A) \[f(a+b)\]
B) \[f(ab)\]
C) \[f\left( \frac{a+b}{1+ab} \right)\]
D) \[0\]
E) \[f\left( \frac{a-b}{1+ab} \right)\]
Correct Answer: C
Solution :
\[f(x)=\log \left( \frac{1-x}{1+x} \right)\] Now, \[f(x)=\log \left( \frac{1-a}{1+a} \right),f(b)=\log \left( \frac{1-b}{1+b} \right)\] \[\therefore \]\[f(a)+f(b)=\log \left( \frac{1-a}{1+a} \right)+\log \left( \frac{1-b}{1+b} \right)\] \[=\log \left( \frac{1-a}{1+a}.\frac{1-b}{a+b} \right)\] \[=\log \left( \frac{1-a-b+ab}{1+a+b+ab} \right)\] \[=\log \left( \frac{1-\left( \frac{a+b}{1+ab} \right)}{1+\left( \frac{a+b}{1+ab} \right)} \right)=f\left( \frac{a+b}{1+ab} \right)\]
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