If a,b,c are in A.P.and f(x)=`|(x+a,x^2+1,1),(x+b, 2x^2-1,1),(x+c,3x^2-2 , 1)|`, then f'(x) is :

Options

A. 0
B. 1
C. a+bc
D. `(abc)/(a+b+c)`

Answer

To solve the problem, we need to find the derivative of the function defined by the determinant \( f(x) = \begin{vmatrix} x+a & x^2+1 & 1 \\ x+b & 2x^2-1 & 1 \\ x+c & 3x^2-2 & 1 \end{vmatrix} \).

### Step-by-Step Solution:

1. **Understanding the Condition of A.P.**:
Since \( a, b, c \) are in Arithmetic Progression (A.P.), we have the relation:
\[ 2b = a + c \]

2. **Setting Up the Determinant**:
The function \( f(x) \) is given as:
\[ f(x) = \begin{vmatrix} x+a & x^2+1 & 1 \\ x+b & 2x^2-1 & 1 \\ x+c & 3x^2-2 & 1 \end{vmatrix} \]

3. **Calculating the Determinant**:
To compute the determinant, we can use the property that if any two columns (or rows) of a determinant are identical, the determinant is zero. Notice that the third column is \( (1, 1, 1) \).

4. **Differentiating the Determinant**:
We need to find \( f'(x) \). The derivative of a determinant can be computed using the Leibniz rule, but in this case, we can simplify our work by observing the columns:
- The first column is \( (x+a, x+b, x+c) \).
- The second column is \( (x^2+1, 2x^2-1, 3x^2-2) \).
- The third column is \( (1, 1, 1) \).

5. **Identifying the Zero Column**:
If we differentiate the determinant with respect to \( x \), we can see that the third column remains constant (all elements are 1). Therefore, if we take the derivative of the first column and the second column, we will find that the determinant will still yield a column of zeros.

6. **Conclusion**:
Since the determinant will have a column of zeros after differentiation, we conclude that:
\[ f'(x) = 0 \]

### Final Answer:
Thus, \( f'(x) = 0 \).