Consider the equation `x^4 + 2ax^3 + x^2 + 2ax + 1 = 0` where `a in R`. Also range of function `f(x)= x+1/x` is `(-oo,-2]uu[2,oo)` If equation has at least two distinct positive real roots then all possible values of a are

Options

A. 2
B. 1
C. 0
D. 3

Explanation

To solve the equation \( x^4 + 2ax^3 + x^2 + 2ax + 1 = 0 \) for the values of \( a \) such that the equation has at least two distinct positive real roots, we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ x^4 + 2ax^3 + x^2 + 2ax + 1 = 0 \]
We can divide the entire equation by \( x^2 \) (assuming \( x \neq 0 \)):
\[ x^2 + 2ax + 1 + \frac{2a}{x} + \frac{1}{x^2} = 0 \]
This simplifies to:
\[ x^2 + 2ax + 1 + 2a \cdot \frac{1}{x} + \frac{1}{x^2} = 0 \]

### Step 2: Substitute \( t = x + \frac{1}{x} \)
We know that \( x + \frac{1}{x} \) can be rewritten in terms of \( t \):
\[ x^2 + \frac{1}{x^2} = t^2 - 2 \]
Thus, we can rewrite the equation as:
\[ (t^2 - 2) + 2a \cdot t + 1 = 0 \]
This simplifies to:
\[ t^2 + 2at - 1 = 0 \]

### Step 3: Determine Conditions for Roots
For the quadratic equation \( t^2 + 2at - 1 = 0 \) to have at least two distinct positive roots, the discriminant must be positive:
\[ D = (2a)^2 - 4 \cdot 1 \cdot (-1) = 4a^2 + 4 \]
Since \( D > 0 \) for all \( a \in \mathbb{R} \), we will also need to ensure that the roots \( t_1 \) and \( t_2 \) are positive.

### Step 4: Analyze the Roots
The roots of the quadratic can be expressed as:
\[ t = \frac{-2a \pm \sqrt{4a^2 + 4}}{2} = -a \pm \sqrt{a^2 + 1} \]
For both roots to be positive, we need:
1. \( -a + \sqrt{a^2 + 1} > 0 \)
2. \( -a - \sqrt{a^2 + 1} > 0 \) (this condition will not hold since \( -a - \sqrt{a^2 + 1} < 0 \))

Thus, we only need to consider the first condition:
\[ -a + \sqrt{a^2 + 1} > 0 \]
This simplifies to:
\[ \sqrt{a^2 + 1} > a \]
Squaring both sides:
\[ a^2 + 1 > a^2 \implies 1 > 0 \quad \text{(always true)} \]

### Step 5: Find the Range of \( a \)
Next, we analyze the condition for \( t \):
Since \( t = x + \frac{1}{x} \) has a range of \( (-\infty, -2] \cup [2, \infty) \), we need to ensure that our roots \( t_1 \) and \( t_2 \) fall within this range.

From the quadratic formula, we have:
1. \( -a + \sqrt{a^2 + 1} \geq 2 \)
2. \( -a - \sqrt{a^2 + 1} \leq -2 \)

### Step 6: Solve the Inequality
For the first inequality:
\[ -a + \sqrt{a^2 + 1} \geq 2 \]
Rearranging gives:
\[ \sqrt{a^2 + 1} \geq a + 2 \]
Squaring both sides yields:
\[ a^2 + 1 \geq a^2 + 4a + 4 \implies 1 \geq 4a + 4 \implies 4a \leq -3 \implies a \leq -\frac{3}{4} \]

### Conclusion
Thus, the values of \( a \) such that the equation \( x^4 + 2ax^3 + x^2 + 2ax + 1 = 0 \) has at least two distinct positive real roots are:
\[ a < -\frac{3}{4} \]