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  • Provide solution for rd sharma maths class 12 chapter Functions exercise 2.4 question 3 sub-question (0)

Provide solution for rd sharma maths class 12 chapter Functions exercise 2.4 question 3 sub-question (0)  

Answers (1)

Answer:

\left ( gof \right )^{-1}=f^{-1}og^{-1}

Given:

f=\left \{ \left ( 1,a \right ),\left ( 2,b \right ),\left ( c,3 \right ) \right \},g=\left \{ \left ( a,apple \right ),\left ( b,ball \right ),\left ( c,cat \right ) \right \}, clearly f and g are bijections.

Hint:

Here, f, g  and fog are invertible.

Solution:

Now, f^{-1}=\left \{ \left ( a,1 \right ),\left ( b,2 \right ),\left ( 3,c \right ) \right \},g^{-1}=\left \{ \left ( apple,a \right ),\left ( ball,b \right ),\left ( cat,c \right ) \right \}

So, f^{-1}og^{-1}=\left \{ \left ( apple,1 \right ),\left ( ball,2 \right ),\left ( cat,3 \right ) \right \}

f:\left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}\: and\: g:\left \{ a,b,c \right \}\rightarrow \left \{ apple,ball,cat \right \}

So, gof:\left \{ 1,2,3 \right \}\rightarrow \left \{ apple,ball,cat \right \}

\Rightarrow \left ( gof \right )\left ( 1 \right )=g\left [ f\left ( 1 \right ) \right ]=g\left ( a \right )=apple

\left ( gof \right )\left ( 2 \right )=g\left [ f\left ( 2 \right ) \right ]=g\left ( b \right )=ball

\left ( gof \right )\left ( 3 \right )=g\left [ f\left ( 3 \right ) \right ]=g\left ( c \right )=cat

\therefore \left ( gof \right )=\left \{ \left ( 1,apple \right ),\left ( 2,ball \right ),\left ( 3,cat \right ) \right \}

Clearly, gof is a bijection.

So, gof is invertible.

\left ( gof \right )^{-1}=\left \{ \left ( apple,1 \right ),\left ( ball,2 \right ),\left ( cat,3 \right ) \right \}

From\: \left ( i \right ),\: \left ( iii \right ),  we get,

\left ( gof \right )^{-1}=f^{-1}og^{-1}

 

Posted by

Gurleen Kaur

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