Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be a function such that \(f(x+y)=f(x)+f(y)\) for all \(x, y \in \mathbb{R}\), and \(g: \mathbb{R} \rightarrow(0, \infty)\) be a function such that \(g(x+y)=g(x) g(y)\) for all \(x, y \in \mathbb{R}\). If \(f\left(\frac{-3}{5}\right)=12\) and \(g\left(\frac{-1}{3}\right)=2\), then the value of \(\left(f\left(\frac{1}{4}\right)+g(-2)-8\right) g(0)\) is . ________.

\(f(x+y)=f(x)+f(y)\) ...(1)
\(\Rightarrow \quad \mathrm{f}(\mathrm{nx})=\operatorname{nf}(\mathrm{x}) \forall \mathrm{n} \in \mathrm{N}\) ...(2)
Now put \(y=-x\) in eq.(1)
\(\begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x})=\mathrm{f}(0) \quad\{\mathrm{f}(0)=0\} \\ \Rightarrow \quad & \mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x})\end{aligned}\)
\(\Rightarrow \quad \mathrm{f}\) is odd function from eq. (2)
from eq. (2)
\(\begin{aligned} & f(-n x)=\operatorname{nf}(-x) \\ & \Rightarrow \quad \mathrm{f}(-\mathrm{nx})=-n f(\mathrm{x}) \\ & \end{aligned}\)
\(\Rightarrow \quad \mathrm{f}(\mathrm{mx})=\operatorname{mf}(\mathrm{x}) \forall \mathrm{m} \in \mathrm{Z}\) ...(3)
from eq. (2) and eq. (3)
\(\mathrm{f}(\mathrm{nx})=\mathrm{nf}(\mathrm{x}) \forall \mathrm{n} \in \mathrm{Z}\) ...(4)
Now put \(x=\frac{p}{q}\) where \(\mathrm{p}, \mathrm{q} \in \mathrm{Z}, \mathrm{q} \neq 0\)
\(\mathrm{f}\left(\frac{\mathrm{np}}{\mathrm{q}}\right)=\operatorname{nf}\left(\frac{\mathrm{p}}{\mathrm{q}}\right) \forall \mathrm{n} \in \mathrm{Z}\)
put \(\mathrm{n}=\mathrm{q}\)
\(f(p)=q f\left(\frac{p}{q}\right)\)
\(\Rightarrow \quad \mathrm{pf}(1)=\mathrm{qf}\left(\frac{\mathrm{p}}{\mathrm{q}}\right) \quad\) {from eq.(4)}
Let \(\mathrm{f}(1)=\mathrm{a}\)
then \(\quad \mathrm{pa}=\mathrm{qf}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)\)
\(\mathrm{f}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)=\frac{\mathrm{ap}}{q}\)
\(\Rightarrow \quad \mathrm{f}(\mathrm{x})=\mathrm{ax} \forall \mathrm{x} \in \mathbb{Q}\)
Now, \(\mathrm{f}\left(\frac{-3}{5}\right)=\mathrm{a}\left(\frac{-3}{5}\right)=12 \Rightarrow \mathrm{a}=-20\)
\(\Rightarrow \quad \mathrm{f}(\mathrm{x})=-20 \mathrm{x} \forall \mathrm{x} \in \mathbb{Q}\) ...(5)
From the given functional equation it is not possible to find a unique function for irrational values of ' \(x\) ', there are infinitely many such functions satisfying given functional equation for irrational values of \(x\), but in this problem we finally need the function at rational values of ' \(x\) ' only. So, for rational values of \(x\) we are getting a unique function mentioned in (5).
Now, \(g(x+y)=g(x) \cdot g(y)\)
\(\Rightarrow \quad \ell \mathrm{n}(\mathrm{g}(\mathrm{x}+\mathrm{y})=\ell \mathrm{n}(\mathrm{g}(\mathrm{x}))+\ell \ln (\mathrm{g}(\mathrm{y}))\)
Let \(\quad \ln (\mathrm{g}(\mathrm{x}))=\mathrm{h}(\mathrm{x})\)
\(\begin{array}{ll}\Rightarrow & \mathrm{h}(\mathrm{x}+\mathrm{y})=\mathrm{h}(\mathrm{x})+\mathrm{h}(\mathrm{y}) \\ \Rightarrow & \mathrm{h}(\mathrm{x})=\mathrm{kx} \forall \mathrm{x} \in \mathbb{Q}\end{array}\)
\(\Rightarrow \quad \mathrm{g}(\mathrm{x})=\mathrm{e}^{\mathrm{kx}} \forall \mathrm{x} \in \mathbb{Q}\) ...(6)
and \(\mathrm{g}\left(\frac{-1}{3}\right)=\mathrm{e}^{-\frac{\mathrm{K}}{3}}=2 \quad \Rightarrow \quad \mathrm{K}=-3 \ell \ln 2\)
\(\Rightarrow \quad \mathrm{K}=\ln \left(\frac{1}{8}\right)\)
\(\Rightarrow \quad \mathrm{g}(\mathrm{x})=\mathrm{e}^{\ln \left(\frac{1}{8}\right) \cdot \mathrm{x}}=\left(\frac{1}{8}\right)^{\mathrm{x}}=2^{-3 \mathrm{x}} \forall \mathrm{x} \in \mathbb{Q}\)
\(\begin{aligned} & \text { Now, } f\left(\frac{1}{4}\right)=-5, g(-2)=2^6=64 \\ & g(0)=1\end{aligned}\)
\(\begin{aligned} & \text { So } \quad\left(\mathrm{f}\left(\frac{1}{4}\right)+\mathrm{g}(-2)-(8) \cdot g(0)\right) \\ & =(-5+64-8)(1)=51 \\ & \end{aligned}\)