Let f: ℝ arrow ℝ be a function such that f(x+y)=f(x)+f(y) for all...

$f(x+y)=f(x)+f(y)$ ...(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}$$$