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Question: If \(f(x) = \log\left\lbrack \frac{1 + x}{1 - x} \right\rbrack,\) then \(f\left\lbrack \frac{2x}{1 +...

If f(x)=log⁡[1+x1−x],f(x) = \log\left\lbrack \frac{1 + x}{1 - x} \right\rbrack, then f[2x1+x2]f\left\lbrack \frac{2x}{1 + x^{2}} \right\rbrack is equal to

A

[f(x)]2\lbrack f(x)\rbrack^{2}

B

[f(x)]3\lbrack f(x)\rbrack^{3}

C

2f(x)2f(x)

D

3f(x)3f(x)

Answer

2f(x)2f(x)

Explanation

Solution

f(x)=log⁡(1+x1−x)f(x) = \log\left( \frac{1 + x}{1 - x} \right)

∴f(2x1+x2)=log⁡[1+2x1+x21−2x1+x2]=log⁡[x2+1+2xx2+1−2x]=log⁡[1+x1−x]2=2log⁡[1+x1−x]=2f(x)\therefore f\left( \frac{2x}{1 + x^{2}} \right) = \log\left\lbrack \frac{1 + \frac{2x}{1 + x^{2}}}{1 - \frac{2x}{1 + x^{2}}} \right\rbrack = \log\left\lbrack \frac{x^{2} + 1 + 2x}{x^{2} + 1 - 2x} \right\rbrack = {\log\left\lbrack \frac{1 + x}{1 - x} \right\rbrack}^{2} = 2\log\left\lbrack \frac{1 + x}{1 - x} \right\rbrack = 2f(x)