Set of values of 'a' for which both roots of the equation `x^(2) - 2x - a^(2) = 0` lie between the roots of the equation `x^(2) - 2x + a^(2) - 11a + 12 = 0`, is

Options

A. (1, 4)
B. (3/2, 4)
C. `(-4, 4)`
D. none of these

Explanation

To solve the problem, we need to find the set of values of \( a \) for which both roots of the equation

\[ x^2 - 2x - a^2 = 0 \]

lie between the roots of the equation

\[ x^2 - 2x + a^2 - 11a + 12 = 0. \]

### Step 1: Find the Discriminant of the First Equation

The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:

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

For the equation \( x^2 - 2x - a^2 = 0 \):

- \( a = 1 \)
- \( b = -2 \)
- \( c = -a^2 \)

Calculating the discriminant:

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

Since \( D > 0 \) for all \( a \in \mathbb{R} \), the roots of this equation are real and distinct.

### Step 2: Find the Roots of the First Equation

The roots \( \alpha_1 \) and \( \alpha_2 \) can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{2 \pm \sqrt{4 + 4a^2}}{2} = 1 \pm \sqrt{1 + a^2}. \]

Thus, the roots are:

\[ \alpha_1 = 1 + \sqrt{1 + a^2}, \quad \alpha_2 = 1 - \sqrt{1 + a^2}. \]

### Step 3: Find the Discriminant of the Second Equation

Next, we find the discriminant of the second equation \( x^2 - 2x + a^2 - 11a + 12 = 0 \):

Here, \( a = 1 \), \( b = -2 \), and \( c = a^2 - 11a + 12 \).

Calculating the discriminant:

\[ D = (-2)^2 - 4(1)(a^2 - 11a + 12) = 4 - 4(a^2 - 11a + 12) = 4 - 4a^2 + 44a - 48 = -4a^2 + 44a - 44. \]

### Step 4: Find the Roots of the Second Equation

The roots of the second equation can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{2 \pm \sqrt{-4a^2 + 44a - 44}}{2} = 1 \pm \frac{\sqrt{-4a^2 + 44a - 44}}{2}. \]

Let \( \beta_1 = 1 + \frac{\sqrt{-4a^2 + 44a - 44}}{2} \) and \( \beta_2 = 1 - \frac{\sqrt{-4a^2 + 44a - 44}}{2} \).

### Step 5: Condition for Roots

We need both roots \( \alpha_1 \) and \( \alpha_2 \) to lie between \( \beta_1 \) and \( \beta_2 \):

\[ \beta_2 < \alpha_2 < \alpha_1 < \beta_1. \]

This leads to the inequalities:

1. \( 1 - \frac{\sqrt{-4a^2 + 44a - 44}}{2} < 1 - \sqrt{1 + a^2} \)
2. \( 1 + \sqrt{1 + a^2} < 1 + \frac{\sqrt{-4a^2 + 44a - 44}}{2} \)

### Step 6: Solve the Inequalities

1. From the first inequality, simplifying gives:

\[ -\frac{\sqrt{-4a^2 + 44a - 44}}{2} < -\sqrt{1 + a^2} \implies \sqrt{-4a^2 + 44a - 44} > 2\sqrt{1 + a^2}. \]

Squaring both sides:

\[ -4a^2 + 44a - 44 > 4(1 + a^2) \implies -4a^2 + 44a - 44 > 4 + 4a^2 \implies -8a^2 + 44a - 48 > 0. \]

Factoring gives:

\[ 2a^2 - 11a + 12 < 0. \]

Finding roots using the quadratic formula:

\[ a = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 2 \cdot 12}}{2 \cdot 2} = \frac{11 \pm \sqrt{121 - 96}}{4} = \frac{11 \pm 5}{4}. \]

Thus, the roots are \( a = 4 \) and \( a = \frac{3}{2} \).

### Step 7: Determine the Interval

The quadratic \( 2a^2 - 11a + 12 < 0 \) opens upwards, so the solution is between the roots:

\[ \frac{3}{2} < a < 4. \]

### Final Answer

The set of values of \( a \) for which both roots of the equation \( x^2 - 2x - a^2 = 0 \) lie between the roots of the equation \( x^2 - 2x + a^2 - 11a + 12 = 0 \) is:

\[ \boxed{\left( \frac{3}{2}, 4 \right)}. \]