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 find the number of negative real roots of the cubic equation \( f(x) = 0 \), where the roots of \( f(x) \) are given by the expressions \( \frac{1+\alpha}{1-\alpha} \), \( \frac{1+\beta}{1-\beta} \), and \( \frac{1+\gamma}{1-\gamma} \), with \( \alpha, \beta, \gamma \) being the roots of the cubic equation \( 3x^3 - 2x + 5 = 0 \).

### Step-by-Step Solution:

**Step 1: Analyze the given cubic equation.**
We start with the cubic equation:
\[ 3x^3 - 2x + 5 = 0 \]
To determine the nature of the roots (whether they are real or complex), we can use the discriminant or evaluate the function at various points.

**Step 2: Evaluate the cubic function at specific points.**
Let's evaluate \( f(x) = 3x^3 - 2x + 5 \) at a few points:
- \( f(-1) = 3(-1)^3 - 2(-1) + 5 = -3 + 2 + 5 = 4 \) (positive)
- \( f(0) = 3(0)^3 - 2(0) + 5 = 5 \) (positive)
- \( f(1) = 3(1)^3 - 2(1) + 5 = 3 - 2 + 5 = 6 \) (positive)

Since \( f(-1) > 0 \), \( f(0) > 0 \), and \( f(1) > 0 \), we can check the derivative to find the critical points.

**Step 3: Find the derivative and critical points.**
The derivative of \( f(x) \) is:
\[ f'(x) = 9x^2 - 2 \]
Setting the derivative to zero to find critical points:
\[ 9x^2 - 2 = 0 \implies x^2 = \frac{2}{9} \implies x = \pm \frac{\sqrt{2}}{3} \]

**Step 4: Evaluate \( f(x) \) at the critical points.**
- \( f\left(\frac{\sqrt{2}}{3}\right) \) and \( f\left(-\frac{\sqrt{2}}{3}\right) \) need to be evaluated to check for local maxima or minima.

Calculating \( f\left(-\frac{\sqrt{2}}{3}\right) \):
\[ f\left(-\frac{\sqrt{2}}{3}\right) = 3\left(-\frac{\sqrt{2}}{3}\right)^3 - 2\left(-\frac{\sqrt{2}}{3}\right) + 5 \]
Calculating this gives a negative value, indicating that there is a local minimum.

**Step 5: Determine the nature of the roots.**
Since \( f(x) \) does not change sign in the intervals we checked and only has one local minimum, we conclude that the cubic equation \( 3x^3 - 2x + 5 = 0 \) has no real roots (all roots are complex).

**Step 6: Analyze the transformation.**
The roots \( \alpha, \beta, \gamma \) are complex, and we need to analyze the transformation \( \frac{1+\alpha}{1-\alpha} \):
- If \( \alpha \) is complex, \( \frac{1+\alpha}{1-\alpha} \) will also be complex.
- Thus, \( f(x) = 0 \) will have no negative real roots.

### Conclusion:
The number of negative real roots of the equation \( f(x) = 0 \) is:
\[ \boxed{0} \]