If $f(a)=\log \left|\frac{1-a}{1+a}\right|$ for $a \neq\{-1,1\}$, then the set of values of all ' $a$ ', for…
If $f(a)=\log \left|\frac{1-a}{1+a}\right|$ for $a \neq\{-1,1\}$, then the set of values of all ' $a$ ', for which $f\left(\frac{2 a}{1+a^2}\right)>0$ is
$(0, \infty)-\{1\}$
$(-\infty, 0)-\{-1\}$
$(-\infty, \infty)-\{-1,1\}$
$(-1,1)$
Solution
Given $f(a)=\log \left|\frac{1-a}{1+a}\right|, a \neq\{-1,1\}$
Now $f\left(\frac{2 a}{1+a^2}\right)>0 \Rightarrow \log \left|\frac{1-\frac{2 a}{1+a^2}}{1+\frac{2 a}{1+a^2}}\right|>0$
$\begin{aligned}
& \Rightarrow\left|\frac{1+a-2 a}{1+a+2 a}\right|>1 \Rightarrow\left|\frac{(1)}{(1)}\right|>1 \\
& \Rightarrow \frac{(1-a)^2-(1+a)^2}{(1+a)^2}>0 \text { or } \frac{\left(1-a^2\right)+\left(1+a^2\right)}{\left(1+a^2\right)} < 0 \\
& \Rightarrow-2 a>0 \text { or } 1+a^2 < 0 \Rightarrow a < 0
\end{aligned}$
Since $\left.\left(1+a^2\right)>0 \forall a \& a \neq \pm 1\right)$
So $a \in(-\infty, 0)-\{-1\}$