Let A is set of all real values of a for which equation `x^(2)-ax+1=0` has no real roots and B is set of al real values of b for which `f(x)=bx^(2)+bx+0.5gt0AAxepsilonR` then `AcapB=`

Options

A. `{x:0ltxlt2}`
B. `{x:0lexlt2}`
C. `{x:0lexle2}`
D. `{x:-2ltxlt2}`

Explanation

To solve the problem, we need to find the sets \( A \) and \( B \) and then determine their intersection \( A \cap B \).

### Step 1: Finding the set \( A \)

The set \( A \) consists of all real values of \( a \) for which the quadratic equation \( x^2 - ax + 1 = 0 \) has no real roots.

To determine when this equation has no real roots, we need to analyze the discriminant \( D \) of the quadratic equation, which is given by:

\[ D = b^2 - 4ac \]

For our equation \( x^2 - ax + 1 = 0 \):
- \( a = -a \)
- \( b = 1 \)
- \( c = 1 \)

Thus, the discriminant is:

\[ D = (-a)^2 - 4 \cdot 1 \cdot 1 = a^2 - 4 \]

For the equation to have no real roots, we require:

\[ D < 0 \]

This gives us:

\[ a^2 - 4 < 0 \]

Rearranging this inequality:

\[ a^2 < 4 \]

Taking the square root of both sides, we find:

\[ -2 < a < 2 \]

Thus, the set \( A \) is:

\[ A = (-2, 2) \]

### Step 2: Finding the set \( B \)

The set \( B \) consists of all real values of \( b \) for which the function \( f(x) = bx^2 + bx + 0.5 > 0 \) for all \( x \in \mathbb{R} \).

Since \( f(x) \) is a quadratic function, it will be positive for all \( x \) if:
1. The leading coefficient \( b > 0 \) (the parabola opens upwards).
2. The discriminant \( D \) of the quadratic must be less than zero.

The discriminant \( D \) for \( f(x) = bx^2 + bx + 0.5 \) is:

\[ D = b^2 - 4 \cdot b \cdot 0.5 = b^2 - 2b \]

We require:

\[ D < 0 \]

This gives us:

\[ b^2 - 2b < 0 \]

Factoring the left-hand side:

\[ b(b - 2) < 0 \]

To solve this inequality, we find the critical points \( b = 0 \) and \( b = 2 \). Testing the intervals:
- For \( b < 0 \): \( b(b - 2) > 0 \)
- For \( 0 < b < 2 \): \( b(b - 2) < 0 \)
- For \( b > 2 \): \( b(b - 2) > 0 \)

Thus, the solution to the inequality is:

\[ 0 < b < 2 \]

Including \( b = 0 \) since \( f(0) = 0.5 > 0 \), we can write:

\[ B = [0, 2) \]

### Step 3: Finding the intersection \( A \cap B \)

Now we need to find the intersection of sets \( A \) and \( B \):

\[ A = (-2, 2) \quad \text{and} \quad B = [0, 2) \]

The intersection \( A \cap B \) consists of values that are in both sets:

\[ A \cap B = [0, 2) \]

### Final Answer

Thus, the intersection \( A \cap B \) is:

\[ A \cap B = [0, 2) \]