If `f(x)=log_(e)((1+x)/(1-x)),g(x)=(3x+x^(3))/(1+3x^(2))andgof(t)=g(f(t))` then what is `g^(@)f((e-1)/(e+1))` equal to?

Options

A. 2
B. 1
C. 0
D. `(1)/(2)`

Answer

To solve the problem step by step, we need to evaluate \( g(f(\frac{e-1}{e+1})) \).

### Step 1: Evaluate \( f\left(\frac{e-1}{e+1}\right) \)

The function \( f(x) \) is defined as:
\[ f(x) = \log_e\left(\frac{1+x}{1-x}\right) \]

Now, substituting \( x = \frac{e-1}{e+1} \):
\[ f\left(\frac{e-1}{e+1}\right) = \log_e\left(\frac{1 + \frac{e-1}{e+1}}{1 - \frac{e-1}{e+1}}\right) \]

Calculating the numerator:
\[ 1 + \frac{e-1}{e+1} = \frac{(e+1) + (e-1)}{e+1} = \frac{2e}{e+1} \]

Calculating the denominator:
\[ 1 - \frac{e-1}{e+1} = \frac{(e+1) - (e-1)}{e+1} = \frac{2}{e+1} \]

Now substituting these into the function:
\[ f\left(\frac{e-1}{e+1}\right) = \log_e\left(\frac{\frac{2e}{e+1}}{\frac{2}{e+1}}\right) = \log_e(e) = 1 \]

### Step 2: Evaluate \( g(f(\frac{e-1}{e+1})) = g(1) \)

Now we need to find \( g(1) \). The function \( g(x) \) is defined as:
\[ g(x) = \frac{3x + x^3}{1 + 3x^2} \]

Substituting \( x = 1 \):
\[ g(1) = \frac{3(1) + (1)^3}{1 + 3(1)^2} = \frac{3 + 1}{1 + 3} = \frac{4}{4} = 1 \]

### Final Result

Thus, we have:
\[ g(f(\frac{e-1}{e+1})) = g(1) = 1 \]

### Summary

The final answer is:
\[ g(f(\frac{e-1}{e+1})) = 1 \]