If exactly one root of the quadratic equation `x ^(2) -(a+1 )x +2a =0` lies in the interval `(0,3)` then the set of value 'a' is given by :

Options

A. `(-oo, 0) uu (6,0)`
B. `(-oo, 0)uu (6, oo)`
C. `(-oo, 0] uu [6,oo)`
D. `(0,6)`

Answer

To solve the problem, we need to analyze the quadratic equation \( x^2 - (a+1)x + 2a = 0 \) and determine the values of \( a \) such that exactly one root lies in the interval \( (0, 3) \).

### Step 1: Determine the Discriminant
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( D = b^2 - 4ac \). For our equation:
- \( a = 1 \)
- \( b = -(a + 1) \)
- \( c = 2a \)

Calculating the discriminant:
\[ D = (-(a + 1))^2 - 4 \cdot 1 \cdot 2a = (a + 1)^2 - 8a \]
\[ D = a^2 + 2a + 1 - 8a = a^2 - 6a + 1 \]
For the quadratic to have exactly one root, the discriminant must be zero:
\[ a^2 - 6a + 1 = 0 \]

### Step 2: Solve the Discriminant Equation
We can solve the quadratic equation \( a^2 - 6a + 1 = 0 \) using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2} \]
Thus, the roots are:
\[ a_1 = 3 + 2\sqrt{2}, \quad a_2 = 3 - 2\sqrt{2} \]

### Step 3: Check the Interval Condition
Next, we need to ensure that exactly one root lies in the interval \( (0, 3) \). We will evaluate the function at the endpoints of the interval.

1. **Evaluate \( f(0) \)**:
\[ f(0) = 2a \]
For \( f(0) < 0 \), we need \( 2a < 0 \) which implies \( a < 0 \).

2. **Evaluate \( f(3) \)**:
\[ f(3) = 3^2 - (a + 1) \cdot 3 + 2a = 9 - 3a - 3 + 2a = 6 - a \]
For \( f(3) > 0 \), we need \( 6 - a > 0 \) which implies \( a < 6 \).

### Step 4: Combine Conditions
Now we combine the conditions:
- From \( f(0) < 0 \): \( a < 0 \)
- From \( f(3) > 0 \): \( a < 6 \)

Thus, we have:
\[ a < 0 \quad \text{and} \quad a < 6 \]
The more restrictive condition is \( a < 0 \).

### Step 5: Identify the Set of Values for \( a \)
Since we also need to consider the roots \( a_1 = 3 + 2\sqrt{2} \) and \( a_2 = 3 - 2\sqrt{2} \):
- \( 3 - 2\sqrt{2} \approx 0.172 \)
- \( 3 + 2\sqrt{2} \approx 5.828 \)

Thus, the values of \( a \) must also satisfy:
\[ a \in (-\infty, 3 - 2\sqrt{2}) \cup (3 + 2\sqrt{2}, \infty) \]

### Final Answer
The set of values for \( a \) such that exactly one root of the quadratic equation lies in the interval \( (0, 3) \) is:
\[ a \in (-\infty, 3 - 2\sqrt{2}) \cup (3 + 2\sqrt{2}, \infty) \]