Question

If $f\left(x\right)=\log \left[\frac{(1+x)}{1-x}\right]$, where $-1<x<1$, then $f\left[\frac{(3x+x^3)}{(1+3x^2)}\right]-f\left[\frac{\left(2x\right)}{(1+x^2)}\right]$ is equal to

• A. ${\left[f\right(x\left)\right]}^3$
• B. $\left[f\right(x){]}^2$
• C. $-f\left(x\right)$
• D. $f\left(x\right)$
• E. $3f\left(x\right)$

Answer 1

The correct option is D

$f\left(x\right)$

Explanation for the correct option:

Find the value of the expression:

Given,

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

Now,

$\begin{array}{rcl}f\left[\frac{(3x+x^3)}{(1+3x^2)}\right]-f\left[\frac{\left(2x\right)}{(1+x^2)}\right]& =& \log \left(\frac{1+{\displaystyle \frac{(3x+x^3)}{(1+3x^2)}}}{1-\left({\displaystyle \frac{3x+x^3}{1+3x^2}}\right)}\right)-\log \left(\frac{1+\left({\displaystyle \frac{2x}{1+x^2}}\right)}{1-\left({\displaystyle \frac{2x}{1+x^2}}\right)}\right)\\ & =& \log \left(\frac{{\displaystyle \frac{1+3x^2+3x+x^3}{1+3x^2}}}{{\displaystyle \frac{1+3x^2-3x-x^3}{1+3x^2}}}\right)-\log \left(\frac{{\displaystyle \frac{1+x^2+2x}{1+x^2}}}{{\displaystyle \frac{1+x^2-2x}{1+x^2}}}\right)\\ & =& \log {\left(\frac{1+x}{1-x}\right)}^3-\log {\left(\frac{1+x}{1-x}\right)}^2\mathbf{[}\mathbf{\because }\mathbf{}{\left(a+b\right)}^{\mathbf{3}}\mathbf{=}{\mathbf{a}}^{\mathbf{3}}\mathbf{+}{\mathbf{b}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{a}}^{\mathbf{2}}\mathbf{b}\mathbf{+}\mathbf{3}{\mathbf{b}}^{\mathbf{2}}\mathbf{a}\mathbf{,}\mathbf{}{\left(a\pm b\right)}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{\pm }\mathbf{2}\mathbf{a}\mathbf{b}\mathbf{]}\\ & =& 3\log \left(\frac{1+x}{1-x}\right)-2\log \left(\frac{1+x}{1-x}\right)\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{l}\mathbf{o}\mathbf{g}{\mathbf{a}}^{\mathbf{m}}\mathbf{=}\mathbf{m}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{a}\mathbf{]}\\ & =& \log \left(\frac{1+x}{1-x}\right)\\ & =& f\left(x\right)\end{array}$

Hence, the correct option is D.