The real number k for which the equation `2x^3 + 3x +k=0` has two distinct real roots in [0, 1]

Options

A. lies between 1 and 2
B. lies between 2 and 3
C. lies between - 1 and 0
D. does not exist

Answer

To find the real number \( k \) for which the equation \( 2x^3 + 3x + k = 0 \) has two distinct real roots in the interval \([0, 1]\), we can follow these steps:

### Step 1: Understand the Function
The function we are dealing with is:
\[ f(x) = 2x^3 + 3x + k \]
We want to find conditions under which this cubic equation has two distinct real roots in the interval \([0, 1]\).

### Step 2: Differentiate the Function
To analyze the behavior of the function, we differentiate it:
\[ f'(x) = \frac{d}{dx}(2x^3 + 3x + k) = 6x^2 + 3 \]
This derivative is always positive since \( 6x^2 + 3 > 0 \) for all \( x \). This means that \( f(x) \) is a strictly increasing function.

### Step 3: Analyze the Roots
Since \( f(x) \) is strictly increasing, it can have at most one real root. Therefore, for the equation to have two distinct real roots in the interval \([0, 1]\), we need to consider the behavior of the function at the endpoints of the interval.

### Step 4: Evaluate the Function at the Endpoints
We evaluate \( f(x) \) at the endpoints \( x = 0 \) and \( x = 1 \):
- At \( x = 0 \):
\[ f(0) = 2(0)^3 + 3(0) + k = k \]
- At \( x = 1 \):
\[ f(1) = 2(1)^3 + 3(1) + k = 2 + 3 + k = 5 + k \]

### Step 5: Set Conditions for Roots
For the function to have two distinct roots in the interval \([0, 1]\), we need:
1. \( f(0) = k < 0 \) (the function must start below the x-axis)
2. \( f(1) = 5 + k > 0 \) (the function must end above the x-axis)

### Step 6: Solve the Inequalities
From the conditions:
1. \( k < 0 \)
2. \( 5 + k > 0 \) implies \( k > -5 \)

Combining these inequalities, we find:
\[ -5 < k < 0 \]

### Conclusion
The values of \( k \) for which the equation \( 2x^3 + 3x + k = 0 \) has two distinct real roots in the interval \([0, 1]\) are:
\[ k \in (-5, 0) \]