If the functions $f$ and $g$ are defined by $f(x)=3 x-4, \quad g(x)=2+3 x \quad$ for $\quad x \in R$…

If the functions $f$ and $g$ are defined by $f(x)=3 x-4, \quad g(x)=2+3 x \quad$ for $\quad x \in R$ respective, then $g^{-1}\left(f^{-1}(5)\right)$ is equal to
  1. $1$
  2. $\frac{1}{2}$
  3. $\frac{1}{3}$
  4. $\frac{1}{4}$

Solution

We have, $ \begin{aligned} & f(x)=3 x-4=y \\ & \Rightarrow \quad 3 x=y+4 \Rightarrow x=\frac{y+4}{3} \\ & \Rightarrow \quad f^{-1}(y)=\frac{y+4}{3} \\ & \text { and } \quad g(x)=2+3 x=z \\ & \Rightarrow \quad 3 x=z-2 \\ & \Rightarrow \quad x=\frac{z-2}{3} \\ & \Rightarrow \quad g^{-1}(z)=\frac{z-2}{3} \\ & \therefore \quad g^{-1}\left(f^{-1}(5)\right)=g^{-1}\left[\frac{5+4}{3}\right]=g^{-1}[3] \\ & =\frac{3-2}{3}=\frac{1}{3} \\ & \end{aligned} $

Asked in: AP EAMCET 2002

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