The value of a for which twice the sum of the cubes of the roots of the equation `a=(x^(2)-3)/(x-2)` attains its minimum value is (where, `a in[0, pi]`)

Options

A. greater than 4
B. less than 2
C. greater than `(7)/(4)`
D. less than 1

Answer

To solve the problem, we will follow these steps:

### Step 1: Rewrite the given equation
The equation given is:
\[ a = \frac{x^2 - 3}{x - 2} \]
We will rearrange this to form a quadratic equation in terms of \(x\).

### Step 2: Rearranging the equation
Multiply both sides by \(x - 2\) to eliminate the fraction:
\[ a(x - 2) = x^2 - 3 \]
This simplifies to:
\[ ax - 2a = x^2 - 3 \]
Rearranging gives:
\[ x^2 - ax + (2a - 3) = 0 \]

### Step 3: Identify the coefficients
From the quadratic equation \(x^2 - ax + (2a - 3) = 0\), we can identify:
- Coefficient of \(x^2\) (denote as \(A\)) = 1
- Coefficient of \(x\) (denote as \(B\)) = -a
- Constant term (denote as \(C\)) = \(2a - 3\)

### Step 4: Calculate the sum and product of the roots
Using Vieta's formulas:
- The sum of the roots \((\alpha + \beta)\) is given by:
\[ \alpha + \beta = -\frac{B}{A} = \frac{a}{1} = a \]
- The product of the roots \((\alpha \beta)\) is given by:
\[ \alpha \beta = \frac{C}{A} = 2a - 3 \]

### Step 5: Find the expression for twice the sum of cubes of the roots
We need to find:
\[ 2(\alpha^3 + \beta^3) \]
Using the identity \(\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2)\), we can express this as:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) \]
Substituting the values we found:
\[ \alpha^3 + \beta^3 = a \left( a^2 - 3(2a - 3) \right) \]
This simplifies to:
\[ \alpha^3 + \beta^3 = a \left( a^2 - 6a + 9 \right) \]
Thus:
\[ 2(\alpha^3 + \beta^3) = 2a(a^2 - 6a + 9) = 2a^3 - 12a^2 + 18a \]

### Step 6: Differentiate the expression to find the minimum
Let:
\[ f(a) = 2a^3 - 12a^2 + 18a \]
To find the minimum, we differentiate \(f(a)\):
\[ f'(a) = 6a^2 - 24a + 18 \]
Setting the derivative to zero to find critical points:
\[ 6a^2 - 24a + 18 = 0 \]
Dividing through by 6:
\[ a^2 - 4a + 3 = 0 \]
Factoring gives:
\[ (a - 3)(a - 1) = 0 \]
Thus, \(a = 3\) or \(a = 1\).

### Step 7: Determine the minimum in the interval \([0, \pi]\)
We need to check the values of \(f(a)\) at \(a = 1\), \(a = 3\), and the endpoints \(a = 0\) and \(a = \pi\):
- \(f(0) = 0\)
- \(f(1) = 2(1)^3 - 12(1)^2 + 18(1) = 8\)
- \(f(3) = 2(3)^3 - 12(3)^2 + 18(3) = 0\)
- \(f(\pi) = 2(\pi)^3 - 12(\pi)^2 + 18(\pi)\) (this value is positive and larger than 0)

The minimum value occurs at \(a = 3\).

### Final Answer
Thus, the value of \(a\) for which twice the sum of the cubes of the roots attains its minimum value is:
\[ \boxed{3} \]