Number of integers in the range of 'c' so that the equation `x^3-3x+c=0` has all its roots real and distinct is

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem of finding the number of integers in the range of \( c \) such that the equation \( x^3 - 3x + c = 0 \) has all its roots real and distinct, we will follow these steps:

### Step 1: Rewrite the Equation
We can rewrite the equation as:
\[ x^3 - 3x = -c \]
Let \( f(x) = x^3 - 3x \). We need to analyze this function to determine the values of \( c \) for which the equation has all roots real and distinct.

### Step 2: Find the Derivative
To find the critical points of \( f(x) \), we compute the first derivative:
\[ f'(x) = 3x^2 - 3 \]
Setting the derivative equal to zero to find critical points:
\[ 3x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1 \]

### Step 3: Determine the Nature of Critical Points
Next, we find the second derivative to determine the nature of these critical points:
\[ f''(x) = 6x \]
- At \( x = 1 \):
\[ f''(1) = 6 \cdot 1 = 6 > 0 \quad \text{(local minimum)} \]
- At \( x = -1 \):
\[ f''(-1) = 6 \cdot (-1) = -6 < 0 \quad \text{(local maximum)} \]

### Step 4: Calculate Maximum and Minimum Values
Now we evaluate \( f(x) \) at the critical points to find the maximum and minimum values:
- For \( x = 1 \):
\[ f(1) = 1^3 - 3 \cdot 1 = 1 - 3 = -2 \]
- For \( x = -1 \):
\[ f(-1) = (-1)^3 - 3 \cdot (-1) = -1 + 3 = 2 \]

### Step 5: Determine the Range of \( c \)
The function \( f(x) = x^3 - 3x \) achieves a maximum value of \( 2 \) and a minimum value of \( -2 \). Therefore, the range of \( c \) is:
\[ -2 \leq c \leq 2 \]

### Step 6: Count the Integer Values of \( c \)
The integers in the range from \( -2 \) to \( 2 \) are:
\[ -2, -1, 0, 1, 2 \]
Thus, there are a total of \( 5 \) integers.

### Final Answer
The number of integers in the range of \( c \) such that the equation \( x^3 - 3x + c = 0 \) has all its roots real and distinct is:
\[ \boxed{5} \]