Mathematics - Functions Question with Solution | TestHub

$f(\alpha )+f(\alpha {)}^2=\ln \left[\left(\frac{1-\alpha }{1+\alpha }\right)\left(\frac{1-{\alpha }^2}{1+{\alpha }^2}\right)\right]=\ln \left[\frac{(1-\alpha {)}^2}{1+{\alpha }^2}\right]$

$f\left(\frac{\alpha }{{\alpha }^2-\alpha +1}\right)=\ln \left[\frac{1-\frac{\alpha }{{\alpha }^2-\alpha +1}}{1+\frac{\alpha }{{\alpha }^2-\alpha +1}}\right]=\ln \left[\frac{(1-\alpha {)}^2}{1+{\alpha }^2}\right]$

$\therefore f(\alpha )+f\left({\alpha }^2\right)=f\left(\frac{\alpha }{{\alpha }^2-\alpha +1}\right)$ for all values of $\alpha$ for which the functions are defined, therefore

(i) $\frac{1-\alpha }{1+\alpha }>0\Rightarrow -1<\alpha <1\dots (1)$

(ii) $\frac{1-{\alpha }^2}{1+{\alpha }^2}>0\Rightarrow 1-{\alpha }^2>0\Rightarrow -1<\alpha <1\dots (2)$

From (1) and (2), we have $-1<\alpha <1$.

$\therefore$ The set of values of $\alpha =(-1,1)$.