If α is a repeated root of a quadratic equation f (x) = 0 and A(x), B(x) and C(x) be polynomials of

degree > 2 then the determinant $(A (x) B (x) C (x) A (α) B (α) C (α) A' (α) B' (α) C' (α))$ is divisible

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A(x)

B(x)

C(x)

f (x)

answer is D.

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Detailed Solution

We assume the given matrix as

(A (x) B (x) C (x) A (α) B (α) C (α) A' (α) B' (α) C' (α))

is equal to g(x).
Then the differential of g(x) can be represented as:
G′(x) =

(A' (x) B' (x) C' (x) A (α) B (α) C (α) A' (α) B' (α) C' (α))

If we are putting α for x then g(α) can be represented as:
g(α) =

(A (x) B (x) C (x) A (α) B (α) C (α) A' (α) B' (α) C' (α))

From the properties of Matrices, we know that when two rows or columns are identical, then the determinant of the matrix becomes equal to zero. Hence, g(α) = 0. Therefore, we can say that (x − α) is the root of g(α) = 0.
Then the differential of g(α) can be represented as:
G′(α) =

(A' (α) B' (α) C' (α) A (α) B (α) C (α) A' (α) B' (α) C' (α))

Now again, since two rows are identical. Hence, the determinant of the matrix becomes equals to zero. Hence, G′(α) = 0. Therefore, we can say that (x−α) is the root of G′(α) = 0.
From the above expressions, we have g(α) = 0 and G′(α) = 0, α is the root and (x−α)
is the factor of the determinant of the g(α) = 0 and G′(α) = 0.
Since, it is given in the question that α is the repeated root of the quadratic equation f(x) = 0. This can be expressed as (x−α)2 is the factor of f(x) = 0.
Since, g(α) = 0, G′(α) = 0and f(x) = 0 has a common root (x−α)2. Hence, we can say that g(x) is divisible by f(x).
Hence, option 4 is correct.

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