Let S be the set of positive integral values of a for which a x ^ 2...

Given:

 $\dfrac{a{x}^{2} + 2 ( a + 1 ) x + 9a + 4}{{x}^{2} - 8x + 32} < 0 \forall x \in R$

For quadratic ${x}^{2} - 8x + 32 = 0, {D}_{1} = {\left(- 8\right)}^{2} - 4\left(32\right) = - 64$

Since the discriminant is less than zero and the leading coefficient is positive, this quadratic will always be positive.

Now, solving $a{x}^{2} + 2 ( a + 1 ) x + 9a + 4 < 0$

We know that, for a quadratic to be always negative, the coefficient of ${x}^{2} < 0, D < 0$.

$⇒ a < 0$ 

But we want positive values.

So, no positive integral value exist.