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MHT CET · Maths · Functions

If \(\mathrm{f}(x)=\log _{\mathrm{c}}\left(\frac{1-x}{1+x}\right),|x| \lt 1\), then \(\mathrm{f}\left(\frac{2 x}{1+x^2}\right)\) is equal to

  1. A \(2 \mathrm{f}\left(x^2\right)\)
  2. B \((\mathrm{f}(x))^2\)
  3. C \(-2 \mathrm{f}(x)\)
  4. D \(2 \mathrm{f}(x)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2 \mathrm{f}(x)\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} f\left(\frac{2 x}{1+x^2}\right) & =\log _{\mathrm{e}}\left(\frac{1-\frac{2 x}{1+x^2}}{1+\frac{2 x}{1+x^2}}\right) \\ & =\log _{\mathrm{e}}\left(\frac{(1-x)^2}{(1+x)^2}\right) \\ & =2 \log _{\mathrm{e}}\left(\frac{1-x}{1+x}\right) \\ & =2 \mathrm{f}(x)\end{aligned}\)