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  • Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 13 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 13 Maths Textbook Solution.

Answers (1)

Answer:

f^{-1}\left ( x \right )=\frac{3x-2}{x-1}

Given:

                A=R-\left \{ 3 \right \},B=R-\left \{ 1 \right \},f:A\rightarrow B    defined by f\left ( x \right )=\frac{x-2}{x-3}.

Hint:

One-one means every domain has a distinct range. One means every image has some preimage in the domain function.

Solution:

We have,A=R-\left \{ 3 \right \},B=R-\left \{ 1 \right \}.

The function f:A\rightarrow B defined by f\left ( x \right )=\frac{x-2}{x-3}.

Let  x,y\epsilon A such that f\left ( x \right )=f\left ( y \right ) then

                \frac{x-2}{x-3}=\frac{y-2}{y-3}

                xy-3x-2y+6=xy-2x-3y+6 

                -x=-y

                x=y

fis one-one.

Let y\epsilon B, then y\neq 1.

The function fis onto if there exists x\epsilon A such that f\left ( x \right )=y.

\Rightarrow \frac{x-2}{x-3}=y

x-2=xy-3y

x-xy=2-3y

x=\frac{2-3y}{1-y}\epsilon A\left ( y\neq 1 \right )

Thus, for any y\epsilon B, there exists \frac{2-3y}{1-y}\epsilon A such that

                f\left ( \frac{2-3y}{1-y} \right )=\frac{\left ( \frac{2-3y}{1-y} \right )-2}{\left ( \frac{2-3y}{1-y} \right )-3}=\frac{2-3y-2+2y}{2-3y-3+3y}=\frac{-y}{-1}=y

f is onto.

So, f is one-one and onto function.

                x=\frac{2-3y}{1-y}

                f^{-1}\left ( x \right )=\frac{3x-2}{x-1}

               

Posted by

infoexpert27

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