Mathematics - Functions Question with Solution | TestHub

We have, $\mathrm{f}(\mathrm{x}+1+\mathrm{f}(\mathrm{x}-1)=\sqrt{3}\mathrm{f}(\mathrm{x})$ for all ...(1) Replacing x by x-1 and x+1 successively in (1), we get $\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{x}-2)=\sqrt{3}\mathrm{f}(\mathrm{x}-1)$ ...(2) and $\mathrm{f}(\mathrm{x}+2)+\mathrm{f}(\mathrm{x})=\sqrt{3}\mathrm{f}(\mathrm{x}+1)$ ...(3) Adding (2) and (3), we get $2\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{x}-2)+\mathrm{f}(\mathrm{x}+2)=\sqrt{3}\{\mathrm{f}(\mathrm{x}-1)\mathrm{f}(\mathrm{x}+1)\}$ $\begin{array}{rlr}& =\sqrt{3}\cdot \sqrt{3}\mathrm{f}(\mathrm{x});\text{ from }(1)& =3\mathrm{f}(\mathrm{x})\end{array}$ $\Rightarrow \mathrm{f}(\mathrm{x}+2)+\mathrm{f}(\mathrm{x}-2)=\mathrm{f}(\mathrm{x})$ Replacing x by x+2 in (4), we get $\mathrm{f}(\mathrm{x}+4)+\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{x}+2)$ ...(5) Adding (4) and (5), we get $\mathrm{f}(\mathrm{x}+4)+\mathrm{f}(\mathrm{x}-2)=0$ ...(6) Again replacing x by x+6 in (6), we get $\mathrm{f}(\mathrm{x}+10)+\mathrm{f}(\mathrm{x}+4)=0$ ...(7) $\begin{array}{c}(6)-(7)\Rightarrow f(x+10)-f(x-2)=0f(x+10)=f(x-2).\end{array}$ Lastly, replacing x by $\mathrm{x}+2$ in (8), we get $\mathrm{f}(\mathrm{x}+12)=\mathrm{f}(\mathrm{x})$.

Hence $\mathrm{f}(\mathrm{x})$ is a periodic function with period 12. This proves part (i).