Login/Sign Up
Home » Let and f: be defined as and Express and as the sets of ordered pairs and verify that (gof) .
Let A=\{1,2,3,4\} ; B=\{3,5,7,9\} ; C=\{7,23,47,79\} and f: A \rightarrow B, g: B \rightarrow C be defined as f(x)=2 x+1 and g(x)=x^{2}-2 . Express (\text { gof })^{-1} and f^{-1} o g^{-1} as the sets of ordered pairs and verify that (gof) ^{-1}=\mathbf{f}^{-1} \mathrm{og}^{-1}.
CBSE | Class 12 | Exercise 2.4 | Function | Maths | RD Sharma
Let A=\{1,2,3,4\} ; B=\{3,5,7,9\} ; C=\{7,23,47,79\} and f: A \rightarrow B, g: B \rightarrow C be defined as f(x)=2 x+1 and g(x)=x^{2}-2 . Express (\text { gof })^{-1} and f^{-1} o g^{-1} as the sets of ordered pairs and verify that (gof) ^{-1}=\mathbf{f}^{-1} \mathrm{og}^{-1}.

\Rightarrow \mathrm{f}=\{(1,2(1)+1),(2,2(2)+1),(3,2(3)+1),(4,2(4)+1)\}

=\{(1,3),(2,5),(3,7),(4,9)\}

Also given that g(x)=x^{2}-2

\Rightarrow g=\left\{\left(3,3^{2}-2\right),\left(5,5^{2}-2\right),\left(7,7^{2}-2\right),\left(9,9^{2}-2\right)\right\}

=\{(3,7),(5,23),(7,47),(9,79)\}

Clearly \mathrm{f} and \mathrm{g} are bijections and, hence, \mathrm{f}^{-1}: \mathrm{B} \rightarrow \mathrm{A} and \mathrm{g}^{-1}: \mathrm{C} \rightarrow \mathrm{B} exist.

So, f^{-1}=\{(3,1),(5,2),(7,3),(9,4)\}

And g^{-1}=\{(7,3),(23,5),(47,7),(79,9)\}

Now, \left(f^{-1} \circ g^{-1}\right): C \rightarrow A

f^{-1} \circ g^{-1}=\{(7,1),(23,2),(47,3),(79,4)\} \ldots \ldots \ldots (1)

Also, f: A \rightarrow B and g: B \rightarrow C,

\Rightarrow gof: A \rightarrow C,(\text { gof })^{-1}: C \rightarrow A

So, f^{-1} \circ g^{-1} and (\text { gof })^{-1} have same domains.

( gof )(x)=g(f(x))

=g(2 x+1)

=(2 x+1)^{2}-2

\Rightarrow( gof )(x)=4 x^{2}+4 x+1-2

\Rightarrow( gof )(x)=4 x^{2}+4 x-1

Then, (gof) (1)=g(f(1))

=4+4-1

=7,

( gof )(2)=g(f(2))

=4(2)^{2}+4(2)-1=23

( gof) (3)=g(f(3))

=4(3)^{2}+4(3)-1=47 and

    \[\begin{array}{*{35}{l}} \left( gof \right)\text{ }\left( 4 \right)\text{ }=\text{ }g~\left( f~\left( 4 \right) \right) \\ =\text{ }4{{\left( 4 \right)}^{2}}~+\text{ }4\left( 4 \right)\text{ }-\text{ }1\text{ }=\text{ }79 \\ So,~gof\text{ }=\text{ }\left\{ \left( 1,~7 \right),~\left( 2,~23 \right),~\left( 3,~47 \right),~\left( 4,~79 \right) \right\} \\ \Rightarrow \text{ }{{\left( gof \right)}^{\text{ }1}}~=\text{ }\left\{ \left( 7,~1 \right),~\left( 23,~2 \right),~\left( 47,~3 \right),~\left( 79,~4 \right) \right\}\ldots \ldots \text{ }\left( 2 \right) \\ \end{array}\begin{array}{*{35}{l}} From~\left( 1 \right)~and~\left( 2 \right),~we~get: \\ {{\left( gof \right)}^{-1~}}=~{{f}^{-1}}~o~{{g}^{-1}} \\ \end{array}\]

i

More Material