Sets, Relations & Functions Question - None

Question

If

f(x) = \log_e \left( \frac{1 - x}{1 + x} \right) |x| < 1

, then

f \left( \frac{2x}{1 + x^2} \right)

is equal to :

Accepted Answer

Step 1: Identify the given function.

We have f(x) = ln((1-x)/(1+x)) where |x| < 1.

Step 2: Substitute the argument into f.

We need to find f(2x/(1+x²)). Let u = 2x/(1+x²).

Then: f(u) = ln((1-u)/(1+u))

Step 3: Calculate (1 - 2x/(1+x²))/(1 + 2x/(1+x²)).

Numerator: 1 - 2x/(1+x²) = (1+x² - 2x)/(1+x²) = (1-x)²/(1+x²)

Denominator: 1 + 2x/(1+x²) = (1+x² + 2x)/(1+x²) = (1+x)²/(1+x²)

Step 4: Form the ratio.

(1 - 2x/(1+x²))/(1 + 2x/(1+x²)) = [(1-x)²/(1+x²)]/[(1+x)²/(1+x²)] = (1-x)²/(1+x)²

Step 5: Apply logarithm.

f(2x/(1+x²)) = ln((1-x)²/(1+x)²) = ln[(1-x)/(1+x)]² = 2ln((1-x)/(1+x))

Step 6: Recognize the pattern.

Since f(x) = ln((1-x)/(1+x)), we have:

f(2x/(1+x²)) = 2ln((1-x)/(1+x)) = 2f(x)

∴ Answer: The answer is A: 2f(x), which equals 1 when properly evaluated.

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