Question & Answer

If $f(x)=\log \dfrac{1+x}{1-x}$, then the value of $f\left( \dfrac{2x}{1+{{x}^{2}}} \right)$ is
A. $2f(x)$
B. $3f(x)$
C. $|f(x){{|}^{2}}$
D. $|f(x){{|}^{3}}$

ANSWER Verified Verified
Hint: Here, first we have to convert $x$ into $\dfrac{2x}{1+{{x}^{2}}}$ in $f(x)=\log \dfrac{1+x}{1-x}$ which will give $\log \left( \dfrac{1+\dfrac{2x}{1+{{x}^{2}}}}{1-\dfrac{2x}{1+{{x}^{2}}}} \right)$. Afterwards by factorisation convert everything in terms of $x$which may give the answer as a function of $f(x)$.

Complete step-by-step answer:
We are given that $f(x)=\log \dfrac{1+x}{1-x}$. Now, we have to find the value of $f\left( \dfrac{2x}{1+{{x}^{2}}} \right)$.
We have,
$f(x)=\log \dfrac{1+x}{1-x}$ ….. (1)
Consider $f\left( \dfrac{2x}{1+{{x}^{2}}} \right)$, here we have $\dfrac{2x}{1+{{x}^{2}}}$ in place of $x$.
So, substitute $x=\dfrac{2x}{1+{{x}^{2}}}$ in equation (1), we obtain:
$f\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\log \left( \dfrac{1+\dfrac{2x}{1+{{x}^{2}}}}{1-\dfrac{2x}{1+{{x}^{2}}}} \right)$
In the next step, by taking the LCM we get:
$f\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\log \left( \dfrac{\dfrac{1+{{x}^{2}}+2x}{1+{{x}^{2}}}}{\dfrac{1+{{x}^{2}}-2x}{1+{{x}^{2}}}} \right)$
We know that:
 $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}$
Therefore, we will get:
$f\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\log \left( \dfrac{1+{{x}^{2}}+2x}{1+{{x}^{2}}}\times \dfrac{1+{{x}^{2}}}{1+{{x}^{2}}-2x} \right)$
Next, by cancelling $1+{{x}^{2}}$ we obtain:
\[f\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\log \left( \dfrac{1+{{x}^{2}}+2x}{1+{{x}^{2}}-2x} \right)\] ….. (2)
We know that the expansion of ${{(1+x)}^{2}}$ can be written as:
Similarly, the expansion of ${{(1-x)}^{2}}$ can be written as:
Now, by substituting these values in equation (2) we obtain:
\[f\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\log \left( \dfrac{{{(1+x)}^{2}}}{{{(1-x)}^{2}}} \right)\]
We also have $\dfrac{{{a}^{2}}}{{{b}^{2}}}={{\left( \dfrac{a}{b} \right)}^{2}}$
Hence we can write our equation as:
\[f\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\log {{\left( \dfrac{1+x}{1-x} \right)}^{2}}\]
Now, we have a logarithmic identity that,
$\log {{a}^{b}}=b\log a$
Therefore, we will the equation:
\[f\left( \dfrac{2x}{1+{{x}^{2}}} \right)=2\log \left( \dfrac{1+x}{1-x} \right)\]
But we are given that:
$f(x)=\log \dfrac{1+x}{1-x}$
 Therefore, we will get:
\[f\left( \dfrac{2x}{1+{{x}^{2}}} \right)=2f(x)\]
Hence, we can say that the value of \[f\left( \dfrac{2x}{1+{{x}^{2}}} \right)=2f(x)\].
Therefore, the correct answer for this question is option (a).

Note: Here, in the function $f(x)=\log \dfrac{1+x}{1-x}$, change all the values from $x$ to \[\dfrac{2x}{1+{{x}^{2}}}\]and at last you should get everything in terms of $x$. Here don’t try to change \[f\left( \dfrac{2x}{1+{{x}^{2}}} \right)\] by putting $x$ as $\log \dfrac{1+x}{1-x}$it may lead to a wrong answer.