Ris a relation from \[\left\{ {11,{\rm{ }}12,{\rm{ }}13} \right\}{\rm{ }}\;to\;{\rm{ }}\left\{ {8,{\rm{ }}10,{\rm{ }}12} \right\}\] defined by $y = x - 3$. Then, ${R^{ - 1}}$ is

$\left\{ {\left( {8,{\rm{ }}11} \right),{\rm{ }}\left( {10,{\rm{ }}13} \right)} \right\}$

$\left\{ {\left( {11,{\rm{ }}18} \right),{\rm{ }}\left( {13,{\rm{ }}10} \right)} \right\}$

$\left\{ {\left( {10,{\rm{ }}13} \right),{\rm{ }}\left( {8,{\rm{ }}11} \right)} \right\}$

JEE MATHEMATICS MCQ: RELATIONS

1 Easy

1 of 1

Application +4 | -1

Ris a relation from

\left\{ {11,\mathrm{ }12,\mathrm{ }13} \right\}\mathrm{ }\;to\;\mathrm{ }\left\{ {8,\mathrm{ }10,\mathrm{ }12} \right\}

defined by

y = x - 3

. Then,

{R^{ - 1}}

View answer and explanation

Correct option: A

\left\{ {\left( {8,\mathrm{ }11} \right),\mathrm{ }\left( {10,\mathrm{ }13} \right)} \right\}

Option A is correct because the inverse relation

{R^{ - 1}}

is obtained by swapping the elements in each ordered pair of the relation R.