Multiple choice

lf $\alpha$ is the repeated root of quadratic equation $f(x)=0$ and $A(x)$ , $B(x)$ and $C(x)$ are polynomials of degree 3,4 and 5 respectively, then $\phi(x)=$ $\begin{vmatrix} A\left ( x \right ) &B\left ( x \right ) &C\left ( x \right ) \ A\left ( \alpha \right )&B\left ( \alpha \right ) & C\left ( \alpha \right )\ A^{'}\left ( \alpha \right ) &B^{'}\left ( \alpha \right ) & C^{'}\left ( \alpha \right ) \end{vmatrix}$ is divisible by

  1. $f\left ( x \right )$
  2. $A\left ( x \right )$
  3. $B\left ( x \right )$
  4. $C\left ( x \right )$
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A Correct answer
Explanation

If alpha is a repeated root of f(x)=0, then f(alpha)=0 and f'(alpha)=0. The determinant phi(x) has a row where all elements are evaluated at alpha, and another row where all derivatives are evaluated at alpha. Since alpha is a repeated root, the determinant will be zero at x=alpha, implying (x-alpha)^2 is a factor, which is f(x).

AI explanation

Let $\phi(x)$ be the given determinant. Differentiating $\phi(x)$ with respect to $x$ yields a determinant with the first row being the derivatives $A'(x)$, $B'(x)$, and $C'(x)$. If we substitute $x = \alpha$ into this derivative, the second row becomes identical to the first row, making the determinant and the derivative equal to zero. Since $\phi(\alpha) = 0$ and $\phi'(\alpha) = 0$, $(x-\alpha)^2$ is a factor of $\phi(x)$. Given that $\alpha$ is the repeated root of $f(x) = 0$, $f(x)$ must be a factor of $\phi(x)$.

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