If a,b,c d `in R` such that `(a+2c)/(b+3d) +4/3=0,` then the equation `ax ^(3) + bx^(3)+ cx+d =0` has

Options

A. atleast one root in `(-1, 0)`
B. atleast one root in `(0,1)`
C. no root in `(-1, 1)`
D. no root in `(0,2)`

Answer

To solve the problem, we need to analyze the given equation and the conditions provided. Let's break it down step by step.

### Given:
\[ \frac{a + 2c}{b + 3d} + \frac{4}{3} = 0 \]

### Step 1: Rearranging the Equation
We can rearrange the equation to isolate the fraction:
\[ \frac{a + 2c}{b + 3d} = -\frac{4}{3} \]

### Step 2: Cross-Multiplying
Cross-multiplying gives us:
\[ 3(a + 2c) = -4(b + 3d) \]
Expanding both sides:
\[ 3a + 6c = -4b - 12d \]

### Step 3: Rearranging the Terms
Rearranging the terms leads to:
\[ 3a + 4b + 6c + 12d = 0 \]

### Step 4: Analyzing the Polynomial
We need to analyze the polynomial:
\[ f(x) = ax^3 + bx^2 + cx + d \]
We want to determine if this polynomial has at least one root.

### Step 5: Applying Rolle's Theorem
To apply Rolle's Theorem, we need to check the values of the polynomial at two points. Let's evaluate \(f(0)\) and \(f(1)\):
- For \(f(0)\):
\[ f(0) = d \]
- For \(f(1)\):
\[ f(1) = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d \]

### Step 6: Setting Up Conditions
From our earlier equation \(3a + 4b + 6c + 12d = 0\), we can express \(d\) in terms of \(a\), \(b\), and \(c\):
\[ d = -\frac{3a + 4b + 6c}{12} \]

### Step 7: Evaluating Roots
Now we can evaluate \(f(0)\) and \(f(1)\):
1. If \(f(0) = d = -\frac{3a + 4b + 6c}{12}\)
2. If \(f(1) = a + b + c + d\)

### Step 8: Conclusion
Since we have established that \(f(0)\) and \(f(1)\) can take values that satisfy the conditions of Rolle's theorem, we conclude that the polynomial \(f(x)\) has at least one root in the interval \([0, 1]\).

### Final Answer
The equation \(ax^3 + bx^2 + cx + d = 0\) has at least one root in the interval \([0, 1]\).

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