$\mathrm{f}: \mathbb{R}-\left(-\frac{3}{5}\right) \rightarrow \mathbb{R}$ is defined by $f(x)=\frac{3 x-2}{5 x+3}$, then $f \circ f(1)$ is

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Aicharya · Accepted Answer

$\begin{aligned} & f(x)=\frac{3 x-2}{5 x+3} \ & \begin{aligned} f(f(x)) & =\frac{3\left(\frac{3 x-2}{5 x+3}ight)-2}{5\left(\frac{3 x-2}{5 x+3}ight)+3} \ & =\frac{3(3 x-2)-2(5 x+3)}{5(3 x-2)+3(5 x+3)} \ & =\frac{9 x-6-10 x-6}{15 x-10+15 x+9} \ & =\frac{-x-12}{30 x-1}\end{aligned} \ & 	ext { fof }(1)=\frac{-1-12}{30-1}=\frac{-13}{29}\end{aligned}$

$\mathrm{f}: \mathbb{R}-\left(-\frac{3}{5}\right) \rightarrow \mathbb{R}$ is defined by $f(x)=\frac{3 x-2}{5…

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