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Provide solution for rd sharma maths class 12 chapter Functions exercise 2.4 question 4 sub-question (0)  

Answers (1)

Answer:

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

Given:

A=\left \{ 1,2,3,4 \right \},B=\left \{ 3,5,7,9 \right \},C=\left \{ 7,23,47,79 \right \}

f:A\rightarrow B,g:B\rightarrow Cdefined\: by\: f\left ( x \right )=2x+1,g\left ( x \right )=x^{2}-2.

Hint:

Clearly f and g are bijections.

Solution:

f\left ( x \right )=2x+1

f=\left \{ \left ( 1,2\left ( 1 \right )+1 \right ),\left ( 2,2\left ( 2 \right ) +1 \right ),\left ( 3,2\left ( 3 \right ) +1\right ),\left ( 4,2\left ( 4 \right ) +1\right )\right \}

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

g\left ( x \right )=x^{2}-2

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 \}

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

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

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

Now, \left ( f^{-1}0g^{-1} \right ):C\rightarrow A

\left ( f^{-1}og^{-1} \right )=\left \{ \left ( 7,1 \right ),\left ( 23,2 \right ),\left ( 47,3 \right ),\left ( 79,4 \right ) \right \}

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

gof:A\rightarrow C,\left ( gof \right )^{-1}:C\rightarrow A

So,  f^{-1}og^{-1}\: \: and\: \: \left ( gof \right )^{-1} have the same domains.

\left ( gof \right )\left ( x \right )=g\left ( x \right )=g\left ( 2x+1 \right )=\left ( 2x+1 \right )^{2}-2

\left ( gof \right )\left ( x \right )=g\left [ f\left ( x \right ) \right ]

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

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

Then, \left ( gof \right )\left ( 1 \right )=g\left ( f\left ( 1 \right ) \right )=4+4-1=7

           \left ( gof \right )\left ( 2 \right )=g\left ( f\left ( 2 \right ) \right )=16+8-1=23

         \left ( gof \right )\left ( 3 \right )=g\left ( f\left ( 3 \right ) \right )=36+12-1=47

         \left ( gof \right )\left ( 4 \right )=g\left ( f\left ( 4 \right ) \right )=64+16-1=79

So,   \left ( gof \right )=\left \{ \left ( 1,7 \right ),\left ( 2,23 \right ),\left ( 3,47 \right ),\left ( 4,79 \right ) \right \}

         \left ( gof \right )^{-1}=\left \{ \left ( 7,1 \right ),\left ( 23,2 \right ),\left ( 47,3 \right ),\left ( 79,4 \right ) \right \}

From \left ( i \right ),\left ( ii \right )

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

\left \{ \left ( 7,1 \right ),\left ( 23,2 \right ),\left ( 47,3 \right ),\left ( 79,4 \right ) \right \}=\left \{ \left ( 7,1 \right ) ,\left ( 23,2 \right ),\left ( 47,3 \right ),\left ( 79,4 \right )\right \}

Posted by

Gurleen Kaur

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