Login/Sign Up
Let \[\mathbf{f}~=\text{ }\left\{ \left( \mathbf{3},\text{ }\mathbf{1} \right),\text{ }\left( \mathbf{9},\text{ }\mathbf{3} \right),\text{ }\left( \mathbf{12},\text{ }\mathbf{4} \right) \right\}\]and \[\mathbf{g}~=\text{ }\left\{ \left( \mathbf{1},\text{ }\mathbf{3} \right),\text{ }\left( \mathbf{3},\text{ }\mathbf{3} \right)\text{ }\left( \mathbf{4},\text{ }\mathbf{9} \right)\text{ }\left( \mathbf{5},\text{ }\mathbf{9} \right) \right\}\]. Show that gof and fog are both defined. Also, find fog and gof.
Let \[\mathbf{f}~=\text{ }\left\{ \left( \mathbf{3},\text{ }\mathbf{1} \right),\text{ }\left( \mathbf{9},\text{ }\mathbf{3} \right),\text{ }\left( \mathbf{12},\text{ }\mathbf{4} \right) \right\}\]and \[\mathbf{g}~=\text{ }\left\{ \left( \mathbf{1},\text{ }\mathbf{3} \right),\text{ }\left( \mathbf{3},\text{ }\mathbf{3} \right)\text{ }\left( \mathbf{4},\text{ }\mathbf{9} \right)\text{ }\left( \mathbf{5},\text{ }\mathbf{9} \right) \right\}\]. Show that gof and fog are both defined. Also, find fog and gof.
CBSE | Class 12 | Exercise 2.2 | Function | Maths | RD Sharma
Let \[\mathbf{f}~=\text{ }\left\{ \left( \mathbf{3},\text{ }\mathbf{1} \right),\text{ }\left( \mathbf{9},\text{ }\mathbf{3} \right),\text{ }\left( \mathbf{12},\text{ }\mathbf{4} \right) \right\}\]and \[\mathbf{g}~=\text{ }\left\{ \left( \mathbf{1},\text{ }\mathbf{3} \right),\text{ }\left( \mathbf{3},\text{ }\mathbf{3} \right)\text{ }\left( \mathbf{4},\text{ }\mathbf{9} \right)\text{ }\left( \mathbf{5},\text{ }\mathbf{9} \right) \right\}\]. Show that gof and fog are both defined. Also, find fog and gof.

Given \[\mathbf{f}~=\text{ }\left\{ \left( \mathbf{3},\text{ }\mathbf{1} \right),\text{ }\left( \mathbf{9},\text{ }\mathbf{3} \right),\text{ }\left( \mathbf{12},\text{ }\mathbf{4} \right) \right\}\]and \[\mathbf{g}~=\text{ }\left\{ \left( \mathbf{1},\text{ }\mathbf{3} \right),\text{ }\left( \mathbf{3},\text{ }\mathbf{3} \right)\text{ }\left( \mathbf{4},\text{ }\mathbf{9} \right)\text{ }\left( \mathbf{5},\text{ }\mathbf{9} \right) \right\}\]

\[f~:\text{ }\left\{ 3,\text{ }9,\text{ }12 \right\}\text{ }\to \text{ }\left\{ 1,\text{ }3,\text{ }4 \right\}\]and \[g~:\text{ }\left\{ 1,\text{ }3,\text{ }4,\text{ }5 \right\}\text{ }\to \text{ }\left\{ 3,\text{ }9 \right\}\]

Co-domain of f is a subset of the domain of g.

So, gof exists and \[gof:\text{ }\left\{ 3,\text{ }9,\text{ }12 \right\}\text{ }\to \text{ }\left\{ 3,\text{ }9 \right\}\]

\[\left( gof \right)~\left( 3 \right)\text{ }=\text{ }g~\left( f~\left( 3 \right) \right)\text{ }=\text{ }g~\left( 1 \right)~=\text{ }3\]

\[\left( gof \right)~\left( 9 \right)\text{ }=\text{ }g~\left( f~\left( 9 \right) \right)\text{ }=\text{ }g~\left( 3 \right)\text{ }=\text{ }3\]

\[\left( gof \right)~\left( 12 \right)\text{ }=\text{ }g~\left( f~\left( 12 \right) \right)\text{ }=\text{ }g~\left( 4 \right)\text{ }=\text{ }9\]

⇒ \[gof~=\text{ }\left\{ \left( 3,~3 \right),~\left( 9,~3 \right),~\left( 12,~9 \right) \right\}\]

Co-domain of g is a subset of the domain of f.

So, fog exists and \[fog:\text{ }\left\{ 1,\text{ }3,\text{ }4,\text{ }5 \right\}\text{ }\to \text{ }\left\{ 3,\text{ }9,\text{ }12 \right\}\]

\[\left( fog \right)~\left( 1 \right)\text{ }=\text{ }f~\left( g~\left( 1 \right) \right)\text{ }=\text{ }f~\left( 3 \right)\text{ }=\text{ }1\]

\[\left( fog \right)~\left( 3 \right)\text{ }=\text{ }f~\left( g~\left( 3 \right) \right)\text{ }=\text{ }f~\left( 3 \right)\text{ }=\text{ }1\]

\[\left( fog \right)~\left( 4 \right)\text{ }=\text{ }f~\left( g~\left( 4 \right) \right)\text{ }=\text{ }f~\left( 9 \right)\text{ }=\text{ }3\]

\[\left( fog \right)~\left( 5 \right)\text{ }=\text{ }f~\left( g~\left( 5 \right) \right)\text{ }=\text{ }f~\left( 9 \right)\text{ }=\text{ }3\]

⇒ \[fog\text{ }=\text{ }\left\{ \left( 1,~1 \right),~\left( 3,~1 \right),~\left( 4,~3 \right),~\left( 5,~3 \right) \right\}\]

i

More Material