The number of integral values of $a$ for which the quadratic expression $(x - a) (x - 10) + 1$ can be factored as a product $(x + α) (x + β)$ of two factors $α, β, \in I$ , is

1

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2

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3

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Solution

The correct option is C $2$
Given quadratic expression $(x - a) (x - 10) + 1$ can be expressed as
$(x - a) (x - 10) + 1 = (x + α) (x + β)$ where $α, β \in I$
Put $x = - α$
$(- α - a) (- α - 10) + 1 = 0$
$(- α - a) (- α - 10) = - 1$
$α, a \in I$
Hence, $α + a$ and $α + 10$ are integers
$α + a = - 1$ and $α + 10 = 1$ or $α + a = 1$ and $α + 10 = - 1$
$a = 8$ or $a = 12$
Hence, number of integral values of $a = 2$

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