If f(x) = log left[ frac{1 + x}{1 - x} right] | vedclass

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(C) Given $f(x) = \log \left( \frac{1 + x}{1 - x} \right)$.We need to find $f\left( \frac{2x}{1 + x^2} \right)$.Substituting $\frac{2x}{1 + x^2}$ for

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$2f(x)$

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(C) Given $f(x) = \log \left( \frac{1 + x}{1 - x} \right)$.We need to find $f\left( \frac{2x}{1 + x^2} \right)$.Substituting $\frac{2x}{1 + x^2}$ for $x$ in the function $f(x)$:$f\left( \frac{2x}{1 + x^2} \right) = \log \left[ \frac{1 + \frac{2x}{1 + x^2}}{1 - \frac{2x}{1 + x^2}} \right]$Simplify the expression inside the logarithm:$= \log \left[ \frac{\frac{1 + x^2 + 2x}{1 + x^2}}{\frac{1 + x^2 - 2x}{1 + x^2}} \right]$$= \log \left[ \frac{1 + x^2 + 2x}{1 + x^2 - 2x} \right]$Recognizing the squares:$= \log \left[ \frac{(1 + x)^2}{(1 - x)^2} \right]$$= \log \left[ \frac{1 + x}{1 - x} \right]^2$Using the property $\log(a^n) = n \log a$:$= 2 \log \left[ \frac{1 + x}{1 - x} \right]$$= 2f(x)$.

If $f(x) = \log \left[ \frac{1 + x}{1 - x} \right]$,then $f\left[ \frac{2x}{1 + x^2} \right]$ is equal to

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