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