If `2^(f(x)) = (2+x)/(2-x), x in (-2, 2)` and `f(x) = lambda f((8x)/(4+ x^(2)))` then value `'lambda'` will be

Options

A. 2
B. `(1)/(2)`
C. 1
D. `-1`

Answer

To solve the problem, we need to find the value of `lambda` given the equations:

1. \( 2^{f(x)} = \frac{2+x}{2-x} \)
2. \( f(x) = \lambda f\left(\frac{8x}{4+x^2}\right) \)

### Step-by-Step Solution:

**Step 1: Take the logarithm of both sides of the first equation.**

We start with the equation:
\[ 2^{f(x)} = \frac{2+x}{2-x} \]

Taking logarithm base 2 on both sides:
\[ f(x) = \log_2\left(\frac{2+x}{2-x}\right) \]

**Step 2: Substitute \( f(x) \) into the second equation.**

Now, we substitute \( f(x) \) into the second equation:
\[ \log_2\left(\frac{2+x}{2-x}\right) = \lambda f\left(\frac{8x}{4+x^2}\right) \]

**Step 3: Find \( f\left(\frac{8x}{4+x^2}\right) \).**

We need to evaluate \( f\left(\frac{8x}{4+x^2}\right) \):
\[ f\left(\frac{8x}{4+x^2}\right) = \log_2\left(\frac{2+\frac{8x}{4+x^2}}{2-\frac{8x}{4+x^2}}\right) \]

**Step 4: Simplify the expression inside the logarithm.**

First, calculate \( 2+\frac{8x}{4+x^2} \) and \( 2-\frac{8x}{4+x^2} \):
\[ 2 + \frac{8x}{4+x^2} = \frac{(2(4+x^2) + 8x)}{4+x^2} = \frac{8 + 2x^2 + 8x}{4+x^2} \]
\[ 2 - \frac{8x}{4+x^2} = \frac{(2(4+x^2) - 8x)}{4+x^2} = \frac{8 + 2x^2 - 8x}{4+x^2} \]

Now, we can write:
\[ f\left(\frac{8x}{4+x^2}\right) = \log_2\left(\frac{8 + 2x^2 + 8x}{8 + 2x^2 - 8x}\right) \]

**Step 5: Substitute back into the equation.**

Now, substitute this back into the equation:
\[ \log_2\left(\frac{2+x}{2-x}\right) = \lambda \log_2\left(\frac{8 + 2x^2 + 8x}{8 + 2x^2 - 8x}\right) \]

**Step 6: Set the logarithmic expressions equal.**

This implies:
\[ \frac{2+x}{2-x} = \left(\frac{8 + 2x^2 + 8x}{8 + 2x^2 - 8x}\right)^{\lambda} \]

**Step 7: Solve for \( \lambda \).**

To find \( \lambda \), we can equate the two expressions. By simplifying, we can find that:
\[ 1 = 2\lambda \]
Thus:
\[ \lambda = \frac{1}{2} \]

### Final Answer:
The value of \( \lambda \) is \( \frac{1}{2} \).