Mathematics - Functions Question with Solution | TestHub

Given $\mathrm{f}(\mathrm{x}+\mathrm{a})=1+{\left\{1-3\mathrm{f}(\mathrm{x})+3(\mathrm{f}(\mathrm{x}){)}^2-(\mathrm{f}(\mathrm{x}){)}^3\right\}}^{1/3}$

$$\Rightarrow \mathrm{f}(\mathrm{x}+\mathrm{a})-1={\left\{(1-\mathrm{f}(\mathrm{x}){)}^3\right\}}^{1/3}$$

$$\Rightarrow \{\mathrm{f}(\mathrm{x}+\mathrm{a})-1{\}}^3=\{1-\mathrm{f}(\mathrm{x}){\}}^3$$

$$\Rightarrow \mathrm{f}(\mathrm{x}+\mathrm{a})-1=1-\mathrm{f}(\mathrm{x})$$

$$\mathrm{f}(\mathrm{x}+\mathrm{a})+\mathrm{f}(\mathrm{x})=2\dots \dots \dots \left(1\right)$$

Replacing x by x-a, the equation (1) becomes $\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{x}-\mathrm{a})=2$........(2)

Subtracting (2) from (1), we get $\mathrm{f}(\mathrm{x}+\mathrm{a})-\mathrm{f}(\mathrm{x}-\mathrm{a})=0$. Finally replacing x by x+a, we get $\mathrm{f}(\mathrm{x}+2\mathrm{a})-\mathrm{f}(\mathrm{x})=0$

$$\Rightarrow \mathrm{f}(\mathrm{x}+2\mathrm{a})=\mathrm{f}(\mathrm{x})$$and hence f is periodic with period 2a.