SATHEE: Functions 2 Question 3

Functions 2 Question 3

4.

If $f (x) = \log_{e} (\frac{1 - x}{1 + x}), | x | < 1$ , then $f (\frac{2 x}{1 + x^{2}})$ is equal to

(a) $2 f (x)$

(c) $(f (x))^{2}$

(b) $2 f (x^{2})$

(d) $- 2 f (x)$

(2019 Main, 8 April I)

Show Answer

Answer:

Correct Answer: 4. (a)

Solution:

Given, $f (x) = \log_{e} \frac{1 - x}{1 + x}, | x | < 1$ , then

$f \frac{2 x}{1 + x^{2}} = \log_{e} \frac{1 - \frac{2 x}{1 + x^{2}}}{1 + \frac{2 x}{1 + x^{2}}} ∵ \frac{2 x}{1 + x^{2}} < 1$

$= \log_{e} \frac{\frac{1 + x^{2} - 2 x}{1 + x^{2}}}{\frac{1 + x^{2} + 2 x}{1 + x^{2}}} = \log_{e} \frac{(1 - x)^{2}}{(1 + x)^{2}} = \log_{e} \frac{1 - x^{2}}{1 + x}$

$= 2 \log_{e} (\frac{1 - x}{1 + x})$ $[∵ \log_{e} | A |^{m} = m \log_{e} | A |]$

$= 2 f (x)$ $∵ f (x) = \log_{e} \frac{1 - x}{1 + x}$

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Functions 2 Question 3

4.

Answer:

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