Let S denote the set of all real values of a for which the roots of the equation `x^(2) - 2ax + a^(2) - 1 = 0` lie between 5 and 10, then S equals

Options

A. `(-1, 2)`
B. (2, 9)
C. (4, 9)
D. (6, 9)

Answer

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

\[ x^2 - 2ax + a^2 - 1 = 0 \]

lie between 5 and 10. We will analyze the conditions required for the roots to lie in this interval.

### Step 1: Sum of the Roots
The sum of the roots of the quadratic equation \( x^2 - 2ax + (a^2 - 1) = 0 \) is given by

\[ \alpha + \beta = 2a. \]

For the roots to lie between 5 and 10, we require:

\[ 5 < \alpha + \beta < 20. \]

This translates to:

\[ 5 < 2a < 20. \]

Dividing the entire inequality by 2 gives:

\[ \frac{5}{2} < a < 10. \]

### Step 2: Product of the Roots
The product of the roots is given by

\[ \alpha \beta = a^2 - 1. \]

For the roots to lie between 5 and 10, we require:

\[ 5 < \alpha \beta < 100. \]

This translates to two inequalities:

1. \( a^2 - 1 > 5 \)
2. \( a^2 - 1 < 100 \)

From the first inequality:

\[ a^2 > 6 \implies a > \sqrt{6} \quad \text{or} \quad a < -\sqrt{6}. \]

From the second inequality:

\[ a^2 < 101 \implies -\sqrt{101} < a < \sqrt{101}. \]

### Step 3: Discriminant Condition
The discriminant \( D \) of the quadratic must be non-negative for the roots to be real:

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

Since \( D = 4 \) is always positive, this condition is satisfied for all \( a \).

### Step 4: Function Values at the Endpoints
Now we need to check the values of the function at \( x = 5 \) and \( x = 10 \):

1. **At \( x = 5 \)**:
\[ f(5) = 5^2 - 2a(5) + (a^2 - 1) = 25 - 10a + a^2 - 1 = a^2 - 10a + 24. \]
We require \( f(5) > 0 \):
\[ a^2 - 10a + 24 > 0. \]
The roots of the equation \( a^2 - 10a + 24 = 0 \) are:
\[ a = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 24}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 96}}{2} = \frac{10 \pm 2}{2} = 5 \pm 1. \]
Thus, the roots are \( a = 4 \) and \( a = 6 \). The quadratic opens upwards, so:

\[ a < 4 \quad \text{or} \quad a > 6. \]

2. **At \( x = 10 \)**:
\[ f(10) = 10^2 - 2a(10) + (a^2 - 1) = 100 - 20a + a^2 - 1 = a^2 - 20a + 99. \]
We require \( f(10) > 0 \):
\[ a^2 - 20a + 99 > 0. \]
The roots of the equation \( a^2 - 20a + 99 = 0 \) are:
\[ a = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 99}}{2 \cdot 1} = \frac{20 \pm \sqrt{400 - 396}}{2} = \frac{20 \pm 2}{2} = 10 \pm 1. \]
Thus, the roots are \( a = 9 \) and \( a = 11 \). The quadratic opens upwards, so:

\[ a < 9 \quad \text{or} \quad a > 11. \]

### Step 5: Combine Conditions
Now we combine all the conditions:

1. From the sum of roots: \( \frac{5}{2} < a < 10 \).
2. From the product of roots: \( a > \sqrt{6} \) and \( a < \sqrt{101} \).
3. From \( f(5) > 0 \): \( a < 4 \) or \( a > 6 \).
4. From \( f(10) > 0 \): \( a < 9 \) or \( a > 11 \).

Combining these intervals, we find:

- The intersection of \( \left(\frac{5}{2}, 10\right) \) and \( (6, 9) \) gives \( (6, 9) \).

Thus, the set \( S \) of all real values of \( a \) for which the roots of the equation lie between 5 and 10 is:

\[ S = (6, 9). \]