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 intersection of two sets:

1. Set A, which contains all real values of \( a \) for which the equation \( x^2 - ax + 1 = 0 \) has no real roots.
2. Set B, which contains all real values of \( b \) for which the function \( f(x) = bx^2 + bx + 0.5 > 0 \) for every \( x \in \mathbb{R} \).

### Step 1: Finding Set A

The quadratic equation \( x^2 - ax + 1 = 0 \) has no real roots when its discriminant is less than zero.

The discriminant \( D \) of the quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:

\[ D = B^2 - 4AC \]

For our equation, \( A = 1 \), \( B = -a \), and \( C = 1 \). Thus, the discriminant is:

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

To ensure there are no real roots, we set the discriminant less than zero:

\[ a^2 - 4 < 0 \]

This simplifies to:

\[ a^2 < 4 \]

Taking the square root of both sides, we find:

\[ -2 < a < 2 \]

Thus, Set A is:

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

### Step 2: Finding Set B

The function \( f(x) = bx^2 + bx + 0.5 \) is positive for all \( x \in \mathbb{R} \) if:

1. The coefficient of \( x^2 \) (which is \( b \)) is greater than zero.
2. The discriminant of the quadratic is less than zero.

First, we require:

\[ b > 0 \]

Next, we calculate the discriminant of \( f(x) \):

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

Setting the discriminant less than zero gives:

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

Factoring this inequality:

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

To solve this inequality, we find the critical points by setting \( b(b - 2) = 0 \):

- \( b = 0 \)
- \( b = 2 \)

Now, we analyze the intervals:

1. For \( b < 0 \): both factors are negative, product is positive.
2. For \( 0 < b < 2 \): the first factor is positive, the second is negative, product is negative.
3. For \( b > 2 \): both factors are positive, product is positive.

Thus, the solution to \( b(b - 2) < 0 \) is:

\[ 0 < b < 2 \]

So, Set B is:

\[ B = (0, 2) \]

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

Now we find the intersection of sets A and B:

- Set A: \( (-2, 2) \)
- Set B: \( (0, 2) \)

The intersection \( A \cap B \) is the set of values that are in both intervals:

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

### Final Answer

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

\[ \boxed{(0, 2)} \]