The range of values of a so that all the roots of the equations `2x^(3)-3x^(2)-12x+a=0` are real and distinct, belongs to

Options

A. (a) (7,20)
B. (b) (-7,20)
C. (c) (-20,7)
D. (d) (-7,7)

Answer

To find the range of values of \( a \) such that all the roots of the equation \( 2x^3 - 3x^2 - 12x + a = 0 \) are real and distinct, we can follow these steps:

### Step 1: Define the function
Let \( f(x) = 2x^3 - 3x^2 - 12x + a \).

### Step 2: Differentiate the function
To find the critical points, we need to differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + a) = 6x^2 - 6x - 12 \]

### Step 3: Set the derivative to zero
Set the derivative equal to zero to find the critical points:
\[ 6x^2 - 6x - 12 = 0 \]
Dividing the entire equation by 6 gives:
\[ x^2 - x - 2 = 0 \]

### Step 4: Factor the quadratic
Factoring the quadratic equation:
\[ (x + 1)(x - 2) = 0 \]
Thus, the critical points are:
\[ x = -1 \quad \text{and} \quad x = 2 \]

### Step 5: Evaluate the function at the critical points
We need to evaluate \( f(x) \) at these critical points to determine the conditions for \( a \).

1. **At \( x = -1 \)**:
\[ f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + a = -2 - 3 + 12 + a = 7 + a \]
For \( f(-1) > 0 \):
\[ 7 + a > 0 \implies a > -7 \]

2. **At \( x = 2 \)**:
\[ f(2) = 2(2)^3 - 3(2)^2 - 12(2) + a = 16 - 12 - 24 + a = -20 + a \]
For \( f(2) < 0 \):
\[ -20 + a < 0 \implies a < 20 \]

### Step 6: Combine the inequalities
From the evaluations, we have:
\[ -7 < a < 20 \]
Thus, the range of values of \( a \) such that all roots of the equation are real and distinct is:
\[ a \in (-7, 20) \]

### Final Answer:
The range of values of \( a \) is \( (-7, 20) \).