Question

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

• A. $f(a+b)$
• B. $f(ab)$
• C. $f\left(\frac{a+b}{1+ab}\right)$
• D. $0$
• E. $f\left(\frac{a-b}{1+ab}\right)$

Answer 1

Answer: (C)

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