Suppose `a in R, a ne - 1//2 . " Let " alpha, beta` be roots of `(2a + 1) x^(2) - ax + a - 2 = 0 If alpha lt 1 lt beta `, then

Options

A. `(1)/(7) (6 - 2 sqrt(23))lt a lt 1`
B. `(-1)/(2) lt a lt (1)/(2)`
C. `(1)/(2) lt a lt (1)/(7) (6 + 2 sqrt(3))`
D. none of these

Answer

To solve the given problem, we need to analyze the quadratic equation and the conditions provided. The equation is:

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

We are given that the roots of this equation, denoted as \(\alpha\) and \(\beta\), satisfy the condition \(\alpha < 1 < \beta\).

### Step 1: Identify the coefficients
The coefficients of the quadratic equation are:
- \(A = 2a + 1\)
- \(B = -a\)
- \(C = a - 2\)

### Step 2: Condition for real roots
For the quadratic equation to have real roots, the discriminant must be greater than zero. The discriminant \(D\) is given by:

\[ D = B^2 - 4AC \]

Substituting the values of \(A\), \(B\), and \(C\):

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

Calculating this step-by-step:

\[ D = a^2 - 4[(2a + 1)(a - 2)] \]
\[ = a^2 - 4[2a^2 - 4a + a - 2] \]
\[ = a^2 - 4[2a^2 - 3a - 2] \]
\[ = a^2 - (8a^2 - 12a + 8) \]
\[ = a^2 - 8a^2 + 12a - 8 \]
\[ = -7a^2 + 12a - 8 \]

Setting the discriminant greater than zero:

\[ -7a^2 + 12a - 8 > 0 \]

### Step 3: Solve the quadratic inequality
To solve this inequality, we first find the roots of the equation:

\[ -7a^2 + 12a - 8 = 0 \]

Using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ a = \frac{-12 \pm \sqrt{12^2 - 4 \cdot (-7) \cdot (-8)}}{2 \cdot (-7)} \]
\[ = \frac{-12 \pm \sqrt{144 - 224}}{-14} \]
\[ = \frac{-12 \pm \sqrt{-80}}{-14} \]

Since the discriminant is negative, the quadratic does not cross the x-axis, and we need to determine the sign of the quadratic expression. Since the coefficient of \(a^2\) is negative, the quadratic opens downwards. Therefore, it is positive between the roots.

### Step 4: Analyze the roots
Next, we need to find the values of \(a\) such that \(\alpha < 1 < \beta\). We can evaluate the function at \(x = 1\):

\[ f(1) = (2a + 1)(1)^2 - a(1) + (a - 2) \]
\[ = 2a + 1 - a + a - 2 \]
\[ = 2a - 1 \]

Since we need \(f(1) < 0\):

\[ 2a - 1 < 0 \implies 2a < 1 \implies a < \frac{1}{2} \]

### Step 5: Combine conditions
Now, we have two conditions:
1. The discriminant condition: \(-7a^2 + 12a - 8 > 0\)
2. The function value condition: \(a < \frac{1}{2}\)

### Step 6: Find the intersection of intervals
To find the valid range for \(a\), we need to solve the discriminant inequality. The roots of \(-7a^2 + 12a - 8 = 0\) can be found as mentioned above, and we can determine the intervals where the quadratic is positive.

### Conclusion
The valid range for \(a\) satisfying both conditions will be the intersection of the intervals derived from the discriminant and the function value condition.