Q82 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
A relation $R$ is defined from $\{2,3,4\}$ to $\{3,6,7,10\}$. If $x R y \Leftrightarrow x$ and $y$ are co prime numbers. Then range of $R$ is
- A$\{3,6,7,10\}$
- B$\{2,3,4\}$
- C$\{3,7\}$
- Correct answerD$\{3,7,10\}$
Solution
The relation $R$ is defined from the set $A = \{2, 3, 4\}$ to the set $B = \{3, 6, 7, 10\}$ such that $x R y$ if and only if $x$ and $y$ are coprime, meaning $\gcd(x, y) = 1$.
We check each element of $A$ against each element of $B$:
For $x = 2$:
$\gcd(2, 3) = 1$ (coprime)
$\gcd(2, 6) = 2$ (not coprime)
$\gcd(2, 7) = 1$ (coprime)
$\gcd(2, 10) = 2$ (not coprime)
So, $2 R 3$ and $2 R 7$.
For $x = 3$:
$\gcd(3, 3) = 3$ (not coprime)
$\gcd(3, 6) = 3$ (not coprime)
$\gcd(3, 7) = 1$ (coprime)
$\gcd(3, 10) = 1$ (coprime)
So, $3 R 7$ and $3 R 10$.
For $x = 4$:
$\gcd(4, 3) = 1$ (coprime)
$\gcd(4, 6) = 2$ (not coprime)
$\gcd(4, 7) = 1$ (coprime)
$\gcd(4, 10) = 2$ (not coprime)
So, $4 R 3$ and $4 R 7$.
The set of all elements $y \in B$ such that there exists $x \in A$ with $x R y$ is the range of $R$.
The elements $y$ that satisfy the condition are $3, 7, 10$.
Therefore, the range of $R$ is $\{3, 7, 10\}$.
Answer: $\{3,7,10\}$
We check each element of $A$ against each element of $B$:
For $x = 2$:
$\gcd(2, 3) = 1$ (coprime)
$\gcd(2, 6) = 2$ (not coprime)
$\gcd(2, 7) = 1$ (coprime)
$\gcd(2, 10) = 2$ (not coprime)
So, $2 R 3$ and $2 R 7$.
For $x = 3$:
$\gcd(3, 3) = 3$ (not coprime)
$\gcd(3, 6) = 3$ (not coprime)
$\gcd(3, 7) = 1$ (coprime)
$\gcd(3, 10) = 1$ (coprime)
So, $3 R 7$ and $3 R 10$.
For $x = 4$:
$\gcd(4, 3) = 1$ (coprime)
$\gcd(4, 6) = 2$ (not coprime)
$\gcd(4, 7) = 1$ (coprime)
$\gcd(4, 10) = 2$ (not coprime)
So, $4 R 3$ and $4 R 7$.
The set of all elements $y \in B$ such that there exists $x \in A$ with $x R y$ is the range of $R$.
The elements $y$ that satisfy the condition are $3, 7, 10$.
Therefore, the range of $R$ is $\{3, 7, 10\}$.
Answer: $\{3,7,10\}$
Hence, the correct answer is option: 'D'.