Question

Mark the correct alternative in the following question:

Let f : R $-$ $\left\{\frac{3}{5}\right\}$ $\to$ R be defined by f(x) = $\frac{3x+2}{5x-3}$. Then,

(a) f ^$-$l(x) = f(x) (b) f ^$-$1(x) = $-$f(x) (c) fof(x) = $-$x (d) f ^$-$1(x) = $\frac{1}{19}$f(x)

Answer 1

We have,

f : R $-$ $\left\{\frac{3}{5}\right\}$ $\to$ R is defined by f(x) = $\frac{3x+2}{5x-3}$

$$\begin{gathered} f\mathrm{o}f\left(x\right)=f\left(f\left(x\right)\right) \\ =f\left(\frac{3x+2}{5x-3}\right) \\ =\frac{3\left({\displaystyle \frac{3x+2}{5x-3}}\right)+2}{5\left({\displaystyle \frac{3x+2}{5x-3}}\right)-3} \\ =\frac{\left({\displaystyle \frac{9x+6}{5x-3}}\right)+2}{\left({\displaystyle \frac{15x+10}{5x-3}}\right)-3} \\ =\frac{\left({\displaystyle \frac{9x+6+10x-6}{5x-3}}\right)}{\left({\displaystyle \frac{15x+10-15x+9}{5x-3}}\right)} \\ =\frac{{\displaystyle 19x}}{19} \\ =x \\ \mathrm{Let}y=\frac{3x+2}{5x-3} \\ \Rightarrow 5xy-3y=3x+2 \\ \Rightarrow 5xy-3x=3y+2 \\ \Rightarrow x\left(5y-3\right)=3y+2 \\ \Rightarrow x=\frac{3y+2}{5y-3} \\ \Rightarrow f^{-1}\left(y\right)=\frac{3y+2}{5y-3} \\ \mathrm{So},f^{-1}\left(x\right)=\frac{3x+2}{5x-3}=f\left(x\right) \end{gathered}$$

Hence, the correct alternative is option (a).