Statement I The equation `3x^(2)+4ax+b=0` has atleast one root in (o,1), if 3+4a=0.
Statement II `f(x)=3x^(2)+4x+b` is continuos and differentiable in (0,1)

Options

A. Statement I is true, Statement II is also true, Statement II is the correct explanation of statement I.
B. Statement I is true, Statement II is also true, Statement II is not the correct explanation of Statement I.
C. Statement I is true, Statement II is false
D. Statement I is false, Statement II is true

Answer

To solve the problem, we need to analyze the two statements provided and determine their validity step by step.

### Step 1: Analyzing Statement I
The first statement claims that the equation \(3x^2 + 4ax + b = 0\) has at least one root in the interval \((0, 1)\) if \(3 + 4a = 0\).

1. **Substituting the condition**: From the condition \(3 + 4a = 0\), we can express \(a\) as:
\[ a = -\frac{3}{4} \]
Thus, the equation becomes:
\[ 3x^2 - 3x + b = 0 \quad \text{(substituting \(a\))} \]

2. **Evaluating the function at the endpoints**: We need to check the values of the quadratic function at \(x = 0\) and \(x = 1\):
- At \(x = 0\):
\[ f(0) = b \]
- At \(x = 1\):
\[ f(1) = 3(1)^2 - 3(1) + b = 3 - 3 + b = b \]

3. **Applying the Intermediate Value Theorem**: For the function \(f(x)\) to have at least one root in the interval \((0, 1)\), we need:
\[ f(0) \cdot f(1) < 0 \]
This means \(b \cdot b < 0\) which implies:
\[ b^2 < 0 \]
This is not possible since the square of a real number cannot be negative.

### Conclusion for Statement I
Since the condition leads to an impossible scenario, **Statement I is false**.

### Step 2: Analyzing Statement II
The second statement claims that the function \(f(x) = 3x^2 + 4x + b\) is continuous and differentiable in the interval \((0, 1)\).

1. **Checking continuity**: The function \(f(x)\) is a polynomial function. Polynomial functions are continuous everywhere, including the interval \((0, 1)\).

2. **Checking differentiability**: Similarly, polynomial functions are differentiable everywhere. Thus, \(f(x)\) is differentiable in the interval \((0, 1)\).

### Conclusion for Statement II
Since \(f(x)\) is both continuous and differentiable in the interval \((0, 1)\), **Statement II is true**.

### Final Conclusion
- Statement I is false.
- Statement II is true.

Thus, the correct option is that Statement I is false and Statement II is true.