Question

If a be a repeated root of the quadratic equation f(x) = 0 and A(x), B(x), C(x) be polynomials of degrees 3, 4 and 5 respectively, then $ \begin{vmatrix}A\left(x\right)&B\left(x\right)&C\left(x\right)\\ A\left(a\right)&B\left(a\right)&C\left(a\right)\\ A'\left(a\right)&B'\left(a\right)&C'\left(a\right)\end{vmatrix} $

• A. is divisible by f(x) for all x
• B. is not divisible by f(x) for all x
• C. is equal to 0
• D. None of these

Answer 1

Answer: (A)

We must have $f (x) = k (x - a)^2$ where k is a constant. Now in order that the given determinant [Say, D(x)] be divisible by f(x) we must show that both D(x) and D'(x) vanish at x = a. Now
$D\left(a\right) = \begin{vmatrix}A\left(a\right)&B\left(a\right)&C\left(a\right)\\ A\left(a\right)&B\left(a\right)&C\left(a\right)\\ A'\left(a\right)&B'\left(a\right)&C'\left(a\right)\end{vmatrix} = 0$
Also
$D'\left(x\right) = \begin{vmatrix}A'\left(x\right)&B'\left(x\right)&C'\left(x\right)\\ A\left(a\right)&B\left(a\right)&C\left(a\right)\\ A'\left(a\right)&B'\left(a\right)&C'\left(a\right)\end{vmatrix}+ \begin{vmatrix}A\left(x\right)&B\left(x\right)&C\left(x\right)\\ 0&0&0\\ A'\left(a\right)&B'\left(a\right)&C'\left(a\right)\end{vmatrix}$
$+\begin{vmatrix}A\left(x\right)&B\left(x\right)&C\left(x\right)\\ A\left(a\right)&B\left(a\right)&C\left(a\right)\\ 0&0&0\end{vmatrix} = \begin{vmatrix}A'\left(x\right)&B'\left(x\right)&C'\left(x\right)\\ A\left(a\right)&B\left(a\right)&C\left(a\right)\\ A'\left(a\right)&B'\left(a\right)&C'\left(a\right)\end{vmatrix}$
which clearly gives D' (a) = 0, since first and third row become identical.