If `3(a+2c)=4(b+3d),` then the equation `a x^3+b x^2+c x+d=0` will have (a) no real solution (b) at least one real root in `(-1,0)` (c) at least one real root in `(0,1)` (d) none of these

Options

A. no real solution
B. at least one real root in `(-1,0)`
C. at least one real root in `(0,1)`
D. none of these

Answer

To solve the problem, we need to analyze the given equation and apply the properties of derivatives and the Intermediate Value Theorem.

### Step-by-Step Solution:

1. **Given Equation**: We start with the equation given in the problem:
\[ 3(a + 2c) = 4(b + 3d) \]
This can be rearranged to:
\[ 3a + 6c = 4b + 12d \]

2. **Define the Polynomial**: We have the polynomial:
\[ f(x) = ax^3 + bx^2 + cx + d \]
We want to find the roots of this polynomial.

3. **Evaluate at Specific Points**: We will evaluate \( f(x) \) at \( x = -1 \) and \( x = 0 \):
- For \( x = -1 \):
\[ f(-1) = a(-1)^3 + b(-1)^2 + c(-1) + d = -a + b - c + d \]
- For \( x = 0 \):
\[ f(0) = d \]

4. **Set Up the Condition**: We need to check the values of \( f(-1) \) and \( f(0) \) based on the rearranged equation:
\[ f(-1) = -a + b - c + d \]
From the rearranged equation \( 3a + 6c = 4b + 12d \), we can express \( b \) in terms of \( a, c, \) and \( d \):
\[ b = \frac{3a + 6c - 12d}{4} \]

5. **Substituting \( b \) into \( f(-1) \)**:
Substituting \( b \) into \( f(-1) \):
\[ f(-1) = -a + \left(\frac{3a + 6c - 12d}{4}\right) - c + d \]
Simplifying this expression:
\[ f(-1) = -a + \frac{3a}{4} + \frac{6c}{4} - \frac{12d}{4} - c + d \]
\[ = -\frac{4a}{4} + \frac{3a}{4} + \frac{6c}{4} - \frac{4c}{4} - \frac{12d}{4} + \frac{4d}{4} \]
\[ = -\frac{a}{4} + \frac{2c}{4} - \frac{8d}{4} \]
\[ = -\frac{a + 2c - 8d}{4} \]

6. **Evaluate \( f(0) \)**:
\[ f(0) = d \]

7. **Apply the Intermediate Value Theorem**: Since \( f(-1) \) and \( f(0) \) are both dependent on the values of \( a, c, \) and \( d \), we can analyze the behavior of the polynomial. If \( f(-1) \) and \( f(0) \) have opposite signs, then by the Intermediate Value Theorem, there exists at least one real root in the interval \((-1, 0)\).

8. **Conclusion**: From the analysis, we find that:
- \( f(-1) = 0 \) and \( f(0) = d \) suggests that there is at least one real root in the interval \((-1, 0)\).

Thus, the answer is:
**(b) at least one real root in (-1, 0)**.