If both roots of `x^(2)-2ax+a^(2)-1=0` lies in `(-2, 1)` then `[a]`, where `[.]` denotes greatest integral function is

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To solve the problem, we need to find the values of \( a \) such that both roots of the quadratic equation \( x^2 - 2ax + (a^2 - 1) = 0 \) lie within the interval \((-2, 1)\).

### Step 1: Find the roots of the quadratic equation
The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation, \( a = 1 \), \( b = -2a \), and \( c = a^2 - 1 \). Thus, the roots are:
\[ x = \frac{2a \pm \sqrt{(-2a)^2 - 4 \cdot 1 \cdot (a^2 - 1)}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{2a \pm \sqrt{4a^2 - 4(a^2 - 1)}}{2} \]
\[ = \frac{2a \pm \sqrt{4a^2 - 4a^2 + 4}}{2} \]
\[ = \frac{2a \pm \sqrt{4}}{2} \]
\[ = \frac{2a \pm 2}{2} \]
\[ = a \pm 1 \]
Thus, the roots are \( r_1 = a + 1 \) and \( r_2 = a - 1 \).

### Step 2: Set the conditions for the roots
We need both roots to lie in the interval \((-2, 1)\):
1. For \( r_1 = a + 1 \):
\[ -2 < a + 1 < 1 \]
This gives us two inequalities:
- From \( -2 < a + 1 \):
\[ a > -3 \]
- From \( a + 1 < 1 \):
\[ a < 0 \]

2. For \( r_2 = a - 1 \):
\[ -2 < a - 1 < 1 \]
This gives us two inequalities:
- From \( -2 < a - 1 \):
\[ a > -1 \]
- From \( a - 1 < 1 \):
\[ a < 2 \]

### Step 3: Combine the inequalities
Now we combine the inequalities:
- From \( r_1 \): \( -3 < a < 0 \)
- From \( r_2 \): \( -1 < a < 2 \)

The intersection of these intervals is:
\[ -1 < a < 0 \]

### Step 4: Find the greatest integer function
The greatest integer function \([a]\) for \( a \) in the interval \((-1, 0)\) is:
\[ [a] = -1 \]

### Final Answer
Thus, the value of \([a]\) is:
\[ \boxed{-1} \]