Question

If a be a repeated roots of the quadratic equation f(x) = 0 and A(x), B(x), C(x) are polynomials of degree 3, 4, 5

respectively then determinant $\left| \begin{matrix} A(x) & B(x) & C(x) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ A'(\alpha) & B'(\alpha) & C'(\alpha) \end{matrix} \right|$is

divisible by (where A¢(a) =$\left( \frac{dA}{dx} \right)_{x = \alpha}$, etc) –

• A. f(x) (x – a)³
• B. (f(x))²
• C. f(x)
• D. None of these

Answer 1

Answer: (C)

f(x)

Answer 2

We know that a is a root of f(x) = 0. This means (x – a) is a factor of f(x). If a is a repeated root of f(x) = 0 then f(x) has a repeated factor (x – a), i.e., (x – a)² is a factor of f(x). But here f(x) is quadratic

\ f(x) = l(x – a)² …(1)

Where l is a constant.

Let D(x) = $\left| \begin{matrix} A(x) & B(x) & C(x) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ A'(\alpha) & B'(\alpha) & C'(\alpha) \end{matrix} \right|$

Which is of the degree 5 at most and 3 at least.

Clearly, D(a) = 0 …(2)

Differentiating D(x) w.r.t. x,

D¢(x)= $\left| \begin{matrix} A'(x) & B'(x) & C'(x) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ A'(\alpha) & B'(\alpha) & C'(\alpha) \end{matrix} \right|$+$\left| \begin{matrix} A(x) & B(x) & C(x) \\ 0 & 0 & 0 \\ A'(\alpha) & B'(\alpha) & C'(\alpha) \end{matrix} \right|$

+ $\left| \begin{matrix} A(x) & B(x) & C(x) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ 0 & 0 & 0 \end{matrix} \right|$

because derivatives of constants = 0

= $\left| \begin{matrix} A'(x) & B'(x) & C'(x) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ A'(\alpha) & B'(\alpha) & C'(\alpha) \end{matrix} \right|$

\D¢(a) =$\left| \begin{matrix} A'(\alpha) & B'(\alpha) & C'(\alpha) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ A'(\alpha) & B'(\alpha) & C'(\alpha) \end{matrix} \right|$= 0…(3)

because R₁ ŗ R₃. We know that

f(a) = 0 Ž (x – a) is a factor of f (x) and f¢(a) = 0

Ž (x – a)² is a factor of f(x)

\ from (2) and (3), D(x) has a factor (x – a)². \ D(x)

= (x – a)² . F(x) = $\frac{1}{\lambda}$l(x – a)² . F(x) = $\frac{1}{\lambda}$f(x).

F(x), using (1) \ D(x) is divisible by f(x).