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Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 16 Maths Textbook Solution.

Answers (1)

Answer:

$f^{-1}\left ( x \right )=\frac{4}{4-3x}$

Given:

$f:R-\left \{ \frac{-4}{3} \right \}\rightarrow R-\left \{ \frac{4}{3} \right \}\:$ is given by $\: f\left ( x \right )=\frac{4x}{3x+4}$

Injectivity,

Let $x,y\: \epsilon \: R-\left \{ \frac{-4}{3} \right \}$ be such that $f\left ( x \right )=f\left ( y \right )$

$\frac{4x}{3x+4}=\frac{4y}{3y+4}$

$4x\left ( 3y+4 \right )=4y\left ( 3x+4 \right )$

$12xy+16x= 12xy+16y$

$16x=16y$

$x=y$

$f$ is a one-one function.

Let $y\:$ be an arbitrary element of $R-\left \{ \frac{4}{3} \right \}$ .

Then $f\left ( x \right )= y$

$\frac{4x}{3x+4}=y$

$4x=3xy+4y$

$4x-3xy=4y$

$x=\frac{4y}{4-3y}$

As $y\: \epsilon \: R-\left \{ \frac{4}{3} \right \},\frac{4y}{4-3y}=\frac{-4}{3}$

Also, $\frac{4y}{4-3y}\neq \frac{-4}{3}$ because, $\frac{4y}{4-3y}= \frac{-4}{3}$

$12y=-16+12y\Rightarrow 0=-16$

Which is not possible.

Thus,

$x=\frac{4y}{4-3y}\: \epsilon \: R-\left \{ \frac{-4}{3} \right \}$

$f\left ( x \right )=f\left ( \frac{4x}{3x+4} \right )=\frac{4\left ( \frac{4y}{4-3y} \right )}{3\left ( \frac{4y}{4-3y} \right )+4}$

$=\frac{16y}{12y+16-12y}=\frac{16y}{16}$

$y$

So every element in $R-\left \{ \frac{4}{3} \right \}$ has pre image in $R-\left \{ \frac{-4}{3} \right \}$

Hence $f$ is onto.

Now, $x=\frac{4y}{4-3y}$ , replacing $x$ by $f^{-1}$ and $y$ by $x$ ,

$f^{-1}\left ( x \right )=\frac{4}{4-3x}$

Posted by

infoexpert27

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