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KCET · Maths · Sets and Relations

Let the relation \(R\) is defined in \(N\) by \(a R b\), if \(3 a+2 b=27\) then \(R\) is

  1. A \(\{(1,12)(3,9)(5,6)(7,3)\}\)
  2. B \(\left\{\left(0, \frac{27}{2}\right)(1,12)(3,9)(5,6)(7,3)\right\}\)
  3. C \(\{(1,12)(3,9)(5,6)(7,3)(9,0)\}\)
  4. D \(\{(2,1)(9,3)(6,5)(3,7)\}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\{(1,12)(3,9)(5,6)(7,3)\}\)

Step-by-step Solution

Detailed explanation

Given relation is \(3 a+2 b=27\), where \(a, b \in N\)
\[
\begin{aligned}
& 3 a+2 b=27 \\
& \Rightarrow \quad b=\frac{27-3 a}{2} \\
& a=1 \Rightarrow b=\frac{27-3}{2}=\frac{24}{2}=12 \\
& a=3 \Rightarrow b=\frac{27-9}{2}=\frac{18}{2}=9 \\
& a=5 \Rightarrow b=\frac{27-15}{2}=\frac{12}{2}=6 \\
& a=7 \Rightarrow b=\frac{27-21}{2}=\frac{6}{2}=3
\end{aligned}
\]
Hence, we can see that all the elements of option (a) satisfies the given relation. Hence, option (a) is correct.