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  4. If f(x)= log ( (1-x/1+x) ), then f(a)+f(b) is equal to:

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Q. If $ f(x)=\log \left( \frac{1-x}{1+x} \right), $ then $ f(a)+f(b) $ is equal to:

KEAMKEAM 2000

Solution:

$ f(x)=\log \left( \frac{1-x}{1+x} \right) $ Now, $ f(x)=\log \left( \frac{1-a}{1+a} \right),f(b)=\log \left( \frac{1-b}{1+b} \right) $ $ \therefore $ $ f(a)+f(b)=\log \left( \frac{1-a}{1+a} \right)+\log \left( \frac{1-b}{1+b} \right) $ $ =\log \left( \frac{1-a}{1+a}.\frac{1-b}{a+b} \right) $ $ =\log \left( \frac{1-a-b+ab}{1+a+b+ab} \right) $ $ =\log \left( \frac{1-\left( \frac{a+b}{1+ab} \right)}{1+\left( \frac{a+b}{1+ab} \right)} \right)=f\left( \frac{a+b}{1+ab} \right) $