If f({x_1}) - f({x_2}) = fleft( {frac{{{x_1} | vedclass

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Question

(a,b,c) When ${x_1} = - 1$ and ${x_2} = 1$ , then

$f( - 1) - f(1) = f\left[ {\frac{{ - 1 - 1}}{{1 + 1(1)}}} \right] = f( - 1) \Rightarrow f(1) = 0$

Vedclass · Accepted Answer

all of these

Vedclass · Answer

(a,b,c) When ${x_1} = - 1$ and ${x_2} = 1$ , then

$f( - 1) - f(1) = f\left[ {\frac{{ - 1 - 1}}{{1 + 1(1)}}} ight] = f( - 1) \Rightarrow f(1) = 0$

Which is satisfied when $f(x) = {	an ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} ight)$

When ${x_1} = {x_2} = 0$, then

$f(0) - f(0) = f\left[ {\frac{{0 - 0}}{{1 - 0}}} ight] = f(0) \Rightarrow f(0) = 0$

When ${x_1} = - 1$ and ${x_2} = 0$ then

$f( - 1) - f(0) = f\left( {\frac{{ - 1 - 0}}{{1 - 0}}} ight) = f( - 1) \Rightarrow f(0) = 0$

Which is satisfied when $f(x) = \log \left( {\frac{{1 - x}}{{1 + x}}} ight)$ and 

$f(x) = \log \left( {\frac{{1 + x}}{{1 - x}}} ight)$.

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

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