R^{-1}$): Let $R$ be a relation from set $A$ to set $B$. The inverse relation $R^{-1}$ is a relation from $B$ to $A$ defined by interchanging the elements of ordered pairs: $R^{-1} = \{(y, x) : (x, y) \in R\}$.
વ્યસ્ત સંબંધ ($R^{-1}$): ધારો કે $R$ એ ગણ $A$ થી ગણ $B$ પરનો સંબંધ છે. $R$ ની દરેક ક્રમયુક્ત જોડના પ્રથમ અને દ્વિતીય ઘટકોની અદલાબદલી કરવાથી મળતા સંબંધને વ્યસ્ત સંબંધ કહે છે: $R^{-1} = \{(y, x) : (x, y) \in R\}$.
Therefore, Domain of $R^{-1}$ = Range of $R$, and Range of $R^{-1}$ = Domain of $R$.
તેથી, સ્પષ્ટ છે કે: $R^{-1}$ નો પ્રદેશ = $R$ નો વિસ્તાર, અને $R^{-1}$ નો વિસ્તાર = $R$ નો પ્રદેશ થાય.
Extra Question (Easy)
Let $A = \{1, 2, 3, 4, 5, 6\}$. Define a relation $R$ from $A$ to $A$ by $R = \{(x, y) : y = x + 1\}$. Write down the domain and range of $R$.
ગણ $A = \{1, 2, 3, 4, 5, 6\}$ આપેલ છે. $A$ થી $A$ પરનો સંબંધ $R = \{(x, y) : y = x + 1\}$ દ્વારા વ્યાખ્યાયિત છે. આ સંબંધનો પ્રદેશ અને વિસ્તાર મેળવો.
Solution: Here, $x \in A$ and $y \in A$.
If we take $x = 1$, then $y = 1 + 1 = 2$. By continuously substituting values of $x$:
$R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$
(Note: We cannot take $x = 6$ because it would result in $y = 7$, which is not in set $A$).
- Domain: The set of first elements = $\{1, 2, 3, 4, 5\}$
- Range: The set of second elements = $\{2, 3, 4, 5, 6\}$
Classwork – 3 (Moderate Logic)
If relations $R_1$ and $R_2$ from set A to set B are defined as $R_1 = \{(1, 2), (3, 4), (5,6)\}$ and $R_2 = \{(2,1), (4,3), (6,5)\}$, then $n(A \times B)$ can be equal to
(A) 35 (B) 53 (C) 91 (D) 55
જો સબંધ $R_1$ અને $R_2$ એ ગણ A થી B માં છે. $R_1 = \{(1,2), (3,4), (5,6)\}$ અને $R_2 = \{(2, 1), (4, 3), (6, 5)\}$, તો $n(A \times B)$ શું હોઈ શકે?
(A) 35 (B) 53 (C) 91 (D) 55
Solution: From the relation $R_1$, the elements in set $A$ must include at least $\{1, 3, 5\}$ and the elements in set $B$ must include at least $\{2, 4, 6\}$.
From the relation $R_2$, the elements in set $A$ must include at least $\{2, 4, 6\}$ and the elements in set $B$ must include at least $\{1, 3, 5\}$.
Combining these, the minimum elements in set $A$ = $\{1, 2, 3, 4, 5, 6\} \Rightarrow n(A) \ge 6$.
Similarly, the minimum elements in set $B$ = $\{1, 2, 3, 4, 5, 6\} \Rightarrow n(B) \ge 6$.
Therefore, $n(A \times B) = n(A) \times n(B) \ge 36$.
Also, $n(A \times B)$ must be a product of two integers, both of which are $\ge 6$.
Looking at the options, $91 = 7 \times 13$ (both factors $\ge 6$). However, $55 = 5 \times 11$, but $5 < 6$.
Correct Answer: (C) 91
Linked Homework: Q1, Q2, Q3, Q12, Q15
Classwork – 6 (Hard / JEE Main)
If $R = \{(x,y) : x,y \in Z, x^2 + 3y^2 \le 8\}$ is a relation on the set of integers Z, then the domain of $R^{-1}$ is:
(A) {-2,-1,1,2} (B) {-1,0,1} (C) {-2,-1,0, 1, 2} (D) {0, 1}
જો $R = \{(x,y) : x,y \in Z, x^2 + 3y^2 \le 8\}$ પૂર્ણાંક સંખ્યા Z પર વ્યાખ્યાયિત છે. તો $R^{-1}$ નો પ્રદેશ મેળવો.
(A) {-2,-1,1,2} (B) {-1,0,1} (C) {-2,-1,0,1,2} (D) {0, 1}
Solution: We know that the domain of $R^{-1}$ is equal to the range of $R$.
Thus, we need to find the possible values of $y$.
Given inequality: $x^2 + 3y^2 \le 8 \Rightarrow 3y^2 \le 8 – x^2$
For any integer $x$, $x^2 \ge 0$. To find the maximum value of $y$, we take $x = 0$:
$$ 3y^2 \le 8 \Rightarrow y^2 \le \frac{8}{3} \Rightarrow y^2 \le 2.66 $$
Since $y \in Z$ (integer), the only possible integer values for $y^2$ are 0 and 1.
If $y^2 = 0 \Rightarrow y = 0$
If $y^2 = 1 \Rightarrow y = \pm 1$
Therefore, $y \in \{-1, 0, 1\}$. This is the range of $R$, which is also the domain of $R^{-1}$.
Correct Answer: (B) {-1, 0, 1}
Linked Homework: Q4, Q5, Q6, Q7 to Q11, Q13, Q14, Q18, Q19, Q20
Theory – 4: Advanced Logical Relations
In JEE level questions, relations are often defined using prime numbers, divisibility tests, or custom conditions. Find combinations manually using logic.
JEE લેવલના પ્રશ્નોમાં સંબંધની શરતો અવિભાજ્ય સંખ્યાઓ (Prime numbers) અથવા વિભાજ્યતા પર આધારિત હોય છે. આવા પ્રશ્નોમાં શરતને સંતોષતી જોડો (Pairs) લોજિકથી ગણવી પડે છે.
Classwork – 5 (Hard)
Let R be a relation from the set {1, 2, 3…….,60} to itself such that $R = \{(a, b) : b = pq$, where p, $q \ge 3$ are prime numbers}. Then, the number of elements in R is:
(A) 600 (B) 660 (C) 540 (D) 720
સંબંધ R એ ગણ {1, 2, 3…….,60} પર વ્યાખ્યાયિત છે. R = {(a, b): b = pq, જ્યાં p, $q \ge 3$ અવિભાજય સંખ્યા છે.} તો R માં રહેલા ઘટકોની સંખ્યા મેળવો.
(A) 600 (B) 660 (C) 540 (D) 720
Solution: Given set $A = \{1, 2, 3, \dots, 60\}$.
Condition: $b = p \times q$, where $p, q \ge 3$ are prime numbers and $b \le 60$.
Possible prime numbers $\ge 3$: $\{3, 5, 7, 11, 13, 17, 19, \dots\}$
Possible values of $b$:
- For $p=3$: $3 \times 3=9, 3 \times 5=15, 3 \times 7=21, 3 \times 11=33, 3 \times 13=39, 3 \times 17=51, 3 \times 19=57$ (Total 7 values)
- For $p=5$: $5 \times 5=25, 5 \times 7=35, 5 \times 11=55$ (Total 3 values)
- For $p=7$: $7 \times 7=49$ (Total 1 value)
Total possible values of $b$ = $7 + 3 + 1 = 11$.
There is no restriction on $a$ in the given condition. Thus, $a$ can take any of the 60 values from the set.
Total elements (ordered pairs) in the relation = (possibilities for $a$) $\times$ (possibilities for $b$) = $60 \times 11 = 660$.
Correct Answer: (B) 660
Linked Homework: Q16, Q21, Q22, Q23, Q24, Q25