Question

If $A=\{0,1,2,3,4,5\}$ and relation $R$ defined by $aRb$ such that $2a+b=10$ then $R^{-1}$ equals

• A. $\left\{\right(4,3),(2,4),(5,0\left)\right\}$
• B. $\left\{\right(3,4),(4,2),(5,0\left)\right\}$
• C. $\left\{\right(4,3),(2,4),(0,5\left)\right\}$
• D. $\left\{\right(4,3),(4,2),(5,0\left)\right\}$

Answer 1

The correct option is C $\left\{\right(4,3),(2,4),(0,5\left)\right\}$

$2a+b=10$, $2a$ is always even and $10$ is even.So, $b$ should also be even.
Hence the solution set is $\left\{\right(0,10),(1,8),(2,6),(3,4),(4,2),(5,0\left)\right\}$
But $A=\{0,1,2,3,4,5\}$

So $R$ becomes $\left\{\right(3,4),(4,2),(5,0\left)\right\}$

Hence $R^{-1}$ is $\left\{\right(4,3),(2,4),(0,5\left)\right\}$