If `f(x)=|(x+a^2,x^4+1,3),(x+b^2,2x^4+2,3),(x+c^2,3x^4+7,3)|` where `x ne 0` and f' (x) = 0 then `a^2,b^2,c^2` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none

Answer

To solve the problem, we need to analyze the determinant given by the function \( f(x) = \begin{vmatrix} x + a^2 & x^4 + 1 & 3 \\ x + b^2 & 2x^4 + 2 & 3 \\ x + c^2 & 3x^4 + 7 & 3 \end{vmatrix} \) and find the conditions under which the derivative \( f'(x) = 0 \) leads to \( a^2, b^2, c^2 \) being in an arithmetic progression (AP).

### Step-by-step Solution:

1. **Evaluate the Determinant**:
We will simplify the determinant by performing row operations. We can subtract the first row from the second and third rows:
\[ R_2 \rightarrow R_2 - R_1 \quad \text{and} \quad R_3 \rightarrow R_3 - R_1 \]
This gives us:
\[ f(x) = \begin{vmatrix} x + a^2 & x^4 + 1 & 3 \\ b^2 - a^2 & x^4 + 1 & 0 \\ c^2 - a^2 & 2x^4 + 6 & 0 \end{vmatrix} \]

2. **Expand the Determinant**:
Since the last column has zeros, we can expand along the third column:
\[ f(x) = 3 \begin{vmatrix} b^2 - a^2 & x^4 + 1 \\ c^2 - a^2 & 2x^4 + 6 \end{vmatrix} \]

3. **Calculate the 2x2 Determinant**:
The 2x2 determinant is calculated as follows:
\[ = (b^2 - a^2)(2x^4 + 6) - (c^2 - a^2)(x^4 + 1) \]
Expanding this gives:
\[ = 2(b^2 - a^2)x^4 + 6(b^2 - a^2) - (c^2 - a^2)x^4 - (c^2 - a^2) \]
Combining like terms:
\[ = (2b^2 - 2a^2 - c^2 + a^2)x^4 + 6(b^2 - a^2) - (c^2 - a^2) \]
Simplifying further:
\[ = (2b^2 - c^2 - a^2)x^4 + 5(b^2 - a^2) + a^2 - c^2 \]

4. **Differentiate \( f(x) \)**:
Now we differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = 3 \left( (2b^2 - c^2 - a^2) \cdot 4x^3 + 5(b^2 - a^2) \cdot 0 \right) \]
Thus,
\[ f'(x) = 12(2b^2 - c^2 - a^2)x^3 \]

5. **Set the Derivative to Zero**:
Since \( f'(x) = 0 \), we have:
\[ 12(2b^2 - c^2 - a^2)x^3 = 0 \]
Given \( x \neq 0 \), we can simplify this to:
\[ 2b^2 - c^2 - a^2 = 0 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 2b^2 = a^2 + c^2 \]
This is the condition for \( a^2, b^2, c^2 \) to be in arithmetic progression (AP).

### Conclusion:
Thus, we conclude that \( a^2, b^2, c^2 \) are in AP.