If f(a) = log left| frac{1-a}{1+a} right| for | vedclass

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(B) Given $f(a) = \log \left| \frac{1-a}{1+a} \right|$.We want to solve $f\left( \frac{2a}{1+a^2} \right) > 0$.Substituting $x = \frac{2a}{1+a^2}$,we

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$(-\infty, 0) - \{-1\}$

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(B) Given $f(a) = \log \left| \frac{1-a}{1+a} ight|$.We want to solve $f\left( \frac{2a}{1+a^2} ight) > 0$.Substituting $x = \frac{2a}{1+a^2}$,we have $\log \left| \frac{1 - \frac{2a}{1+a^2}}{1 + \frac{2a}{1+a^2}} ight| > 0$.This implies $\left| \frac{1+a^2-2a}{1+a^2+2a} ight| > 1$,which simplifies to $\left| \frac{(1-a)^2}{(1+a)^2} ight| > 1$.Since both numerator and denominator are squares,we have $\left| \frac{1-a}{1+a} ight| > 1$.Squaring both sides,$\frac{(1-a)^2}{(1+a)^2} > 1 \Rightarrow (1-a)^2 > (1+a)^2$.$1 - 2a + a^2 > 1 + 2a + a^2$.$-2a > 2a \Rightarrow 4a Also,we must satisfy the domain constraints $a 
eq \pm 1$ and $\frac{2a}{1+a^2} 
eq \pm 1$.For $a Thus,the solution is $a \in (-\infty, 0) - \{-1\}$.

List-$I$	List-$II$
$A$. Range of $\sec ^{-1}\left[1+\cos ^2 x\right]$, where $[.]$ denotes the greatest integer function	$I$. Odd function
$B$. Domain of $f(x)$ where $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$	$II$. $\left\{0, \frac{1}{2}\right\}$
$C$. $f(x+y)=f(x)+f(y) ; f(1)=5$	$III$. $\left\{\sec ^{-1} 5, \sec ^{-1} 4\right\}$
$D$. $\sin ^{-1} x-\cos ^{-1} x+\sin ^{-1}(1-x)=0 \Rightarrow x \in$	$IV$. $R$
	$V$. $\left\{\sec ^{-1} 1, \sec ^{-1} 2\right\}$

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 $f\left( \frac{2a}{1+a^2} \right) > 0$ is

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