Key Concepts and Formulas

• Quadratic Discriminant: For a quadratic equation $ax^2 + bx + c = 0$, the discriminant is given by $\Delta = b^2 - 4ac$. The sign of the discriminant determines the nature of the roots.
• Quadratic Inequality: For a quadratic expression $ax^2 + bx + c$ to be always positive (or negative) for all real $x$, two conditions must hold: (1) the leading coefficient $a$ must be positive (or negative) and (2) the discriminant $\Delta$ must be negative.
• Rational Inequalities: When solving rational inequalities, pay attention to the sign of the denominator. If the denominator is always positive (or negative), the sign of the rational expression depends only on the numerator.

Step-by-Step Solution

Step 1: Analyze the Denominator

• What: Determine the sign of the denominator $x^2 - 8x + 32$ for all real $x$.
• Why: Knowing the sign of the denominator simplifies the inequality.
• Math: Let $D(x) = x^2 - 8x + 32$. The discriminant of $D(x)$ is: $\Delta_D = (-8)^2 - 4(1)(32) = 64 - 128 = -64$ Since $\Delta_D < 0$ and the leading coefficient is positive (1), $D(x) > 0$ for all $x \in \mathbb{R}$.
• Explanation: A negative discriminant means the quadratic has no real roots. Since the leading coefficient is positive, the parabola opens upwards and is always above the x-axis.

Step 2: Simplify the Inequality

• What: Simplify the given inequality using the fact that the denominator is always positive.
• Why: This reduces the problem to analyzing the numerator.
• Math: Since $x^2 - 8x + 32 > 0$ for all $x \in \mathbb{R}$, the inequality $\frac{a x^2+2(a+1) x+9 a+4}{x^2-8 x+32} < 0$ is equivalent to $a x^2 + 2(a+1) x + 9 a + 4 < 0 \quad \forall x \in \mathbb{R}$
• Explanation: Multiplying both sides of the inequality by a positive quantity ($x^2 - 8x + 32$) preserves the inequality sign.

Step 3: Apply Conditions for a Quadratic to be Always Negative

• What: State the conditions for the quadratic expression $a x^2 + 2(a+1) x + 9 a + 4$ to be negative for all real $x$.
• Why: These conditions will give us inequalities to solve for $a$.
• Math: Let $P(x) = a x^2 + 2(a+1) x + 9 a + 4$. For $P(x) < 0$ for all $x \in \mathbb{R}$, we need:
  1. $a < 0$ (leading coefficient must be negative)
  2. $\Delta_P < 0$ (discriminant must be negative)
• Explanation: $a<0$ ensures the parabola opens downwards. $\Delta_P < 0$ ensures the parabola does not intersect the x-axis.

Step 4: Calculate the Discriminant and Set Up the Inequality

• What: Calculate the discriminant $\Delta_P$ and set up the inequality $\Delta_P < 0$.
• Why: This sets up the inequality needed to solve for $a$.
• Math: $\Delta_P = [2(a+1)]^2 - 4(a)(9a + 4) = 4(a^2 + 2a + 1) - 36a^2 - 16a = 4a^2 + 8a + 4 - 36a^2 - 16a = -32a^2 - 8a + 4$ We need $\Delta_P < 0$, so $-32a^2 - 8a + 4 < 0$
• Explanation: This is a straightforward calculation of the discriminant.

Step 5: Solve the Discriminant Inequality

• What: Solve the inequality $-32a^2 - 8a + 4 < 0$.
• Why: This finds the range of values for $a$ that satisfy the discriminant condition.
• Math: Divide by -4 (and reverse the inequality): $8a^2 + 2a - 1 > 0$ Find the roots of $8a^2 + 2a - 1 = 0$: $a = \frac{-2 \pm \sqrt{2^2 - 4(8)(-1)}}{2(8)} = \frac{-2 \pm \sqrt{36}}{16} = \frac{-2 \pm 6}{16}$ So, $a_1 = \frac{4}{16} = \frac{1}{4}$ and $a_2 = \frac{-8}{16} = -\frac{1}{2}$. Since the parabola $8a^2 + 2a - 1$ opens upwards, $8a^2 + 2a - 1 > 0$ when $a < -\frac{1}{2}$ or $a > \frac{1}{4}$.
• Explanation: Using the quadratic formula to find roots, and then using the shape of the parabola to determine the intervals where the quadratic is positive.

Step 6: Combine the Conditions

• What: Combine the conditions $a < 0$ and ($a < -\frac{1}{2}$ or $a > \frac{1}{4}$).
• Why: This finds the values of $a$ that satisfy both conditions.
• Math: We need $a < 0$ AND ($a < -\frac{1}{2}$ OR $a > \frac{1}{4}$).
  • $a < 0$ AND $a < -\frac{1}{2}$ implies $a < -\frac{1}{2}$.
  • $a < 0$ AND $a > \frac{1}{4}$ is impossible. Therefore, we must have $a < -\frac{1}{2}$.
• Explanation: Basic logic and set intersection.

Step 7: Determine the Set S

• What: Determine the set $S$ of positive integral values of $a$ that satisfy $a < -\frac{1}{2}$.
• Why: This finds the solution to the original problem.
• Math: $S = \{a \in \mathbb{Z}^+ \mid a < -\frac{1}{2} \}$. Since there are no positive integers less than $-\frac{1}{2}$, $S = \emptyset$.
• Explanation: By definition, positive integers are greater than 0. Therefore, no positive integer can satisfy the inequality $a < -1/2$.

Step 8: Find the Number of Elements in S

• What: Find the number of elements in the set $S$.
• Why: This answers the question asked in the problem.
• Math: $n(S) = 0$.
• Explanation: The number of elements in the empty set is 0.

Common Mistakes & Tips

• Forgetting to analyze the denominator: Always check the sign of the denominator in rational inequalities.
• Incorrectly applying the quadratic conditions: Remember that for a quadratic to be always positive/negative, the leading coefficient and discriminant must satisfy specific conditions.
• Not considering the domain of the variable: Pay close attention to whether the variable must be an integer, positive, etc.

Summary

We analyzed the given rational inequality by first determining that the denominator is always positive. This reduced the problem to finding when the numerator is always negative. We applied the conditions for a quadratic to be always negative (negative leading coefficient and negative discriminant) and solved the resulting inequalities. Finally, we considered the requirement that '$a$' must be a positive integer. Since no positive integer satisfies the derived inequality, the set $S$ is empty, and the number of elements in $S$ is 0.

Final Answer

The number of elements in $S$ is $\boxed{0}$, which corresponds to option (A).