The real number k for which the equation, `2x^3+""3x""+""k""=""0` has two distinct real roots in [0, 1] (1) lies between 2 and 3 (2) lies between -1 and 0 (3) does not exist (4) lies between 1 and 2

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 solve the problem, we need to analyze the cubic equation \( f(x) = 2x^3 + 3x + k \) and determine the conditions under which it has two distinct real roots in the interval \([0, 1]\).

### Step 1: Define the function
We start with the function:
\[ f(x) = 2x^3 + 3x + k \]

### Step 2: Find the derivative
Next, we find the derivative of the function to analyze its behavior:
\[ f'(x) = \frac{d}{dx}(2x^3 + 3x + k) = 6x^2 + 3 \]

### Step 3: Analyze the derivative
The derivative \( f'(x) = 6x^2 + 3 \) is always positive for all \( x \) because:
- \( 6x^2 \geq 0 \) for all \( x \)
- Adding 3 ensures that \( f'(x) > 0 \)

Since \( f'(x) > 0 \) for all \( x \), it indicates that \( f(x) \) is a strictly increasing function.

### Step 4: Determine the nature of roots
For a strictly increasing function, it can have at most one real root. Therefore, for the equation \( f(x) = 0 \) to have two distinct real roots, it is impossible because a strictly increasing function cannot cross the x-axis more than once.

### Step 5: Conclusion
Since the function \( f(x) \) cannot have two distinct real roots for any value of \( k \), we conclude that the correct answer is:
(3) does not exist.