To solve the problem, we need to analyze the given quadratic equation \(x^2 - 2ax + (a^2 + a - 3) = 0\) under different conditions related to its roots. Let's break down the solution step by step.
### Step 1: Determine the conditions for real roots
For the roots of the quadratic equation to be real, the discriminant must be non-negative. The discriminant \(D\) is given by:
\[
D = b^2 - 4ac
\]
Here, \(a = 1\), \(b = -2a\), and \(c = a^2 + a - 3\).
Calculating the discriminant:
\[
D = (-2a)^2 - 4 \cdot 1 \cdot (a^2 + a - 3) = 4a^2 - 4(a^2 + a - 3)
\]
\[
D = 4a^2 - 4a^2 - 4a + 12 = -4a + 12
\]
For the roots to be real:
\[
-4a + 12 \geq 0 \implies a \leq 3
\]
### Step 2: Analyze the conditions for the roots to be less than 3
We want both roots to be less than 3. The roots of the quadratic can be expressed as:
\[
x = \frac{2a \pm \sqrt{D}}{2}
\]
To ensure both roots are less than 3, we need:
\[
\frac{2a + \sqrt{D}}{2} < 3
\]
This simplifies to:
\[
2a + \sqrt{D} < 6 \implies \sqrt{D} < 6 - 2a
\]
Squaring both sides gives:
\[
D < (6 - 2a)^2
\]
Substituting \(D\):
\[
-4a + 12 < (6 - 2a)^2
\]
Expanding the right side:
\[
-4a + 12 < 36 - 24a + 4a^2
\]
Rearranging gives:
\[
4a^2 - 20a + 24 > 0
\]
Factoring or using the quadratic formula:
\[
a^2 - 5a + 6 > 0 \implies (a - 2)(a - 3) > 0
\]
This inequality holds when:
\[
a < 2 \quad \text{or} \quad a > 3
\]
Since we also have \(a \leq 3\), we conclude:
\[
a < 2
\]
### Step 3: Combine conditions
From Step 1, we have \(a \leq 3\) and from Step 2, \(a < 2\). Therefore, the final condition for part (A) is:
\[
a \in (-\infty, 2)
\]
### Step 4: Analyze the conditions for the roots to be greater than 3
For part (B), we want both roots to be greater than 3. Following a similar process:
\[
\frac{2a - \sqrt{D}}{2} > 3
\]
This leads to:
\[
2a - \sqrt{D} > 6 \implies \sqrt{D} < 2a - 6
\]
Squaring gives:
\[
D < (2a - 6)^2
\]
Substituting \(D\):
\[
-4a + 12 < (2a - 6)^2
\]
Expanding:
\[
-4a + 12 < 4a^2 - 24a + 36
\]
Rearranging yields:
\[
4a^2 - 20a + 24 > 0
\]
This is the same quadratic as before, which gives:
\[
a < 2 \quad \text{or} \quad a > 3
\]
However, since we want \(a > 3\) and \(a \leq 3\), there are no values of \(a\) that satisfy this condition. Thus, the answer for part (B) is:
\[
\text{(p) } a \text{ does not exist.}
\]
### Step 5: Analyze the condition for exactly one root in (1,3)
For part (C), we need \(f(1) \cdot f(3) < 0\):
\[
f(1) = 1 - 2a + (a^2 + a - 3) = a^2 - a - 2
\]
\[
f(3) = 9 - 6a + (a^2 + a - 3) = a^2 - 5a + 6
\]
The product:
\[
(a^2 - a - 2)(a^2 - 5a + 6) < 0
\]
Factoring gives:
\[
(a - 2)(a + 1)(a - 2)(a - 3) < 0
\]
The intervals are:
\[
(-\infty, -1) \cup (2, 3)
\]
Thus, the answer for part (C) is:
\[
\text{(r) } (-1, 2) \cup (2, 3)
\]
### Step 6: Analyze the condition for one root greater than 3 and the other less than 1
For part (D):
\[
f(1) < 0 \quad \text{and} \quad f(3) < 0
\]
This leads to:
\[
a \in (-1, 2) \quad \text{and} \quad a \in (2, 3)
\]
The intersection is empty, so the answer for part (D) is:
\[
\text{(p) } a \text{ does not exist.}
\]
### Final Answers
- (A) corresponds to (r): \(a \in (-\infty, 2)\)
- (B) corresponds to (p): \(a \text{ does not exist.}\)
- (C) corresponds to (r): \(a \in (-1, 2) \cup (2, 3)\)
- (D) corresponds to (p): \(a \text{ does not exist.}\)
