If `Delta(x)=|((e^x,sin2x,tanx^2)),((In(1+x),cosx,sinx)),((cosx^2,e^x-1,sinx^2))|=A+Bx+Cx^2+....` then B is equal to

Options

A. 0
B. 1
C. 2
D. none of these

Answer

To solve the question, we need to find the value of \( B \) in the polynomial expansion of the determinant \( \Delta(x) \).

### Step-by-Step Solution:

1. **Set Up the Determinant**:
We have the determinant defined as:
\[ \Delta(x) = \begin{vmatrix} e^x & \sin(2x) & \tan(x^2) \\ \ln(1+x) & \cos(x) & \sin(x) \\ \cos(x^2) & e^x - 1 & \sin(x^2) \end{vmatrix} \]

2. **Differentiate the Determinant**:
To find the coefficient \( B \), we can differentiate \( \Delta(x) \) with respect to \( x \):
\[ \Delta'(x) = \frac{d}{dx} \Delta(x) \]

3. **Evaluate at \( x = 0 \)**:
We need to evaluate \( \Delta'(0) \) to find \( B \):
\[ B = \Delta'(0) \]

4. **Differentiate Each Column**:
We can differentiate the determinant by differentiating each column one at a time. For example, if we differentiate the first column:
- The derivative of \( e^x \) is \( e^x \).
- The derivative of \( \ln(1+x) \) is \( \frac{1}{1+x} \).
- The derivative of \( \cos(x^2) \) is \( -2x \sin(x^2) \).

We will repeat this for the other columns as well.

5. **Construct the New Determinant**:
After differentiating the columns, we will construct a new determinant:
\[ \Delta'(x) = \begin{vmatrix} e^x & \sin(2x) & \tan(x^2) \\ \frac{1}{1+x} & -\sin(x) & \cos(x) \\ -2x \sin(x^2) & e^x & 2x \cos(x^2) \end{vmatrix} \]

6. **Evaluate the New Determinant at \( x = 0 \)**:
Now we substitute \( x = 0 \) into the new determinant:
\[ \Delta'(0) = \begin{vmatrix} e^0 & \sin(0) & \tan(0) \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{vmatrix} = \begin{vmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{vmatrix} \]

7. **Calculate the Determinant**:
The determinant simplifies to:
\[ \Delta'(0) = 1 \cdot (0 - 0) - 0 \cdot (1 - 0) + 0 \cdot (1 - 0) = 0 \]

8. **Final Result**:
Since we have \( \Delta'(0) = 0 \), we conclude that \( B = 0 \).

### Conclusion:
The value of \( B \) is \( 0 \).