Mathematics - Functions Question with Solution | TestHub

$f(x+y)=f(x)+f(y)$ ...(1) $\Rightarrow \mathrm{f}(\mathrm{nx})=\mathrm{nf}(\mathrm{x})\forall \mathrm{n}\in \mathrm{N}$ ...(2) Now put $y=-x$ in eq.(1) $\begin{array}{rl}& \mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x})=\mathrm{f}(0)\{\mathrm{f}(0)=0\}\\ \Rightarrow & \mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x})\end{array}$ $\Rightarrow \mathrm{f}$ is odd function from eq. (2) from eq. (2) $\begin{array}{rl}& f(-nx)=\mathrm{nf}(-x)\\ & \Rightarrow \mathrm{f}(-\mathrm{nx})=-nf(\mathrm{x})\\ & \end{array}$ $\Rightarrow \mathrm{f}(\mathrm{mx})=\mathrm{mf}(\mathrm{x})\forall \mathrm{m}\in \mathrm{Z}$ ...(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 $\mathrm{p},\mathrm{q}\in \mathrm{Z},\mathrm{q}\ne 0$ $\mathrm{f}\left(\frac{\mathrm{np}}{\mathrm{q}}\right)=\mathrm{nf}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)\forall \mathrm{n}\in \mathrm{Z}$ put $\mathrm{n}=\mathrm{q}$ $f(p)=qf\left(\frac{p}{q}\right)$ $\Rightarrow \mathrm{pf}(1)=\mathrm{qf}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)$ {from eq.(4)} Let $\mathrm{f}(1)=\mathrm{a}$ then $\mathrm{pa}=\mathrm{qf}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)$ $\mathrm{f}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)=\frac{\mathrm{ap}}{q}$ $\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{ax}\forall \mathrm{x}\in \mathbb{Q}$ Now, $\mathrm{f}\left(\frac{-3}{5}\right)=\mathrm{a}\left(\frac{-3}{5}\right)=12\Rightarrow \mathrm{a}=-20$ $\Rightarrow \mathrm{f}(\mathrm{x})=-20\mathrm{x}\forall \mathrm{x}\in \mathbb{Q}$ ...(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, $g(x+y)=g(x)\cdot g(y)$ $\Rightarrow \ell \mathrm{n}(\mathrm{g}(\mathrm{x}+\mathrm{y})=\ell \mathrm{n}(\mathrm{g}(\mathrm{x}))+\ell \ln (\mathrm{g}(\mathrm{y}))$ Let $\ln (\mathrm{g}(\mathrm{x}))=\mathrm{h}(\mathrm{x})$ $\begin{array}{ll}\Rightarrow & \mathrm{h}(\mathrm{x}+\mathrm{y})=\mathrm{h}(\mathrm{x})+\mathrm{h}(\mathrm{y})\\ \Rightarrow & \mathrm{h}(\mathrm{x})=\mathrm{kx}\forall \mathrm{x}\in \mathbb{Q}\end{array}$ $\Rightarrow \mathrm{g}(\mathrm{x})={\mathrm{e}}^{\mathrm{kx}}\forall \mathrm{x}\in \mathbb{Q}$ ...(6) and $\mathrm{g}\left(\frac{-1}{3}\right)={\mathrm{e}}^{-\frac{\mathrm{K}}{3}}=2\Rightarrow \mathrm{K}=-3\ell \ln 2$ $\Rightarrow \mathrm{K}=\ln \left(\frac{1}{8}\right)$ $\Rightarrow \mathrm{g}(\mathrm{x})={\mathrm{e}}^{\ln \left(\frac{1}{8}\right)\cdot \mathrm{x}}={\left(\frac{1}{8}\right)}^{\mathrm{x}}=2^{-3\mathrm{x}}\forall \mathrm{x}\in \mathbb{Q}$ $\begin{array}{rl}& \text{ Now, }f\left(\frac{1}{4}\right)=-5,g(-2)=2^6=64\\ & g(0)=1\end{array}$ $\begin{array}{rl}& \text{ So }\left(\mathrm{f}\left(\frac{1}{4}\right)+\mathrm{g}(-2)-(8)\cdot g(0)\right)\\ & =(-5+64-8)(1)=51\\ & \end{array}$