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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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