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Provide solution for RD Sharma maths Class 12 Chapter Functions Exercise 2.2 Question 2.

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Answer : f\; o\; g= \left \{ (1,1),(3,1),(4,3),(5,3) \right \}

               g\; o\; f= \left \{ (3,3),(9,3),(12,9) \right \}

Hint : The set which contains all the elements of all the ordered pairs of relation R is known as the domain of the relation. The set            which contains all the second element, is known as the range of the relation.

Given : f=\left \{ (3,1),(9,3),(12,4) \right \}

              g=\left \{ (-1,-2),(-2,-4),(-3,-6),(4,8) \right \}

Prove : g\; o\; f and f\; o\; g are both defined

Solution :  f:\left \{ 3,9,12 \right \}\rightarrow \left \{ 1,3,4 \right \}\; \text {and}\; g:\left \{ 1,3,4,5 \right \}\rightarrow \left \{ 3,9 \right \}

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

So, g\; o\; f exist and g\; o\; f:\left \{ 3,9,12 \right \}\rightarrow \left \{ 3,9 \right \}

      (g\; o\; f)(3)=g(f(3))=g(1)=3

      (g\; o\; f)(9)=g(f(9))=g(3)=3

      (g\; o\; f)(12)=g(f(12))=g(4)=9

       g\; o\; f =\left \{ \left ( 3,3 \right ),\left ( 9,3 \right ),\left ( 12,9 \right ) \right \}

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

So, f\; o\; g exist and f\; o\; g:\left \{ 1,3,4,5 \right \}\rightarrow \left \{ 3,9,12 \right \}

      (f\; o\; g)(1)=f(g(1))=f(3)=1

       (f\; o\; g)(3)=f(g(3))=f(3)=1

      (f\; o\; g)(4)=f(g(4))=f(9)=3

      (f\; o\; g)(5)=f(g(5))=f(9)=3

     f\; o\; g=\left \{ (1,1),(3,1),(4,3),(5,3) \right \}

Hence proved, g\; o\; f and f\; o\; g are both defined.

             f\; o\; g=\left \{ (1,1),(3,1),(4,3),(5,3) \right \} and

             g\; o\; f=\left \{ (3,3),(9,3),(12,9) \right \}

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