Let f(a)= a-1 a+1 . Consider the following : 1) f(2 a)=f(a)+1 2) (f...

$$\begin{aligned} & \mathrm{f}(\mathrm{a})=\frac{\mathrm{a}-1}{\mathrm{a}+1} \\ & \mathrm{f}(2 \mathrm{a})=\frac{2 \mathrm{a}-1}{2 \mathrm{a}+1} \end{aligned}$$\mathrm{f}(\mathrm{a})+1=\frac{\mathrm{a}-1}{\mathrm{a}+1}+1=\frac{2 \mathrm{a}}{\mathrm{a}+1}$So,$\mathrm{f}(2 \mathrm{a}) \neq \mathrm{f}(\mathrm{a})+1$Now,$f\left(\frac{1}{\mathrm{a}}\right)=\frac{\frac{1}{\mathrm{a}}-1}{\frac{1}{\mathrm{a}}+1}=\frac{1-\mathrm{a}}{1+\mathrm{a}}=-\left(\frac{\mathrm{a}-1}{\mathrm{a}+1}\right)=-\mathrm{f}(\mathrm{a})$