If p(x) ,q(x) and r(x) are 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 given problem, we need to evaluate the determinant

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

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

### Step 1: Write the polynomials
Since \( p(x), q(x), r(x) \) are polynomials of degree 2, we can express them in the general form:

\[ p(x) = a_2 x^2 + a_1 x + a_0 \]
\[ q(x) = b_2 x^2 + b_1 x + b_0 \]
\[ r(x) = c_2 x^2 + c_1 x + c_0 \]

### Step 2: Compute the derivatives
Next, we compute the first and second derivatives of these polynomials:

\[ p'(x) = 2a_2 x + a_1 \]
\[ q'(x) = 2b_2 x + b_1 \]
\[ r'(x) = 2c_2 x + c_1 \]

\[ p''(x) = 2a_2 \]
\[ q''(x) = 2b_2 \]
\[ r''(x) = 2c_2 \]

### Step 3: Substitute into the determinant
Now we can substitute these expressions into the determinant:

\[ D = \begin{vmatrix} a_2 x^2 + a_1 x + a_0 & b_2 x^2 + b_1 x + b_0 & c_2 x^2 + c_1 x + c_0 \\ 2a_2 x + a_1 & 2b_2 x + b_1 & 2c_2 x + c_1 \\ 2a_2 & 2b_2 & 2c_2 \end{vmatrix} \]

### Step 4: Factor out common terms
We can factor out \( x^2 \) from the first row, \( 2x \) from the second row, and \( 2 \) from the third row:

\[ D = x^2 \cdot 2x \cdot 2 \cdot \begin{vmatrix} a_2 & b_2 & c_2 \\ 2a_2 & 2b_2 & 2c_2 \\ 2a_2 & 2b_2 & 2c_2 \end{vmatrix} \]

### Step 5: Analyze the determinant
Notice that the second and third rows of the determinant are identical:

\[ \begin{vmatrix} a_2 & b_2 & c_2 \\ 2a_2 & 2b_2 & 2c_2 \\ 2a_2 & 2b_2 & 2c_2 \end{vmatrix} \]

Since two rows of the determinant are the same, the value of the determinant is 0.

### Conclusion
Thus, we conclude that:

\[ D = 0 \]

### Final Answer
The determinant is independent of \( x \) and equals 0.

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