Question

If \(f(x), g(x)\) and \(h(x)\) are three polynomials of degree 2 and
\[ \Delta(x)= \begin{vmatrix} f(x) & g(x) & h(x)\\ f'(x) & g'(x) & h'(x)\\ f''(x) & g''(x) & h''(x) \end{vmatrix} \] then \(\Delta(x)\) is a polynomial of degree

• A. 2
• B. 3
• C. 0
• D. at most 3

Hint:

For three quadratic polynomials, the determinant \(\begin{vmatrix} f & g & h \\ f' & g' & h' \\ f'' & g'' & h'' \end{vmatrix}\) becomes constant (degree 0), similar to Wronskian determinant behaviour.

Answer 1

The Correct Option is C

Step 1: Write general forms of degree 2 polynomials.
Let:
\[ f(x)=a_1x^2+b_1x+c_1,\quad g(x)=a_2x^2+b_2x+c_2,\quad h(x)=a_3x^2+b_3x+c_3 \] Step 2: Compute derivatives.
\[ f'(x)=2a_1x+b_1,\quad f''(x)=2a_1 \] Similarly:
\[ g'(x)=2a_2x+b_2,\quad g''(x)=2a_2 \] \[ h'(x)=2a_3x+b_3,\quad h''(x)=2a_3 \] Step 3: Form determinant structure.
\[ \Delta(x)= \begin{vmatrix} a_1x^2+b_1x+c_1 & a_2x^2+b_2x+c_2 & a_3x^2+b_3x+c_3\\ 2a_1x+b_1 & 2a_2x+b_2 & 2a_3x+b_3\\ 2a_1 & 2a_2 & 2a_3 \end{vmatrix} \] Step 4: Observe degree of determinant.
The third row is constant (degree 0).
The second row is linear (degree 1).
The first row is quadratic (degree 2).
When expanded, highest power terms cancel because first row is a linear combination of the derivatives structure.
This determinant is a Wronskian-type determinant for degree 2 polynomials, which becomes a constant.
Hence \(\Delta(x)\) has degree 0.
Final Answer: \[ \boxed{0} \]