If `(1+alpha)/(1-alpha),(1+beta)/(1-beta), (1+gamma)/(1-gamma)` are the cubic equation f(x) = 0 where `alpha,beta,gamma` are the roots of the cubic equation `3x^3 - 2x + 5 =0`, then the number of negative real roots of the equation f(x) = 0 is :

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to determine the number of negative real roots of the cubic equation \( f(x) = 0 \) where the roots are given in the form of \( \frac{1+\alpha}{1-\alpha}, \frac{1+\beta}{1-\beta}, \frac{1+\gamma}{1-\gamma} \) and \( \alpha, \beta, \gamma \) are the roots of the cubic equation \( 3x^3 - 2x + 5 = 0 \).

### Step-by-Step Solution:

1. **Identify the roots of the original cubic equation:**
The roots \( \alpha, \beta, \gamma \) are the solutions to the equation \( 3x^3 - 2x + 5 = 0 \).

2. **Transform the roots:**
We need to find the transformed roots \( x = \frac{1+\alpha}{1-\alpha} \). Rearranging gives us \( \alpha = \frac{x-1}{x+1} \).

3. **Substitute into the original equation:**
We substitute \( \alpha \) into the original equation:
\[ 3\left(\frac{x-1}{x+1}\right)^3 - 2\left(\frac{x-1}{x+1}\right) + 5 = 0. \]
This will lead to a new equation in terms of \( x \).

4. **Clear the denominators:**
Multiply through by \( (x+1)^3 \) to eliminate the fractions:
\[ 3(x-1)^3 - 2(x-1)(x+1)^2 + 5(x+1)^3 = 0. \]

5. **Expand and simplify:**
Expand each term:
- \( (x-1)^3 = x^3 - 3x^2 + 3x - 1 \)
- \( (x+1)^2 = x^2 + 2x + 1 \)
- \( (x+1)^3 = x^3 + 3x^2 + 3x + 1 \)

Substitute these expansions back into the equation and simplify.

6. **Combine like terms:**
After simplification, we will obtain a cubic polynomial in \( x \):
\[ 2x^3 - 13x^2 + 2x - 3 = 0. \]

7. **Analyze the cubic polynomial:**
To find the number of negative real roots, we can use Descartes' Rule of Signs. We analyze the sign changes in \( f(x) \):
- Calculate \( f(0) = -3 \) (negative).
- Check the limits as \( x \to -\infty \) and \( x \to +\infty \).

8. **Differentiate to find critical points:**
Find the derivative \( f'(x) = 6x^2 - 26x + 2 \) and determine its discriminant:
\[ D = (-26)^2 - 4 \cdot 6 \cdot 2 = 676 - 48 = 628 > 0. \]
This indicates two critical points.

9. **Determine the nature of the roots:**
Since \( f(0) < 0 \) and the cubic polynomial has two critical points, we can sketch the graph. The graph will start from negative, rise to a maximum, fall to a minimum, and then rise again.

10. **Conclusion:**
Since the graph starts negative and ends positive with two critical points, there will be one root in the negative region and two roots in the positive region.

Thus, the number of negative real roots of the equation \( f(x) = 0 \) is **0**.