If `4a+2b+c=0` , then the equation `3ax^(2)+2bx+c=0` has at least one real lying in the interval

Options

A. (0,1)
B. (1,2)
C. (0,2)
D. none of these

Answer

To solve the problem step by step, we will analyze the given equation and apply the Mean Value Theorem.

### Step 1: Understand the Given Information
We have the equation:
\[ 4a + 2b + c = 0 \]
and we need to analyze the quadratic equation:
\[ 3ax^2 + 2bx + c = 0 \]

### Step 2: Define the Function
Let's define the function:
\[ f(x) = 3ax^2 + 2bx + c \]

### Step 3: Evaluate the Function at Specific Points
We will evaluate \( f(x) \) at \( x = 0 \) and \( x = 2 \).

1. **Calculate \( f(0) \)**:
\[ f(0) = 3a(0)^2 + 2b(0) + c = c \]

2. **Calculate \( f(2) \)**:
\[ f(2) = 3a(2)^2 + 2b(2) + c = 3a(4) + 2b(2) + c = 12a + 4b + c \]

### Step 4: Use the Given Condition
From the condition \( 4a + 2b + c = 0 \), we can express \( c \):
\[ c = - (4a + 2b) \]

### Step 5: Substitute \( c \) in \( f(2) \)
Now substitute \( c \) into \( f(2) \):
\[ f(2) = 12a + 4b - (4a + 2b) = 12a + 4b - 4a - 2b = 8a + 2b \]

### Step 6: Set Up the Equations
Now we have:
- \( f(0) = c = - (4a + 2b) \)
- \( f(2) = 8a + 2b \)

### Step 7: Check for Roots
We need to check if there is at least one real root in the interval [0, 2].

1. **Evaluate \( f(0) \)**:
\[ f(0) = - (4a + 2b) \]

2. **Evaluate \( f(2) \)**:
\[ f(2) = 8a + 2b \]

### Step 8: Analyze the Signs
From the above evaluations:
- If \( 4a + 2b + c = 0 \), then \( f(0) = 0 \).
- If \( 8a + 2b \) is positive, then \( f(2) > 0 \).

### Step 9: Apply the Intermediate Value Theorem
Since \( f(0) = 0 \) and \( f(2) = 8a + 2b \) (which can be positive), by the Intermediate Value Theorem, there must be at least one real root in the interval [0, 2].

### Conclusion
Thus, the equation \( 3ax^2 + 2bx + c = 0 \) has at least one real root in the interval [0, 2].

### Final Answer
The correct option is:
**Option 3: 0 to 2**

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