Question

If f(x)=log$\left[\frac{1+x}{1-x}\right]$ and -1<x${}_1$,x${}_2$<1, then f(x${}_1$)-f(x${}_2$) is equal to:

• A. $f\left[\right(x_1-x_2)/1+x_1{x}_2]$
• B. $f\left[\right(x_1-x_2)/1-x_1{x}_2]$
• C. $f\left[\right(x_1+x_2)/1-x_1{x}_2]$
• D. $f\left[\right(x_1+x_2)/1+x_1{x}_2]$

Answer 1

The correct option is B $f\left[\right(x_1-x_2)/1-x_1{x}_2]$

If $f\left(x\right)=\log \left|\begin{array}{c}1+x\\ 1-x\end{array}\right|$ and $-1<x_1,x_2<1$, then

$f\left(x_1\right)=\log \left(\frac{1+x_1}{1-x_1}\right)$

$f\left(x_2\right)=\log \left(\frac{1+x_2}{1-x_2}\right)$

$f\left(x_1\right)-f\left(x_2\right)=\log \left(\frac{1+x_1}{1-x_1}\right)-\log \left(\frac{1+x_2}{1-x_1}\right)$

$=\log \left[\frac{\frac{1+x_1}{1-x_1}}{\frac{1+x_2}{1-x_2}}\right]$

$=\log \left[\frac{(1+x_1)(1-x_2)}{(1-x_1)(1+x_2)}\right]$

$=\log \left[\frac{1+x_1-x_2-x_1{x}_2}{1+x_2-x_1-x_1{x}_2}\right]$

$=\log \left[\frac{1-x_1{x}_2+(x_1-x_2)}{1-x_2{x}_2-(x_1-x_2)}\right]$

Divide $N^r$ and $D^r$ by

$1-x_1{x}_2$

$=\log \left[\frac{1+\left(\frac{x_1-x_2}{1-x_1{x}_2}\right)}{1-\left(\frac{x_1-x_2}{1-x_1{x}_2}\right)}\right]$

$=f\left(\frac{x_1-x_2}{1-x_1{x}_2}\right)$