If f ( x ) = log ( 1 + x 1 - x ) , - 1 x 1 , then f ( 3 x + x 3 1 + 3 x 2 ) - f ( 2 x 1 + x 2 ) is

Question

If $f (x) = \log (\frac{1 + x}{1 - x})$ , $- 1 < x < 1$ , then $f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$ is

Smart Achievers · Accepted Answer

(4)

$f (x) = \log (\frac{1 + x}{1 - x})$

$f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$

$= \log (\frac{1 + \frac{3 x + x^{3}}{1 + 3 x^{2}}}{1 - \frac{3 x + x^{3}}{1 + 3 x^{2}}}) - \log (\frac{1 + \frac{2 x}{1 + x^{2}}}{1 - \frac{2 x}{1 + x^{2}}})$

$= \log (\frac{1 + 3 x^{2} + 3 x + x^{3}}{1 + 3 x^{2} - 3 x - x^{3}}) - \log (\frac{1 + x^{2} + 2 x}{1 + x^{2} - 2 x})$

$= \log {(\frac{1 + x}{1 - x})}^{3} - \log {(\frac{1 + x}{1 - x})}^{2}$

$= \log (\frac{{(1 + x)}^{3} \times {(1 - x)}^{2}}{{(1 - x)}^{3} \times {(1 + x)}^{2}}) = \log (\frac{1 + x}{1 - x}) = f (x)$

Solution

Q.
If $f (x) = \log (\frac{1 + x}{1 - x})$ , $- 1 < x < 1$ , then $f (\frac{3 x + x^{3}}{1 + 3 x^{2}}) - f (\frac{2 x}{1 + x^{2}})$ is

1 ${[f (x)]}^{3}$

2 ${[f (x)]}^{2}$

3 $- f (x)$

4 $f (x)$

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Solution

Q. If f(x)=log(1+x1-x), -1<x<1, then f(3x+x31+3x2)-f(2x1+x2) is

1 [f(x)]3

2 [f(x)]2

3 -f(x)

4 f(x)