Taking the set of natural numbers as the universal set,write down the complement of the following set:
$A = \{ x : x \in N \text{ and } 2x + 1 > 10 \}$
$A = \{ x : x \in N \text{ and } 2x + 1 > 10 \}$
The universal set is $U = N = \{1, 2, 3, 4, 5, \dots \}$.
Given set $A = \{ x : x \in N \text{ and } 2x + 1 > 10 \}$.
Solving the inequality: $2x > 9 \implies x > 4.5$.
Since $x \in N$,the set $A = \{5, 6, 7, 8, \dots \}$.
The complement of $A$ is $A' = U - A$.
$A' = \{ x : x \in N \text{ and } x \le 4.5 \}$.
Since $x \in N$,$A' = \{1, 2, 3, 4 \}$.
Given set $A = \{ x : x \in N \text{ and } 2x + 1 > 10 \}$.
Solving the inequality: $2x > 9 \implies x > 4.5$.
Since $x \in N$,the set $A = \{5, 6, 7, 8, \dots \}$.
The complement of $A$ is $A' = U - A$.
$A' = \{ x : x \in N \text{ and } x \le 4.5 \}$.
Since $x \in N$,$A' = \{1, 2, 3, 4 \}$.
Explore More
Similar Questions
If $P(A) = 0.65$ and $P(B) = 0.15$,then $P(\bar{A}) + P(\bar{B}) = $
Medium
View SolutionLet $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,$A = \{1, 2, 3, 4\}$,$B = \{2, 4, 6, 8\}$,and $C = \{3, 4, 5, 6\}$. Find $(A \cup C)'$.
Easy
View SolutionLet $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$,$A = \{1, 2, 5\}$,and $B = \{6, 7\}$. Find $A \cap B'$.
Easy
View SolutionTaking the set of natural numbers as the universal set,write down the complement of the following set:
$A = \{ x : x \text{ is a positive multiple of } 3 \}$
$A = \{ x : x \text{ is a positive multiple of } 3 \}$
Easy
View SolutionLet $U$ be the universal set and $A \cup B \cup C = U$. Then $\{ (A - B) \cup (B - C) \cup (C - A)\} '$ is equal to
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo