Question

Let $f\left(x\right)=\ln \left(\frac{1-x}{1+x}\right)$. The set of values of '$\alpha$' for which $f\left(\alpha \right)+f\left({\alpha }^2\right)=f\left(\frac{\alpha }{{\alpha }^2-\alpha +1}\right)$ is satisfied are

• A. $\left(-\infty ,-1\right)\cup \left(1,\infty \right)$
• B. $(-1,1)$
• C. $(0,1)$
• D. $(1,2)$

Answer 1

Answer: (B)

$(-1,1)$

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

Since LHS=RHS

$\alpha$ $ϵ$ $R$

But $\frac{1-\alpha }{1+\alpha }>0$ and $\frac{1-{\alpha }^2}{1+{\alpha }^2}>0$ and $\frac{{\alpha }^2-2\alpha +1}{{\alpha }^2+1}>0$

$\frac{1-\alpha }{1+\alpha }>0⟹\alpha ϵ(-1,1)$

$\frac{1-{\alpha }^2}{1+{\alpha }^2}>0$

$(\alpha -1)(\alpha +1)<0$

$\alpha ϵ(-1,1)$

So, $\alpha ϵ(-1,1)$