If two roots of the equation `(p-1)(x^2 +x +1)^2 -(p+1)(x^4+x^2+1)=0` are real and distinct and `f(x)=(1-x)/(1+x)` then `f(f(x))+f(f(1/x))` is equal to

Options

A. (a) `p`
B. (b) `-p`
C. (c) `2p`
D. (d) `-2p`

Answer

To solve the problem step by step, we will start with the given equation and the function \( f(x) \).

### Step 1: Analyze the given equation
The equation is:
\[ (p-1)(x^2 + x + 1)^2 - (p+1)(x^4 + x^2 + 1) = 0 \]
We need to find conditions under which this equation has two real and distinct roots.

### Step 2: Simplify the equation
Rearranging the equation gives:
\[ (p-1)(x^2 + x + 1)^2 = (p+1)(x^4 + x^2 + 1) \]
This can be simplified further by substituting \( u = x + \frac{1}{x} \), which leads to \( x^2 + \frac{1}{x^2} = u^2 - 2 \).

### Step 3: Substitute and analyze the roots
We can express \( x^2 + x + 1 \) and \( x^4 + x^2 + 1 \) in terms of \( u \):
- \( x^2 + x + 1 = \frac{u^2 + 3}{2} \)
- \( x^4 + x^2 + 1 = \frac{u^4 + 2u^2 + 3}{2} \)

Substituting these into the equation will allow us to analyze the roots in terms of \( u \).

### Step 4: Determine conditions for real and distinct roots
To ensure that the roots are real and distinct, we will need to analyze the discriminant of the resulting polynomial in \( u \).

### Step 5: Evaluate the function \( f(x) \)
The function given is:
\[ f(x) = \frac{1-x}{1+x} \]
We need to compute \( f(f(x)) + f(f(1/x)) \).

### Step 6: Compute \( f(f(x)) \)
First, we compute \( f(f(x)) \):
1. Substitute \( f(x) \) into itself:
\[ f(f(x)) = f\left(\frac{1-x}{1+x}\right) = \frac{1 - \frac{1-x}{1+x}}{1 + \frac{1-x}{1+x}} \]
2. Simplifying this expression gives:
\[ = \frac{\frac{(1+x) - (1-x)}{1+x}}{\frac{(1+x) + (1-x)}{1+x}} = \frac{\frac{2x}{1+x}}{\frac{2}{1+x}} = x \]

### Step 7: Compute \( f(f(1/x)) \)
Next, we compute \( f(f(1/x)) \):
1. Substitute \( 1/x \) into \( f(x) \):
\[ f(1/x) = \frac{1 - 1/x}{1 + 1/x} = \frac{x-1}{x+1} \]
2. Now compute \( f(f(1/x)) \):
\[ f(f(1/x)) = f\left(\frac{x-1}{x+1}\right) = \frac{1 - \frac{x-1}{x+1}}{1 + \frac{x-1}{x+1}} \]
3. Simplifying this expression gives:
\[ = \frac{\frac{(x+1) - (x-1)}{x+1}}{\frac{(x+1) + (x-1)}{x+1}} = \frac{\frac{2}{x+1}}{\frac{2x}{x+1}} = \frac{1}{x} \]

### Step 8: Combine results
Now we combine the results:
\[ f(f(x)) + f(f(1/x)) = x + \frac{1}{x} \]

### Step 9: Conclusion
Thus, the final result is:
\[ f(f(x)) + f(f(1/x)) = x + \frac{1}{x} \]

### Final Answer
The value of \( f(f(x)) + f(f(1/x)) \) is \( x + \frac{1}{x} \).

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