If $x^{2} + a x - 3 x - (a + 2) = 0$ has real and distinct roots, then the minimum value of $\frac{(a^{2} + 1)}{(a^{2} + 2)}$ is

1

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0

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\frac{1}{2}

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\frac{1}{4}

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Solution

The correct option is C $\frac{1}{2}$
$x^{2} + a x - 3 x - (a + 2) = 0$
$D = {(a - 3)}^{2} + 4 (a + 2)$
$\Rightarrow D = a^{2} - 2 a + 17$
$D_{1} = 4 - 4 (17) < 0$
Therefore, $a^{2} - 2 a + 17 > 0$ for all $a \in R$
Now, $\frac{a^{2} + 1}{a^{2} + 2} = 1 - \frac{1}{a^{2} + 2} > 1 - \frac{1}{2} = \frac{1}{2}$
Ans: C

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