1. Key Concepts and Formulas

• Condition for a Quadratic to be Always Non-Negative: For a quadratic expression $Ax^2 + Bx + C$ to be always non-negative (i.e., $Ax^2 + Bx + C \ge 0$) for all real values of $x \in \mathbb{R}$, two fundamental conditions must be met:
  1. The leading coefficient must be positive ($A > 0$). This ensures that the parabola represented by the quadratic opens upwards.
  2. The discriminant must be non-positive ($D \le 0$). The discriminant, $D = B^2 - 4AC$, determines the nature of the roots. If $D < 0$, there are no real roots, and the parabola is entirely above the x-axis. If $D = 0$, there is exactly one real root (a repeated root), and the parabola touches the x-axis at that point, remaining above it otherwise. Given the options and common interpretations in competitive exams, "always positive" in such contexts can sometimes imply "always non-negative" to include the case where the quadratic touches the x-axis.
• Counting Integers in a Range: The number of integers in an inclusive range $[p, q]$ (i.e., from $p$ to $q$ including both $p$ and $q$) is given by $q - p + 1$.
• Probability: The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes. $P(\text{Event}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$

2. Step-by-Step Solution

Step 1: Identify the coefficients of the quadratic and apply the condition for $A > 0$. The given quadratic expression is $x^2 + 2(a + 4)x - 5a + 64 > 0$. Comparing this to the standard form $Ax^2 + Bx + C > 0$, we identify the coefficients: $A = 1$ $B = 2(a + 4)$ $C = -5a + 64$

The first condition for the quadratic to be always non-negative is $A > 0$. Here, $A = 1$, which is indeed greater than 0. This condition is satisfied, meaning the parabola opens upwards.

Step 2: Apply the discriminant condition ($D \le 0$). The second condition is that the discriminant $D$ must be less than or equal to zero. The discriminant is given by $D = B^2 - 4AC$. Substitute the coefficients into the discriminant formula: $D = (2(a + 4))^2 - 4(1)(-5a + 64)$ Expand the terms: $D = 4(a^2 + 8a + 16) - 4(-5a + 64)$ $D = 4a^2 + 32a + 64 + 20a - 256$ Combine like terms to simplify the expression for $D$: $D = 4a^2 + 52a - 192$ Now, set the condition $D \le 0$: $4a^2 + 52a - 192 \le 0$ Divide the entire inequality by 4 (since 4 is a positive number, the inequality sign remains unchanged): $a^2 + 13a - 48 \le 0$

Step 3: Solve the quadratic inequality for 'a'. To solve $a^2 + 13a - 48 \le 0$, we first find the roots of the corresponding quadratic equation $a^2 + 13a - 48 = 0$. We can use the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $a = \frac{-13 \pm \sqrt{13^2 - 4(1)(-48)}}{2(1)}$ $a = \frac{-13 \pm \sqrt{169 + 192}}{2}$ $a = \frac{-13 \pm \sqrt{361}}{2}$ Since $\sqrt{361} = 19$: $a = \frac{-13 \pm 19}{2}$ This gives two roots: $a_1 = \frac{-13 - 19}{2} = \frac{-32}{2} = -16$ $a_2 = \frac{-13 + 19}{2} = \frac{6}{2} = 3$ Since the quadratic $a^2 + 13a - 48$ has a positive leading coefficient (1), its parabola opens upwards. For the expression to be less than or equal to zero, $a$ must lie between or be equal to its roots. Therefore, the solution to the inequality is: $-16 \le a \le 3$

Step 4: Determine the favorable integer values of 'a'. The problem states that $a$ is an integer selected from the range $[-5, 30]$. This means the possible values for $a$ are integers such that $-5 \le a \le 30$. We need to find the integers $a$ that satisfy both conditions:

1. $-16 \le a \le 3$ (from the quadratic condition)
2. $-5 \le a \le 30$ (from the given range for $a$) To find the intersection of these two ranges, we take the maximum of the lower bounds and the minimum of the upper bounds: $\max(-16, -5) \le a \le \min(3, 30)$ So, the favorable integer values for $a$ are those satisfying: $-5 \le a \le 3$ The integers in this range are $\{-5, -4, -3, -2, -1, 0, 1, 2, 3\}$.

Step 5: Count the number of favorable outcomes. The number of favorable integer values for $a$ is the count of integers in the range $[-5, 3]$. Number of favorable outcomes = $3 - (-5) + 1 = 3 + 5 + 1 = 9$.

Step 6: Count the total number of possible outcomes. The total range for $a$ is $[-5, 30]$. The integers in this range are $\{-5, -4, \dots, 29, 30\}$. Total number of possible outcomes = $30 - (-5) + 1 = 30 + 5 + 1 = 36$.

Step 7: Calculate the probability. The probability of selecting an integer $a$ that satisfies the given conditions is the ratio of the number of favorable outcomes to the total number of possible outcomes: $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{9}{36}$ Simplify the fraction: $P = \frac{1}{4}$

3. Common Mistakes & Tips

• Misinterpreting "always positive": A common mistake is to interpret "always positive" as strictly $Ax^2+Bx+C > 0$, which would imply $D < 0$. However, in many competitive exams, "always positive" can include the case where the quadratic touches the x-axis (i.e., $D=0$), meaning it's "always non-negative." Always check the options; if $D<0$ leads to none of the options, consider $D \le 0$.
• Errors in solving quadratic inequalities: Remember that for $ax^2+bx+c \le 0$ (with $a>0$), the solution lies between or at the roots. For $ax^2+bx+c \ge 0$ (with $a>0$), the solution lies outside or at the roots.
• Incorrectly counting integers: When counting integers in an inclusive range $[p, q]$, use the formula $q - p + 1$. Forgetting to add 1 is a common error.
• Ignoring the domain of 'a': Always ensure to intersect the derived range for 'a' with the given domain for 'a' (in this case, $a \in [-5, 30]$).

4. Summary

To determine the probability, we first established the conditions for a quadratic expression to be always non-negative, which are a positive leading coefficient and a non-positive discriminant ($D \le 0$). The leading coefficient was already positive. We then calculated the discriminant in terms of $a$ and set it less than or equal to zero, resulting in a quadratic inequality for $a$. Solving this inequality, we found the range of $a$ that satisfies the quadratic condition. Finally, by intersecting this range with the given domain for $a$ and counting the favorable integers, we calculated the probability.

5. Final Answer

The final answer is $\boxed{{1 \over 4}}$, which corresponds to option (D).