If quadratic equation `f(x)=x^(2)+ax+1=0` has two positive distinct roots then

Options

A. `f(1)gt0`
B. `f^(')(1)lt0`
C. `f(0)-f^(')(0)=2`
D. `f^(')(0)lt-2`

Explanation

To determine the conditions under which the quadratic equation \( f(x) = x^2 + ax + 1 = 0 \) has two positive distinct roots, we can follow these steps:

### Step 1: Identify the roots
Let the roots of the quadratic equation be \( \alpha \) and \( \beta \). According to Vieta's formulas, we have:
- The sum of the roots: \( \alpha + \beta = -a \)
- The product of the roots: \( \alpha \beta = 1 \)

### Step 2: Express one root in terms of the other
From the product of the roots, we can express \( \beta \) in terms of \( \alpha \):
\[ \beta = \frac{1}{\alpha} \]

### Step 3: Substitute into the sum of roots
Substituting \( \beta \) into the sum of roots gives us:
\[ \alpha + \frac{1}{\alpha} = -a \]

### Step 4: Analyze the expression
Since \( \alpha \) is positive, we can analyze the expression \( \alpha + \frac{1}{\alpha} \). By the AM-GM inequality, we know:
\[ \alpha + \frac{1}{\alpha} \geq 2 \]
This implies:
\[ -a \geq 2 \quad \Rightarrow \quad a \leq -2 \]

### Step 5: Ensure distinct roots
For the roots to be distinct, the discriminant of the quadratic must be positive. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac = a^2 - 4 \cdot 1 \cdot 1 = a^2 - 4 \]
For the roots to be distinct, we require:
\[ D > 0 \quad \Rightarrow \quad a^2 - 4 > 0 \quad \Rightarrow \quad a < -2 \quad \text{or} \quad a > 2 \]

### Step 6: Combine conditions
From the analysis, we have:
1. \( a \leq -2 \) (from the positivity of the roots)
2. \( a < -2 \) or \( a > 2 \) (from the distinctness of the roots)

The only condition that satisfies both is:
\[ a < -2 \]

### Conclusion
Thus, the condition for the quadratic equation \( f(x) = x^2 + ax + 1 = 0 \) to have two positive distinct roots is:
\[ a < -2 \]