Let `x_(1),x_(2),x_(3)` bet the roots of equation `x^(3) - x^(2) +betax + gamma = 0 ` If ` x_(1),x_(2),x_(3)` are in A. P ., then

Options

A. `beta in (- oo,1/3]`
B. `beta in [ 1/3 ,oo)`
C. `gamma in ( - oo,-1/27]`
D. `gamma in [ -1/27 ,oo)`

Answer

To solve the problem, we need to analyze the given cubic equation \( x^3 - x^2 + \beta x + \gamma = 0 \) and the condition that its roots \( x_1, x_2, x_3 \) are in arithmetic progression (A.P.).

### Step 1: Express the roots in A.P.
Since the roots are in A.P., we can denote them as:
- \( x_1 = a - d \)
- \( x_2 = a \)
- \( x_3 = a + d \)

### Step 2: Use Vieta's formulas
According to Vieta's formulas for a cubic polynomial \( ax^3 + bx^2 + cx + d = 0 \):
1. The sum of the roots \( x_1 + x_2 + x_3 = -\frac{b}{a} \)
2. The sum of the products of the roots taken two at a time \( x_1x_2 + x_2x_3 + x_3x_1 = \frac{c}{a} \)
3. The product of the roots \( x_1x_2x_3 = -\frac{d}{a} \)

For our equation:
- \( a = 1 \)
- \( b = -1 \)
- \( c = \beta \)
- \( d = \gamma \)

### Step 3: Calculate the sum of the roots
Using the first formula:
\[ x_1 + x_2 + x_3 = (a - d) + a + (a + d) = 3a \]
Setting this equal to \( -\frac{-1}{1} = 1 \):
\[ 3a = 1 \implies a = \frac{1}{3} \]

### Step 4: Calculate the sum of the products of the roots
Using the second formula:
\[ x_1x_2 + x_2x_3 + x_3x_1 = (a - d)a + a(a + d) + (a - d)(a + d) \]
This simplifies to:
\[ = a^2 - ad + a^2 + ad + (a^2 - d^2) = 3a^2 - d^2 \]
Setting this equal to \( \beta \):
\[ 3a^2 - d^2 = \beta \]
Substituting \( a = \frac{1}{3} \):
\[ 3\left(\frac{1}{3}\right)^2 - d^2 = \beta \implies 3 \cdot \frac{1}{9} - d^2 = \beta \implies \frac{1}{3} - d^2 = \beta \]

### Step 5: Calculate the product of the roots
Using the third formula:
\[ x_1x_2x_3 = (a - d)a(a + d) = a(a^2 - d^2) \]
Setting this equal to \( -\gamma \):
\[ a(a^2 - d^2) = -\gamma \]
Substituting \( a = \frac{1}{3} \):
\[ \frac{1}{3}\left(\left(\frac{1}{3}\right)^2 - d^2\right) = -\gamma \implies \frac{1}{3}\left(\frac{1}{9} - d^2\right) = -\gamma \]
This simplifies to:
\[ \frac{1}{27} - \frac{d^2}{3} = -\gamma \implies \gamma = -\frac{1}{27} + \frac{d^2}{3} \]

### Step 6: Determine the conditions for \( \beta \) and \( \gamma \)
From \( \beta = \frac{1}{3} - d^2 \):
- Since \( d^2 \geq 0 \), we have \( \beta \leq \frac{1}{3} \).

From \( \gamma = -\frac{1}{27} + \frac{d^2}{3} \):
- Since \( d^2 \geq 0 \), \( \gamma \geq -\frac{1}{27} \).

### Conclusion
Thus, we conclude:
- \( \beta \) can take values from \( -\infty \) to \( \frac{1}{3} \) (inclusive).
- \( \gamma \) can take values from \( -\frac{1}{27} \) to \( +\infty \).