Question

If \(f(x)=\log _{e}\left(\frac{1-x}{1+x}\right)\), \(|x|\lt 1\), then \(f\left(\frac{4x}{4+x^{2}}\right)\) is equal to

• A. \(2f(x)\)
• B. \(-2f(x)\)
• C. \(2f\left(\frac{x}{2}\right)\)
• D. \(-2f\left(\frac{x}{2}\right)\)

Hint:

Substitute the argument into f, simplify to a perfect square ratio and compare with f(x/2).

Answer 1

The Correct Option is C

Step 1: Try a substitution
Put $x=2t$, so $\dfrac{4x}{4+x^{2}}=\dfrac{8t}{4+4t^{2}}=\dfrac{2t}{1+t^{2}}$.

Step 2: Use the known identity
$\log\dfrac{1-\frac{2t}{1+t^{2}}}{1+\frac{2t}{1+t^{2}}}=\log\dfrac{(1-t)^{2}}{(1+t)^{2}}=2\log\dfrac{1-t}{1+t}$

Step 3: Return to x
Since $t=\dfrac{x}{2}$, the result is $2f\left(\dfrac{x}{2}\right)$.

Final Answer:
Using x = 2t shows the value equals 2f(x/2). \[ \boxed{\text{(C) }2f\left(\dfrac{x}{2}\right)} \]