If p(x) ,q(x) and r(x) and three polynomials of degree 2, then `|{:(p(x),q(x),r(x)),(p'(x),q'(x),r'(x)),(p''(x),q''(x),r''(x)):}|` is ………….of x .

Answer

To solve the problem, we need to evaluate the determinant given by:

\[ D = \begin{vmatrix} p(x) & q(x) & r(x) \\ p'(x) & q'(x) & r'(x) \\ p''(x) & q''(x) & r''(x) \end{vmatrix} \]

where \( p(x), q(x), r(x) \) are polynomials of degree 2.

### Step 1: Define the Polynomials
Let’s assume the polynomials are of the following forms:
- \( p(x) = ax^2 + bx + c \)
- \( q(x) = dx^2 + ex + f \)
- \( r(x) = gx^2 + hx + i \)

### Step 2: Calculate the Derivatives
Next, we compute the first and second derivatives of these polynomials:
- \( p'(x) = 2ax + b \)
- \( q'(x) = 2dx + e \)
- \( r'(x) = 2gx + h \)

- \( p''(x) = 2a \)
- \( q''(x) = 2d \)
- \( r''(x) = 2g \)

### Step 3: Set Up the Determinant
Now we can set up the determinant:

\[ D = \begin{vmatrix} ax^2 + bx + c & dx^2 + ex + f & gx^2 + hx + i \\ 2ax + b & 2dx + e & 2gx + h \\ 2a & 2d & 2g \end{vmatrix} \]

### Step 4: Factor Out Common Terms
From the first row, we can factor out \( x^2 \), from the second row, we can factor out \( 2x \), and from the third row, we can factor out \( 2 \):

\[ D = x^2 \cdot 2x \cdot 2 \cdot \begin{vmatrix} a & d & g \\ 2a & 2d & 2g \\ 0 & 0 & 0 \end{vmatrix} \]

### Step 5: Evaluate the Inner Determinant
The inner determinant now is:

\[ \begin{vmatrix} a & d & g \\ 2a & 2d & 2g \\ 0 & 0 & 0 \end{vmatrix} \]

This determinant evaluates to zero because the third row consists entirely of zeros. Therefore, we have:

\[ D = x^2 \cdot 2x \cdot 2 \cdot 0 = 0 \]

### Conclusion
Since the determinant evaluates to zero, we conclude that:

\[ D \text{ is independent of } x. \]

### Final Answer
Thus, the answer is **independent of x**.

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