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  • Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 12 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 12 Maths Textbook Solution.

Answers (1)

Answer: 

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

Given:

                f:Q\rightarrow Q,g:Q\rightarrow Q defined by f\left ( x \right )=2x,g\left ( x \right )=x+2.

Hint:

                Bijective function should be fulfil the injective and surjective function.

Solution:

x,y be two elements of domain\left ( Q \right ), such

                \! \! \! \! \! \! \! \! \! f\left (x \right )=f\left ( y \right )\\\\2x=2y\\\\x=y

                f is one-one.

Let y be in the co-domain\left ( Q \right ), such that

                f\left ( x \right )=y

                2x =y

                x=\frac{y}{2}\epsilon \: Q.

f is onto.

So, f is a bijection and hence, it’s invertible
Let          f^{-1}\left ( x \right )=y                                                                                                                                           …(i)                                                                       

       \\\\x=f\left ( y \right )\\\\y=\frac{x}{2}\\\\f^{-1}\left ( x \right )=\frac{x}{2}                                                                                                                                                                                                             Injectivity of g

Let  x,y be two elements of domain\left ( Q \right ).

                \! \! \! \! \! \! \! \! g\left ( x \right )=g\left ( y \right )\\\\x+2=y+2\\\\x=y

           So, g is one-one.

Subjectivity of g:

Let y be in the co-domain\left ( Q \right ), such that g\left ( x \right )=y.

                x+2=y

                x=2-y\: \epsilon \: (domain)

g is onto.

So, g is a bijection, hence it’s invertible.

Let          g^{-1}\left ( x \right )=y

                x=g\left ( y \right )                                                                                                                                                         …(ii)

                x=y+2

                y=x-2

So,          g^{-1}\left ( x \right )=x-2                                                                                                                                   [from (ii)]

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

f\left ( x \right )=2x,g\left ( x \right )=x+2,f^{-1}\left ( x \right )=\frac{x}{2},g^{-1}\left ( x \right )=x-2

\left ( f^{-1} og^{-1}\right )\left ( x \right )=f^{-1}\left ( x-2 \right )                                                   \left [ \because \left ( f^{-1} og^{-1}\right )\left ( x \right )=f^{-1}\left ( g^{-1}\left ( x \right ) \right ) \right ]

\left ( f^{-1}og^{-1} \right )\left ( x \right )=f^{-1}\left ( \frac{x-2}{2} \right )                                                                                                              …(iii)

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

                   =g\left (2 x \right )

                   =2x+1

Let          \left ( gof \right )\left ( x \right )=y                                                                                                                                 …(vi)

                x=\left ( gof \right )\left ( y \right )

                x=2y+2

                y=\frac{x-2}{2}

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

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