Let Z be the set of all integers,A = {(x, y) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) First,find the set $A \cap B$:$A = \{(x, y) \in Z 	imes Z : (x-2)^{2} + y^{2} \leq 4\}$$B = \{(x, y) \in Z 	imes Z : x^{2} + y^{2} \leq 4\}$The

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) First,find the set $A \cap B$:$A = \{(x, y) \in Z 	imes Z : (x-2)^{2} + y^{2} \leq 4\}$$B = \{(x, y) \in Z 	imes Z : x^{2} + y^{2} \leq 4\}$The integer points $(x, y)$ satisfying both inequalities are $(1, 0), (1, 1), (1, -1), (2, 0), (0, 0)$.Thus,$n(A \cap B) = 5$.Next,find the set $A \cap C$:$A = \{(x, y) \in Z 	imes Z : (x-2)^{2} + y^{2} \leq 4\}$$C = \{(x, y) \in Z 	imes Z : (x-2)^{2} + (y-2)^{2} \leq 4\}$The integer points $(x, y)$ satisfying both inequalities are $(2, 0), (2, 1), (2, 2), (1, 1), (3, 1)$.Thus,$n(A \cap C) = 5$.The total number of relations from $A \cap B$ to $A \cap C$ is given by $2^{n(A \cap B) 	imes n(A \cap C)} = 2^{5 	imes 5} = 2^{25}$.Comparing this with $2^{p}$,we get $p = 25$.

Similar Questions

If $R$ is a relation from set $A$ to set $B$ and $S$ is a relation from set $B$ to set $C$,then the relation $S \circ R$ is:

Let $R$ be the real line. Let the relations $S$ and $T$ on $R$ be defined by $S = \{(x, y) : y = x + 1, 0 < x < 2\}$ and $T = \{(x, y) : (x - y) \text{ is an integer}\}$. Then:

Let $T$ be the set of all triangles in a plane with $R$ a relation in $T$ given by $R = \{(T_1, T_2) : T_1 \text{ is congruent to } T_2\}$. Show that $R$ is an equivalence relation.

Let $X$ be a family of sets and $R$ be a relation on $X$ defined by '$A$ is disjoint from $B$'. Then $R$ is

Let $A = \{1, 2, 3\}$. The number of relations on $A$ containing $(1, 2)$ and $(2, 3)$ which are reflexive and transitive but not symmetric is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module