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If $f({x_1}) - f({x_2}) = f\left( {\frac{{{x_1} - {x_2}}}{{1 - {x_1}{x_2}}}} \right)$ for ${x_1},{x_2} \in [ - 1,\,1]$, then $f(x)$ is
$\log \frac{{(1 - x)}}{{(1 + x)}}$
${\tan ^{ - 1}}\frac{{(1 - x)}}{{(1 + x)}}$
$\log \frac{{(1 + x)}}{{(1 - x)}}$
all of these
Solution
(a,b,c) When ${x_1} = – 1$ and ${x_2} = 1$ , then
$f( – 1) – f(1) = f\left[ {\frac{{ – 1 – 1}}{{1 + 1(1)}}} \right] = f( – 1) \Rightarrow f(1) = 0$
Which is satisfied when $f(x) = {\tan ^{ – 1}}\left( {\frac{{1 – x}}{{1 + x}}} \right)$
When ${x_1} = {x_2} = 0$, then
$f(0) – f(0) = f\left[ {\frac{{0 – 0}}{{1 – 0}}} \right] = f(0) \Rightarrow f(0) = 0$
When ${x_1} = – 1$ and ${x_2} = 0$ then
$f( – 1) – f(0) = f\left( {\frac{{ – 1 – 0}}{{1 – 0}}} \right) = f( – 1) \Rightarrow f(0) = 0$
Which is satisfied when $f(x) = \log \left( {\frac{{1 – x}}{{1 + x}}} \right)$ and
$f(x) = \log \left( {\frac{{1 + x}}{{1 – x}}} \right)$.
