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 $f(x)+f(x-2)=\sqrt{3}f(x-1)$ .........(2)

and $\mathrm{f}(\mathrm{x}+2)+\mathrm{f}(\mathrm{x})=\sqrt{3}\mathrm{f}(\mathrm{x}+1)$.......(3)

$$\begin{array}{rlrl}\text{ Adding (2) and (3), we get }2\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{x}-2)+\mathrm{f}(\mathrm{x}+2)& =\sqrt{3}\{\mathrm{f}(\mathrm{x}-1)\mathrm{f}(\mathrm{x}+1)\}& =\sqrt{3}\cdot \sqrt{3}\mathrm{f}(\mathrm{x});\text{ from (1) }& =3\mathrm{f}(\mathrm{x})\end{array}$$

$$\Rightarrow f(x+2)+f(x-2)=f(x)\dots \dots \left(4\right)$$

Replacing x by $\mathrm{x}+2$ 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 $f(x+10)+f(x+4)=0$ ...........(7)

$$\begin{array}{c}(6)-(7)\Rightarrow f(x+10)-f(x-2)=0\end{array}$$

$$f(x+10)=f(x-2)..\dots \left(8\right)$$

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).

$$\text{ (ii) }\begin{array}{rlrl}\sum _{r=0}^{9}f(2+12r)& =f(2)+f(2+12)+f(2+24)+\dots +f(2+12\times 9)& & \end{array}$$

$$=f(2)+f(2)+f(2)+\dots to10terms,sincef(x)isperiodicwithperiod12.=10\times f(2)=10\times 9=90.$$