KCET · Maths · Functions
Let the relation \(R\) be defined in \(N\) by \(a R b\), if \(3 a+2 b=27\), then \(R\) is
- A \(\left\{\left(0, \frac{27}{2}\right),(1,12),(3,9),(5,6),(7,3)\right\}\)
- B \(\{(1,12),(3,9),(5,6),(7,3),(9,0)\}\)
- C \(\{(2,1),(9,3),(6,5),(3,7)\}\)
- D \(\{(1,12),(3,9),(5,6),(7,3)\}\)
Answer & Solution
Correct Answer
(D) \(\{(1,12),(3,9),(5,6),(7,3)\}\)
Step-by-step Solution
Detailed explanation
Given that \(3 a+2 b=27\)
\(\begin{array}{ll}\Rightarrow & 2 b=27-3 a \\ \Rightarrow & b=\frac{27-3 a}{2}\end{array}\)
For \(a=1, b=12\)
\(\begin{aligned} & a=3, b=9 \\ & a=5, b=6 \\ & a=7, b=3 \\ & R=\{(1,12),(3,9),(5,6),(7,3)\} \\ & \end{aligned}\)
\(\begin{array}{ll}\Rightarrow & 2 b=27-3 a \\ \Rightarrow & b=\frac{27-3 a}{2}\end{array}\)
For \(a=1, b=12\)
\(\begin{aligned} & a=3, b=9 \\ & a=5, b=6 \\ & a=7, b=3 \\ & R=\{(1,12),(3,9),(5,6),(7,3)\} \\ & \end{aligned}\)
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