Question

If $f\left( x \right) = \log \left( {\dfrac{{1 + x}}{{1 - x}}} \right)$, then find $f\left( {\dfrac{{3x + {x^3}}}{{1 + 3{x^2}}}} \right) - f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$. (Here x<1 and x>-1)
A) ${\left( {f\left( x \right)} \right)^3}$
B) ${\left( {f\left( x \right)} \right)^2}$
C) $- f\left( x \right)$
D) $f\left( x \right)$
E) $3f\left( x \right)$

Answer 1

Given is one function in its original form and the rest two are replacing the value of x. What we will do is replace the x in the $f\left( x \right)$ with the x given in the second given equation and then take the LCM solve further. We will get the numerator and denominator of both the terms as perfect squares or cubes. According to that, make simplifications.

Complete step by step answer:
Given that,
$f\left( x \right) = \log \left( {\dfrac{{1 + x}}{{1 - x}}} \right)$
Now we are asked to find,
$f\left( {\dfrac{{3x + {x^3}}}{{1 + 3{x^2}}}} \right) - f\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$
Just replace the x from $f\left( x \right)$ by the x given in the above respective brackets,
$= \log \left( {\dfrac{{1 + \dfrac{{3x + {x^3}}}{{1 + 3{x^2}}}}}{{1 - \dfrac{{3x + {x^3}}}{{1 + 3{x^2}}}}}} \right) - \log \left( {\dfrac{{1 + \dfrac{{2x}}{{1 + {x^2}}}}}{{1 - \dfrac{{2x}}{{1 + {x^2}}}}}} \right)$
We need to use the concept of LCM.
$= \log \left( {\dfrac{{1 + 3{x^2} + 3x + {x^3}}}{{1 + 3{x^2} - 3x - {x^3}}}} \right) - \log \left( {\dfrac{{1 + {x^2} + 2x}}{{1 + {x^2} - 2x}}} \right)$
Now as we can see, the numerator and denominator is the expansion of the perfect cube and that of the second bracket is the expansion of the perfect square. Just by rearranging them,
$= f\left( {\dfrac{{1 + 3{x^2} + 3x + {x^3}}}{{1 + 3{x^2} - 3x - {x^3}}}} \right) - f\left( {\dfrac{{1 + 2x + {x^2}}}{{1 - 2x + {x^2}}}} \right)$
$= \log \left( {\dfrac{{{{\left( {1 + x} \right)}^3}}}{{{{\left( {1 - x} \right)}^3}}}} \right) - \log \left( {\dfrac{{{{\left( {1 + x} \right)}^2}}}{{{{\left( {1 - x} \right)}^2}}}} \right)$
Now this can be further simplified as,
$= \log {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^3} - \log {\left( {\dfrac{{1 + x}}{{1 - x}}} \right)^2}$
This can be written as, $\log {a^n} = n\log a$
$= 3\log \left( {\dfrac{{1 + x}}{{1 - x}}} \right) - 2\log \left( {\dfrac{{1 + x}}{{1 - x}}} \right)$
$= \log \left( {\dfrac{{1 + x}}{{1 - x}}} \right)$
This is the same as the given function.
$= f\left( x \right)$
Thus option (D) is the correct answer.

Note:
Note that, if $f\left( x \right)\& g\left( x \right)$ are two function given then $f\left( {g\left( x \right)} \right)$ is nothing but expressing the $f\left( x \right)$ in $g\left( x \right)$ form. Thus this is what we have done here. Just for two functions. Then also note that the positive expansions are not affected but the negative expansions might be changed because generally we write x terms first but since here 1 is written first the cube will be ${\left( {1 - x} \right)^3}$ and same for ${\left( {1 - x} \right)^2}$.