Answer:
Let's calculate \( f(2x/1+x^2) \) using the given function \( f(x) = \log(1-x) \):
Given \( f(x) = \log(1-x) \),
Substitute \( x = \frac{2x}{1+x^2} \) into \( f(x) \):
\[ f\left(\frac{2x}{1+x^2}\right) = \log\left(1 - \frac{2x}{1+x^2}\right) \]
Now, we simplify the expression inside the logarithm:
\[ 1 - \frac{2x}{1+x^2} = \frac{1+x^2-2x}{1+x^2} = \frac{x^2 - 2x + 1}{1 + x^2} = \frac{(x-1)^2}{(1+x)(1-x)} \]
Therefore,
\[ f\left(\frac{2x}{1+x^2}\right) = \log\left(\frac{(x-1)^2}{(1+x)(1-x)}\right) \]
Now, using properties of logarithms, we can rewrite this as:
\[ f\left(\frac{2x}{1+x^2}\right) = \log((x-1)^2) - \log((1+x)(1-x)) \]
\[ f\left(\frac{2x}{1+x^2}\right) = 2\log|x-1| - (\log(1+x) + \log(1-x)) \]
\[ f\left(\frac{2x}{1+x^2}\right) = 2\log|x-1| - f(x) \]
So, \( f\left(\frac{2x}{1+x^2}\right) \) is equal to \( 2f(x) \).
Therefore, the correct option is:
(2) 2f(x)
Explanation:
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