If f(a)=log⁡1−a1+a for a≠{−1,1}, then the set of values of all ' a ', for which f2a1+a2>0 is

f2a1+a2=log⁡1+2a1+a21−2a1+a2=log⁡1+a2+2a1+a21+a2−2a1+a2=log⁡a2+2a+1a2−2a+1=log⁡(a+1)2(a−1)2=log⁡a+1a−12=2f(a)

If f(a)=log⁡1−a1+a for a≠{−1,1}, then the set of values of all ' a ', for which f2a1+a2>0 is

f2a1+a2=log⁡1+2a1+a21−2a1+a2=log⁡1+a2+2a1+a21+a2−2a1+a2=log⁡a2+2a+1a2−2a+1=log⁡(a+1)2(a−1)2=log⁡a+1a−12=2f(a)

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If $f (a) = \log |\frac{1 - a}{1 + a}|$ for $a \neq {- 1, 1}$ , then the set of values of all ' a ', for which $f (\frac{2 a}{1 + a^{2}}) > 0$ is

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

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

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

$(- 1, 1)$

answer is B.

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Detailed Solution

$f (\frac{2 a}{1 + a^{2}}) = \log \frac{1 + (\frac{2 a}{1 + a^{2}})}{1 - (\frac{2 a}{1 + a^{2}})}$

$= \log \frac{(\frac{1 + a^{2} + 2 a}{1 + a^{2}})}{(\frac{1 + a^{2} - 2 a}{1 + a^{2}})}$

$= \log (\frac{a^{2} + 2 a + 1}{a^{2} - 2 a + 1})$

$\begin{aligned} = \log (\frac{(a + 1)^{2}}{(a - 1)^{2}}) \\ = \log {(\frac{a + 1}{a - 1})}^{2} \end{aligned}$

$= 2 f (a)$

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If f(a)=log⁡1−a1+a for a≠{−1,1}, then the set of values of all ' a ', for which f2a1+a2>0 is

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If $f (a) = \log |\frac{1 - a}{1 + a}|$ for $a \neq {- 1, 1}$ , then the set of values of all ' a ', for which $f (\frac{2 a}{1 + a^{2}}) > 0$ is