The number of integral values of $a$ for which $x^{2} - (a - 1) x + 3 = 0$ has both roots positive and $x^{2} + 3 x + 6 - a = 0$ has both roots negative is

0

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1

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2

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infinite

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Solution

The correct option is D $1$
Given that the $x^{2} + 3 x + 6 - a = 0$ has both roots negative then
$6 - a > 0$
$\Rightarrow a < 6$
Let consider $x^{2} - (a - 1) x + 3 = 0$ have roots $α$ and $β$
So $α + β = a - 1, α β = 3$
Only for $a = 5$ , there exists only two values found as $(3, 1)$ and $(1, 3)$ .
So, only one value of $a$ exists.
Hence, option 'B' is correct.

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