Let us consider an equation `f(x) = x^(3)-3x +k=0`.then the values of k for which the equation has
Exactly one root which positive , then k belongs to

Options

A. `(-oo,-2)`
B. `(2,oo)`
C. (0,2)
D. `(-2,0)`

Answer

To find the values of \( k \) for which the equation \( f(x) = x^3 - 3x + k = 0 \) has exactly one positive root, we can follow these steps:

### Step 1: Analyze the function
We start with the function:
\[ f(x) = x^3 - 3x + k \]
To understand the behavior of this function, we need to find its critical points by taking the derivative.

### Step 2: Find the derivative
The derivative of \( f(x) \) is:
\[ f'(x) = 3x^2 - 3 \]
Setting the derivative equal to zero to find critical points:
\[ 3x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1 \]

### Step 3: Evaluate the function at critical points
Now, we evaluate \( f(x) \) at the critical points \( x = -1 \) and \( x = 1 \):
\[ f(-1) = (-1)^3 - 3(-1) + k = -1 + 3 + k = k + 2 \]
\[ f(1) = (1)^3 - 3(1) + k = 1 - 3 + k = k - 2 \]

### Step 4: Determine the conditions for exactly one positive root
For the function to have exactly one positive root, we need:
1. \( f(1) = k - 2 \) should be equal to 0 (the function must touch the x-axis at this point).
2. \( f(-1) = k + 2 \) should be greater than 0 (the function must be above the x-axis at this point).

From \( f(1) = 0 \):
\[ k - 2 = 0 \implies k = 2 \]

From \( f(-1) > 0 \):
\[ k + 2 > 0 \implies k > -2 \]

### Step 5: Combine the conditions
We have:
- \( k = 2 \) for the function to have a root at \( x = 1 \).
- \( k > -2 \) to ensure the function is above the x-axis at \( x = -1 \).

Thus, the values of \( k \) for which the equation has exactly one positive root are:
\[ k \in (-2, 2] \]

### Final Answer
The values of \( k \) for which the equation \( f(x) = x^3 - 3x + k = 0 \) has exactly one positive root are:
\[ k \in (-2, 2] \]