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Question: If \(f(x)=\log \dfrac{1+x}{1-x}\), then the value of \(f\left( \dfrac{2x}{1+{{x}^{2}}} \right)\) is ...

If f(x)=log⁡1+x1−xf(x)=\log \dfrac{1+x}{1-x}, then the value of f(2x1+x2)f\left( \dfrac{2x}{1+{{x}^{2}}} \right) is
A. 2f(x)2f(x)
B. 3f(x)3f(x)
C. ∣f(x)∣2|f(x){{|}^{2}}
D. ∣f(x)∣3|f(x){{|}^{3}}

Explanation

Solution

Hint: Here, first we have to convert xx into 2x1+x2\dfrac{2x}{1+{{x}^{2}}} in f(x)=log⁡1+x1−xf(x)=\log \dfrac{1+x}{1-x} which will give log⁡(1+2x1+x21−2x1+x2)\log \left( \dfrac{1+\dfrac{2x}{1+{{x}^{2}}}}{1-\dfrac{2x}{1+{{x}^{2}}}} \right). Afterwards by factorisation convert everything in terms of xxwhich may give the answer as a function of f(x)f(x).

Complete step-by-step answer:
We are given that f(x)=log⁡1+x1−xf(x)=\log \dfrac{1+x}{1-x}. Now, we have to find the value of f(2x1+x2)f\left( \dfrac{2x}{1+{{x}^{2}}} \right).
We have,
f(x)=log⁡1+x1−xf(x)=\log \dfrac{1+x}{1-x} ….. (1)
Consider f(2x1+x2)f\left( \dfrac{2x}{1+{{x}^{2}}} \right), here we have 2x1+x2\dfrac{2x}{1+{{x}^{2}}} in place of xx.
So, substitute x=2x1+x2x=\dfrac{2x}{1+{{x}^{2}}} in equation (1), we obtain:
f(2x1+x2)=log⁡(1+2x1+x21−2x1+x2)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(2x1+x2)=log⁡(1+x2+2x1+x21+x2−2x1+x2)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:
abcd=ab×dc\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}
Therefore, we will get:
f(2x1+x2)=log⁡(1+x2+2x1+x2×1+x21+x2−2x)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+x21+{{x}^{2}} we obtain:
f(2x1+x2)=log⁡(1+x2+2x1+x2−2x)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{{(1+x)}^{2}} can be written as:
(1+x)2=1+2x+x2{{(1+x)}^{2}}=1+2x+{{x}^{2}}
Similarly, the expansion of (1−x)2{{(1-x)}^{2}} can be written as:
(1−x)2=1−2x+x2{{(1-x)}^{2}}=1-2x+{{x}^{2}}
Now, by substituting these values in equation (2) we obtain:
f(2x1+x2)=log⁡((1+x)2(1−x)2)f\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\log \left( \dfrac{{{(1+x)}^{2}}}{{{(1-x)}^{2}}} \right)
We also have a2b2=(ab)2\dfrac{{{a}^{2}}}{{{b}^{2}}}={{\left( \dfrac{a}{b} \right)}^{2}}
Hence we can write our equation as:
f(2x1+x2)=log⁡(1+x1−x)2f\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\log {{\left( \dfrac{1+x}{1-x} \right)}^{2}}
Now, we have a logarithmic identity that,
log⁡ab=blog⁡a\log {{a}^{b}}=b\log a
Therefore, we will the equation:
f(2x1+x2)=2log⁡(1+x1−x)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⁡1+x1−xf(x)=\log \dfrac{1+x}{1-x}
Therefore, we will get:
f(2x1+x2)=2f(x)f\left( \dfrac{2x}{1+{{x}^{2}}} \right)=2f(x)
Hence, we can say that the value of f(2x1+x2)=2f(x)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⁡1+x1−xf(x)=\log \dfrac{1+x}{1-x}, change all the values from xx to 2x1+x2\dfrac{2x}{1+{{x}^{2}}}and at last you should get everything in terms of xx. Here don’t try to change f(2x1+x2)f\left( \dfrac{2x}{1+{{x}^{2}}} \right) by putting xx as log⁡1+x1−x\log \dfrac{1+x}{1-x}it may lead to a wrong answer.